8-1

��������� 8.1

a) For step response,

input is Mtu =′ )( , s

MsU =′ )(

)(1

1)( sU

s

sKsY

D

Dca ′

+ατ

+τ=′ =

+ατ

+τ)1(

1

ss

sMK

D

Dc

=′ )(sYa )1(1 +ατ+

+αττ

ss

MK

s

MK

D

c

D

Dc

Taking inverse Laplace transform

)1()( )/()/( DD tc

tca eMKe

MKty ατ−ατ− −+

α=′

As α →0

MKdte

tMKty c

t

t

ca

D

+α

δ=′ ∫∞

=

ατ−

0

)/(

)()(

MKtMKty cDca +τδ=′ )()(

Ideal response,

)()()( sUsGsY ii ′=′ =

+τ

s

sMK D

c

1 = KcMτD +

s

MKc

MKtMKty cDci +δτ=′ )()(

Hence )()( tyty ia ′→′ as 0→α

For ramp response,

input is Mttu =′ )( , 2

)(s

MsU =′

Solution Manual for Process Dynamics and Control, 2nd edition,

Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.

8-2

)(1

1)( sU

s

sKsY

D

Dca ′

+ατ

+τ=′ =

+ατ

+τ)1(

12 ss

sMK

D

Dc

=′ )(sYa )1()1( 2 +ατ+

+αττ

ss

MK

ss

MK

D

c

D

Dc

+ατ

ατ++

ατ−+

+ατ

ατ−τ=

1

)(1

1

1 2

2 sssMK

ssMK

D

DDc

D

DDc

Taking inverse Laplace transform [ ] [ ])1(1)( )/()/( −ατ++−τ=′ ατ−ατ− DD t

Dct

Dca etMKeMKty

As α →0 MtKMKty cDca +τ=′ )(

Ideal response,

=′ )(sYi

+τ

2

1

s

sMK D

c = 2s

MK

s

MK cDc +τ

MtKMKty cDci +=′ τ)(

Hence )()( tyty ia ′→′ as 0→α

b) It may be difficult to obtain an accurate estimate of the derivative for use

in the ideal transfer function.

c) Yes. The ideal transfer function amplifies the noise in the measurement by taking its derivative. The approximate transfer function reduces this amplification by filtering the measurement.

8.2

a) 11)(

)(

1

21212

1

1

+τ+τ+

=++τ

=′

s

KsKKK

s

K

sE

sP

+τ

++τ

+=1

1

)(1

21

12

21 s

sKK

K

KK

8-3

b) Kc = K1 + K2 → K2 = Kc − K1

Dατ=τ1

21

2

21

12

KK

K

KK

K DD +

ατ=+τ=τ

or 21

21KK

K

+α=

α=+ 221 KKK )1(2221 −α=−α= KKKK Substituting, 111 )1()1()1)(( KKKKK cc −α−−α=−α−=

Then,

cKK

α−α= 1

1

c) If Kc = 3 , τD = 2 , α = 0.1 then,

2731.0

9.01 −=×−=K

30)27(32 =−−=K τ1 = 0.1 × 2 = 0.2

Hence

K1 + K2 = -27 + 30 = 3

23

2.030

21

12 =×=+τKK

K

++=

12.0

123)(

s

ssGc

8-4

8.3

a) From Eq. 8-14, the parallel form of the PID controller is :

τ′+

τ′+′= s

sKsG D

Ici

11)(

From Eq. 8-15, for α →0, the series form of the PID controller is:

[ ]11

1)( +τ

τ

+= ss

KsG DI

ca

τ+

τ+

ττ

+= ss

K DII

Dc

11

ττ

+

τ+

τ

ττ

++

ττ

+=

I

D

D

II

DI

Dc

s

s

K

11

111

Comparing Ga(s) with Gi(s)

ττ

+=′I

Dcc KK 1

ττ

+τ=τ′I

DII 1

I

D

DD

ττ

+

τ=τ′

1

b) Since

+

I

D

ττ

1 ≥ 1 for all τD, τI, therefore

cc KK ′≤ , II τ′≤τ and DD τ′≥τ

c) For Kc = 4, τI=10 min , τD =2 min

8.4=′cK , min12=τ′I , min67.1=τ′D

d) Considering only first-order effects, a non-zero α will dampen all responses, making them slower.

8-5

8.4 Note that parts a), d), and e) require material from Chapter 9 to work.

a) System I (air-to-open valve) : Kv is positive.

System II (air-to-close valve) : Kv is negative.

b) System I : Flowrate too high → need to close valve →decrease controller output → reverse acting

System II: Flow rate too high →need to close valve → increase controller output → direct acting.

c) System I : Kc is positive

System II : Kc is negative d) Kc Kv Kp Km

System I : + + + + System II : − − + + Kc and Kv must have same signs

e) Any negative gain must have a counterpart that "cancels" its effect. Thus,

the rule:

# of negative gains to have negative feedback = 0 , 2 or 4. # of negative gains to have positive feedback = 1 or 3.

8.5

a) From Eqs. 8-1 and 8-2,

[ ])()()( tytyKptp mspc −+= (1)

The liquid-level transmitter characteristic is ym(t) = KT h(t) (2) where h is the liquid level

KT > 0 is the gain of the direct acting transmitter.

8-6

The control-valve characteristic is q(t) = Kvp(t) (3) where q is the manipulated flow rate Kv is the gain of the control valve. From Eqs. 1, 2, and 3

[ ] [ ])()()()( thKtyKKptpKqtq TspcVv −=−=−

)(

)(

thKy

qtqKK

TspcV −

−=

For inflow manipulation configuration, q(t)> q when ysp(t)>KTh(t). Hence KvKc > 0 then for "air-to-open" valve (Kv>0), Kc>0 : reverse acting controller and for "air-to-close" valve (Kv<0), Kc<0 : direct acting controller For outflow manipulation configuration, KvKc <0 then for "air-to-open" valve, Kc<0 : direct acting controller and for "air-to-close" valve, Kc>0 : reverse acting controller

b) See part(a) above 8.6 For PI control

τ

++= ∫t

Ic dtteteKptp

0

**)(1

)()(

τ

+=′ ∫t

Ic dtteteKtp

0

**)(1

)()(

Since

e(t) = ysp – ym and ym= 2

8-7

Then e(t)= -2

τ

−−=

−

τ+−=′ ∫ tKdtKtp

Ic

t

Ic

22*)2(

12)(

0

Initial response = − 2 Kc

Slope of early response = I

cK

τ−

2

− 2 Kc = 6 → Kc = -3

I

cK

τ−

2= 1.2 min-1 → τI = 5 min

8.7

a) To include a process noise filter within a PI controller, it would be placed in the feedback path

b)

f

1

� ��s

τ

+s

KI

c

11

c) The TF between controller output )(sP′ and feedback signal Ym(s) would be

8-8

)1(

)1(

)(

)(

+ττ+τ−

=′

ss

sK

sY

sP

fI

Ic

m

Negative sign comes from comparator

For s

MsYm =)(

+τ

++τ

−=

+τ+τ

τ−

=′1)1(

1)(

22 s

C

s

B

s

AMK

ss

sMKsP

fI

c

f

I

I

c

The 1+τ s

C

f

term gives rise to an exponential.

To see the details of the response, we need to obtain B (= τI - τf) and A(=1) by partial fraction expansion. The response, shown for a negative change in Ym, would be

Note that as 0→ττ

I

f , the two responses become the same.

d) If the measured level signal is quite noisy, then these changes might still be large enough to cause the controller output to jump around even after filtering.

One way to make the digital filter more effective is to filter the process output at a higher sampling rate (e.g., 0.1 sec) while implementing the controller algorithm at the slower rate (e.g., 1 sec). A well-designed digital computer system will do this, thus eliminating the need for analog (continuous) filtering.

time

y

-KcM

-KcM(1-τf/τI)

"Ideal" PI

Filtered PI

Slope = KcM/τI

8-9

8.8

a) From inspection of Eq. 8-26, the derivative kick = rt

K Dc ∆

∆τ

b) Proportional kick = rK c∆

c) e1 = e2 = e3 = …. = ek-2 = ek-1 = 0

ek = ek+1 = ek+2 = …= ∆r

ppk =−1

∆

∆τ

+∆τ∆+∆+= r

tr

trKpp D

Ick

∆

τ∆++∆+=+ r

tirKpp

Icik )1( , i = 1, 2, …

c) To eliminate derivative kick, replace (ek – ek-1) in Eq. 8-26 by (yk-yk-1).

k k+1 k+2 k+3

rt

K Dc ∆

∆τ

rKc∆ rt

KI

c ∆τ∆

kp

p

k-1

8-10

8.9

a) The digital velocity P algorithm is obtained by setting 1/τI = τD = 0 in Eq. 8-28 as

∆pk = Kc(ek – ek-1)

= ( )[ ]1)( −−−− kspkspc yyyyK

= [ ]kkc yyK −−1

The digital velocity PD algorithm is obtained by setting 1/τI = 0 in Eq. 8-28 as

∆pk = Kc [(ek – ek-1) + t

D

∆τ

(ek – 2ek-1 + ek-2)]

= Kc [ (-yk + yk-1) + t

D

∆τ

(-yk – 2yk-1 + yk-2) ]

In both cases, ∆pk does not depend on spy .

b) For both these algorithms ∆pk = 0 if yk-2 = yk-1 = yk. Hence steady state is

reached with a value of y that is independent of the value of spy . Use of

these algorithms is inadvisable if offset is a concern.

c) If the integral mode is present, then ∆pk contains the term Kc )( kspI

yyt −

τ∆

.

Thus, at steady state when ∆pk = 0 and yk-2 = yk-1 = yk , yk = spy and the

offset problem is eliminated. 8.10

a)

+ατ

τ+

τ+=

′1

11

)(

)(

s

s

sK

sE

sP

D

D

Ic

( )( 1) 1

( 1)I D D D I

cI D

s s s s sK

s s

τ ατ + + ατ + + τ τ=

τ ατ +

+αττ

ττα++ατ+τ+=

)1(

)1()(1 2

ss

ssK

DI

DIDIc

8-11

Cross- multiplying

( ) )()1()(1)()( 22 sEssKsPss DIDIcIDI ττα++ατ+τ+=′τ+τατ

ατ+τ+=

′τ+

′τατ

dt

tdeteK

dt

tpd

dt

tpdDIcIDI

)()()(

)()(2

2

ττα++

2

2 )()1(

dt

tedDI

b)

+ατ

τ

τ

+τ=

′1

1

)(

)(

s

s

s

sK

sE

sP

D

D

I

Ic

Cross-multiplying

( ) )()1)(1()()1(2 sEssKsPss DIcDI +τ+τ=′+αττ

τ+τ+=

′τ+

′τατ

dt

tdeteK

dt

tpd

dt

tpdDIcIDI

)()()(

)()(2

2

ττ+

2

2 )(

dt

tedDI

c) We need to choose parameters in order to simulate:

e.g., 2=cK , 3=τ I , 5.0=τ D , 1.0=α , M = 1

By using Simulink-MATLAB

Step Response

Time 0 2 4 6 8 10

2

4

6

8

10

12

14

16

18

20

22Parallel PID with a derivative filterSeries PID with a derivative filter

p'(t)

Figure S8.10. Step responses for both parallel and series PID controllers with derivative filter.

8-12

8.11

a) ( )11

)(

)( +τ

τ

+τ=

′s

s

sK

sE

sPD

I

Ic

( ) )()1)(1()( sEssKsPs DIcI +τ+τ=′τ

( ) ( )

( ) ( )cI D

I

Kdp t de te t

dt dt

′ = + τ + ττ

ττ+

2

2 )(

dt

tedDI

b) With the derivative mode active, an impulse response will occur at t = 0. Afterwards, for a unit step change in e(t), the response will be a ramp with slope = ( ) /c I D IK τ + τ τ and intercept = /c IK τ for 0>t .

t

p '

slope =

Impulse at t=0

c

I

K

τ

( )c I D

I

K τ + ττ