dtft & zt the z transform

T. Y. Choi, ECE, Ajou University DTFT & zT 47

DTFT & zT zT

| | 1 ( ) ( ) | j

j

X z ez R X e X z

C is a closed counterclockwise

contour in the complex z-plane

lying entirely within the ROC RX

The z Transform

z-Transform

2

( ) [ ]

1[ ] ( )

2

j j n

n

j j n

X e x n e

x n X e e d

1

( ) [ ] ,

1[ ] ( )

2

n

X

n

n

c

X z x n z R

x n X z z dzj

DTFT

jz e jdz je d


DTFT & zT zT

( ) [ ]

[ ]( )

[ ]

jj

n

z re nz re

j n

n

n j n

n

X z x n z

x n re

x n r e

1. Relationship between the DTFT and the z Transform

Proof:

0 2

DTFT [ ] nx n r

( ) ( ) [ ]j

j n

z reX z X re x n r

jIm

Re

jz re

(1) If a circle of radius r is included within the ROC RX in the z

plane, the DTFT of exists along the circle, i.e., [ ] nx n r


DTFT & zT zT

( [ ] ) [ ] [ ]( )n n j n j n

n nx n r x n r e x n re

Proof:

( ) ( [ ] ) , | |j

n

re zX z x n r z r

0 2

[ ] nx n r

( [ ] ) [ ] ( ), | |j

n n

nre zx n r x n z X z z r

( ), | |X z z r

(2) Reversely, if the DTFT of exists, the zT of

exists along a circle of radius r and that is equivalent to

the substitution of in the DTFT of , i.e.,

[ ] nx n r [ ]x n

jre z

( )X z

( )X z

[ ] nx n r

jIm

Re

jz re


DTFT & zT zT

If (| | 1) , ( ) ( ) | j

j

X z ez R X e X z

jIm

Re

jz e

| | 1z

1

( ) ( ) | , | | 1j

j

e zX z X e z

(3) If the unit circle (|z|=1) is included within the ROC RX in the

z plane, the DTFT of x[n] exists along the circle, i.e., ( )jX e

Reversely, if the DTFT exists, must exist along

the unit circle and , i.e.,

( )jX e ( )X z

( ) ( ) | j

j

e zX z X e


DTFT & zT zT

Example: DTFT and zT of [ ] 2 [ ]nx n u n

0 2

10

1( ) 2 , | | 2

1 2

n n

nX z z z

z

(2) zT:

0( ) 2j n j n

nX e e

(1) DTFT: diverges

[ ] 2 [ ]n n nx n r r u n (3) DTFT of

1

( ) ( [ ] )

1, | | 2

1 2

j

n

re zX z x n r

z rz

[ ] nx n r ( ), | |X z z r

If r>2, the DTFT of exists. Therefore,

[ ] nx n r

jIm

Re

2 r1

1

10

1

1 (2 )( [ ] ) (2 )

1 2

1, 2

1 2

jn j n

jn

j

r ex n r r e

r e

rr e

jz e


DTFT & zT zT

1

10

1

1 ( )( )

1

1, | | 1

1

n n

n

azX z a z

a z

a

a z z

jIm

Re

| | | |z a

z a

( | | 1)a

-1 0 1 2 n

1

-1 0 1 2 n

( | | 1 )a 1

If |a|<1, ROC includes |z|=1 and then we have the DTFT from the zT as

1( ) ( ) |

1j

j

jz eX e X z

ae

Proof:

elsewhere (|a|>1), |z|=1 is not included in ROC and the DTFT diverges.

If a=1, i.e., x[n]=u[n], X(z) = ?

1

1[ ] , | | | |

1

zna u n z aa z

(1) Right sided signal: [ ] [ ]nx n a u n

RX is the open exterior of a circle lying a pole

2. zTs of right, left, and double sided sequences


DTFT & zT

-3 -2 -1 0 1 2 3 n

1

2 n

-3 -2 -1 0 1 2 3 n

1

2n

1

1 1

12 [ 1] , | | 2

1 2

zn u n zz

1

1 1

12 [ ] , | | 2

1 2

zn u n zz

1

12 [ 1] , | | 2

1 2

zn u n zz

1

12 [ ] , | | 2

1 2

zn u n zz

Right/left sided

Right/left sided

± sign, ROC(outer/inner)


DTFT & zT zT

2 [ ] 3 [ 1]n nu n u n Ex-1.

2 [ ] 3 [ 1]n nu n u n

jIm

Re

| | | | | |a z b

z a

z b

Common ROC is an open ring

between two adjacent poles.

1/2<|z|<3

2<|z|<3

0 n

2n

3n

1

Ex-2.

0 n

1 ROC

Ex-3. 3 3 ( [ ] [ 1])n n u n u n 3<|z|<3 NO common ROC ?

[ ] [ ] [ 1]n ny n a u n b u n (3) Double sided signal:

1 1

1 1( ) , | | | | | |

1 1Y z a z b

az bz


DTFT & zT zT

3. Confliction between DTFT and zT on the circle lying a pole

It is not obvious whether the zT can be expressed on the circle lying a pole z=1.

Which is the superset? The DTFT or the ZT ?

The DTFT of u[n] is expressed even with an impulse train.

In the same way, its z transform could be expressed.

jz e

jz e

is not allowed since |z|>1.

Substitution with

DTFT 1[ ] ( 2 )

1 jku n k

e

1

1[ ] ( 1) , | | 1

1

zu n z zz

?

( 1)je

See Property of Delta function

1

1[ ] , | | 1

1

zu n zz


DTFT & zT

( ) diric( , 2 1)N

j j n

n N

X e e N

DTFT: 1 0 1N N n

1

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

1 1/2 1/2

(1 )( ) ,

1 1

N N N N N NNn

n N

z z z z z zX z z z

z z z z

z-T:

The limiting value 2N+1 of X(z) as z approaches 1 is finite unless N is infinite. Thus if N is finite, we regard there is a pole-zero cancellation at z=1.

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

/2 /2

sin((2 1) / 2)( ) diric( , 2 1)

sin( / 2)j

j n j n

j jz e

e e NX z N

e e

(0<|z|<∞)

DTFT and zT of [ ] [ ]N

k Nx n n k

lim ( ) diric( , ) 2 ( 2 ) 2 ( 1)j j

odd nNX e n e

lim ( ) lim ( ) 2 ( 1), | | 1j

j

N N e z

X z X e z z

(Around the unit circle)

As N→∞, lim [ ] 1N

x n

(dc sequence 1)


DTFT & zT zT

1( )

2X f

a j f

1 1( )

2 1

aT jj

aT j

e eX e

e e

4. Summary of X(f), Xs(f), DTFT, and zT

1( ) ( ), 0atx t e u t a

t0

12

[ ] ( )

[ ] [ ]

anT

anT

x n e u n

e u n n

12

0

( ) ( ) ( )anT

s

n

x t t e t nT

n

12

0 t nT1 2 3 4

( 2 )

0

2

2

1( )

2

1 1

2 1

aT j Tf n

s

n

aT j Tf

aT j Tf

X f e

e e

e e

1

1

1 1( ) , | | 1

2 1

aTaT

aT

e zX z z e

e z

1

1[ ] , | | 1

1

zaTn aT

aTe u n z e

e z

2j Tfe z

2 Tf

je z

n

12

0 1 2 3 4

Red dots

+ 1/2

With repetition of X(f)


DTFT & zT zT

Red

[ ]anTe u n

1

10

1

1 ( )[ ]

1

1, | |

1

aTzaTn aTn n

aTn

aT

aT

e ze u n e z

e z

z ee z

n

12

0 1 2 3 4

( ) ( ), 0atx t e u t a

( )anTe u n

10

1

1

1 1 1( )

2 2 1

1 1, | |

2 1

zaTn aTn n

aTn

aTaT

aT

e u n e ze z

e zz e

e z

Blue


DTFT & zT zT

x[n - N]

- a

Xs(f)

xs(t)

s=j2f

(1)

Xs(s)

xs(t)

esT = z

(4)

X(ej)

x[n]

z = ei

(2)

X(z)

x[n]

2 fT =

(3)

ej2 fT = z(6)

sT=j

(5)

( ) [ ] ( )s

n

x t x n t nT

( ) [ ] sTn

s

n

X s x n e

2( ) [ ] j nTf

s

n

X f x n e

( ) [ ]j j n

n

X e x n e

( ) [ ]n

n

X z x n z

[ ] ( )x n x nT

( )X fIf exists, all of do too. ( ), ( ), ( ), ( ), ( )j

s sX s X f X s X e X z


DTFT & zT zT

Basically, the discrete-time system does a weighted sum of shifted input sequences. Thus the time shift is the fundamental algorithm in the discrete-time system analysis.

( )[ ] [ ] ( )o on n nn

o

n n

x n n z x n z X z z Proof:

Verification:

(1) Time shift [ ] ( ),onz

o Xx n n z X z R

[ ] ( ) ojnj

ox n n X e e

[ ] ( )on

ox n n z X z

5. Properties of the z transform Pole-zero cancellation

je z

1

1

1

1[ ] ( 1) , | | 1

1

[ 1] ( 1) , | | 11

u n z zz

zu n z z

z

1[ ] ( 1)

1

j

ju n e

e

[ 1] ( 1) ( 1)1 1

j jj j j

j j

e eu n e e e

e e


DTFT & zT zT

1

1

[ ] [ ] ( )

( ), | |

n n n

o o

n n

o o X

z x n z x n z z

X z z z R

(2) Modulation 1[ ] ( ), | |zn

o o o Xz x n X z z z R

( )[ ] ( )o oj n j wDTFTe x n X e

1[ ] ( ), | |zn


oj

oe z

je z

Proof:

Verification:

( )

( 1) [ ] ( ),

( 1) [ ] ( ) ( )

n

X

n j j

x n X z R

x n X e X e

( 1)n j n n

oz e


DTFT & zT

1[ ] ( ), | |zn


1

1

1

1

1

1 2

1

1 2

1[ ] , | | 1

1

1[ ] , | | | |

1

1[ ] , | | 1

1

1[ ] , | | 1

1

1 cos( )cos( ) [ ] , | | 1

1 2cos( )

sin( )sin( ) [ ] , | | 1

1 2cos( )

o

o

o

o

n

j n

j

j n

j

oo

o

oo

o

u n zz

a u n z aaz

e u n ze z

e u n ze z

zn u n z

z z

zn u n z

z z


DTFT & zT zT

(3) Convolution sum [ ]* [ ] ( ) ( ),z

X Yx n y n X z Y z R R R

[ ]* [ ] ( ) ( )DTFT j jx n y n X e Y e

[ ]* [ ] ( ) ( ),z

X Yx n y n X z Y z R R R je z

Proof:

Verification:

[ ] [ ]* [ ] [ ] [ ]

( ) [ ] [ ] [ ] [ ]

( ) [ ] ( ) ( )

m

n n

n m m n

m

m

c n x n y n x m y n m

C z x m y n m z x m y n m z

Y z x m z Y z X z


DTFT & zT zT

( ) [ ]n

n

X z x n z

1' 1( ) [ ] [ ]n n

n n

X z n x n z z n x n z

'( ) [ ]n

n

z X z n x n z

Differentiating the above defining sum with respect to z on both sides,

( )[ ]

jDTFT dX e

n x n jd

'[ ] ( ),z

Xn x n z X z R

( ) ( ) ( )

/

jdX e dX z dX zjz

d dz jz dz

(4) Differentiation: '[ ] ( ),z

Xn x n z X z R

je z

Proof:

Verification:

,jje d dz jzd dz

' 1

1 1 2

1[ ] , | | | |

1 (1 )

n azna u n z z a

az az


DTFT & zT zT

2

1

( ) [ ] , 1 | | 2n

n

n n

X z x n z r z r

[ ] ( )DTFT jx n X e

1 1[ ] ( ),z

Xx n X z R

1 21 1

2 1

1

[ ] [ ] ( ), 1 | | 2

1 1( ), | |

2 1

n nn n

zn n n n

x n z x n z X z r r

X z zr r

(5) Folding: 1 1[ ] ( ),z

Xx n X z R

je z

1

1

1

[ 1] ( 1) , | | 11

1[ 1] ( 1) , | | 1

1

zu n z z

z

u n z zz

Proof:

Verification:


DTFT & zT zT

Useful z transform pairs

[ ]N

k N

n k

1

1

1

1 2

1

1 2

1

1 2

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

2 1 1/2

1[ ] , | | | |

1

1[ ] , | | 1

1

1 cos( )cos( ) [ ] , | | 1

1 2 cos( )

sin( )sin( ) [ ] , | | 1

1 2 cos( )

[ ] , | | | |(1 )

o

o

n

j n

j

oo

o

oo

o

n

N N

nN

a u n z aaz

e u n ze z

zn u n z

z z

zn u n z

z z

azna u n z a

az

z z

z z

1/2, 0 | |z

or z


DTFT & zT zT

6. The Inverse z Transform

( ) [ ] , {| | }n

X

n

X z x n z z r R

When |z| = r is contained in the ROC, i.e.,

( ) ( ) [ ] DTFT( [ ] )i

i n j n n

z ren

X z X re x n r e x n r

2

| |

1[ ] ( )

2

1( )

2

n j j n

n n

z r

x n r X re e d

dzX z z r

jz

, (0 2 )

,

j

j

re z

dzjre d dz d

jz

Taking the IDTFT,

Therefore

1

| |

1[ ] ( )

2

n

z r

x n X z z dzj

|z|=r is a closed counterclockwise contour in the complex z-plane lying entirely within the ROC RX


DTFT & zT zT

7. DLTI System analysis in the z transform domain

( )jX e DLTI h[n]

( ) ( ) ( )j j jY e H e X e

[ ]x n [ ] [ ]* [ ]y n x n h n

( )X z ( ) ( ) ( )Y z H z X z

Causality: [ ] 0 for 0h n n

20

[1] [2]( ) [ ] [0]n

n

h hH z h n z h

z z

Right-sided signal

H(z) does not contain any positive power of z. The system is causal if and only if

Stability: Absolutely summable h[n] leads to the convergence of H(ejω). The system is stable if and only if

the RH is the exterior of a circle including infinity.

the RH contains the unit circle.

|z|>r


DTFT & zT zT

Causality and Stability depending on possible ROCs

{| | 1} Xz R

jIm

Re1

{| | 1} Xz R

jIm

Re1

jIm

Re1

jIm

Re1

| | 0.7z 0.7 | | 2z

| | 0.7z | | 1.5z

Stable, causal Stable, anticausal

Unstable, anticausal Unstable, causal

Causal: ROC contains

a unit circle including infinity


DTFT & zT zT

0

1 0

[ ] [ ] [ ], ( 0)N M

k k

k k

y n a y n k b x n k b

Transfer Function Characterization of LTI Difference Systems

1 0

( )(1 ) ( )N M

j jk j jk

k k

k k

Y e a e X e b e

0

1

( )

1

Mjk

kj k

Njk

k

k

b e

H e

a e

For a given difference equation:

Taking the DTFT or the zT,

The transfer function is then given by the DTFT (zT) of the output divided by the DTFT (zT) of the input:

1 0

( )(1 ) ( )N M

k k

k k

k k

Y z a z X z b z

0

1

( )

1

Mk

k

k

Nk

k

k

b z

H z

a z


DTFT & zT zT

x[n] y[n]

y[n-1]

z-1

-a

+

+

한 표본지연

[ ] [ 1] [ ]y n ay n x n

Design of the first-order system

1

1( ) , | | | |

1H z z a

az

[ ] ( ) [ ]nh n a u n

This is the IIR (infinite impulse response) system.

or, the unit circle is not contained in the ROC and therefore the system is unstable

If |a|>1,

the system may be saturated and so the system is unstable.


DTFT & zT zT

1 2[ ] [ 1] [ 2] [ ]y n a y n a y n x n

1 2

1 2

1( )

1H z

a z a z

Design of the second-order system

[ ]y n[ ]x n

1z

1z

-a1

-a2

[ 1]y n

[ 2]y n


DTFT & zT zT

1 2

1 2

1 2

1 2

( )1

M

o M

N

N

b b z b z b zH z

a z a z a z

1

1( ) ( )M

o MY z b b z b z W z

1

1( ) 1 ( )N

NW z a z a z X z

1[ ] [ ] [ 1] [ ]o My n b w n b w n b w n M

1[ ] [ 1] [ ] [ ]Nw n a w n a w n N x n

Canonical form [ ]y n[ ]x n

1z

-a1

[ ]w n

1z

-a2

1z

-aN

-aN-1

b0

b1

b2

bM

(1) y[n] with w[n] (Forward):

(2) x[n] with w[n] (Feedback):

---- (A)

---- (B)

H(z)=(A)/(B)

Numerator polynomial of H(z)

Denominator polynomial of H(z)


DTFT & zT zT

[ ] 0.9 [ 4] [ ]y n y n x n

41/ 4

4 4

1( ) , | | 0.9

1 0.9 0.9

zH z z

z z

Example:

[ ]y n[ ]x n

4z

1/4 /2 40.9 ( 0,1,2,3) 0.9jk

kz e k z

4

4( )

0.9

jj

j

eH e

e

0.9

Comb filter (빗살여파기)

H(z) has 4 poles located at intervals of π/2 on the circle of radius 0.91/4 such as


DTFT & zT zT

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

12Comb filter: y[n] - 0.9 y[n-4] = x[n]

in units

4

.974r 1r

[ ] 0.9 [ 4] [ ]y n y n x n

Pole-zero diagram

| ( )|jH e

1/40.9 0.974


DTFT & zT

Impulse train of period N (N=3)

8. Sampling Techniques

Upsampling

Downsampling

Impulse Train Sampling

… … n

3[ ]x n

… …

[ ]x n

n

… …

3[ ]x n

n

… …

3[ ] [ ]x n i n

n

…

0 3 6 9

… 3 3[ ] [ ]i n i n

n

3[ ] [ ]x n i n

3[ ]x n

3[ ]x n

3 3[ ] [ ]i n i n


DTFT & zT

(1) Upsampling

[ ] [ ] [ ] ( )j n j Nm jN

Nn mN m

nx n x e x m e X e

N

[ ] [ ] [ ] ( )n Nm N

Nn mN m

nx n x z x m z X z

N

Expansion in Time Insertion of (N-1) zeros (No loss in signal) Compression in Frequency

… … n

[ ]x n

… … n

3[ ]x n

2 0

A )( jeX

max

3

3( ) ( )j jX e X e

2

0

A

max

3

2

3

4

3

,[ / ],

[ ] ( ), ( )0,

jN N

N

x n N n mNx n X e X z

n mN


DTFT & zT

Example of upsampling: Periodic impulse train of period N

3

3

2 2( ) ( ) ( )

3 3

j j

k

kI e I e

3 3[ ] [ ] [ 3 ]

k

i n i n n k

[ ] [ ]k

i n n k

2

3/43/2

2

3

0

… …

3( )jI e

… n

0 1 2 3 4 5 6

…

… n

0 3 6 9

… 3[ ]i n

[ ]i n

( ) [ ]

diric( , ) 2 ( 2 )

j j n j n

n k n

odd

k

I e n k e e

k

2 2[ ] [ ] ( ) ( )j

N N

k k

ki n n kN I e

N N

Impulse train

3 3

3( ) diric(3 , ) ( )j j k j

odd

k

I e e I e

Or, direct evaluation:


DTFT & zT

(2) Impulse training sampling

Shift 2k/3, k=0,1,2

Gain 1/3

… … n

[ ]x n

… … n

3[ ] [ ]x n i n

21

( )

0

1[ ] [ ] ( )

k

N

Nj

N

k

x n i n X eN

23

23

( )132 2

2 212 3 32

2( ) 21

3 320

2( )1

3

0

( ) ( ) ( )

( ) ( )

( ) ( )

( )

k

k

j j j

j k

k

j k

k

j

k

Y e X e I e d

X e d

X e d

X e

Let 3[ ] [ ] [ ]y n x n i n

Periodic convolution

For every k=0,1,2, unit area


DTFT & zT

(3) Downsampling (decimation):

Expansion in Frequency

Upsampling

3

3

[ ] [ ]

( ) ( )j j

y n d n

Y e D e

Compression in Time 21

( )

0

1[ ] [ ] ( )

k

N

Nj

Nk

x n x Nn X eN

… … n

3[ ] [ ] [3 ]d n x n x n

… … n

[ ]x n

… … n

3[ ] [ ] [ ]y n x n i n Impulse train sampling 2

3

2( )

0

1( ) ( )

3

kjj

k

Y e X e

2 2

3 3

2 1( ) ( )

0 0

1 1( ) ( ) ( ) ( )

3

k k

N

Nj j jj

k k

D e Y e X e X eN


DTFT & zT

Spectrum of downsampled sequence

2 0

A )( jeX

21

0

1( ) ( )

kN

Njj

Nk

X e X eN

20

A/N

max

maxN

max

N

To prevent aliasing,

21( )

0

1[ ] [ ] ( )

k

N

Nj

Nk

x n x Nn X eN


DTFT & zT

Impulse sampling, upsampling, downsampling

… …

n

… …

… …

3[ ]x n

[ ]x n

… …

3[ ]x n

3[ ] [ ] [ ]y n x n i n

2 3/43/2

/ 3A

0

BW

2

/ 3A

0

3BW

2

)( jeXA

0

BW

21( )

0

1( ) ( )

kN j wj N

k

Y e X eN

2 3/43/2

A

0

/ 3BW( ) ( )j jN

NX e X e

21

0

1( ) ( ) ( )

k

N N

Nj jj

Nk

X e Y e X eN

Bandwidth


DTFT & zT zT

21

( ) ( )k

jj N

Nk

X e X eN

(4) Reconstruction of decimated signals [ ]

Nx n

( )jH e N N

[ ]N

x n

[ ]x n [ ]y n[ ]uy n

2( )1

( ) ( ) ( )k

jj jN N

u Nk

Y e X e X eN

( ) ( )* ( 2 )2

j

k

NH e N k

N

N

N

( )jH e

( ) ( ) ( ) ( )j j j j

uY e Y e H e X e

Assume that a signal x[n] was downsampled to without aliasing.

That is, , or x[n] was N times over sampled.

Then x[n] can be recovered from by first upsampling by N followed by an ILPF of cutoff frequency .

N max

[ ]N

x n

c N

[ ]N

x n

N max

Perfect reconstruction


DTFT & zT

2

)( jeX

3BW ( )j

NX e

0

A

2

0

max 4

maxN

/A N

No aliasing.

max4 3

f

( )X f

0

AT

1khzmf

1 8khzs Tf

max 42 mTf

( )j

UY e

2

0

/A N

max 2

N

4

N

2

N

2

( )jH e

0max

N

N

(N=3)


DTFT & zT zT

[ ] [ ] [ ]y n ay n N x n

1/1( ) , | | | |

1

N

N NH z z a

az

[ ]y n[ ]x n

Nz

a

1 1

1( ) ( ) , | | | |

1H z H z z a

az

(N=1):

1[ ] [ ] [ ]nh n h n a u n

1( ) ( ) ( )N N

NH z H z H z (N>1):

1[ ] [ ] [ ]N N N

h n h n h n

Upsampled h1[n] by N

(insert N-1 zeros between samples)

(1) Unit sample response

→ hN = 1, z, a, z, a2, z, a3, . . .

z= (N-1) zeros: z= 0,0,. . .,0

→ h = 1, a, a2, a3, . . .

9. Feedback comb filter analysis


DTFT & zT

(2) Unit step response (N=1)

n 0 1 2 3 …

h[n] 1 a a2 a3 …

h[n-1] 1 a a2 …

h[n-2] 1 a …

… …

s[n] 1 1+a 1+a+a2 1+a+a2+a3 …

Running sum

Sum of geometric series

10 0

1 0 0

2

1[ ] [ ] [ ]* [ ] [ ] [ ]

1

[ ] [ 1] [ 2]

(1 ) [ ]

nn n k

k k

n

as n s n h n u n h k a u n

a

h n h n h n

a a a u n


DTFT & zT

Unit step response (N>1)

Ex.: (N=2)

n 0 1 2 3 4 5 …

h2[n] 1 0 a 0 a2 0 …

h2[n-1] 1 0 a 0 a2 …

h2[n-2] 1 0 a 0 …

h2[n-3] 1 0 a …

h2[n-4] 1 0 …

…

sN[n] 1 1 1+a 1+a 1+a+a2 … …

[ ] [ ]*ones(1, )Ns k s k N : N times repetition of s[k]

0

0

[ ] [ ]* [ ] [ ] [ 1] [ 2]

[ ]

[0] [1] [2] [ ]

N N N N N

n

Nk

N N N N

s n h n u n h n h n h n

h k

s s s s n


DTFT & zT zT

1 1

1( ) ( ) , | | | |

1H z H z z a

az

(N=1):

(3) Unit step response in zT domain

Sum of geometric series

1

1 0

1[ ] [ ] [ ] [ ]

1

nn k

k

as n s n u n a u n

a

1 1 1

11 1 1

1 2

1 1( ) ( )

1 1

1

1 1

1

1 (1 )

a

S z S zz az

a

z az

a z az

n 0 1 2 3 …

s[n] 1 1+a 1+a+a2 1+a+a2 +a3 …

1[ ]s n[ ]n

1z

1z

1+a

-a


DTFT & zT zT 1/1

( ) , | | | |1

N

N NH z z a

az

(N>1):

[ ] [ ] [ 1] [ 2] [ 1]N N N N Ns n s n s n s n s n N

1 1

1 2 ( 1)

1

1 1 1 1 1( )

1 1 1 1 1

1( ) 1 ( )

1

N

N N N N

NN N N

zS z

z az z z az

zS z z z z S z

z

Unit step response in zT domain

1. S↑N[n]: Upsampling s[n] by N

2. Sum of all shifted S↑N[n-k] for k=0:N-1, i.e.,

Replacing N-1 zeros occurred by upsampling with S↑N[n].

n 0 1 2 3 …

s[n] 1 1+a 1+a+a2 1+a+a2 +a3 …

s↑2[n] 1 0 1+a 0 1+a+a2 0 …

s↑2[n-1] 1 0 1+a 0 1+a+a2 …

s2[n] 1 1 1+a 1+a 1+a+a2 1+a+a2 …

EX.: (N=2)


DTFT & zT

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

Unit sample response h[n]

(1) y[n] - 0.8y[n - 1] = x[n]

0 2 4 6 8 100

2

4

6

5

Unit step response uo[n]

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

Unit sample response h[n]

(2) y[n] - 0.8y[n - 4] = x[n]

0 5 10 15 20 25 30 35 400

2

4

6

5

Unit step response uo[n]

Unit sample and step responses

(the 1st and 4 th comb filters): class_zt05.m


DTFT & zT

Frequency responses

(the 1st and 4 th comb filters)

-1 -0.5 0 0.5 10

2

4

6

in units

(1) |H(e j)|: H(z)=1/(1-0.8z-1)

-1 -0.5 0 0.5 10

2

4

6

in units

(2) |H(e j)|: H(z)=1/(1-0.8z-4)

1 1 1

1 1(1) | ( ) | | |: ( )

1 0.8 1 0.8

j

jH e H z

e z

4 4

4 1 4 14 4

1 1(2) | ( ) | | | | ( ) |: ( ) ( )

1 0.8 1 0.8

j j

jH e H e H z H z

e z


DTFT & zT

= 0

|H(ej

)|

= -/2

0.8<r1

Mesh surface of X(rej

) for 0.8<r1

unit circle |z| = 1

= /2

|H(rej

)|: H(z) = 1/(1 + 0.8z-1

)


DTFT & zT zT

1 0

[ ] [ ] [ ]N M

k k

k k

y n a y n k b x n k

1 2

1 2

1 2

1 2

( )1

M

o M

N

N

b b z b z b zH z

a z a z a z

10. DLTI system Analysis with MATLAB

y = filter(b, a, x)

1D filtering of x, column by column

r = filter( a, b, y)

1D inverse filtering of y, column by column

x[n] r[n] = x [n] y[n]

(b, a) (a, b)

Cascaded system

1 2

1 2

[ , , , , ]

[1, , , , ]

o M

N

b b b b b

a a a a

Coefficient vectors


DTFT & zT zT

(1) Pole-zero plot

zplane(b,a)

(2) Frequency response H(ejw) for |w|pi

H = freqz(b,a, pi*[-100:100]/100)

(3) Impulse response h[n] for n=0:49

h = impz(b,a,50)

h = filter(b,a,[1, zeros(1,49)] ) % size of h: 1x50

h = filter(b,a,[1, zeros(1,49)]’ ) % size of h: 50x1

(4) Step response s[n] for n=0:49

s = stepz(b,a,50);

s = filter(b,a,ones(1,50) ) % size of s: 1x50

s = filter(b,a,ones(1,50)’ ) % size of s: 50x1

Plotting the pole-zero, frequency response, h[n], s[n]


DTFT & zT zT

(1) Image reading

f = imread('flower.jpg’), s = size(f), % (502x753x3): 3D color

(2) Color components

fr = f(:,:,1); size(fr), % Red component (502x753)

fg = f(:,:,2); % Green

fb = f(:,:,3); % Blue

(3) 1D filtering of each color component (Column by column)

yr = filter(b,a,fr)

yg = filter(b,a,fg)

yb = filter(b,a,fb)

(4) Concatenate each component.

y = cat(3,yr,yg,yb)

Image processing

y = filter(b,a,f)


DTFT & zT zT

save class y % Store y in the file class.mat

load class % Take y from class.mat again

whos % List current variables

imshow(y) % Display y

Image saving and loading


DTFT & zT

Given image x


DTFT & zT

Columnwise 1D filtering result y = filter(b,a,x): b = [1 -1 1 -1 1]/5, a = [1-0.8]

dtft & zt the z transform

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