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abor

Linear Feedback Loops

Prof Dr Martin J W Schubert

Electronics Laboratory

Ostbayerische Technische Hochschule Regensburg

Regensburg Bavaria Germany

Martin JW Schubert Linear Feedback Loops OTH Regensburg

- 2 -


Abstract Linear feedback loop models are introduced

1 Introduction

Fig 1 (a) basic model of linear feedback loop (b) operational amplifier application example The feedback loop model in Fig 1(a) with F and B being linear and time-invariant (LTI) models in s or z features signal transfer function (STF) and noise transfer function (NTF)

1

0 1FB

E

Y FSTF B

X FB

(11)

0

10

1FB

X

YNTF

E FB

(12)

To derive STF according to (11) set E=0 (corresponding to a short circuit) so that

( ) ( )Y F F X V F X BY FX FBY rarr (1 )Y FB FX 1

Y F

X FB

Exercise To derive (12) do the same derivation with input signal E and X=0 Y F E Exercise for electronics engineers (i) Relate quantities of Fig 1(a) and (b) Assume that E results from non-linearity in the operational amplifier and noise feed-through from the power supply (ii) Which errors can and which cannot be suppressed by high loop gain |FB| rarr infin X corresponds to helliphellip ε corresponds to hellip Y corresponds to hellip V corresponds to helliphellip F corresponds to hellip B corresponds to hellip


- 3 -

Solutions (i) X rarr Uip ε rarr Uin Y rarr Uout V rarr Uim F rarr Av B rarr R1 ( R1 + R2 ) (ii) |FB| rarr infin suppresses non-linear behavior of Av not errors in B or input noise eg offset The important features of this loop are High loop amplification |FB| given then Loop inverts transfer function of feedback network B Error entries behind forward network F are suppressible Errors of feedback network B are not suppressible (loop can never be better than B) Input errors such as input noise of input offset of the op-amp are unsurpassable ldquoinputrdquo Note that we have calculated with frequency-domain functions Expressions like Y=STFꞏX in frequency domain correspond to a convolution (deutsch Faltung) in time-domain according to y(t) = hSTF(t)x(t) with hSTF(t) being the impulse response of the STF and the lsquorsquo operator standing for convolution Exercise 1 Compute the amplification STF=YX of the System in Fig 1(a) for F=10000 B=110 Compute the noise suppression NTF=YE o f the System in Fig 1(a) for F=10000 B=110 Solution STF = 104(1+01ꞏ104) = 999 NTF = 1(1+01ꞏ104) = 999ꞏ10-4 Exercise 2 Using Matlabrsquos LTI systems [1] created by tf(hellip) the behavior of a feedback loop as illustrated in Fig 1(a) is modeled with command feedback Example Listing 1 Computing STF and NTF with Matlab for the system in Fig 1

Use Matlabs LTI functions to compute STF and NTF as feedback loops F = tf(1E4) Forward network amplification B = tf(01) Backward network amplification STF = feedback(FB) build STF as LTI system NTF = feedback(1FB) build NTF as LTI system The organization of this document is as follows

Section 1 presents in a very short form the basic feedback loop configuration and equations Section 2 presents 4 axioms of linear and time invariant (LTI) signal processing and appl Section 3 extends loop topologies to higher order systems Section 4 explains filters canonic direct structures Section 5 illustrates translation from s rarr z domain Section 6 is concerned with stability Section 7 details theory in several practical application examples Section 8 draws relevant conclusion and Section 9 offers references Section 10 is an appendix used eg for detailed mathematical proofs


- 4 -

2 Linearity and Time Invariance (LTI) 21 The 4 Axioms of Signal Processing

211 Linearity Axiom

y[ c1x1(t) + c2x2(t) ] = c1y[ x1(t) ] + c2y[x2(t)] (21)

H

c1

c2

x1(t)

x2(t)

y(t)

(a) (b)

x1(t)

x2(t)

c1

c2

y(t)H

H

Fig 211 (a) linear superposition of two signals (b) equivalent system Linearity for signal processing systems is defined according equation (21) illustrated by Fig 211 Proportionality Implication

Setting c2=0 in equation (21) shows Linearity implies proportionality y[ cx(t)] = cy[x(t)] (22) as illustrated in Fig 212 Proportionality allows to shift constants over LTI systems and therefore to combine several constants within the circuit mathematically to a single constant

Hcx(t) y(t)

(a) (b)

cx(t) y(t)H

Fig 212 Proportionality Systems (a) and (b) are equivalent for linear circuits Zero-Offset Implication

Setting c=0 in equation (22) shows Proportionality implies zero offset y[0] = 0y[x(t)] = 0 (23) Conclusion The resistive divider in Fig 213(a) is linear because U2 = constant U1 The circuit with operational amplifier in Fig 213(b) is non-linear as U20 when U1=0


- 5 -

R1

R2

Uoff

R1 R2(a) (b)

U1

U2

U2U1

Fig 213 (a) resistive divider (b) circuit using OpAmp with offset voltage Uoff0 Remark Linearity according to Eq (21) is a signal processing definition From a mathematical point of view a system Uout = aUin + b with constants a b is linear 212 Time Invariance Axiom

A system is time invariant when its impulse response h(t) is not a function of time h(t) = h(t-) (24) Fig 214 Time-variant system when Uctrl varies with time

UctrlUin Uout

CvLCk1 Ck2

Most systems we use are time-invariant An example for a time-variant system is shown in Fig 214 where response Uout to the impulses at Uin depends on control voltage Uctrl which varies with time

213 Causality Axiom y(t) = f(x()) with t

The present state of a system y(t) is a function of the past and present state of its input but not of future inputs 214 Stability Definition Bounded Input Bounded Output (BIBO)

Definition There exist constants M and K so that from |x(t)|M follows |y[x(t)]| KM Question Is an ideal integrator BIBO stable (Hint consider f0)


- 6 -

22 Application LTI Axioms

Fig 22 Evaluation of the feedback loop equations In Fig 22(a) the transfer function of the system is Y = QI + FX - FBL In Fig 22(b) from L=Y follows (1 + FB)ꞏY = FꞏX + QꞏI and consequently

1 1

F QY X I

FB FB

= STF X NTF I (25)

The so-called signal and noise transfer functions are

1

0 1FB

E

Y FSTF B

X FB

(26)

1

0 1FB

X

Y QNTF B

I FB

(27)

In Fig 22(c) FB was split in BꞏF Proportionality allows moving factor -1 over network F In Fig 22(d) Linearity allows to unify the 2 networks F without changing (26) (27) Opening the Loop at ldquocutrdquo in Fig 22(d) allows to measure Q=EI F=YX and FB=YL


- 7 -

23 Application ΔΣ Modulator

Fig 23 ΔΣ system with (a) quantizer and (b) ADA conversion (c) quantization noise density unmodulated is flat (ldquowhiterdquo) and (d) shaped by modulatorrsquos NTF

With our notation the ΔΣ modulator would have to be called ε int modulator with

Forward network 1

11

zF

z

is a time-discrete integrator

Backward network B = 1 Input signal X = input signal (eg sound) Interferer I = inferring Quantizer with minimum step Δ Quantization noise power 2 2 12qE total noise power is constant

Quantization noise density Eq= 6 sf flat (ldquowhiterdquo) distributed over bandwidth 0hellipfs2

The signal transfer function is flat

1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF

The NTF is a Differentiator with F = f fs

1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F

Eq ~| NTF| ΔΣ modulator shapes quantization noise power density to high frequencies


- 8 -

3 Network Topologies Higher order systems often come in one of two topologies

Topology 1 distributed feed-in common network concentrated feed-out of it

Topology 2 distributed feed-out of common network concentrated feed-in to it

31 Topology 1 Distributed Feed-In

311 Derivation

Fig 311

Distributed feed-in network

Open the loop at its concentrated point (cut) as shown in Fig 311 Then determine the feed forward network F for L=0 Identify any possible path from X to Y In linear systems solutions superpose linearly Feed Forward network of order R

0 1 1 2 1 2 1 20 0

1 R R

L E

YF a a F a F F a F F F

X b

(311)

Feed Back loop network of order R

1 1 2 1 2 1 20 0

1 R R

X E

YFB b F b F F b F F F

L b

(312)

Signal and noise transfer functions are obtained by closing the loop setting L=Y

0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R

a a F a F F a F F FY FSTF

X FB b b F b F F b F F F

(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF

E FB b b F b F F b F F F

(314)

In this network type input signal X passes the feed-forward filter defined by coefficients ak k=0hellipR before it enters the feedback loop We have the chance to cancel or attenuate critical frequencies before they enter the feedback loop


- 9 -

312 Single Feedback Topology with Feedforward Path

Fig 312 Single Feedback Topology

with Feedforward Path

Set in Fig 311 bR = aR = a0 = 1 and all other coefficients to zero To obtain Fig 312 Let G = F1ꞏhellipꞏFR-1 Then

0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)

The signal transfer function is always ideally flat over frequency and the G can be used to shape noise by the NTF The particular advantage of this topology is in the fact that the accuracy of the loop depends on the accuracy of its feedback network B This is realized in Fig 312 as single coefficient bR=1 which is in most cases relatively easy and accurate to realize


- 10 -

32 Topology 2 Distributed Feed-Out

321 Derivation

Fig 321 (a) Two networks that are combined to (b typical distributed feed-out topology Figs 321(a) and (b) show two linear networks that behave identical for L=W Opening the feedback loop in Fig 321(a) at (cut) and yields

0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E

WFB b F b F F b F F F

L b

(322)

Closing the loop by setting L=Y delivers the transfer functions for the feedback network as

10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)

The upper network part in Fig 321(a) contains feed forward only

0 1 1 2 1 2 1 2 R R


W (325)

As feedforward network F is in series with STF1 and NTF1 we get

0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R

a a F a F F a F F FW GSTF STF F


(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion

Feeding forward network F and feedback-loop network FB are identical in the topology 1 of Fig 311 and topology 2 of Fig 321 Consequently signal transfer functions in (313) and (326) are identical at the first glance In topology 1 the signal passes fist the feedforward network F and then the feedback loop

network FB Hence network F has the chance to cancel or attenuate critical frequencies before entering the loop

In topology 2 the signal stimulates at first the feedback loop and is then shaped ty forward network F which cannot protect the loop from critical frequencies any more

The noise transfer functions are different as we cannot prevent the noise from transferring the forward network F The cut in Fig 321(a) separates L and W for the lower series of transfer functions Fk only while the upper row of Fk remains connected Therefore an equivalent cut is not possible in Fig 3(b) Exercise We need an integrator and a differentiator in series (eg sinc filter) Which sequence should be preferred Solution Placing the integrator after the diffentiator is strongly recommended because the differentiator removes the DC component of the signal A differentiator after an integrator cannot protect the integrator from overflow due to DC input Fig 322 sinc filterDifference xn-xn-R avoidsDC frequencies at theintegratorrsquos input


- 12 -

4 Canonic Direct Structures 41 Time-Continuous Modeling

Fig 41 Time-continuous filter of order R in 1st canonic direct structure LTI models of time-continuous systems are modeled as

( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R

R Ra x t a x t a x t b y t b y t b y t (411)

1 ( ) ( )

0 0 1 1( ) ( ) ( ) ( ) ( ) ( )R RR Ry t b a x t a x t a x t b y t b y t (412)

For frequency domain modeling we translate ( ) ( )x t X s and ( ) ( )x t s X s

0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R RR Ra X s a sX s a s X s b Y s b sY s b s Y s (413)

yielding transfer function 0 1

0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)

Using s=jω=j2πf 0 0( 0) TF f a b ( ) R RTF f a b (415)


- 13 -

42 Time-Discrete Modeling

Fig 42 Time-discrete filter of order R in (a) 1st and (b) 2nd canonic direct structure Canonic number of delays (z-1) is R with z=exp(j2πF) with F=f fs at sampling rate fs =1Ts Direct structure taps of impulse hn=h(n) response directly set by coefficients whereas index n denotes time point tn = nꞏTs

0 1 1 0 1 1 n n R n R n n R n Rd x d x d x c y c y c y (421)

10 0 1 1 1 1 n n n R n R n R n Ry c d x d x d x c y c y

(rarr typical 0 1c ) (412)

For frequency domain modeling we translate ( )nx X z and ( )kn kx z X z

1 2 1 20 1 2 0 1 2( ) ( ) ( ) ( ) ( ) ( )c Y z c z Y z c z Y z d X z d z X z d z X z (413)


0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R

d d z d z d z d z dTF z

c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -

5 Translation between s rarr z The accurate relationship behaves difficult in formulae

sTz e 1

ln( )s zT

(501)

Using S sT (502) obtains ln( )S z (503)

51 Linear Interpolation

1( ) n nx xx t

T

translates to 1 11X z X z

s X XT T

Cancelling X delivers

1

11(1 )s

zs f z

T

(511)

which results in the back and forth symmetrical transformation

11S z 1 1z S (512) Translate a transfer function in s to a transfer function in z order is R 1 k

k k sa a f kk k sb b f k=0hellipR

2 replace 11s z 3 compute H(z) 4 Divide all coefficients of H(z) by denominator coefficient c0

20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2

First order transfer functions are obtained from the solution above setting a2 = b2 = 0 Check From a2 = b2 = 0 should result c2 = d2 = 0


- 15 -

52 Bilinear (Tustin) Approximation

From series expansion of ln(x) we approximate 1

1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)

results in transformation 1

1

1

1

zs

z

1

1

1

sz

s

(523)

suffers from no amplitude error as

1

11

1

jz

j

(524)

but frequency shift arctan( )z s 2

arctan2z s

T

T

(525)

with ωs being the original frequency axis of H(s) and ωz the compressed axis of H(z) To get a particualr frequnceny ωx to the right point after translation to z predistort it according to but frequency shift

tan( )s predistorded s (526)

Translate a transfer function in s to a transfer function in z order is R 1 (2 )k

k k sa a f (2 )kk k sb b f k=0hellipR

2 replace 11s z 3 compute H(z) 4 Divide all coefficients of H(z) by denominator coefficient c0 Application for 2nd order

20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2

1st order system | 2nd order system

0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -

53 Impulse-Invariant Time-Discretization Basic principle

1 Translate the analog frequency response Hc(s) to a time-continuous impulse response hc(t)

2 Find a time-discrete model Hd(z) that generates an impulse response hd(n) which samples hc(t) so that hd(n) = hc(nT) where T=1fs is the sampling interval

Example

A RC lowpass features the impulse response

0

00)( tife

tifth RCtc shown in Fig 44(b)

From Fig 53(a) we derivebz

a

zb

zazH d

1

1

1)( It is easily found by trial and error that

its impulse response is 0 1 b b2 b3 b4 As illustrated in Fig 44(b) the taps of y(n)

ldquosamplerdquo the impulse response of an RC lowpass (dashed lined) when RCTT eeb g and ba 1

yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0

Fig 53 (a) recursive IIR filter (b) impulse response time-discrete versus RC lowpass

54 Matlab Matlabrsquos function c2d translates Matlabrsquos time-continuous LTI systems to Matlabrsquos time- discrete LTI systems The conversion method can be selected The reverse function is d2c


- 17 -

6 Stability 61 Considering Poles and Zeros

611 Poles Describing the Momentum of a System

Computing poles (ie zeros of the denominator) and zeros (of the numerator) of STF and NTF is often complicated and requires computer-aided tools If we have these poles and zeros we have deep system insight Zeros describe the stimulation (dt Anregung) and poles the momentum (dt Eigendynamik) of a system For X=0 both STF and NTF deliver Y(1+FB) = 0 (621) Consequently the output signal must follow the poles that fulfill (621)

Time-continuous ( ) pks t

kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)

With Ck being constants and index pk indicating pole numper k Fig 52 illustrates the relation between the location of poles sp=σp+jp in the Laplace domain s=σ+j and the respective transient behavior For a polynomial with real coefficients complex poles will always come as

complex pairs sp=σpjp that can be combined to )cos( te ptp

or )sin( te ptp

because

ejx+e-jx=2cos(x) and ejx-e-jx=2jsin(x) It is obvious that these functions increase with σpgt0 and decay with negative σplt0

The conclusion from a time-continuous to time-discrete is obvious from z=esT and Ts pe with

T=1fs being the sampling interval Increasing time corresponds to increasing n for npz

Consequently stability requires σplt0 or |zp|lt1 for all poles spi or zp1 respectively


- 18 -

7 Applications 71 Second-Order Lowpass

711 Poles of Second Order Lowpass

Fig 711 Time-continuous system of 2nd order characteristic curves

A technically particularly important system is the 2nd order lowpass frequently modeld as

200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)

with A0 being the DC amplification D stability parameter ω0 bandwith and S = sω0

1

11

111

202021 DjD

DDs p (713)


- 19 -

712 Characteristic Lowpasses

The Butterworth lowpass is the mathematically flattest characteristics in frequency domain and has always ndash3dB attenuation in its bandwith freuency regardless ist order It can behaviorally modeled by

N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)

Note that this is a behavioral model obtaining always ndash3dB at f=f0 Butterworth coefficients for order N=123 taken from [httpsdewikipediaorgwikiButterworth-Filter] are Table 712 Poles and their complex phases for Butterworth lowpasses of 1st 2nd 3rd order 45deg-phae-margin technique and aperiodic borderline case (AP)

Lowpass Type

Denominator Poly with S=sω0

D Poles represented as σpplusmnjωp

Φp pj

pe step resp overshoot

AP1=BW1 S+1 - Sp = 1 sp = ω0 0deg 0 AP2 S2+2S+1 = (S+1)2 1 Sp12 = 1 sp12 = ω0 0deg 0 BW2 S2 + 2 S + 1 21 Sp12 = j 121 plusmn45deg 4

BW3 (S2 + S + 1) (S+1) - Sp12 =frac12 (1+j 3 ) Sp3=1 plusmn60deg 0deg 8

PM45 S2 + S + 1 12 Sp12 = frac12 (1 + j 3 ) plusmn60deg 16

Fig 712 Characteristics of 1st (bl) 2nd (red) 3rd (ye) Butter-worth and a 45deg-phase-margin lowpasses (pink)

(a) Bode diagram

(b) Step responses Listing 712 Matlab code for Fig 71

Filter 2 poles Hs_BW1=tf([1][1 1]) Hs_BW2=tf([1][1 sqrt(2) 1]) Hs_BW3=tf([1][1 2 2 1]) Hs_M45=tf([1][1 1 1]) figure(1) step(Hs_BW1Hs_BW2Hs_BW3Hs_M45) grid on figure(2) h=bodeplot(Hs_BW1Hs_BW2Hs_BW3Hs_M45) grid on setoptions(hFreqUnitsHzPhaseVisibleon)


- 20 -

713 Example RLC Lowpass

An RLC lowpass according to Fig 713 is modeld as

21

1)(

sLCsRCsH LP (715)

R is a sum of both voltage-source output impedance and inductorrsquos copper resistor

Combining (711) and (714) yields 10 A LC

f2

10 and

L

CRD

2

(a) RLC lowpass

OSC D=01 oscillating case M45 D=frac12 phase marg 45deg

BW2 D= 21 Butterworth AP2 D=1 aperiodic borderline

(b) Bode diagram

(c) Step responses

Fig 713 Characteristics of an RLC lowpass Listing 713 Matlab code generating screen shots of Fig 713 (b) and (c)

RLC lowpass as filter with 2 poles ---------------------------------- setting cut-off frequency f0 and restor R f0 = 10000 set cut-off frequency in Hz R = 1 set resistor in Ohms Stability settings D_OSC = 01 2nd order oscillating D_BW2 = sqrt(12) Butterworth 2nd order D_M45 = 12 phase margin 45deg 2n order D_AP2 = 10001 aperiodic borderline case 2nd order conclusions w0 = 2pif0 LC = 1w0^2 L_OSC = R(2D_OSCw0) C_OSC = 2D_OSC(Rw0) L_M45 = R(2D_M45w0) C_M45 = 2D_M45(Rw0) L_BW2 = R(2D_BW2w0) C_BW2 = 2D_BW2(Rw0) L_AP2 = R(2D_AP2w0) C_AP2 = 2D_AP2(Rw0) set up LTI systems Hs_AP2=tf([1][LC RC_AP2 1]) aperiodic borderline case 2nd order Hs_BW2=tf([1][LC RC_BW2 1]) Butterworth 2nd order Hs_M45=tf([1][LC RC_M45 1]) phase margin 45deg 2n order Hs_OSC=tf([1][LC RC_OSC 1]) 2nd order oscillating Graphical postprocessing figure(1) step(Hs_AP2k-Hs_BW2gHs_M45bHs_OSCr) grid on figure(2) h=bodeplot(Hs_AP2k-Hs_BW2gHs_M45bHs_OSCr) grid on p = bodeoptions setoptions(hFreqUnitsHz)


- 21 -

72 Placing a Pole on a 1st Order Integrator According to Fig 72(a) a integrating open loop amplification is modeled as

11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)

with A1 being the 1st order feed-forward and k the feedback network In Fig 72 we use ωrsquo=ωꞏsec as both a log-functoinrsquos argument and a closed-loop amplification must be dimensionless The Signal transfer function of the loop will be

01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s

STF s B B B BFB FB s s

(722)

The system has a phase of -90deg and is stable due to a rest of 90deg phase margin against -180deg In the model above the low pole is spl = 0 but a finite pole ωpl as indicatd in Fig part (c) might be reasonable or necessary to respect limited amplification of an operational amplifier compute an operatin point in LTspice (divide by zero at s=jω=0 ) or avoid digital number overflow A low pole ωpl ltlt ω1 has few impact on the closed loop behavior To model the a second pole ωp in the open loop we model

1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)

Comparing it with the general system (711) we find A0=k -1 (725)

150 pD (726)

120 p (727)

Fig 72 (a) ideal integrator ωpl=0 (b) including second pole ωp (c) low pole 0ltωplltltωp


- 22 -

PM45 ωp = ω1 The most usual strategy to set the pole to ωp = ω1 which is typically said to optimize to 45deg phase margin because it results into a total phase shift of -135deg at ω1 This corresponds to D=frac12 and poles at sp12 = -Dꞏω1ꞏexp(plusmn60deg) They cause 125dB peking in frequency domain and 16 step-response voltage overshoot in time-domain BW3 3rd order Butterworth PM45 + ωp3=ω1 The PM45 polynomial is contained in a 3rd order Butterworth polynomial Appending a simple 1st order lowpass to a PM45-system obtains a 3rd order Butterworth characteristics ie no peaking and 8 voltage overshoot in time domain

BW2 2nd order Butterworth ωp = 2ꞏω1 This yields 21D and poles at sp12 = -Dꞏω0ꞏexp(plusmn45deg) No peking in frequency domain ca 4 step-response voltage overshoot Typically considered optimal with respect to settling time AP2 ωp = 4ꞏω1 This yields D=1 and consequently the aperiodic borderline case corresponding to two 1st order lowpasses in series -6dB at ω1 and poles sp12 = -ω0 No peaking and no voltage overshoot but slow


- 23 -

73 Placing a Zero on a 2nd Order Integrator According to Fig 73(a) a integrating loop amplification is modeled as

2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)

with A2 being the 2nd order feed-forward and k the feedback network In Fig 73 we use ωrsquo=ωꞏsec as both a log-functoinrsquos argument and a closed-loop amplification must be dimensionless The Signal transfer function of the loop will be

2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)

The system has a phase of -180deg and is an ideal oscillator 0deg phase margin against -180 To regain positive phase margin we have to place a zero ωn in the open loop we model

2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)

Fig 73 (a) ideal integrator ωpl=0 (b) including zero ωn


- 24 -

PM45 ωn = ω2 The most usual strategy is to place the zero at ωn = ω2 This corresponds to D=frac12 and consequently poles at sp12 = -Dꞏω1ꞏexp(plusmn60deg) Together with the zero this obtains 33dB peking in the frequency domain and 30 step-response voltage overshoot

BW2 2nd order Butterworth ωn = ꞏω2 2 This yields 21D and poles at sp12 = -Dꞏω0ꞏexp(plusmn45deg) Together with the zero this obtains 21dB peking in the frequency domain and 21 step-response voltage overshoot in time-domain AP2 ωn = ω2 2 This yields D=1 and consequently two poles at sp12 = -ω2 However together with the zero the system is not aperiodic but features 13dB peking in the frequency domain and 14 step-response voltage overshoot in time-domain

(a) Bode diagram (b) step response

Photo 73 Matlab screen yellow ωn= ω2 PM=45deg red Butterworth poles ωn= ω2 2D blue ωn= ω22 poles of aperiodic borderline case

Listing 73 Matlab code generating Fig 73

Filter 2 poles 1 zero Hs_M45=tf([11 1][1 1 1]) Hs_BW2=tf([1sqrt(05) 1][1 sqrt(2) 1]) Hs_AP=tf([105 1][1 2 1]) figure(1) step(Hs_APHs_BW2Hs_M45) grid on figure(2) h=bodeplot(Hs_APHs_BW2Hs_M45) grid on setoptions(hFreqUnitsHzPhaseVisibleon)


- 25 -

74 RLC Lowpass as 2nd Order System

741 System Setup

Fig 741 (a) RLC lowpass (b) corresponding Bode diagram The RLC lowpass shown in Fig 741 is important for DCDC conversters It is good-natured with respect to both low and high frequencies but may show instabilities around f0 and within control loops RC is the equivalent series resistance (ESR) of capacitor C and RD is the wire resistance of the inductor in series with the voltage source output impedance In the two subsections below will compute LTI model of Fig 741

1 In the polynomial coefficients of 2

0 1 22

0 1 2

a a s a s

b b s b s

2 as standard 2nd order model 2

0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s

Both has to be done for process transfer function PTF (or short HP) coputing output voltage Uout as function of input voltage Uin

0

( )L

out

in I

UPTF s

U

(741)

and inference (quarrel) transfer function respectings the impact of inference IL on Uout as

0

( )in

out

L U

UQTF s

I

(742)

so that

( ) ( ) ( ) ( ) ( )out in LU s PTF s U s QTF s I s (743)


- 26 -

According to the appendix we get

( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s

R G R R R R G C G L s R G LC s

(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)

742 Modelling the RLC Lowpass with 2nd Order Polynomials Ansatz

Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)

0 1pa 1p Ca C R 2 0pa

0 1p D Lb R G 1 ( )p C D D C L Lb R R R R G C G L 2 (1 )p C Lb R G LC

Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)

0q Da R 1 ( )q C Da R R C L 2q Ca R LC

0 0q pb b 1 1q pb b 2 2q pb b

743 Modelling the RLC Lowpass as Standard 2nd Order System

Process transfer function in the standard 2nd order system ansatz

0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)

with a zero in 1n and poles ndash depending on |D|gt1 or lt1 ndash in

2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)

D LD D D 0C D LR R G 2

D CR R C

L

+

2LG L

C (7417)

Inference (quarrel) transfer function modelling the impact of inference IL on Uout with ansatz

0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)

has the same denominator and ωn1 as PTF(s) but a second zero at

2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -

75 PID Controller as 2nd order System

Fig 75 (a) PID controller and (b) its Bode diagram amplifiy both low and high frequencies General bi-quadratic polynomial dexcription is

2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid

751 Controller Transfer Function Defined by KP KI KD ωp2

KP KI KD Controller engineers model PID controllers typically as

( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)

with KP KI KD being proportional integral and derivative control parameters Additional Pole ωp1 In practical applications infinite integrator amplification is often not possible so the we have an unintentional pole ωp1 in the integral part like KI(s+ωp1) limiting its maximum amplification to KI ꞏωp2 ωp1 as detailed in (754) Additional Pole ωp2 For control stability we frequently need also a pole for the derivative part so that KDꞏs becomes

2

221 1

pD D

pp

sK K

ss

(752)

The total controller transfer function evaluates with ωp1=0 to

22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p

s K K K s K K sKCTF s K K

s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers

0 2c I pa K 1 2c I P pa K K 2 2c P D pa K K (753)

0 0cb 1 2c pb 2 1pb

DC amplification according to (753) is CTF(s=0) = ac0bc0 rarr infin In reality the maximum controller amplification is limited eg by the DC amplification of an operational amplifier Parameter bc0 can be used to model a realistic DC amplification by bc0 = ac0 CTF(s=0) = KI ꞏωp2 ωp1 (754) Table 752 Matlabrsquos designations and symbols for controller parameters

Matlab parameter designations Matlab symbol Authorrsquos symbol Proportional controller parameter KP KP Integral controller parameter I KI Derivative controller parameter D KD Filter coefficient N ωp2

752 Controller Transfer Function Defined by KP ωn1 ωn2 ωp1 ωp2

Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -

753 Translation to Time-Discrete Transfer Function

Translation to the time-discrete domain yields

221

2210

1)(

zczc

zdzddzH PID

(759)

Listing 753 Function f_c2d_bilin translating s rarr z using bi-linear substitution acc to chap 5

function Hz = f_c2d_bilin_order2(HsT) fs2=2T a0=Hs(11) a1=Hs(12) a2=Hs(13) if size(Hs1)==1 b0=1 b1=0 b2=0 else b0=Hs(21) b1=Hs(22) b2=Hs(23) end as0=a0 as1=a1fs2 as2=a2fs2fs2 bs0=b0 bs1=b1fs2 bs2=b2fs2fs2 cs0=bs0+bs1+bs2 cs1=2(bs0-bs2) cs2=bs0-bs1+bs2 ds0=as0+as1+as2 ds1=2(as0-as2) ds2=as0-as1+as2 Hz = [[ds0 ds1 ds2][cs0 cs1 cs2]] if cs0~=1 Hz=Hzcs0 end 754 Exercise

How does Matlabrsquos filter coefficient N in the PID controller equation depend on the PID parametes detailed above -------------------------------------------- Solution exercise 75 N = ωp2


- 31 -

76 Time-Continuous Filter using Integrotors

761 Linearization of a Time-Continuous Integrator Offset Removal

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy

Fig 761 (a) Real Situation (b) circuit linearized (c) equivalent signal-flow model First of all we have to linearize the circuit Assume a 3-input integrator as shown in part (a) of the figure above Its behavior is described by

xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)

where the virtual ground voltage Uvg is given by VoutoffBvg AUUUU (762)

with AV being the OpAmprsquos amplification To linearize (761) we have to remove the constant term ie UB-Uoff The amplifierrsquos offset voltage Uoff is either compensated for or neglected assuming Uoff0V To remove UB we remember that any voltage in the circuit can be defined to be reference potential ie 0V and define U = U ndash UB (763) U = U + UB (764) This yields the circuit in Fig part (b) which is linear in the sense of (21) Things are facilitated assuming also AVrarr so that Uvg=UB so that Uvg=0V Then (761) facilitates to

xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)

It is always possible to factor out coefficients such that the weight of one input is one eg

sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -

762 Application to a Time-Continuous 1st-Order System

R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)

Fig 762 (a) Real Circuit (b) general model (c) particular model for circuit in (a) Fig 762 (a) shows the analog circuit under consideration Fig part (b) illustrates the very general form of the topology To get forward network A set b1=E1=0 Identify all paths a signal can take from X to Y Assuming a linear system the sum of all those partial transfer function delivers the STF Open-loop network B is found by summing all paths from Y to Y

110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)

Adapting the general topology in part (b) to the particular model in Fig part (c) we can select either ak or bk free Letrsquos set ak=1 Here with order k=1 we get

11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)

Signal and noise transfer functions can then be computed from

11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -

763 Application to a Time-Continuous 2nd-Order System

F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)

Fig 763 (a) Real Circuit (b) general model (c) particular model for circuit in (a) Fig 763 (a) shows the analog circuit under consideration Fig part (b) illustrates the very general form of the topology To get forward network A set b1=b2=E1=0 Identify all paths a signal can take from X to Y Assuming a linear system the sum of all those partial transfer function delivers the STF Open-loop network B is found by summing all paths from Y to Y

2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)

The signal transfer function can then be computed from


- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)

We compare this result to the general 2nd-order model 200

2

200

2 2

sDs

AH ndOrder (7619)

with DC amplification A0 cutoff frequency ω0 and damping constant D (7618)=(7619) delivers

ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2

Comparing coefficients of s delivers for DC amplification cutoff frequency damping factor

2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)

The noise transfer function is computed from

212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)

8 Conclusions A feedback loop model valid for time-continuous and time-discrete domain modeling was derived discussed and applied to different circuit topologies Translation methods from s rarr z domain were introduced stability issues addressed At the several practical applications are presented

9 References [1] Matlab available httpsdemathworkscom


- 35 -

10 Appendices 101

102

103


- 36 -


Fig 741 (a) RLC circuit (b) corresponding Bode diagram The RLC lowpass shown in Fig 741 is important for DCDC conversters It is good-natured with respect to both low and high frequencies but may show instabilities in its oscillation frequency and within control loops RC is the equivalent series resistance (ESR) of capacitor C and RD is the wire resistance of the inductor in series with the voltage source output impedance In the two subsections below will compute LTI model of Fig 741


0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that

( ) ( ) ( ) ( ) ( )out in LU s PTF s U s QTF s I s


- 37 -

1041 Computing the Laplace Transform of the Oscillator

We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z

sCRG G Z sC G sCG R G sC G RGsC

Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR

sCR R G sCR G R sLG s LC G R

2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C

R G sR C sR C R G sLG s LC R G

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


The impact of load current IL is modelled easily as QTF = STFZD

out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -

1042 Modelling the Oscillator with 2nd Order Polynomials

Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s

for the process transfer function delivers

0 1pa 1p Ca R C 2 0pa


Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s

For the inference (quarrel) transfer function delivers

0osc Wa R 1osc C Da CR R L 2osc Ca R LC


1043 Modelling the Oscillator as 2nd Order System

Process transfer function modelling the impact of input voltage Uin on Uout

0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

with a zero in 1n and poles in

212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G

G L R R R R G C R Gs LC s

R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G

R R R R G C G L R G R GLC s LC s

R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G

R R R R G C GC L s sL CR G R G R G R G

Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C

To obtain the quarrel transfer function wa correct DC amplification and add a second zero

0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s

with second pole in 2 n DR L and DC impedance 0 0DR R A


- 2 -


Abstract Linear feedback loop models are introduced

1 Introduction

Fig 1 (a) basic model of linear feedback loop (b) operational amplifier application example The feedback loop model in Fig 1(a) with F and B being linear and time-invariant (LTI) models in s or z features signal transfer function (STF) and noise transfer function (NTF)

1

0 1FB

E

Y FSTF B

X FB

(11)

0

10

1FB

X

YNTF

E FB

(12)

To derive STF according to (11) set E=0 (corresponding to a short circuit) so that

( ) ( )Y F F X V F X BY FX FBY rarr (1 )Y FB FX 1

Y F

X FB

Exercise To derive (12) do the same derivation with input signal E and X=0 Y F E Exercise for electronics engineers (i) Relate quantities of Fig 1(a) and (b) Assume that E results from non-linearity in the operational amplifier and noise feed-through from the power supply (ii) Which errors can and which cannot be suppressed by high loop gain |FB| rarr infin X corresponds to helliphellip ε corresponds to hellip Y corresponds to hellip V corresponds to helliphellip F corresponds to hellip B corresponds to hellip


- 3 -





- 4 -


211 Linearity Axiom


H

c1

c2

x1(t)

x2(t)

y(t)

(a) (b)

x1(t)

x2(t)

c1

c2

y(t)H

H



Hcx(t) y(t)

(a) (b)

cx(t) y(t)H




- 5 -

R1

R2

Uoff

R1 R2(a) (b)

U1

U2

U2U1



UctrlUin Uout

CvLCk1 Ck2






- 6 -



1 1

F QY X I

FB FB

= STF X NTF I (25)


1

0 1FB

E

Y FSTF B

X FB

(26)

1

0 1FB

X

Y QNTF B

I FB

(27)



- 7 -




Forward network 1

11

zF

z





1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF


1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 3 -





- 4 -


211 Linearity Axiom


H

c1

c2

x1(t)

x2(t)

y(t)

(a) (b)

x1(t)

x2(t)

c1

c2

y(t)H

H



Hcx(t) y(t)

(a) (b)

cx(t) y(t)H




- 5 -

R1

R2

Uoff

R1 R2(a) (b)

U1

U2

U2U1



UctrlUin Uout

CvLCk1 Ck2






- 6 -



1 1

F QY X I

FB FB

= STF X NTF I (25)


1

0 1FB

E

Y FSTF B

X FB

(26)

1

0 1FB

X

Y QNTF B

I FB

(27)



- 7 -




Forward network 1

11

zF

z





1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF


1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 4 -


211 Linearity Axiom


H

c1

c2

x1(t)

x2(t)

y(t)

(a) (b)

x1(t)

x2(t)

c1

c2

y(t)H

H



Hcx(t) y(t)

(a) (b)

cx(t) y(t)H




- 5 -

R1

R2

Uoff

R1 R2(a) (b)

U1

U2

U2U1



UctrlUin Uout

CvLCk1 Ck2






- 6 -



1 1

F QY X I

FB FB

= STF X NTF I (25)


1

0 1FB

E

Y FSTF B

X FB

(26)

1

0 1FB

X

Y QNTF B

I FB

(27)



- 7 -




Forward network 1

11

zF

z





1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF


1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 5 -

R1

R2

Uoff

R1 R2(a) (b)

U1

U2

U2U1



UctrlUin Uout

CvLCk1 Ck2






- 6 -



1 1

F QY X I

FB FB

= STF X NTF I (25)


1

0 1FB

E

Y FSTF B

X FB

(26)

1

0 1FB

X

Y QNTF B

I FB

(27)



- 7 -




Forward network 1

11

zF

z





1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF


1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 6 -



1 1

F QY X I

FB FB

= STF X NTF I (25)


1

0 1FB

E

Y FSTF B

X FB

(26)

1

0 1FB

X

Y QNTF B

I FB

(27)



- 7 -




Forward network 1

11

zF

z





1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF


1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 7 -




Forward network 1

11

zF

z





1

11

1

1

1

1 11

zzSTF zz

z

rarr 1STF


1 121

1

1 21 sin

1 11

s

NTF z z fz f

z

rarr sin 2NTF F



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 8 -





311 Derivation

Fig 311



0 1 1 2 1 2 1 20 0

1 R R

L E


X b

(311)


1 1 2 1 2 1 20 0

1 R R

X E


L b

(312)


0 1 1 2 1 2 1 2

0 0 1 1 2 1 2 1 2

1 R R

E R R



(313)

0

0 0 1 1 2 1 2 1 2

1

1 X R R

bYNTF


(314)



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 9 -





0 1 11

1 1 1R

R

d G d GSTF

b G G

(315)

1

1NTF

G

(316)



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 10 -


321 Derivation


0 0

1

L E

WG

X b

(321)

1 1 2 1 2 1 20 0

1 R R

X E


L b

(322)


10 0 1 1 2 1 2 1 2

1

1 E R R

W GSTF


(323)


- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 11 -

1 0 10

1

1X

WNTF b STF

E FB

(324)


0 1 1 2 1 2 1 2 R R


W (325)


0 1 1 2 1 2 1 21

0 0 1 1 2 1 2 1 2

1 R R

E R R



(326)

2 1 00E

WNTF NTF F b STF

X

(327)

322 Discussion






- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 12 -



( ) ( )0 1 0 1( ) ( ) ( ) ( ) ( ) ( )R R


1 ( ) ( )





0 1

( )( )

( )

RR

RR

a a s a sY sTF s

X s b b s b s

(414)



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 13 -









0 1 0 11 1

0 1 0 1

( )

R R RR R

R R RR R


c c z c z c z c z c

(414)

0 1

0 1

0 1

R

R

d d dTF F TF

c c c

0 1 2 3

0 1 2 3

05

d d d dTF F

c c c c

0

00

dh

c (415)


- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 14 -


sTz e 1

ln( )s zT

(501)



1( ) n nx xx t

T


s X XT T


1

11(1 )s

zs f z

T

(511)





20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

2 1 10 1 2

2 1 10 1 2

1 1( )

1 1

a a z a zH z

b b z b z

rarr

1 20 1 2

1 20 1 2

z

d d z d zH

c c z c z

kk k sa a f k

k k sb b f k=0 1 2

0 0 1 2xd a a a

1 1 22xd a a 2 2xd a

0 0 1 2xc b b b

1 1 22xc b b 2 2xc b

0x xk kd d c 0x x

k kc c c k=0 1 2



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 15 -



1

2 1

1

zs

T z

(521)

Using

2

T

2

Ts j s (522)


1

1

1

zs

z

1

1

1

sz

s

(523)


1

11

1

jz

j

(524)


arctan2z s

T

T

(525)






20 1 2

20 1 2

( )s

a a s a sH s

b b s b s

rarr

21 1 0 1 21 1

21 1

0 1 21 1

1 11 1

( )1 11 1

z za a a

z zH z

z zb b b

z z

rarr 1 2

0 1 21 2

0 1 2z

d d z d zH

c c z c z

(2 )kk k sa a f (2 )k

k k sb b f k=0 1 2


0 0 1xd a a

1 0 1xd a a |

0 0 1 2xd a a a

1 0 22 2xd a a 2 0 1 2xd a a a

0 0 1xc b b

1 0 1xc b b |

0 0 1 2xc b b b

1 0 22 2xc b b 2 0 1 2xc b b b

0x xk kd d c 0x x

k kc c c k=0 1 2


- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 16 -




Example


0

00)( tife



a

zb

zazH d

1

1




yn

Y(z)a

b

z-1xn

X(z)

tn=nT

h(n)

h(t)

(a) (b)

0




- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 17 -





kk

y t C e (622)

Time-discrete p

k

s nT nn k k p

k k

y C e C z (623)



or )sin( te ptp

because






- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 18 -





200

2

200

20

212)(

sDs

A

DSS

AsH LP (711)


1

11

111

202021 DjD

DDs p (713)


- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 19 -



N

js

NBW

ssH

2

0

2

0

1

1

1

1)(

NBW

f

ffH

2

0

1

1)(

(714)


Lowpass Type



Φp pj






(a) Bode diagram




- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 20 -



21

1)(

sLCsRCsH LP (715)



f2

10 and

L

CRD

2

(a) RLC lowpass



(b) Bode diagram

(c) Step responses




- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 21 -


11FB

s

rarr 1

1( )FB jj

rarr 11( )

fFB f

f (721)


01 1 1 11 1 1 1

1 1 1 1

( )

1 1 1 sF FB s


(722)


1112 ( )

(1 ) ( )p

p p

FB ss s s s

(723)

1 011 1 112 122

12 12 1

( )1 1

sp

p p

F FBSTF s B B B

FB FB s s

(724)


150 pD (726)

120 p (727)



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 22 -




- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 23 -


2

22FB

s

rarr 22

2 2( )FB j

rarr 2

22 2( )

fFB f

f (731)


2 2 201 1 1 12 2 2 2

2 2 2 22 2 2 2

( )

1 1 1 sF FB s


(732)


2 22 2

21 2 2( ) (1 ) n

nn

sFB s s

s s

(733)

1

2 201 1 12 221 21

2 2 221 21 2 2

( )( )

1 1 ( )sn

n

sF FBSTF s B B B

FB FB s s

(734)


n

D2

2 (736)

20 (737)



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 24 -








- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 25 -


741 System Setup



0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U

(741)


0

( )in

out

L U

UQTF s

I

(742)

so that



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 26 -


( ) out

in

UPTF s

U

2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s


(744)

out

L

UQTF

I 2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R


(745)


Ansatz

0

( )L

out

in I

UPTF s

U

2

0 1 2

20 1 2

p p p

p p p

a a s a s

b b s b s

(746)

Delivers (747)



Ansatz

0

( )in

out

L U

UQTF s

I

20 1 2

20 1 2

q q q

q q q

a a s a s

b b s b s

(748)

Delivers (749)





0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s

(7410)


2 212 0 01 1ps D D D j D (7411)


- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 27 -

delivers

0

1

1 D L

AR G

0D LR G 1 (7412)

1

1n

CR C (7413)

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC (7414)

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

(7415)

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C (7416)


D CR R C

L

+

2LG L

C (7417)


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2D n nR A s s

s D s

(7418)


2 n DR L (7419)

and DC impedance

0 0 DR A R (7420)

It is seen that

2( ) (1 ) ( )D nQTF s R s PTF s (7421)


- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 28 -



2210

2210)(

sbsbb

sasaasH

pidpidpid

pidpidpidspid

rarr 2

21

1

22

110

1)(

zczc

zdzddzH

pidpid

pidpidpidzpid



( )

( )

C sQTF

E s

2I I P I

P D

K K K s K sK K s

s s

(751)


2

221 1

pD D

pp

sK K

ss

(752)


22 2 2 2

22 2

( ) ( )( )

0p I p I P p P D pI

P Dp p


s s s s


- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 29 -

Ansatz 2

0 1 22

0 1 2

( ) c c c

c c c

a a s a sQTF s

b b s b s

delivers


0 0cb 1 2c pb 2 1pb




Ansatz

2

2121

21

2

2121

21

21

21

1

1

))((

))(()(

ss

ss

Kpss

ssconstsH

pppp

pp

nnnn

nn

pp

nnPID

(755)

Comparing it to

2210

2210)(sbsbb

sasaasH PID

(756)

delivers

21

210

nn

nnpKa

pKa 1 21

2nn

pKa

(757)

21

210

pp

ppb

11 b 21

2

1

pp

b

(758)


- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 30 -



221

2210

1)(

zczc

zdzddzH PID

(759)





- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 31 -



Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux

Uvg

UB

Uy

0V

(c)

-xs

ax

bx

Rx

Cx

OAx

Rb

Ra

(a)Ua

Ub

Ux Uy

0VUa

Ub

Ux Uy


xb

vgb

xx

vgx

xa

vgavgout CsR

UU

CsR

UU

CsR

UUUU

(761)



xb

b

xx

x

xa

aout CsR

U

CsR

U

CsR

UU

(765)


sbUUaUU x

bxaout

(767) with kx

x CR

1 (767)

a

x

R

Ra (768) and

b

x

R

Rb (769)


- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 32 -


R1

C1

OA1

Rb1

UoutUin

0V

Uerr1

Uout1

F1

a0

Y

E1a1

b1

X(a) (b)

1

1s

Uin

Uout

E1

b1

(c)


110011

FaaX

YA

bE

(7610) 110

FbY

YB

X

(7611)


11 a 00 a 1

11

bR

Rb

111

1

CR

sF 1

1

(7612)


11

1

11

1

1

1

11

110

)(1

)(

111

bssb

s

F

F

Fb

Faa

B

ASTF

Hence 1

1

1

0

111

1

11

1 1

1

1

R

R

bCsRR

R

bsSTF bs

b

b

(7613)

)(1

1

1

1

1

1

1111 sbFbBNTF

=gt 0

10

11

11

11

s

b

b

CsR

CsR

bs

sNTF

(7614)


- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 33 -


F2 F1

a0

Y

E1a2

-b2

a1

b1

X

Y

E1

-b2

1

b1

X

1s

2s

R2Uin

C2 R1C1

OP2 OP1

Rb1Rb2

Uout

Uerr1

-1

Uerr2

0V 0V

Uout2 Uout1

(a)

(b)(b) (c)


2121100211

FFaFaaX

YA

bbE

(7615) 212110

FFbFbY

YB

X

(7617)


12 a 001 aa 2

22

bR

Rb

1

11

bR

Rb

222

1

CR

111

1

CR

sF 2

2

sF 1

1

(7617)



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 34 -

ssb

sb

ssFFbFb

FF

FFbFb

FFaFaa

B

ASTF

212

11

21

12211

12

12211

122110

11)(11

=gt

212112

21

bsbsSTF

(7618)


2

200

2 2

sDs

AH ndOrder (7619)


ndOrderHsDs

A

bsbsSTF 22

002

200

212112

21

2


2

2

20

1

R

R

bA b (30)

2211

2120

1

CRCRb

b

(31)1

2

1

21

0

11

22 C

C

R

RRbD

b

b

(7620)


212112

2

212

11

1221112211 1

1

1

1

)(1

1

1

1

bsbs

s

ssb

sbFFbFbFFbFbB

NTF

Hence 20200

2

2

212112

2

02

s

sDs

s

bsbs

sNTF

(7621)




- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 35 -

10 Appendices 101

102

103


- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 36 -




0 1 22

0 1 2

a a s a s

b b s b s


0 0 1 22 2

0 0

(1 )(1 )

2n nA s s

s D s


0

( )L

out

in I

UPTF s

U


0

( )in

out

L U

UQTF s

I

so that



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 37 -


We define

D DZ R sL

11 C

C C

sCRZ R

sC sC

1

1 11||

11 (1 )1

C

C C CCL C

CL L C L L C L L CL

sCRZ sCR sCRsCZ Z


Then

0L

out CL

in D CLI

U ZPTF

U Z Z

0

||in

outD CL

out U

UQTF Z Z

I

Consequently

( ) out

in

UPTF s

U CL

CL D

Z

Z Z

1

(1 )1

( )(1 )

C

L L C

CD

L L C

sCR

G sC G RsCR

R sLG sC G R

1

1 ( ) (1 )C

C D L L C

sCR

sCR R sL G sC G R

2

1

1 (1 ) (1 )C

C D L D L C L L C

sCR


2

1

1 (1 ) (1 )C

D L C D C L L C L

sR C


2

1

1 ( ) (1 )C

D L C D D C L L C L

CR s



out

L

UQTF

I

CP D

CP D

Z Z

Z Z

2

(1 )(1 )

1 ( ) (1 )D C D

D L C D D C L L C L

R sCR sL R



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 38 -


Ansatz

20 1 2

20 1 2

( )( )

( )p p pout

in p p p

a a s a sU sPTF s

U s b b s b s




Ansatz

20 1 2

20 1 2

( ) ( ) q q qosc

q q q

a a s a sQTF s Z s

b b s b s






0

( )L

out

in I

UPTF s

U

2

0 0 12 2

0 0

(1 )

2nA s

s D s


212 0 1ps D j D

delivers

0

1

1 D L

AR G

0D LR G 1

1

1n

CR C


- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s



- 39 -

0L

out

in I

UPTF

U

2

11

1( ) 1

11 1

CD L

L C D D C L C L

D L D L

sCRR G


R G R G

2

11

1

( ) 1 11

1 1(1 )(1 )

CW L

C D D C L L C L C L

D L D LD L C L

sCRR G


R G R GR G R G

2

0 0

11

1

)1 2

2 (1 )(1 ) 2 (1 )(1 )

CD L

C D D C L L

D L C L D L C L

sCRR G


Comparison with

0 1 0 12 2

0 0 0 0

1 1

1 2 1 2

n n

C L

A s A sPTF

s s s sD D D

delivers

0

0

11

1W LR G

W L

AR G

0W LR G 1

1

1n

CR C

0

1 1

1D L

C L

R G

R G LC

0C D LR R G 1

LC

2 (1 )(1 )D C D C L

D

D L C L

R R R R G CD

LR G R G

0C D LR R G

2D CR R C

L

2 (1 )(1 )L

L

D L C L

G LD

CR G R G

0C D LR R G

2LG L

C


D CR R C

L

+

2LG L

C


0

( )in

out

L U

UQTF s

I

2

0 0 1 22 2

0 0

(1 )(1 )

2n nR s s

s D s


Linear Feedback Loops - hps.hs-regensburg.de › scm39115 › homepage › ... · Martin J.W....

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Transcript of Linear Feedback Loops - hps.hs-regensburg.de › scm39115 › homepage › ... · Martin J.W....