07 Dft Fft Release

8/13/2019 07 Dft Fft Release

1/55

Curtin University, Australia 7-1YH Leung (2005, 2006, 2007, 2008, 2012)

THE DFT AND FFT

1 The Discrete Fourier Transform (DFT)

Recall discrete-time Fourier transform DTFT of a sequence [ ]x n defined by

q q

-

=-

= ( ) [ ]j jnn

X e x n e (1)

In practice, record length of [ ]x n usually finite. Suppose [ ]x n has length N,

and with no loss of generality, suppose = -0, , 1n N . It then follows

q q-

-

=

= 1

0

( ) [ ]N

j jn

n

X e x n e (2)

Next, suppose we know q( )jX e but not [ ]x n . It can be shown, since [ ]x n

has length N, it is sufficientto know only q( )jX e , at Ndistinct frequency points

q q -0 1, , N in the interval [ )0, 2q p , to determine [ ]x n1

Discrete Fourier transform(DFT) defined by choosing q p= 2k k N, i.e.,

12

0

[ ] [ ] , 0, , 1N

j nk N

n

X k x n e k Np-

-

=

= = - (3)

Inverse DFT(IDFT) given by

12

0

1[ ] [ ] , 0, , 1

Nj nk N

k

x n X k e n NN

p-

+

=

= = - (4)

DFT is simply DTFT of a length Nsequence, evaluated at

Nequally-spaced frequency points

1Basically, we need N equations to find N unknowns where the unknowns are [ ]x n . The resulting

system of equations is assured to be consistent if q q -0 1, , N are distinct


2/55


DFT is just another sequence of numbers. We put its argument in square

brackets

Appendix I shows DTFT q( )jX e can be recovered from DFT [ ]X k by

interpolation:

( )( )

2 1212 2

20 2

sin1( ) [ ]

sin

k NN kN jj N

N kk N

X e X k eN

pq p qq

q p

- - - - -

-=

= (5)

WARNINGA fundamental result in signal theory states a signal cannot be bandlimited

and time-limited at the same time

\ If ( )x t is time limited, then its sampling rate should be infinite. Alternatively,

if ( )x t is bandlimited, then its sample sequence must have infinite length

Great care must be exercised in using and interpreting the DFT

The DFT Eq. (3) can be written in vector-matrix notation as follows

2 2 2

2 2 2

2 2 2

2 ( 1)

2 4 2( 1)

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

1 1 1 1[0] [0]

1[1] [1]

[2] [2]1

[ 1] [ 1]1

N N N

N N N

N N N

j j j N

j j j N

j N j N j N N

X x

e e eX x

X xe e e

X N x Ne e e

p p p

p p p

p p p

- - - -

- - - -

- - - - - - -

= - -

(6)

Matrix above called DFT matrix

Above representation useful in analysis of signal processing systems which

involve on-line calculation of the DFT (e.g. filterbanks and OFDM)

Exercise. Express the IDFT Eq. (4) in vector-matrix notation. Corresponding

matrix is called IDFT matrix


3/55


Recall definition of DFT Eq. (3)

p-

-

=

= = - 1

2

0

[ ] [ ] , 0, , 1N

j nk N

n

X k x n e k N

Suppose [ ]x n is specified for , ( 1), , ( 1)o o on n n n N = + + - . Definition of

N-point DFT can be generalised as follows

p

+ --

=

= = - 1

2[ ] [ ] , 0, , 1o

o

n Nj nk N

n n

X k x n e k N (7)

But

p p ++ - -

- -

= == +

( )1 12 2

0[ ] [ ]

n m ko onkN N

o

n N Nj j

on n mx n e x n m e

p p p+

-- -

--

= =

= +1

21

0

2 2[ ] [ ]

o nk n ko nN

o

kN N

Nj j

on

n Nj

n n

e x n n ex n e

Therefore2

[ ] [ ]n koN

jX k e X k

p-= (8)

where { }1

2

0

[ ] DFT [0], , [ 1] [ ]nkN

Nj

n

X k x x N x n e p

--

=

= - = (9)

and = + = -[ ] [ ], 0, , 1ox n x n n n N (10)

It can be readily verified inverse of generalised DFT given by

{ }1

2

0

1[ ] [ ] [ ]

Nj nk N

ok

x n n x n X k eN

p-

+

=

+ = =

1 2 2

0

[ ] [1

[ ] , 0] , , 1n koN

N j jo

nk N

k

X k e e Nx n nx nN

n p p

-+

=

= = -

+ =

(11)

In view of Eq. (8), in the sequel we shall assume, with no loss of generality, that

all sequences start at 0n=


4/55


2 Properties of the DFT

DFT is a finite sum, therefore always converge

Since [ ]x n is finite-valued by assumption, [ ]X k also finite-valued

DFT of a sequence is unique, i.e., given a length Nsequence [ ]x n , there is

one and only one N-point DFT [ ]X k

Since DTFT is periodic, so [ ]X k also periodic, with period N

+ = [ ] [ ],X k mN X k m (12)

In practice, range of [ ]X k restricted to { } -0,1, , 1k N

Inverse DFT also periodic with period N. Define

p-

+

=

= 1

2

0

1[ ] [ ] ,

Nj nk N

k

x n X k e nN

(13)

Can readily verify that

+ = [ ] [ ],x n mN x n m (14)

and by uniqueness

= = - [ ] [ ], 0, , 1x n x n n N (15)

[ ]x n isperiodic extension of [ ]x n

Fig. 1 Periodic extension

0 N-1

0 N-1

[ ]x n

[ ]x n


5/55


Denote moda b byb

a such that

=[ ] [ ]N

x n x n (16)

Suppose [ ]x n and [ ]y n both have length N with DFTs [ ]X k and [ ]Y k

respectively

DFT

[ ] [ ]x n X k

DFT

[ ] [ ]y n Y k

Linearity + +DFT

[ ] [ ] [ ] [ ]ax n by n aX k bY k

Time shifting p-- DFT 2

[ ] [ ] oj kn N

o Nx n n X k e

Frequency shiftingp -

DFT2[ ] [ ]oj k n N o Nx n e X k k

Conjugation * * -DFT

[ ] [ ]N

x n X k

Time reversal * *- DFT

[ ] [ ]N

x n X k

Circular convolution-

=

- 1

DFT

0

[ ] [ ] [ ] [ ]N

Nm

x m y n m X k Y k

Multiplication-

=

-1

DFT

0

1[ ] [ ] [ ] [ ]

N

Nm

x n y n X m Y k mN

x n[ ] real *= -[ ] [ ]N

X k X k

x n[ ] real { } { }= -Re [ ] Re [ ]NX k X k

x n[ ] real { } { }= - -Im [ ] Im [ ]NX k X k

Parsevals relation- -

= =

= 1 1

2 2

0 0

1[ ] [ ]

N N

n k

x n X kN

Duality -DFT[ ] [ ]N

X n Nx k

Note, for -0 ( 1)n N , - = -Nn N n


6/55


3 Circular Convolution

Suppose [ ]x n and [ ]y n both have length N

The N-point circular convolution (or cyclicconvolution) of [ ]x n and [ ]y n is

defined by

1

0

[ ] [ ] [ ] [ ] [ ]NN

Nm

w n x n y n x m y n m-

=

= = - (17)

where N denotes N-point circular convolution

Example 1 { }3 01, 1, 0, 2 nx == - { } == 3 01, 0, 3, 2 ny

First note that: 4 4 4 4 4 4 4 4 4 4 45 4 3 2 1 0 1 2 3 4 5

3 0 1 2 3 0 1 2 3 0 1

- - - - -

Therefore

0n= 4 4 4 4 4

4

0 1 2 3

[ ] 1 1 0 2

[ ] [ 0 ] [0] 1 [ 1 ] [3] 2 [ 2 ] [2] 3 [ 3

[

] [1] 0

[ ] [ ] 1 2 0 0

0

0

0

]

1

m

x m

y m y y y y y y y y

x m y m

w

-

- = = - = = - = = = =

- - -

-

1n = 4 4 4 4 4

4

0 1 2 3

[ ] 1 1 0 2

[ ] [ 1 ] [1] 0 [ 0 ] [0] 1 [ 1 ] [3] 2 [ 2

[

]

1

[2] 3

[ ] [

1

] 0 1 0 6

]

1 5

m

x m

y m y y y y y y y y

x m y m

w

-

- = = = = - = = - = =

- -

2n= -

- = = = = = = - = =

-4 4 4 4 4

4

0 1 2 3

[ ] 1 1 0 2

[ ] [ 2 ] [2] 3 [ 1 ] [1] 0 [ 0 ] [0] 1 [ 1 ]

2

2 [3] 2

[ ] [ ] 3 0 0

]

4

[

2 7

m

x m

y

w

m y y y y y y y y

x m y m

3n= 4 4 4 4 4

4

0 1 2 3

[ ] 1 1 0 2

[ ] [ 3 ] [3] 2 [ 2 ] [2] 3 [ 1 ] [1] 0

[

[ 0 ] [0] 1

[ ] [ ] 2 3 0 2

3

1

3

3

]m

x m

y m y y y y y y y y

x

w

m y m

-

- = = = = = = = =

- -

4n= -

- = = = = = = = =

- - -4 4 4 4 4

4

0 1 2 3

[ ] 1 1 0 2

[ ] [ 4 ] [0] 1 [ 3 ] [3

4

4 ] 2 [ 2 ] [2] 3 [ 1 ] [1] 0

[ ] [ ] 14 0

[ ]

0 12

m

x m

y m y y y y y y y y

m y m

w

x

Observe circular shiftingof { }3

4 0[ ]

my n m

=- as nincreases


7/55


Example 1 shows [ ]w n is periodic with period N. Follows since, for any k

1 1

0 0

[ ] [ ] [ ( ) ] [ ] [ ] [ ]N N

N Nm m

w n kN x m y n kN m x m y n m w n- -

= =

+ = + - = - = (18)2

By convention, range of [ ] [ ] [ ]Nw n x n y n= is restricted to { }0,1, , 1n N -

Circular convolution has another interpretation

1

0

[ ] [ ] [ ] [ ] [ ] , 0, , 1NN

m

w n x n y n x m y n m n N -

=

= = - = - (19)

where [ ]y n is periodic extension of [ ]y n

Circular convolution equivalent to finding linear convolution of [ ]x n with periodic

extension of [ ]y n and then keeping only first Npoints

Fig. 2 Circular convolution as linear convolution with periodic extension

As can be seen, for any given n, [ ]N

y n m- and [ ]y n m- have same data

points in summation region (yellow), thus explaining equivalence

2 Alternatively, from circular convolution property of DFTs, since [ ]w n is inverse DFT of N-point DFT

[ ] [ ] [ ]W k X k Y k = , therefore, by Eq. (13), [ ]w n is periodic with period N

[ ]x m

[ ]y m

[ 1 ]N

y m-

[ ]y m

[1 ]y m-

[ ]N

y m- [ ]y m-

0n=

1n =

m

m

m

m

m

m

m

[ 2 ]N

y m- [2 ]y m-

2n=

m m

0

0 0

0

0

00

0

0 1N-

1N- 1N-

1N-

1N-

1N-

1N-1N-

1N-


8/55


We next derive direct relationship between linear and circular convolutions

Suppose [ ]x n and [ ]y n both have length N. Denote by [ ]w n and [ ]cw n theirlinear and N-point circular convolutions, respectively. Then

[ ] [ ]cr

w n w n rN

=-

= - (20)

Proof

Let ( )jX e q

and ( )jY e q

be DTFTs of [ ]x n and [ ]y n , and [ ]X k and [ ]Y k be their N-point DFTs

From properties of DTFTs, we have

( ) ( ) ( )j j jW e X e Y eq q q= (21)

Suppose we sample ( )jW e q

at N equally-spaced points from 0q= to 2p and denote this

N-point DFT by [ ]oW k

Clearly,

2 2 2[ ] ( ) ( ) ( ) [ ] [ ]j k N j k N j k NoW k W e X e Y e X k Y k p p p= = = (22)

and so IDFT of [ ]oW k is just N-point circular convolution of [ ]x n with [ ]y n , i.e.

{ }IDFT [ ] [ ]o cW k w n= (23)

But, linear convolution of [ ]x n and [ ]y n has length 2 1N- . Therefore, points of [ ]oW k represent a sub-sampling of ( )

jW e q

Proof completes after application of the Time Aliasingresult Eq. (40), to be discussed later

Alternatively, it can be shown directly (Problem 18), that Eqs. (19) and (20) are

equivalent

i.e.

1

0[ ] [ ] [ ]

N

m rx m y n m w n rN

-

= =-- = -

(24)


9/55


4 Zero-Padding

Suppose 1[ ]x n is a 1N-point sequence. DTFT of 1[ ]x n given by, see Eq. (2)

q q

--

==

1 1

1 10( ) [ ]

Nj jn

nX e x n e (25)

Next extend 1[ ]x n to 2N points by appending (padding) -2 1N N zeros to its

end

= -= = -

1 1

2

1 2

[ ] , 0, , 1[ ]

0 , , , 1

x n n Nx n

n N N (26)

Fig. 3 Zero padding

DTFT of 2[ ]x n given by

q q q q

- - -- - -

= = == = =

2 1 11 1 1

2 2 2 10 0 0

( ) [ ] [ ] [ ]

N N Nj jn jn jn

n n nX e x n e x n e x n e (27)

Thus q q=2 1( ) ( )j jX e X e (28)

i.e. 1[ ]x n and 2[ ]x n both have same underlying DTFT. Call this DTFTq( )jX e

DFT of 1[ ]x n ( 2[ ]x n ) is simplyq( )jX e evaluated at 1N ( 2N ) equi-angular

points about unit circle\ Since >2 1N N , so frequency resolution of 2[ ]X k greater than 1[ ]X k

Can increasefrequency resolution byzero padding

Note, by frequency resolution, we mean how closely spaced in q we compute

DFT. Therefore, for an N-point DFT, its frequency resolution is given by

2N

pqD = (29)

0

x2[n]x1[n]

N1-1 N2-10


10/55


Do notconfuse frequency resolution with spectral resolution, to discuss later

Example 2

7( ) 0.0083cX j w p

w=

= . Therefore, can sample ( )cx t at w p p= =2 8 16s with acceptable aliasing.

With p p= =2 16 0.125sT , we get =2 0.125 16 sample points

16-point DFT plotted below on left. Right plot is 51-point DFT obtained by zero-padding [ ]x n

Remarks

(i) As discussed, 1[ ]x n and its zero-padded version 2[ ]x n both have same

DTFT, i.e. 1 2( ) ( )j jX e X eq q=

In contrast, 1 2[ ] [ ]X k X k , as can be seen from above example. In any

event, 1[ ]X k has length 1N while 2[ ]X k has length 2N

(ii) Zero-padding as described above also called post-pending. This differs

from pre-pendingwhich appends zeros to the front of a sequence and

then shifts the extended sequence right such that it starts at 0n=

Suppose N is length of extended sequence and on is number of zeros

pre-pended or post-pended. By time-shifting property of DFTs, one gets

2pre-pend post-pend[ ] [ ]

n koNjX k e X kp-= (30)

0 t21

1

xc(t)

ww

w

=

2sin( 2)

( )2

cX j

0 0.25 0.5 0.75 1 1.25 1.5 1.75 20

1

2

3

4

5

6

7

8

(x) 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2

0

1

2

3

4

5

6

7

8

(x)


11/55


5 Zero-Padding the DFT

A powerful concept in the Fourier analysis of signals is duality

In simple terms, duality means: if we perform an operation in the time domain

to get a certain result in the frequency domain, then performing the sameoperation in frequency will yield a mirrored result in time

An example of duality is following DFT property:

DFT[ ] [ ]x n X k

DFT[ ] [ ]

N

X n Nx k -

In this section, we consider the dual of zero-padding

Let ( )cx t be a finite-duration continuous-time signal that is being sampled,

and [ ]x n the resulting N-point sequence with DFT [ ]X k

Suppose we have no access to ( )cx t , except its samples in [ ]x n , but wish

to increase the sampling rate in order to get a closer look at ( )cx t . That is,

we wish to interpolatebetween the samples of [ ]x n

zero pad [ ]X k

We coin above process zero-padding in frequency. Its dual relationship to

zero-padding in timeis summarised below

Zero pad [ ]x n to 1N N> points

sample ( )jX e q with finer freq spacing1

2 21 N N

p pqD = <

Zero pad [ ]X k to 2N N> points

sample ( )cx t with finer sampling period2

2NTN

T T= <

Question: How to zero-pad in frequency?


12/55


First recall following result from Sampling

Suppose ( )cx t is a continuous-time signal with Fourier transform ( )cX jw , and

it is sampled with sampling period T. DTFT of sampled sequence [ ] ( )cx n x nT=

given by

21( ) ( )j c T Tk

X e X j jkT

q q p

=-

= - (31)

Suppose next ( )cx t is sampled with sampling periods 1T and 2 1T T<

Let 1[ ]x n and 2[ ]x n be resulting sample sequences, and 1( )jX e q and 2( )

jX e q

their respective DTFTs

Figure below compares the DTFTs

Fig. 4 DTFTs of sequences obtained with two different sampling periods

Now, suppose ( )cx t is being sampled over the time interval 0 t t . It then

follows that

1 1 2 2N T N T t= = (32)

where 1N is length of 1[ ]x n and 2N is length of 2[ ]x n

( )cX jw

w0

1( )jX e q

0

1

11T

0

2( )jX e q

2 1T T see Eq. (32) so 2[ ]X k has more zeros than 1[ ]X k

Indeed, together with Eq. (35), we see that:

To go from 1[ ]X k to the higher sampling rate 2[ ]X k , we scale 1[ ]X k by

1 2 2 1T T N N = and zero-pad it in the middle mid-pending3

We next consider situation when 1[ ]X k does not have zeros in its middle

3 Mid-pending also ensures the expanded DFT satisfies [ ]X k = [ ]

NX k* - , which is required by an

N-point real sequence


14/55


We begin by noting that since 1[ ]X k has 1N points between 0q= and 2q p= ,

its kth point corresponds to

11

2, 0,1, , ( 1)k

kk N

N

pq = = - (36)

1N is odd

Solving Eq. (36) for kq p= , the mid-point of the interval 0 2q p , we get

1 2k N= which is not an integer, i.e. 1[ ]X k does not have a point at q p=

Therefore, to mid-pend 1[ ]X k , we simply insert zeros between 1 1[( 1) 2]X N -

and 1 1[( 1) 2]X N + which are DFT points immediately on either side of q p=

Example 3 1( 5)N = { }4

10

, , , ,k

X a b c c b* *=

=

{ } 221

20

, , , 0, , 0, ,NN

N kX a b c c b* *

==

1

N is even

Solving Eq. (36) for kq p= , we get 1 2k N= which is now an integer, i.e.

1[ ]X k has a point at q p=

A method to mid-pend 1[ ]X k is to first observe that 1 1[ 2]X N is always real,

whereupon we define 1 1 1 1[ 2] [ 2]X N X N- += = 1 1 12 [ 2]X N , and insert zeros

between 1 1[ 2]X N- and 1 1[ 2]X N

+ 4

Observe above technique preserves requirement that2

2 2[ ] [ ]NX k X k*

= -

Example 4 1( 6)N = { }5

10

, , , , ,k

X a b c d c b* *=

=

{ } 221

2 2 2 0, , , , 0, , 0, , ,

NN d dN k

X a b c c b* *=

=

4 Because of the splitting of 1 1[ 2]X N into 1 1[ 2]X N

-and 1 1[ 2]X N

+, the number of zeros we actually

insert is one less than the number we originally wanted to insert


15/55


Example 5 2 23

2( ) ( ) ( )

(2 )

tc cx t t e u t X j

jw

w

-= =+

As can be seen, duration of ( )c

x t is about 5 sec and its bandwidth is about 2.5 Hz

Accordingly, we set sampling frequency to 5 Hz, or a sampling period of 0.2 sec, and we acquire

(5 0.2) 1 26+ = samples. Denote these samples by 1[ ]x n

In the following figure, we zero pad the DFT of 1[ ]x n to 80 points, and take the IDFT of the scaled

and zero-padded DFT

As can be seen, except for a slight erratic behaviour near 5t= , the interpolated sequence2

[ ]x n does

give us a closer look at ( )cx t

The erratic behaviour near 5t= can be explained in terms of the fact that strictly, ( )cx t is not time-

limited and what we have done is effectively apply a rectangular window to ( )cx t . More will be said

about this temporal windowing in Section 9

Finally, to conform with common usage, unless otherwise stated, in the sequel

zero-padding shall mean zero-padding in time

0 1 2 3 4 5

0

0.03

0.06

0.09

0.12

0.15

t(sec)

xc

(t)

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

f(Hz)

|X

c(j2f)|

0 1 2 3 4 5

0

0.03

0.06

0.09

0.12

0.15

t(sec)

x1[n]

x2[n]


16/55


6 Time Aliasing

Suppose [ ]x n is an N-point sequence with DFT [ ]X k and DTFT q( )jX e

Suppose we zero pad [ ]x n to 1N points and denote zero-padded sequence

by 1[ ]x n and its DFT by 1[ ]X k

We have seen 1[ ]X k can be thought of as DFT obtained by samplingq( )jX e at

>1N Nequally-spaced frequency points

Question What happens if we under-sampleq( )jX e at


17/55


p

---

= =

=

2

2

112 ( )

220 0

1[ ] [ ]

NNj n m k N

m k

x n x m eN

Now, p d-

= =-

= = = -

2

2

12 2

2

2 0

1, ,1[ ]

0, otherwise

Nj n k N

k r

n rN r e n rN

N (38)

d d

=- =- =-

\ = - - = * -

2 2 2[ ] [ ] [( ) ] [ ] [ ]m r r

x n x m n m rN x n n rN (39)

i.e.

=-

= -2 2[ ] [ ]r

x n x n rN (40)

An application of Eq. (40) is shown below

Fig. 5 Time aliasing

As can be seen

(i) If [ ]x n has (non-zero padded) finite length N and


18/55


Example 6 { }5

0[ ]

nx x n

==

0 1 5( ) [0] [1] [5]j j j jX e x e x e x eq q q q- - -\ = + + +

Suppose we sample ( )jx e q

at four frequency points to obtain the DFT:

2 0 42[0] ( ) [0] [1] [2] [3] [4] [5]

jX X e x x x x x xp= = + + + + +

2 1 4 2 4 2 2 4 5 2 42[1] ( ) [0] [1] [2] [5]

[0] [1] [2] [3] [4] [5]

j j j jX X e x x e x e x e

x jx x jx x jx

p p p p - - - = = + + + +

= - - + + -

2 2 4 4 4 2 4 4 5 4 42[2] ( ) [0] [1] [2] [5]

[0] [1] [2] [3] [4] [5]


x x x x x x

p p p p - - - = = + + + +

= - + - + -

2 3 4 6 4 2 6 4 5 6 42[3] ( ) [0] [1] [2] [5]

[0] [1] [2] [3] [4] [5]


x jx x jx x jx

p p p p - - - = = + + + +

= + - - + +

IDFT of the resulting 4-point DFT is given by the 4 4 IDFT matrix

2 2

2 4 4 4 6 42 2

4 4 8 4 12 42 2

6 4 12 4 18 42 2

[0] [0] 1 1 1 11 1 1 1

[1] [1] 1 111 1

4 4[2] [2] 1 1 1 11

[3] [3] 1 11

j j j

j j j

j j j

x X

x X j je e e

x Xe e e

x X j je e e

p p p

p p p

p p p

- - = = - - - -

2

2

2

2

[0]

[1]

[2]

[3]

X

X

X

X

Thus

{ }

{ }{ }{ }

2 2 2 2 24 [0] [0] [1] [2] [3] [0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5][0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

4 [0] 4 [4]

x X X X X x x x x x x

x jx x jx x jxx x x x x x

x jx x jx x jx

x x

= + + + = + + + + +

+ - - + + -+ - + - + -

+ + - - + +

= +

{ }{ }

{ }{ }

2 2 2 2 24 [1] [0] [1] [2] [3] [0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

4 [1] 4 [5]

x X jX X jX x x x x x x

j x jx x jx x jx

x x x x x x

j x jx x jx x jx

x x

= + - - = + + + + +

+ - - + + -

- - + - + -

- + - - + +

= +

{ }{ }{ }{ }

2 2 2 2 2

4 [2] [0] [1] [2] [3] [0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

4 [2]

x X X X X x x x x x x

x jx x jx x jx

x x x x x x

x jx x jx x jx

x

= - + - = + + + + +

- - - + + -

+ - + - + -

- + - - + +

=

{ }{ }

{ }{ }

2 2 2 2 24 [3] [0] [1] [2] [3] [0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

[0] [1] [2] [3] [4] [5]

4 [3]

x X jX X jX x x x x x x

j x jx x jx x jx

x x x x x x

j x jx x jx x jx

x

= - - + = + + + + +

- - - + + -

- - + - + -

+ + - - + +

=


19/55


Remark

Time aliasing can be viewed as another consequence of the duality that

exists between the time and frequency domains in Fourier analysis

Frequency Aliasing Time Aliasing

( )cX jw is bandlimited to Nw w [ ]x n is time limited to 0 ( 1)n N -

Freq aliasing if ( )c

x t sampled at

2 (2 )NT p w>

Time aliasing if ( )jX e q sampled at

2 Nq pD >

1( ) ( )s c s

k

X j X j jkT

w w w

=-

= -

2s Tw p=

2 2[ ] [ ]r

x n x n rN

=-

= -

2 2N p q= D


20/55


7 The Fast Fourier Transform (FFT)

As noted earlier, DFT does not give true spectrum. So why bother with DFT?

Ans. The Fast Fourier Transform (FFT)

FFT is just a computationally efficient way of calculating the DFT. Also, com-

paring Eqs. (3) and (4), we see it can be easily modified to compute the IDFT

It is the speed of the FFT that makes it a popular and powerful signal

processing tool

Recall N-point DFT Eq. (3)

p-

-

=

= 1

2

0

[ ] [ ]N

j nk N

n

X k x n e , = -0, , 1k N

To compute above directly will require

2N complex multiplications

-( 1)N N complex additions

where 1 complex multiply = 4 real multiplies and 2 real adds

1 complex add = 2 real adds

In contrast, Cooley-Tukey radix-2 FFT algorithm, as implemented in MATLAB

and many other similar software packages, requires only

2( 2)logN N complex multiplications

2logN N complex additions

DFT FFT

N real real real real

8 256 240 48 72

32 4,096 4,032 320 480

128 65,536 65,280 1,792 2,688

512 1,048,576 1,047,552 9,216 13,824

2048 16,777,216 16,773,120 45,056 67,584


21/55


Cooley-Tukey requires Nto be an integer power of 2

There exist many other FFT algorithms, some even faster than Cooley-Tukey,but Cooley-Tukey remains most popular because of its ease in programming

DFT/FFT (i) evaluates DTFT at Nfrequency points simultaneously

(ii) requires all of [0]x to -[ 1]x N to be available before computation

can start

If interested in only a few frequency points, or if want to spread computational

load over time interval when [0]x to -[ 1]x N are acquired, use Goertzel

algorithm(see Appendix II)


22/55


8 Linear Convolution with the DFT/FFT

Question: Can we use DFT/FFT to compute, efficiently, the linear convolution of

two sequences [ ]x n and [ ]y n ?

8.1 Full Linear Convolution with Circular Convolution

Figure below compares linear convolution with circular convolution

Fig. 6 Linear vs circular convolution

As can be seen, circular shifting in circular convolution introduces time aliasing

(dashed green ovals)

Time aliasing can be overcome by zero-padding(see Fig. 7 next page)

Idea is to force a gap between the repeated patterns of [ ]y m in [ ]y m

[ ]x m

[ ]y m

[1 ]y m-

[ ]y m

[1 ]y m-

[ ]y m- [ ]y m-

0n=

1n=

m

m

m

m

m

m

m

[ ]y N m- [ ]y N m-

n =

m

[ ]x m

m

1

0

[ ] [ ] [ ] [ ]N

m

x n y n x m y n m-

=

* = -1

0

[ ] [ ] [ ] [ ]NN

m

x n y n x m y n m-

=

= -

m

0

0 0

0

0

00

0

0 1N-

1N- 1N-

1N-

1N-

1N-

1N-1N-

1N- 0 1N-


23/55


Fig. 7 Linear convolution from circular convolution with zero-padding

It follows from Fig. 7 that

If [ ]x n and [ ]y n both have length N, then to compute their linear convolution

using DFT/FFT, we must zero pad them to at least (2 1)N- points

[ ]x m

[ ]y m

[1 ]y m-

zp[ ]y m

zp[1 ]y m-

[ ]y m- zp[ ]y m-

0n=

1n=

m

m

m

m

m

m

m

[ ]y N m- zp[ ]y N m-

n N=

m

zp[ ]x m

m

1

0

[ ] [ ] [ ] [ ]N

m

x n y n x m y n m-

=

* = -2 2

zp zp zp zp0

[ ] [ ] [ ] [ ]

2 1

N

N

m

x n y n x m y n m

N N

-

=

= -

= -

m

0

0 0 1N-1N-

1N- 0 1N- 2 2N-

0 0 1N-1N-

0 0 1N-1N-

0 0 1N-1N-

2 2N-

2 2N-

2 2N-

2 2N-

2 2n = -

[2 2 ]y N m- - zp[2 2 ]y N m- -

m m0 0 1N-1N- 2 2N-


24/55


Key points to explain why zero-padding works:

(i) It follows from Discrete-Time Signals and Systems, Problem 2, that:

{ } { } { }1 1 2( 1)

0 0 0[ ] , [ ] [ ]

N N N

n n nx x n y y n w x y w n

- - -

= = == = = * =

i.e. if xand yboth have length N, then x y* has length (2 1)N-

(ii) Circular convolution of two length (2 1)N- sequences will result in a length

(2 1)N- sequence

(iii) Points of zp[ ]y n m- for , , (2 2)m N N= - do not enter convolution sum

since corresponding points of zp[ ]x m are zero (green regions)6

(iv) Because of zero-padding, when 0 (2 2)n N - , zp[ ] [ ]y n m y n m- = -

for 0 ( 1)m N - (yellow regions)7. More specifically, number of zerosappended to [ ]y n must cover length of [ ]x n , less 1

In view of above discussion, it is not difficult to see that, more generally

If [ ]x n has length P and [ ]y n has length L, then if we are to compute

their linear convolution using the DFT, we must zero pad them to at least

( 1)L P+ - points

6 Thus in Fig. 7,

1 1 1

zp zp zp zp zp zp zp0 0 0

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]NN N N

m m m

x n y n x m y n m x m y n m x m y n m- - -

= = =

= - = - = -

7 Thus in Fig. 7,

1 1

zp0 0

[ ] [ ] [ ] [ ], 0 2 2

[ ] [ ]

N N

m m

x m y n m x m y n m n N

x n y n

- -

= =

- = - -

= *


25/55


Example 7 Linear Convolution with DFT

{ }3

01, 1, 0, 2

nx

== - { }

==

3

01, 0, 3, 2

ny

{ }6

zp 01, 1, 0, 2, 0, 0, 0

nx

== - { }

6zp 0

1, 0, 3, 2, 0, 0, 0n

y=

=

It can be readily verified that

{ }6

0[ ] [ ] 1, 1, 3,1, 2, 6, 4

nx n y n

=* = - -

Now

{

}

zp

6

0

[ ] 2, 1.4254 0.0859, 2.4695 2.5386, 1.4559 1.5160,

1.4559 1.5160, 2.4695 2.5386, 1.4254 0.0859k

X k j j j

j j j=

= - - + -

+ - - -

{

}

zp

6

0

[ ] 6, 1.4695 3.7926, 0.4559 2.8653, 2.4254 0.3956,

2.4254 0.3956, 0.4559 2.8653, 1.4695 3.7926k

Y k j j j

j j j=

= - - - + +

- - - - +

{

}

zp zp

6

0

[ ] [ ] [ ]

12, 1.7687 5.5323, 8.3998 5.9185, 4.1310 3.1009,

4.1310 3.1009, 8.3998 5.9185, 1.7687 5.5323k

W k X k Y k

j j j

j j j=

\ =

= + - + -

+ - - -

and { } { }6

0[ ] IDFT [ ] 1, 1, 3, 1, 2, 6, 4

nw n W k

== = - -

In above example, [ ]x n and [ ]y n were both zero-padded to 7 points. This is

not commensurate with FFT. However, result in pp. 4 says we can also zero

pad them to32 8 7= > points

Example 8 { }3

01, 1, 0, 2

nx

== - { }

==

3

01, 0, 3, 2

ny

{ }7

zp 01, 1, 0, 2, 0, 0, 0, 0

nx

== - { }

7zp 0

1, 0, 3, 2, 0, 0, 0, 0n

y=

=

{

}zp 7

0

[ ] 2, 1.1213 0.7071, 1 3, 3.1213 0.7071,

0, 3.1213 0.7071, 1 3, 1.1213 0.7071k

X k j j j

j j j=

= - - + -

+ - - +

{

}

zp

7

0

[ ] 6, 0.4142 4.4142, 2 2, 2.4142 1.5858,

2, 2.4142 1.5858, 2 2, 0.4142 4.4142k

Y k j j j

j j j=

= - - - + +

- - - - +

{

}

zp zp

7

0

[ ] [ ] [ ]

12, 2.6569 5.2426, 8 4, 8.6569 3.2426,

0, 8.6569 3.2426, 8 4, 2.6569 5.2426k

W k X k Y k

j j j

j j j=

=

= - + - - +

- - + - -

{ } { }7

0[ ] IDFT [ ] 1, 1, 3,1, 2, 6, 4, 0

nw n W k

= = = - -


26/55


Next example illustrate more general result

Example 9 { }1

01, 1 nx == - { } ==3

01, 0, 3, 2 ny

Thus 2P= , 4L= , and we zero pad [ ]x n and [ ]y n to 1 5L P+ - = points:

{ }4

zp 01, 1,0, 0,0

nx

== - { }

4zp 0

1, 0, 3, 2, 0n

y=

=

Next, following the procedure

{

}

zp

4

0

[ ] 0, 0.6910 0.9511, 1.8090 0.5878,

1.8090 0.5878, 0.6910 0.9511k

X k j j

j j=

= + +

- +

{}

zp

4

0

[ ] 6, 3.0451 0.5878, 2.5451 0.9511,

2.5451 0.9511, 3.0451 0.5878k

Y k j j

j j=

= - - +

- - +

{

}

zp zp

4

0

[ ] [ ] [ ] 0, 1.5451 3.3022, 4.0451 3.2164,

4.0451 3.2164, 1.5451 3.3022k

W k X k Y k j j

j j=

= = - - +

- - +

{ } { }4

0[ ] IDFT [ ] 1, 1, 3, 1, 2

nw n W k

= = = - - -


[ ] [ ] [ ]x n y n w n* =


27/55


8.2 Partial Linear Convolution with Circular Convolution

Suppose [ ]x n has length P and [ ]y n has length L P>

Figure below compares linear with L-point circular convolution of [ ]x n and

[ ]y n where, for circular convolution, [ ]x n has been zero-padded to Lpoints

Fig. 8 Linear convolution vs L-point circular convolution

(i) Zero-padding [ ]x n sum in [ ]cw n applies only over first Ppoints

(ii) For 0, , ( 2)n P= - , [ ] [ ]cw n w n because of time aliasing (dashed

green oval) in summation region (yellow)

(iii) For ( 1), , ( 1)n P L= - - , [ ] [ ]cw n w n= since [ ]y n m- and [ ]y n m-

have same data points in (truncated) summation region

(iv) For n L= , [ ]cw n again suffers from time aliasing but we do not compute

[ ]cw L

01P-

[ ]x m

m

[ ]y m

0

m

1

0

[ ] [ ] [ ]P

m

w n x m y n m-

=

= -

1L -

1 1

zp zp0 0

[ ] [ ] [ ] [ ] [ ]L P

cm m

w n x m y n m x m y n m- -

= =

= - = -

01P-

zp[ ]x m

m0

m1L -

[ ]y m

m

[ ]y m-

0 m0

[ ]y m-

0=n

m

[( 1) ]y P m- -

0 m0

[( 1) ]y P m- -

1= -n P

m

[ ]y P m-

0 m0

[ ]y P m-

=n P

m

[( 1) ]y L m- -

0 m0

[( 1) ]y L m- -

1= -n L

1L - 1L -1P- 1P-


28/55


In summary

First ( 1)P- points of circular convolution do not equal linear convolution

Remaining ( 1)L P- + points of circular convolution equal linear convolution

Partial linear convolution method does not find last ( 1) 1L P L P + - - = - points of linear convolution

Example 10 { }1

01, 1

nx

== - { }

==

3

01, 0, 3, 2

ny

Thus 2P= and 4L= . Zero-padding [ ]x n to 4L= points:

{ }3zp 01, 1, 0, 0 nx == -

Accordingly

{ }3

zp 0[ ] 0, 1 1, 2, 1 1

kX k j j

== + -

{ }3

0[ ] 6, 2 2, 2, 2 2

kY k j j

== - + - -

{ }3

zp 0[ ] [ ] [ ] 0, 4, 4, 4

kW k X k Y k

== = - -

{ } { }3

0[ ] IDFT [ ] 1, 1, 3, 1

nw n W k

= = = - - -


{ }4

0[ ] [ ] 1, 1, 3, 1, 2

nx n y n

=* = - - -

Thus:

1. Frst ( 1) 2 1 1P- = - = point of [ ]w n and [ ] [ ]x n y n* are not equal

2. Next ( 1) 4 2 1 3L P- + = - + = points of [ ]w n and [ ] [ ]x n y n* are equal

3. Last ( 1) 2 1 1P- = - = point of [ ] [ ]x n y n* is not found


29/55


9 Implementation of LTI Systems with DFT/FFT

Let [ ]x n be input to an LTI system with impulse response [ ]h n . Suppose [ ]h n

has length Pand [ ]x n has length Q P

May not be possible to use full linear convolution method to find output [ ]y n :

(i) Computation can start only when all of [ ]x n has been gathered. This may

lead to unacceptable delay

(ii) DFT length can be very large if Q is large and such DFTs may be

impractical to compute

(iii) [ ]x n may have infinite length

Two basic methods: Overlap-Addand Overlap-Save

Overlap-add discussed below

Overlap-save discussed in Appendix IV


30/55


Overlap-Add Method

First partition [ ]x n into Rsegments of length Leach

1

0[ ] [ ]

R

rrx n x n rL

-

== - (42)

where [ ]x n has been zero-padded to RL points, [ ]rx n is rth segment

[ ], 0 ( 1)[ ]

0, otherwiser

x n rL n Lx n

+ -=

(43)

P L Q (44)

and = R Q L (45)

It follows that

{ }1 1

0 0

[ ] [ ] [ ] [ ] [ ] [ ] [ ]R R

r rr r

y n x n h n x n rL h n x n rL h n- -

= =

= * = - * = - * (46)

Or1

0

[ ] [ ]R

rr

y n y n rL-

=

= - (47)

where [ ] [ ] [ ]r ry n x n h n= * (48)

and linear convolution of Eq. (48) can be computed using full linear con-

volution method

(i) rth input segment [ ]rx n starts at 0n= while [ ]rx n rL- starts at n rL=

(ii) rth output segment [ ]ry n starts at 0n= while [ ]ry n rL- starts at n rL= (iii) [ ]rx n , and so [ ]rx n rL- , has length L

(iv) [ ]ry n , and so [ ]ry n rL- , has length ( 1)L P+ - where Pis length of [ ]h n

Relationship between [ ]x n , [ ]rx n rL- , [ ]ry n rL- and [ ]y n illustrated in Fig. 9


31/55


1

0

[ ] [ ]R

rr

x n x n rL-

=

= - [ ], 0 ( 1)

[ ]0, otherwise

r

x n rL n Lx n

+ -=

1

0

[ ] [ ]R

rr

y n y n rL-

=

= - [ ] [ ] [ ]r ry n x n h n= *

Fig. 9 Overlap-add method

Observe output segments longer than input segments and they overlap. They

are added together as per Eq. (47) to form [ ]y n

01P-

[ ]h n

n

[ ]x n

nL0 2L 3L

0 [ ]x n 1[ ]x n L- 2[ 2 ]x n L-

n0

0[ ]y n

2P L+ -

n0 L

1[ ]y n L-

n

2[ 2 ]y n L-

2L0

0

[ ]y n

L 2L

n

3L

2 2P L+ -


32/55


Example 11

n 0 1 2

[ ]h n 1 0 -1

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

[ ]x n 1 1 -1 0 -2 2 0 -1 1 0 2 -1 -1 0 1

Thus, length of [ ]h n is 3P= ,

length of [ ]x n is 15Q= ,

and length of [ ]y n is ( 1)P Q+ - = 3 15 1 17+ - =

Suppose segment length is 6L = . Then,

number of segments is R Q L = = 15 6 3 =

length of output segments is ( 1) 6 3 1 8L P+ - = + - =

and we zero pad [ ]x n to 3 6 18RL= = points.

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

[ ]x n 1 1 -1 0 -2 2 0 -1 1 0 2 -1 -1 0 1 0 0 0

0 [ ]x n 1 1 -1 0 -2 2

0[ ]ny 1 1 -2 -1 -1 2 2 -2

-1[ ]x n L 0 -1 1 0 2 -1

-1[ ]n Ly 0 -1 1 1 1 -1 -2 1

-2 [ 2 ]x n L -1 0 1 0 0 0

-2 [ 2 ]n Ly -1 0 2 0 -1 0 0 0

[ ]y n 1 1 -2 -1 -1 2 2 -3 1 1 1 -1 -3 1 2 0 -1


33/55


Appendix I Derivation of DFT to DTFT Interpolation Formula

Recall definition of DTFT Eq. (1)

q q

-

=-= ( ) [ ]j jn

nX e x n e

If [ ]x n has finite length N, then

q q-

-

=

= 1

0

( ) [ ]N

j jn

n

X e x n e (49)

Now, from Eq. (4)

p

p

q q q

q

- - -- -

= = =

- --

= =

= =

=

2

2

1 1 1

0 0 0

1 1

0 0

1( ) [ ] [ ]

1[ ]

nkN

nkN

N N Njj jn jn

n n k

N Nj jn

k n

X e x n e X k e eN

X k e eN

(50)

But

( ) ( )

( )

( )

( )

( ) ( )

( ) ( )

( )

( )

( )( )

( ) ( ){ }

p

pp

p p q

p q p q p q

p q p q p q

p q p q

q p qqq

q

p q

p q

-

- - -

- - -

- -

- -- ---

-= =

-

-

--

-

- -= = =

- -

-=

-

-=

-

2

22

2 2

2 2 22 2 2

2 2 22 2 2

2 22 2

21 1

0 0

22

2

2

1 1

1 1

sin

sin

kNknk

NNk k N

N N

k N k N k N

k N k N k N N N N

k N k N N

j N j k NN Nj nj jn

j jn n

j j j

j j j

k Nj

k N

N

e ee e e

e e

e e e

e e e

je

j

(51)

Substituting Eq. (51) into Eq. (50), yields Eq. (5)

( )( )

2 1212 2

20 2

sin1( ) [ ]

sin

k NN kN jj N

N kk N

X e X k eN

pq p qq

q p

- - - - -

-=

=


34/55


Appendix II Goertzel Algorithm

A.1 Basic Principle

Consider following single-pole (complex) IIR filter (Goertzel cell)

Fig. 10 Goertzel cell

p= + -(2 )[ ] [ ] [ 1]oj k Ny n x n e y n (52)

p -=

- (2 ) 11

( )1 o

j k NH z

e z (53)

p= (2 )[ ] [ ]oj k N nh n e u n (54)

Appendix III shows that if (i) filter is initially at rest, i.e., - =[ 1] 0y

(ii) [ ]x n is causal, i.e., =[ ] 0x n for < 0n

and (iii) =[ ] 0x N

thenp= 2[ ] [ ]oj k oy N e X k (55)

i.e., above filter can be used to compute [ ]oX k (and only ok ) which is not

surprising given its pole-zero structure

Remarks

(i) Goertzel cell spits outp2

[ ]oj k

oe X k at time =n N, and not = -1n N

(ii) At =n N, we must set =[ ] 0x N

(iii) Derivation in Appendix III indicates ok doesnot have to be an integer

(iv) If ok , thenp =2 1oj ke and =[ ] [ ]oy N X k

y[n]x[n]

p2 oj k Ne

p2 ok

N

z-plane

-1z


35/55


A.2 Efficient Implementation

Recall Eq. (52)

p= + -(2 )[ ] [ ] [ 1]oj k Ny n x n e y n

1 real add and 1 complex multiply, or 3 real adds and 4 real multiplies, persampling period

Now, Eq. (53) gives transfer function of Goertzel cell

p -=

- (2 ) 11

( )1 o

j k NH z

e z

\

{ }

p p

p p p

- -- -

- - - - -

- - = - - + +

(2 ) (2 )1 1

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

1 1( )

1 1

o o

o o o

j k N j k N

j k N j k N j k N

e z e z H z

e z e e z z

orp

p

- -

- -

-=

- +

(2 ) 1

2 1 2

1( )

1 2cos

o

o

j k N

k

N

e zH z

z z (56)

y[n]x[n]

p2 oj k Ne

-1z


36/55


p

p

- -

- -

-=

- +

(2 ) 1

2 1 2

1( )

1 2cos

o

o

j k N

k

N

e zH z

z z

Fig. 11 Goertzel cell as 2nd order IIR filter

LCCDE

AR:p

= + - - -2

[ ] [ ] 2cos [ 1] [ 2]ok

Nw n x n w n w n (57)

MA:p

-= - -

2

[ ] [ ] [ 1]ko

Nj

y n w n e w n p p

= - - + -2 2

[ ] cos [ 1] sin [ 1][ ] o ok k

N Nw n w n j w ny n (58)

Sampling periods 0 to N1:

Goertzel cell in real form requires only 2 real adds/subtracts and 1 real

multiply per sampling period to maintain AR section ( [ ]w n is real)

Sampling period N:

AR section now requires 1 less real add since =[ ] 0x N but MA section kicks

in and requires 1 real add and 2 real multiplies

2 real adds and 3 real multiplies

y[n]x[n]

p22cos o

k

N

p-- 2 oj k Ne

-1z

-1z

w[n]


37/55


A.3 Continuous Operation

Block Mode [ ]oX k every Nth sampling period

Output response of Goertzel cell given by Eq. (65) of Appendix III

p -

=

= (2 )( )0

[ ] [ ] on

j k N n m

m

y n x m e

\ p p p- -

- - - -

= =

- = = 2 2 21 1

( 1 ) ( 1)

0 0

[ 1] [ ] [ ]k k ko o o

N N N

N Nj N m j N j m

m m

y N x m e e x m e

i.e.,p

- -- =

2( 1)

[ 1] [ ]ko

Nj N

oe y N X k (59)

Can extract [ ]oX k at 1n N= - by scaling output

Fig. 12 Goertzel cell continuous operation, block mode

At =n N, we reset filter, wait for [ ]x N , and proceed as before

Next valid [ ]oX k available at 2 1, 3 1, ,n N N= - - etc.

n = N1, 2N1, 3N1,

x[n]

n = 0, 1, 2, X[ko]

p- -

2( 1)

koN

j Ne

N

Np22cos o

k

N

-1z

-1z

p-- 2 oj k Ne

n = N1, 2N1, 3N1,

x[n]

n = 0, 1, 2, p2 oj k NeN

Np22cos o

k

N

-1z

-1z

X[ko]p- 2 oj ke


38/55


Sampling periods 0 to N2: 2 real adds/subtracts and 1 real multiply

Sampling period N1: 3 real adds/subtracts and 3 real multiplies ( ok )

5 real adds/subtracts and 7 real multiplies ( ok )

Sliding Mode [ ]oX k every sampling period

Definep-

-

=

= - - +21

0

[ , ] [ ( 1) ]ko

N

Nj m

o o om

X k n x n N m e (60)

i.e. [ , ]o oX k n is DFT of the Nsuccessive samples - - [ ( 1)], , [ ]o ox n N x n

Fig. 13 Grouping of data samples in sliding mode operation

Now, from Eq. (65) of Appendix III

p -

=

= (2 )( )0

[ ] [ ] on

j k N n m

m

y n x m e

Therefore

p p

p p

p

p

p p-

- - -

= =

-- -

= =

-

= - +

-- -

=

- - = -

= -

=

= - + + = - - +

2 2

2 2

2

2

( ) ( )2 2

0 0

( ) ( )

0 0

( )

1

1( 1 )

0

[ ] [ ] [ ] [ ]

[ ] [ ]

[ ]

[ 1 ] , ( 1)

k ko oo oN N

k ko oN N

koN

koN

n n Nj n m j n N mj k j k

m m

n n Nj n m j n m

m m

nj n m

m n N

Nj N k

k

y n e y n N x m e e x m e

x m e x m e

x m e

x n N k e k m n N

i.e.,

p p

p-

- -

=- - = - - +

2 21( 1)2

0[ ] [ ] [ ( 1) ]

k ko oo N N

Nj N j kj k

ky n e y n N e x n N k e (61)

nno-(N-1)

Npoints

x[n]

no

X[ko,no]

X[ko,no-1]


39/55


Substituting Eq. (60) into Eq. (61), we get

p

p -- - =2

( 1)2[ ] [ ] [ , ]

koo N

j Nj koy n e y n N e X k n (62)

i.e.

p p- -

- - =

2 2( 1)

[ ] [ ] [ , ]

k ko oN N

j N j

oe y n e y n N X k n (63)

Fig. 14 Goertzel cell continuous operation, sliding mode

Remarks

(i) Sliding mode structure is basically FIR since it simply computes an N-point

DFT

\ Stable in theory

But Goertzel cell embeds a recursive filter with a pole on unit circle, so can

be unstable if pole actually outside unit circle due to coefficient round-offs

A remedy is to put pole atp= 2 oj k Nz r e where ris just less than 1

(ii) Block mode structure always stable because of periodic resets

(iii) If ok , such thatp =2 1oj ke , then Eq. (62) yields the more efficient

structure shown in Fig. 15

p- -

2( 1)

koN

j Ne

p22cos o

k

N

-1z

-1z

- +2Nz

x[n] X[ko,n]

-1z

p2 oj k Ne

p- 2 oj ke


40/55


Fig. 15 Goertzel cell continuous operation, sliding mode, ok

x[n]

p2 oj k Ne

p22cos o

k

N

-1z

-1z

X[ko,n]

Nz-


41/55


Appendix III Derivation of Goertzel Algorithm

Recallp= + -(2 )[ ] [ ] [ 1]oj k Ny n x n e y n

p -=

- (2 ) 11

( )1 o

j k NH z

e z

andp= (2 )[ ] [ ]oj k N nh n e u n

Suppose filter is initially at rest. Then

p

-

=- =-

= - = - (2 )( )[ ] [ ] [ ] [ ] [ ]oj k N n mm m

y n x m h n m x m e u n m (64)

But [ ]x n iscausal, i.e., =[ ] 0x m for < 0m . Also - =[ ] 0u n m for >m n

Therefore p -

=

= (2 )( )0

[ ] [ ] on j k N n m

m

y n x m e (65)

and p -

=

= (2 )( )0

[ ] [ ] oN

j k N N m

m

y N x m e

p p-

=

= 2 (2 )0

[ ][ ] o oN

j k j k N m

m

e x m ey N (66)

Now, if =[ ] 0x N (67)

then p p p-

-

=

= =1

2 (2 ) 2

0

[ ] [ ] [ ]o o oN

j k j k N m j ko

m

y N e x m e e X k (68)

where [ ]oX k is thok output of an N-point DFT

y[n]x[n]

p2 oj k Ne

-1z


42/55


Appendix IV Overlap-Save Method

Method based on partial linear convolution method

Idea is to partition [ ]x n into overlapping equal-length segments and use partial

linear convolution method to determine from these segments, contiguous blocksof correct linear convolution terms

The method: (Recall [ ]h n has length P and [ ]x n has length Q P )

1. Choose segment length Lsuch that

P L Q (69)

2. Pre- and post-pend [ ]x n with ( 1)P- and { }( 1)R L P Q- + - zeros,respectively, where Ris number of segments

( 1) ( 1)R Q L L P = + - - + (70)

i.e. construct the extended sequence ext[ ]x n where

= = - =ext ext[0] [ 2] 0x x P (71)

+ - = -ext[ 1] [ ], 0 ( 1)x n P x n n Q (72)

and + - = = - + + - =ext ext[ 1] [ ( 1) 2] 0x Q P x R L P P (73)

3. Partition ext[ ]x n into overlapping L-point segments where successive

segments overlap by ( 1)P- points. Thus, rth segment is

+ - + -=

ext [ ( 1)], 0 ( 1)[ ]

0, otherwiser

x n r L P n Lx n (74)

4. Compute L-point circular convolution of [ ]h n (zero-padded to L points)

with [ ]rx n , discard first ( 1)P- points of convolution and retain last

( 1)L P- + points. Denote retained points by [ ]ry n , 0 ( )n L P -

5. Abut [ ]ry n s to form output

1

0

[ ] [ ( 1)]R

rr

y n y n r L P -

=

= - - + (75)


43/55


Fig. 16 Overlap-save method

Method called overlap-save since we overlap input segments and we save

last ( 1)P- points of input segment [ ]rx n to use them as first ( 1)P- points

of next input segment 1[ ]rx n+

[ ]x n

n0

0[ ]x n- - +1[ ( 1)]x n L P

- - +2[ 2( 1)]x n L P

ext [ ]x n

nL0

n0

0y

01P-

[ ]h n

n

- +1L P - +2( 1)L P

n0

1y

n0

2y

n0

[ ]y n

-2P


44/55


Example

n 0 1 2

[ ]h n 1 0 -1

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

[ ]x n 1 1 -1 0 -2 2 0 -1 1 0 2 -1 -1 0 1

Thus, length of [ ]h n is 3P= ,

length of [ ]x n is 15Q= ,

and length of [ ]y n is ( 1)P Q+ - = 3 15 1 17+ - =

Suppose segment length is 6L = . Then,

number of segments is ( 1) ( 1) 20 4 5R Q L L P = + - - + = =

segments overlap by ( 1) 3 1 2P- = - = points

length of output segments is 6L =

we retain last ( 1) 6 3 1 4L P- + = - + = points of each output segment

we pre-pend [ ]x n with ( 1) 3 1 2P- = - = zeros

and we post-pend [ ]x n with { }( 1) 5(6 3 1) 15 5R L P Q- + - = - + - = zeros

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

ext [ ]x n 0 0 1 1 -1 0 -2 2 0 -1 1 0 2 -1 -1 0 1 0 0 0 0 0

0 [ ]x n 0 0 1 1 -1 0

0h x 1 0 1 1 -2 -1

0 [ ]y n 1 1 -2 -1

-1[ 4]x n -1 0 -2 2 0 -1

1h x -1 1 -1 2 2 -3

-1[ 4]y n -1 2 2 -3

-2 [ 8]x n 0 -1 1 0 2 -1

2h x -2 0 1 1 1 -1

-2 [ 8]y n 1 1 1 -1

-3 [ 12]x n 2 -1 -1 0 1 0

3h x 1 -1 -3 1 2 0

-3 [ 12]y n -3 1 2 0

-4 [ 16]x n 1 0 0 0 0 0 4h x 1 0 -1 0 0 0

-4 [ 16]y n -1 0 0 0

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

[ ]y n 1 1 -2 -1 -1 2 2 -3 1 1 1 -1 -3 1 2 0 -1


45/55


Problems

1. Compute the DFT of each of the following finite-length sequences considered to be of

length N(where Nis even):

(a) [ ] [ ]d=n n

(b) [ ] [ ], 0 1d= - -o ox n n n n N

(c)1, even, 0 1

[ ]0, odd, 0 1

-= -

n n Nx n

n n N

(d) 2

2

1, 0 1[ ]

0, 1

-= -

N

N

nx n

n N

(e), 0 1

[ ]0, otherwise

-=

na n Nx n

2. Consider the finite-lengthsequence { }3 0[ ] 6, 5, 4, 3 nx n == . Determine the sequences

(a) 1 4[ ] [ 2 ], 0 3x n x n n= -

(b) 2 4[ ] [ ], 0 3x n x n n= -

3. A finite-duration sequence [ ]x n of length 8 has the 8-point DFT [ ]X k shown below.

A new sequence of length 16 is defined by

[ 2], even[ ]

0, odd

x n ny n

n

=

Sketch the 16-point DFT of [ ]y n .

k

[ ]X k

0 1 2 3 4 5 6 70 1 2 3 4 5 6 7


46/55


4. Two 8-point sequences 1[ ]x n and 2[ ]x n shown below have DFTs 1[ ]X k and 2[ ]X k

respectively. Determine the relationship between 1[ ]X k and 2[ ]X k .

5. Two finite-length sequences [ ]n and 1[ ]x n are shown below.

TheN-point DFTs of these sequences, [ ]X k and 1[ ]X k , respectively, are related by

2 2

1[ ] [ ] Nj k

X k X k ep

=

where N is an unknown constant. Can you determine a value of N consistent with

the figures drawn for [ ]x n and 1[ ]n ? Is your choice for N unique? If so, justify

your answer. If not, find another choice of Nconsistent with the information given.

6. Consider the real finite-length sequence [ ]x n shown below.

(a) Determine the finite-length sequence [ ]y n whose 6-point DFT is

26

4[ ] [ ]

j kY k X k e

p- =

where [ ]X k is the 6-point DFT of [ ]x n

(b) Determine the finite-length sequence [ ]w n whose 6-point DFT is

{ }[ ] Re [ ]W k X k =

(c) Determine the finite-length sequence [ ]q n whose 3-point DFT is

[ ] [2 ], 0,1, 2Q k X k k = =

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

a

b

c

d

e

n n

e

d

a

b

cx1[n] x2[n]

0

1

2 3 4 5 6 7 0 1 2 3

4

5 6 7n n

x [n] x1[n]1 1

1- 1-

1 1

n

[ ]x n

0 1 2 3 4 5 6 7-1

43

21


47/55


7. [ ]x n is a real-valued finite-length sequence of length 10 and is nonzero in the interval

from 0 to 9, i.e.,

[ ] 0, 0, 10

[ ] 0, 0 9

x n n n

x n n

= <

( )jX e q denotes the DTFT of [ ]x n , and [ ]X k denotes the 10-point DFT of [ ]x n .

Determine a choice for [ ]n so that [ ]X k is real valued for all kand

( ) ( ) ,j jX e A eq aqq q p= <

where ( )A q is real and a is a nonzero real constant.

8. Let [ ]x n be a real-valued 5-point sequence whose 7-point DFT is denoted by [ ]X k .

If { }Re [ ]X k is the 7-point DFT of [ ]g n , show that [0] [0]g x= , and determine therelationship between [1]g and [1]x .

9. Suppose we have two 4-point sequences [ ]n and [ ]h n as follows:

( )2[ ] cos , 0,1, 2, 3

[ ] 2 , 0,1, 2, 3

n

n

x n n

h n n

p= =

= =

(a) Calculate the 4-point DFT [ ]X k .

(b) Calculate the 4-point DFT [ ]H k .

(c) Calculate 4[ ] [ ] [ ]y n x n h n= by doing the circular convolution directly.

(d) Calculate [ ]y n of Part (c) by multiplying the DFTs of [ ]n and [ ]h n and

performing an inverse DFT.

10. Consider the two 4-point sequences { }31 0[ ] 1, 0, 2, 1 nx n == and { }3

2 0[ ] 0, 1, 0,

nx n a ==

where a is unknown. Suppose the 4-point circular convolution of 1[ ]n and 2[ ]x n is

given by

{ }43

1 2 0[ ] [ ] [ ] 1, 1, 1, 1

ny n x n x n

== = - -

Can you specify auniquely? If so, what is a? If not, give two possible values of a.


48/55


11. In this exercise, we investigate the plausibility of the zero-insertion method described in

the Lecture Notes for zero-padding in frequency, particularly when the original DFT

has an even number of points.

Consider the continuous-time signal

( ) sin 2 cos 4p p= +cx t A t B t

Clearly, ( )cx t is periodic with period 1 sec and bandwidth 2 Hz. Suppose ( )cx t is

sampled over 1 sec at 4, 5, 6 and 7 Hz with the first sample taken at 0=t to yield

the 4-, 5-, 6- and 7-point sequences 4[ ]n , 5[ ]x n , 6[ ]n and 7[ ]n , respectively. Show

that

{ }3

4 0[ ] 0, 2 , 4 , 2

== -

kX k j A B j A

{ }45

5 4 0[ ] 0, 2 , 2 , 2 , 2

== -

kX k j A B B j A

{ }56

6 4 0[ ] 0, 2 , 2 , 0, 2 , 2

== -

kX k j A B B j A

{ }67

7 4 0[ ] 0, 2 , 2 , 0, 0, 2 , 2

== -

kX k j A B B j A

12. Consider the finite-length sequence { }5 0[ ] 1,1,1,1,1,1 == nx n . Let ( )X z be the z-

transform of [ ]n . If we sample ( )X z at(2 4)p= j kz e , 0,1, 2, 3=k , we obtain

(2 4)1[ ] ( ) p=

= j kz e

X k X z , 0,1, 2, 3=k .

Sketch the sequence 1[ ]n obtained as the inverse DFT of 1[ ]X k

13. Suppose 1[ ]x n is an infinite-length, stable (i.e., absolutely summable) sequence with

z-transform given by

1 113

1( )

1X z

z-=

-

Suppose 2[ ]x n is a finite-length sequence of lengthN, and itsN-point DFT is given by22 1[ ] ( ) , 0,1, , 1j k Nz eX k X z k Np== = -

Determine 2[ ]x n .

14. Consider the two finite-length sequences { }31 0[ ] 1, 2, 1, 3 nx n == - - and 2[ ]x n =

{ }5 00, 2, 0, 0, 1, 1 n=- . What is the smallest N such that the N-point circularconvolution of 1[ ]n and 2[ ]n equals the linear convolution of these sequences, i.e.,

such that 1 2 1 2[ ] [ ] [ ] [ ]Nn x n x n x n= * ?


49/55


15. Determine a real sequence [ ]n that satisfies all of the following three conditions:

Condition 1: The DTFT of [ ]n has the form

1 2( ) 1 cos cos 2jX e A Aq q q= + +

where 1 and 2 are some unknown constants.

Condition 2: The sequence [ ] [ 3]x n nd* - evaluated at 2n= is 5.

Condition 3: For the 3-point sequence { }2 0[ ] 1, 2, 3 nw n == , the result of the 8-point

circular convolution of [ ]w n and [ 3]x n - is 11 when 2n= ; i.e.,

7

80 2

[ ] [ 3 ] 11

m n

w m x n m

= =

- - =

16. Two finite-length sequences 1[ ]x n and 2[ ]x n , which are zero outside the interval

0 99n , are circularly convolved to form a new sequence [ ]y n .

100

99

1 2 1 2 1000

[ ] [ ] [ ] [ ] [ ] , 0 99k

y n x n x n x k x n k n=

= = -

If, in fact, 1[ ]n is nonzero only for 10 39n , determine the set of values of n for

which [ ]y n is guaranteed to be identical to the linearconvolution of 1[ ]x n and 2[ ]n .

17. Consider two finite-duration sequences [ ]x n and [ ]y n . [ ]n is zero for 0n < , 40n ,

and 9 30n< < , and [ ]y n is zero for 10n < and 19n > , as shown below.

Let [ ]w n denote the linear convolution of [ ]x n and [ ]y n . Let [ ]g n denote the 40-point

circular convolution of [ ]x n and [ ]y n :

[ ] [ ] [ ] [ ] [ ]k

w n x n y n x k y n k

=-

= * = -

40

39

400

[ ] [ ] [ ] [ ] [ ]k

g n x n y n x k y n k== = -

0

0 9

10 19

30 39 n

n

[ ]y n

[ ]x n


50/55


(a) Determine the values of nfor which [ ]w n can be nonzero.

(b) Determine the values of nfor which [ ]w n can be obtained from [ ]g n . Explicitly

specify at what index values nin [ ]g n these values of [ ]w n appear.

18. Suppose [ ]n and [ ]y n are two finite-duration sequences, both with the same length N.

Show, directly, that

1

0

[ ] [ ] [ ]N

m r

m y n m w n rN -

= =-

- = -

where [ ]y n is the periodic extension of [ ]y n , and [ ]w n is the linear convolution of

[ ]x n with [ ]y n .

19. We want to implement the linear convolution of a 10,000-point sequence with an FIR

impulse response that is 100 points long. The convolution is to be implemented by using

DFTs and IDFTs of length 256.

(a) If the overlap-add method is used, what is the minimum number of 256-point

DFTs and the minimum number of 256-point IDFTs needed to implement the

convolution of the entire 10,000-point sequence? Justify your answer.

(b) If the overlap-save method is used, what is the minimum number of 256-point

DFTs and the minimum number of 256-point IDFTs needed to implement the

convolution of the entire 10,000-point sequence? Justify your answer.

(c) Suppose the DFTs and IDFTs of Parts (s) and (b) are implemented using FFTs

and IFFTs. Compare the number of arithmetic operations (multiplications and

additions) required in the overlap-add method, the overlap-save method, and

direct convolution.

Other problems to attempt from Oppenheim and Schafer, Chapter 8:

8, 11, 13, 14, 16, 18, 19, 26, 31, 32, 38, 53


51/55


Answers

1. (a) [ ] 1=X k (b) 2[ ] p-= oj kn NX k e

(c)2, 0, 2

[ ]0, otherwise

==

N k NX k (d) 2

2

1

2 , 0

, odd[ ]

0 , even

p--

==

j k Ne

N k

kX k

k

(e) 21

1[ ]

p-

-

-=

N

j k N

a

aeX k

2. (a) { }3

10

[ ] 4, 3, 6, 5n

x n

=

=

(b) { }3

2 0[ ] 6, 3, 4, 5

nx n

==

3.

4. 2 1[ ] ( 1) [ ]kX k X k= -

5. 5N =

6. (a) { }5

0[ ] 2, 1, 0, 0, 4, 3

ny n

==

(b) { }5

3 32 2 0[ ] 4, , 1, 1, 1, nw n ==

(c) { }2

0[ ] 5, 3, 2

nq n

==

k

[ ]Y k

0 1 2 3 4 5 6 7 15


52/55


7. We first establish the following result. Let [ ]Y k be the DFT of an arbitrary realN-point

sequence. Define

[ ] [ ] [ ]X k Y k Y k*= +

Clearly, [ ]X k is real for all k. Now, from the properties of DFTs, the IDFT of [ ]X k is given by

[ ] [ ] [ ] [ ] [ ] [ ] [ ]N N

n y n y n y n y n y n y N n*= + - = + - = + -

Accordingly, if [ ]X k is an N-point DFT and is real for 0 ( 1)k N - , then its

IDFT [ ]n must satisfy

[ ], 1 1[ ]

arbitrary real , 0

x N n n Nx n

n

- -= =

In the context of this Problem, the above result means

[1] [9]

[2] [8]

[3] [7]

[4] [6]

x

x x

x x

x x

=

=

=

=

and [0] and [5]are arbitrary but real.

Next, recall our earlier work on linear phase FIR filters. The condition on ( )j

X e q

implies [ ]x n is linear phase. Therefore, [ ]x n must satisfy

[ ] [ 1 ], 0 ( 1)x n x N n n N= - - -

or in the context of this Problem,

[0] [9]

[1] [8]

[2] [7]

[3] [6]

[4] [5]

x

x

x x

x x

x x

=

=

=

=

=

Combining both sets of conditions on [ ]n , we see that [ ]x n must be a constant for

0 ( 1)n N - .

8. [0] [0]g x= ; 12

[1] [1]g x=

9. (a) { }4

43

0[ ] 1 0, 2, 0, 2

kj

kX k e

p-

== - =

(b) { }2 4 6

4 4 43

0[ ] 1 2 4 8 15, 3 6, 5, 3 6

k k kj j j

kH k e e e j j

p p p- - -

== + + + = - + - - -

(c)(d) { }3

0[ ] 3, 6, 3, 6

ny n

== - -


53/55


10. 1a = -

11. Sampling ( )cx t at 4, 5 and 6 Hz, we get

{ }34 0[ ] , ( ), , ( ) == - - - nx n B A B B A B

{}

2 25 5 5 5 5

42 25 5 5 5 0

[ ] , sin cos , sin cos ,

sin cos , sin cos

p p p p

p p p p

=

= - +

- + - -n

x n B A B A B

A B A B

{}

1 16 2 2

51 12 2 0

[ ] , ( 3 ), ( 3 ),

, ( 3 ), ( 3 )=

= - -

- - - -n

x n B A B A B

B A B A B

DFTs can be found, after some algebra, through the DFT matrices

4

1 1 1 1

1 1

1 1 1 1

1 1

FFT - -

- -

- -

=

j j

j j

2 5 4 5 4 5 2 5

4 5 2 5 2 5 4 55

4 5 2 5 2 5 4 5

2 5 4 5 4 5 2 5

1 1 1 1 1

1

1

1

1

FFT

p p p p

p p p p

p p p p

p p p p

- -

- -

- -

- -

=

j j j j

j j j j

j j j j

j j j j

e e e e

e e e e

e e e e

e e e e

3 2 3 2 3 3

2 3 2 3 2 3 2 3

2 3 2 3 2 3 2 3

3 2 3 2 3 3

6

1 1 1 1 1 1

1 1

1 1

1 1 1 1 1 1

1 1

1 1

FFT

p p p p

p p p p

p p p p

p p p p

- -

- -

- -

- -

-

- - -

-

=

j j j j

j j j j

j j j j

j j j j

e e e e

e e e e

e e e e

e e e e

12. { }3

1 0[ ] 2, 2, 1, 1

nx n

==

13. ( ) ( )131 12 3

1[ ] Nnx n

-=

14. 9N=


54/55


15. From Condition 1, we can write

1 2 2 21 2 2 2

( ) 1 cos cos 2 1 ( ) ( )A Aj j j j jX e A A e e e eq q q q qq q

- -= + + = + + + + (1)

Comparing the above expression with the definition of the DTFT

( ) [ ]j jn

n

X e x n eq q

-

=-

= (2)

we see that [ ]x n is the non-causal finite-length sequence

{ }2 1 1 22

2 2 2 2 2[ ] , , 1, ,

A A A A

nx n

=-= (3)

Next, from Condition 2

( ) ( )2 2

[ ] [ 3] [ 3] [ 1] 5n n

x n n x n xd= =

* - = - = - = (4)

Combining (3) and (4), we get

[ 1] [1] 5x x- = = (5)

and 1 10A = (6)

Finally, with reference to Condition 3, we first remark that [ 3]x n- is a causal 6-point

sequence

{ }2 1 1 25

2 2 2 2 0[ 3] 0, , , 1, ,

A A A A

nx n

=- = (7)

Therefore, the linearconvolution of [ 3]x n- with the causal 3-point sequence [ ]w n will

yield a causal 3 6 1 8+ - = point causal sequence. Moreover, from our study on

the relationship between linear and circular convolutions, we see the 8-point circular

convolution of [ 3]x n- with [ ]w n will equal their linear convolution. In other words,

8[ ] [ 3] [ ] [ 3]w n x n w n x n* - = - (8)

It can be shown, using the table method to calculate linear convolutions, that

{ }7

0[ ] [ 3] 0, , 5 2 , 11 3 , 22, 13 , 15 2 , 3

nw n x n a a a a a a

=* - = + + + + (9)

where 2[ 2] [2] 2a x x A= - = = . Hence, from Condition 3

2[ ] [ 3] 5 2 11

nw n x n a

=* - = + = (10)

23 or 6a A = = (11)

Summarising

{ }2 2[ ] 3, 5, 1, 5, 3 nx n =-=

16. 39 99n


55/55

17. (a) 10 28n and 40 58n

(b) [ ] [ ]g n w n= for 19 39n , and [ ] [ 40]g n w n= + for 0 9n

19.Multiplications Addtions

Overlap-add 528,384 798,813

Overlap-save 536,576 804,864

Direct convolution 1,009,900 999,801

07 Dft Fft Release

Documents

Transcript of 07 Dft Fft Release