Z Transform When the transform is identical to DFT (11)

z Transformz Transform

X z x n z n

n

( ) ( )

When the transform is identical to DFT

(11)

X z x n z

X e x n e x n e

n

z en

j j n

n

j kn N

n

j k

k k

( ) ( )

( ) ( ) ( ) /

2

z e j k

z Transformz Transform

(12)

X z x n z

X e x n e x n e

n

z en

j j n

n

Nj kn N

n

N

j k

k k

( ) ( )

( ) ( ) ( ) /

0

12

0

1

The sequence x(n) is non-zero within [0,N-1], hence

The transform is identical to DFT

Some basic properties of z TransformSome basic properties of z Transform

H z h n z n

n1( ) ( )

1. Basic definition1. Basic definition

h n ah n bh n H z aH z bH z( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2

2. Linearity2. Linearity

x n k z X zk( ) ( )

3. Delay3. Delay

y n h n x n Y z H z X z( ) ( ) ( ) ( ) ( ) ( )

4. Convolution4. Convolution

Essential properties of z TransformEssential properties of z Transform

x n z z X z x n z X z

x n z x n

( ) ( ) ( ) ( )

( ) ( )

1 1 1

1

1

1

and

5. Multiplication by ‘z-1’5. Multiplication by ‘z-1’

x n X z( ) ( )

6. Relation between X(z) and X(-z)6. Relation between X(z) and X(-z)

( ) ( ) ( ) 1 n x n X z

x n X z( ) ( )

7. Relation between X(z) and X(z-1)7. Relation between X(z) and X(z-1)

x n X z( ) ( ) 1

Why z Transform?Why z Transform?

1. z transform can be used to calculate DFT.

2. Filter architecture can be deduced directly from the transfer function in the z domain.

Specify filter characteristics(LP, HP, BP...)

Determine transfer

function H(z)

Determine filter sturcture

Filter structure can be inferred from H(z)

Figure 21

y (n) = ak y(n-k) +

k=1

Mbk x(n-k)

k= -NF

NF

(13)

Generalised finite order LTI system

Making use of the delay property, equation (4) can be rewritten as

x n z x n( ) ( ) 1 1

y (n) = ak z-k y(n) +

k=1

Mbk z-k

x(n)k= -NF

NF

(14)

y (n) = ak z-k y(n) +

k=1

Mbk z-k

x(n)k= -NF

NF

(14)

z

b-NF

z-1

bNF

b0 + +

z-1

a1

z-1

aM

x(n) y(n)

Figure 22

y (n) = ak z-k y(n) +

k=1

Mbk z-k

x(n)k= -NF

NF

(14)

z transform

Y (z) = ak z-k Y(z) +

k=1

Mbk z-k

X(z)k= -NF

NF

(15)

Note the similarity between the time and z domain

Digital Filter DesignDigital Filter Design

Specify filter characteristics(LP, HP, BP...)

Determine transfer

function H(z)

Construct filter sturcture

Y (z) = ak z-k Y(z) +

k=1

Mbk z-k

X(z)k= -NF

NF

Rarrange H(z) to the form

Figure 21

Example ( ) =

( )

( )

1- cos

cos

0

0

H zY z

X z

r z

r z r z

1

1 2 21 2

(z) cos ( ) - cos0 0Y r Y z z r Y z z X z r X z z 2 1 2 2 1 ( ) ( ) ( )

(z) ( ) - 1Y a Y z z a Y z z b X z b X z z 1

12

20

1( ) ( ) ( )

z-1

b1

b0 + +

z-1

a1

z-1

a2

x(n) y(n)

Figure 23

Poles and Zeros of Transfer FunctionPoles and Zeros of Transfer Function

(z)H

b z

a z

kk N

Nk

kk

Mk

F

P

11

Az

c z

d z

Nk

k

N N

kk

MF

P F

1

1

1

1

1

1

= (16)

Poles and zeros are values of ‘z’ which results in H(z) = infinity and zero, respectively

H(z) can be divided into 3 groups for NF > 0 :

Az N F 1 1

1

c zkk

N NP F 1 1

1

d zkk

M

1. 2. 3.

Poles and Zeros of Transfer FunctionPoles and Zeros of Transfer Function

Gourp 1 Az N F NF poles at z = and NF zeros at z = 0

Gourp 2 a zero at z = ck and a pole at z = 0 for each term from k=1 to NP+NF

1 1

1

c zkk

N NP F

Gourp 3 a zero at z = 0 and a pole at z = dk for each term from k=1 to M

1 1

1

d zkk

M

Example ( ) =H z z z1

311

( ) =H z Az c z c z1 111

21

The transfer function can be expressed as

= 1

3 and A c

jc

j, ,1 2

1 3

2

1 3

2

Term Group Pole Zero

z 1 0 2 0 c1

1 11 c z

2 0 c2 1 21 c z

Is a filter useful?Is a filter useful?

A filter transfer function H(z) is only useful if :

1. It is stable

2. It is finite

Stability of Transfer FunctionStability of Transfer Function

Given a system with unit-sample response

h(n) = [h(0), h(1), ....., h(N-1)] with z transform given by

H(z).

The system is stable if ( )k=-

h k

Finite sequence is generally stable as the absolute sum of finite sample values is always finite

(17)

Example ( ) = 1

0

h n

for n

otherwise

0

( )k=-

h kk

10

The system is unstable


H z

b z

a z

kk

k N

N

kk

k

MF

P

11

Generalized LTI transfer function

b zkk

k N

N

F

P

The numerator is < for finite NP and NF

1

11

a zk

k

k

MThe denominator can lead to unstability


Stability of a digital system depends on the pole locations that are contained in

k

kk

M

d z1 11

M

k

kk za

zH

1

1

1

after partial fraction decomposition

M

k

nkk

M

kk nudnh)n(h

11

It can be easily shown that


M

k

nkk

M

kk nudnh)n(h

11

For a stable system,

h n d u nkn

k kn

n

0 0

k kn

n

d0

For finite summation result, |dk| < 1.



h n d u nkn

k kn

n

0 0

k kn

n

d0


dk are the poles of H’(z)

M

k

nkk

M

kk nudnh)n(h

11



h n d u nkn

k kn

n

0 0

k kn

n

d0


dk are the poles of H’(z)

|dk| < 1 means that the poles of a stable system must lies within the unit circle in the z plane.

M

k

nkk

M

kk nudnh)n(h

11

Region of Convergence (ROC) of Transfer FunctionRegion of Convergence (ROC) of Transfer Function

A transfer function H(z) is only useful if it is finite, i.e.,

H z h n z n

n

Two classes of digital filters

y (n) = ak y(n-k) +

k=1

Mbk x(n-k)

k= -NF

NF

(13)

The generalised finite order LTI system

1. Finite Impulse Response (FIR) Filter

2. Infinite Impulse Response (IIR) Filter

formulates an IIR filter

Two classes of digital filters

y (n) = bk x(n-k)k= 0

N-1(18)

When ak = 0 for all values of k,

formulates an FIR filter

h(n) = [h(0), h(1), ....., h(N-1)] = [b0, b1, .......bN-1]

As N is finite, according to eqn. (17), FIR filter is inherently stable

FIR filters

z-1 z-1 z-1

h(N-1)

x(n)

y(n)

Figure 24

h(2)h(1)h(0)

H z h k z

h z h z h N

z

k

Nk

N N

N

0

1

1 2

1

0 1 1..... (19)

FIR filters

Finite Impulse Response (FIR) Filter can guarantee linear phase

With linear phase, all input sinsuoidal components are delayed by the same amount. Consider

x n X z x n k X z z k

x n k X e ej j k In the frequency domain,

Phase delay for frequency = k

Phase Distortion - example

x n cos n cos n1 0 02( ) Given:

Y e H e X e X e ej j j j jk

According to previous analysis,

y n cos n k cos n k( ) ( ) ( ) 0 02

y n x n h n( ) ( ) ( ) 1

and

H e ej j( )

x n k X e ej j k

(a linear phase X-function)

(Same signal as before, only delay added to each sample)

Phase Distortion - example

x n cos n cos n1 0 02( ) Given:

y n cos n cos n( )

0 04

2

y n x n h n( ) ( ) ( ) 1

and

H efor

forj( )

/

/

4

0 3 2

3 20

0

y(n) is not the same as x1(n)

(a non-linear phase X-function)

Analogue Transfer Function H(s)

Inverse Fourier Transform h(t)

Sample Impulse Response h[n]

Corresponding Transfer Function H(z)

H(z) = H(s)?

h(t)

t

Analogue filter: y(t) = x(t) * h(t)

Y(s) = X(s)H(s)

Analogue filter: y(t) = x(t) * h(t)

Y(s) = X(s)H(s)

as

sH

1

Given

Applying inverse Laplace Transform

ateth

h(t)

t

1 unit

Digital filter: Sampled and Digitized x(t) and h(t)

x(t) -> x(n) , h(t) -> h(n)

y(n) = x(n) * h(n)

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

h(t)

t

1 unit

H z h n z n

( )0

If h(t) is sampled at unit interval, we have

h(t)

t

1 unit

H e h n ej jn

( )0

e ean jn

0

1

1 e ea j

h(t) = e-an for t > 0

However, h(t) is sampled at interval of TS instead of the following,

h(t)

t

1 unit

0

0

)(

)(

m

njj

n

n

enheH

znhzH

Does sampling rate affects the above transfer function?

If h(t) is sampled at interval of TS,

h(t)

t

TS unit

1. h(n) will be replace with h(nTS) 2. Frequency will be scaled by 1/TS

Answer this question by computing the transfer function again based on the new sampling rate


h(t)

t

TS unit


0

// )(n

TjnS

Tj SS enTheH


h(t)

t

TS unit


0

// )(n

TjnS

Tj SS enTheH

0

''

)(n

jnS

j enTheH


h(t)

tTS unit


0

// )(n

TjnS

Tj SS enTheH

0

''

)(n

jnS

j enTheH

0

'

)(n

nS znThzH

The transfer function has similar form as before, may not need to re-compute the transfer function again.

Given (sampling period of 1 unit)

0n

nz)n(hzH

If sampling period changes to TS , then

0

'

)(n

nS znThzH

Suppose dtethjH tjAA

0

0

'

)(n

nS znThzH

If the analogue unit response is sampled by a period TS ,

Represents the analogue transfer function.

Is the digital response similar to the analogue response?

Suppose dtethjH tjAA

0

If the analogue unit response is sampled by a period TS ,

Represents the analogue transfer function.

Digital and Analogue responses are equal if the maximum frequency of the signal is restricted to

k

SSATj TkjTjHeH S )/2/(/

ST

M

(otherwise the images start to overlap each other)

(a single spectrum becomes an infinite string of replicas)

Given

1

1 e eaTjTS S

STjeH /

'

'

1

1

jaT

j

eeeH

S

11

1

ze

zHSaT

jaj

eeeH

1

1 11

1

ze

zHa

Given a desire response HD(), find hD(n)

h n H e e dD Dj j n( ) ( )

1

2

Applying inverse Fourier Transform, we have,

(20)

Noted that: n is extended to These kind of filter is not available in practice

with n being infinite and assuming Impulse Invariant, the digital transfer function will be identical to the Analogue ones.

In practice, the FIR structure in figure 24 cannot be infinite, hence n is restricted by a window function wR(n)

2/1-Nn1)/2-N-

for

nhnwnh DRN (21)

jR

jD

jN eWeHeH (22)

circular convolution

Rectangular Window

Hanning Window

Hamming Window

Blackman Window

1

2cos5.05.0

N

nnw

otherwise

Nnnw

0

2/11

1

2cos46.054.0

N

nnw

1

4cos08.0

1

2cos5.042.0

N

n

N

nnw

Analogue Transfer Function H(s)

Inverse Fourier Transform h(t)

Sample Impulse Response h[n]

Apply window function w[n]h[n]

Modified Transfer Function H’(z)

Assume Impulse Invariant

Consider a low pass response H(s)

|H ()|

c-c 0 -

Without window

c-c 0 -

|H ()|

With window

Side lobes

1. Side lobes decreases stop band attenuation A

c-c 0 -

|H ()|

A

Window A (dB)

Rectangular 21

Hanning 44

Hamming 55

Blackman 75

2. Window determines the length of the FIR filter

Consider a Low Pass Frequency Response

P0

|H ()|

S

Desired Pass Band Edge Frequency

Stop Band Edge Frequency

Transition Width (TW)

Actual Pass Band Edge Frequency

Relations between the Window, Filter length and A

z-1 z-1 z-1

h(N-1)

x(n)

y(n)

h(2)h(1)h(0)

N

75Blackman

55Hamming

44Hanning

21Rectangular

A (dB)Window

TW

f s91.0

TW

f s32.3

TW

f s44.3

TW

f s98.5

fs is the sampling frequency

|HD()|1

c-c 0 -

Figure 25

H efor

oDj c

1

0

therwise c

An ideal LPF

h n e d e dDj n j n( )

1

21

1

2

n

nsin c

n -n

Figure 26

Due to windowing, the pass band edge frequency will be shifted to higher frequency end.

Usually the revised pass band edge is taken to be the middle point of the Transition Region, i.e. 2

TWc

hD(n) will be revised as n

nsin

Next select the window that complies with the stop band attenuation A.

Determine the length of the filter N from fS and TW

The Prolate Spheroidal Wave Sequence (Slepian 1978)

The Prolate sequence is a real sequence of lengh N+1 and unit energy.

Aims at mininizing ripple energy

2jeV

0

Optimal energy concentration at low frequency

Assuming the sequence is casual and of length N+1, i.e.,

N

n

nznvzV0

deV j

S

2

Since the sequence is unit energy, we have

110

2

0

2 d

eVnv jN

Compute the energy contained in the sequence

Define an objective function to denote the pass band energy as

deV j

2

0

Maximizing is equivalent to minimizing the stop band energy

i.e., Optimal energy concentration at low frequency

jj eneV evT

veve !* jjjjj eeeVeVeV 2

Let TNvvvv .....v 210 TNzzzz .....e 211

vRv

ddeV j

0

T2

0

R is a (N+1)x(N+1) matrix with the (m,n) element equals to

nmjnme nmj sincos

0T vQv

Splitting R() into real and imaginary parts as

QPR j

Nnmmn nm

nmdnmp

,

sincos 00

vPv T

Each entry in

vvvv

dP

dR

deV j

0

T

0

T2

0We have

is real, hence

and

dP0

Maximization based on Rayleigh’s principle

1 kk vPv

econveniencfor dropping TT PvvvPv

The objective function is maximized if v is the eigenvector corresponding to the maximum eigenvalue of P.

This can be found by the “power method” which is an ilterative process starting with an arbitrary value for v0.

vvvvPvv TTT

maxmax

Search for v in a vast space!

Step 1 0 kSet

Step 2 kvCompute P

Step 3 kk vves eigenvaluthe set ofFind P :

Step 4 kgenvaluemaximum eiFind :

Step 5 kkk PvvCompute /1

Step 6 2 1 StepGotokk ,

Method does not guarantee the best (optimal) solution and depends on the initial choice of vk

Better optimization methods:

Simulated Annealing. Binary Genetic Algorithm. Real Coded Genetic Algorithm. Particle Swarm Optimization.

0v

kv

Stuck in comfort zone!

Drive to higher value, but not the other way round

maxvBest position, needs to leave the comfort zone first

x

Optimal window does not imply optimal filter.

nN

n

znhzH

0

A LPF is optimized if the passband and stopband error are minimum. Consider a linear phase FIR filter. N is even and h(n)=h(N-n)

R

Mjj HeeH

cbcos/

TNM

nnR nbH

2

0

The amplitude response is:

and

Pbbbccb TTTjS

ddeHE

ss

2

The stopband energy is

bcbc Tj

RHeH 22

dnmp

s

mn coscos

The m,n entry of P is

ES should be zero, otherwise it will become the “stopband error”.

cbcbb 1TTT

The stopband error is

TTRH bcb 00

d

where

EP

T

TP

0

11 ccQ

Qb,b

The error energy at passband =

The pass band should be flat from DC onwards as

RbbQbbbPb TTT 1

Which can be rewritten as

PS EE 1

QPR 1

Where

An objective function can now be defined as

The minimum value corresponding to the eigenvector of R with the minimum eigenvalue.

However the transfer functions of FIR filters are restricted in certain form (as shown below). IIR filters are more flexible and have a counterpart in analog filters.

FIR filters are to design, optimize, and implement. It is also possible to select different window functions to adjust the stop band attenuation.

H z h k z

h z h z h N

z

k

Nk

N N

N

0

1

1 2

1

0 1 1.....

Eqn. (19) shows that the transfer function of FIR filters are restricted in certain form. IIR filters are more flexible and have a counterpart in analog filters.

Time signal Filter Output

Analog Transfer Function

Digital Transfer Function

DigitalFilter Output

Figure 27

Stability of filtersAssigning poles at s=-a to H(s) and s=a to H(-s) results in a stable system, i.e.,

H sa s

1

General form of Transfer Function

H s

s z s z s z s z

s p s p s p s p

i N

j N

1 2

1 2

.... ....

.... ....(25)

Stability of filtersAssigning poles at s=-a to H(s) and s=a to H(-s) results in a stable system, i.e.,

H sa s

1

A pole at s = -pj results in a -20db per decade drop at f = pi

A zero at s = -zi results in a +20db per decade rise at f = zj

H s

s z s z s z s z

s p s p s p s p

i N

j N

1 2

1 2

.... ....

.... ....

General form of Transfer Function

(25)

f

db p1 z1 z2 p2

Figure 28

Img

Re

z plane

j

s plane

z=1z=-1

TS

TS

s=0

The problem is obvious: define on the unit circleis limited, extends to infinity!

fs

20

fs

16

2fs

16

4fs

16

3fs

16

5fs

16

6fs

16

7fs

16

fs

160

2fs

16

3fs

16

7fs

16

6fs

16

5fs

16

4fs

16

- Digital Frequency - Analogue Frequency

For a digital N point sampling lattice, the maximum frequency it can represent is fs/2 (i.e. 1/2TS or rads/s)

The frequency resolution is 2/N

N=16

extends to infinity!

fs

20

fs

16

2fs

16

4fs

16

3fs

16

5fs

16

6fs

16

7fs

16

fs

160

2fs

16

3fs

16

7fs

16

6fs

16

5fs

16

4fs

16




N=16fs

16 2fs

16

3fs

16

4fs

16

fs

2= /TS


fs

20

fs

16

2fs

16

4fs

16

3fs

16

5fs

16

6fs

16

7fs

16

fs

160

2fs

16

3fs

16

7fs

16

6fs

16

5fs

16

4fs

16




N=16fs

16 2fs

16

3fs

16

4fs

16

fs

2

9fs

16

10fs

16?


Z Transform When the transform is identical to DFT (11)

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Transcript of Z Transform When the transform is identical to DFT (11)