3

1Digital Signal Processing A.S.Kayhan

DIGITAL SIGNAL

PROCESSING

Part 2

Digital Signal Processing A.S.Kayhan

Frequency Analysis of LTI Systems:the Input/Output (I/O) relation of a LTI system is given by convolution:

k

knxkhnhnxny ][][][*][][

In z-domain)()()( zHzXzY

jez

jjj eHeXeY

)()()(

On the unit circle

where is the frequency response of the system.

)( jeH


)()()( jjj eHeXeY

Magnitude is

where is the magnitude response of the system.

)( jeH

Phase is )()()( jjj eHeXeY

where is the phase response of the system.

)( jeH


Discrete-time ideal filters: LPF, HPF, BPF:

These are not realizable. Why?


Phase distortion and delay:Assume that ][][ did nnnh

thendnjj

id eeH )(

ord

jid

jid neHeH

)(,1)(

Linear phase distortion causes simple delay.

Group delay: It measures linearity of the phase response.Consider nnsnx ocos][][ where is a narrowband (slowly varying) signal.][ ns


Assume around o

dojj neHeH )(,1)(

Then doood nnnnsny cos][][

Thus, group delay of a system is

)(arg)]([

jj eHd

deHgrd

where is the continuous phase. )(arg jeH



Example: Consider the following filter


Input is the following signal


Pulses are at 85.0,5.0,25.0


Systems with Difference Equation Models:Assume that the Input/Output (I/O) relation of a system is given by a constant coefficient difference equation:

M

kk

N

kk knxbknya

00

][][

Applying z-transform, using linearity and shifting properties, we obtain

M

k

kk

N

k

kk zXzbzYza

00

)()(

)()(00

zXzbzYzaM

k

kk

N

k

kk


Then, the transfer function (or system function) is

N

k

kk

M

k

kk

za

zb

zX

zYzH

0

0

)(

)()(

We can write the transfer function in terms of its poles, dk, and zeros, ck, as

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(


Example: Consider system function

)4

31)(

2

11(

)1()(

11

21

zz

zzH

)(

)(

83

41

1

21)(

21

21

zX

zY

zz

zzzH

then

)()8

3

4

11()()21( 2121 zYzzzXzz

]2[8

3]1[

4

1][

]2[]1[2][

nynyny

nxnxnx


Stability and Causality: A system is causal if ROC for the transfer function H(z) is outward.A sytem is stable if ROC for the transfer function H(z) includes the unit circle.


Inverse Systems :Let be the inverse system of , then )( zH)( zH i

1)()()( zHzHzG i

then

)(

1)(,

)(

1)(

j

jii eH

eHzH

zH

Not all systems have an inverse. Ideal LPF does not have an inverse, we can not recover high frequency components.Now, consider

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(

nnhnhng i *Eq.(1)


Then the inverse system has

M

kk

N

kk

o

oi

zc

zd

b

azH

1

1

1

1

)1(

)1()(

Poles become zeros of the inverse system, zeros become poles. For Eq.(1) to hold, ROC of and must overlap. If is causal, ROC is

)( zH i )( zH)( zH

kdz max

Some part of ROC of must overlap with this.)( zH i


Example: Suppose

9.0,9.01

5.0)(

1

1

zz

zzH

The inverse system is

1

1

1

1

21

8.12

5.0

9.01)(

z

z

z

zzH i

Two possible ROC:

2z nununh nni 11 28.1121which is stable but noncausal.

2z 128.12 112

nununh nni

unstable but causal.


Example: Suppose

9.0,9.01

5.01)(

1

1

zz

zzH

1

1

5.01

9.01)(

z

zzH i

The inverse system is

The only choice for ROC is overlaps with is 9.0z

5.0z

then

15.09.05.0 1 nununh nni

which is both stable and causal.

10


Impulse Response for Rational System:Assume that a stable LTI system has a rational transfer function.

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(

N

k k

kNM

NMifr

rr zd

AzBzH

11

00 1

)(

then

N

k

nkk

NM

rr nudArnBnh

10

1

Impulse response is of infinite length, called Infinite Impulse Response (IIR) system.


If the system has no poles, then

M

k

kk zbzH

0

)(

M

kk knbnh

1

Impulse response is of finite length, called Finite Impulse Response (FIR) system.

Example: Consider a causal system with][]1[][ nxnayny

If then stable and the impulse response is

1a

.nuanh n

11


Example: Consider .

otherwise,0

0,

Mna

nhn

Then

1

11

0 1

1)(

az

zazazH

MMM

n

nn

Zeros at

.,,1,0,12

Mkaez Mk

j

k

Difference equation is

However, we can also write

1][]1[][ 1 Mnxanxnayny M

M

k

k knxany1


Frequency Response for Rational System Functions:Assume that a stable LTI system has a rational transfer function. Then frequency response is obtained by evaluating it on the unit circle:

N

k

kjk

M

k

kjk

j

ea

eb

eH

0

0)(

N

k

jk

M

k

jk

o

oj

ed

ec

a

beH

1

1

)1(

)1()(

12


The magnitude response of the system is

N

k

jk

M

k

jk

o

oj

ed

ec

a

beH

1

1

1

1)(

The magnitude-squared response of the system is

)()()( *2 jjj eHeHeH

N

k

jk

jk

M

k

jk

jk

o

oj

eded

ecec

a

beH

1

*

1

*2

2

11

11)(


Log magnitude or Gain in decibels(dB) :

N

k

jk

M

k

jk

o

oj

ed

eca

beH

110

1101010

1log20

1log20log20)(log20

Attenuation in dB = - Gain in dB

Note that :

)(log20

)(log20)(log20

10

1010

j

jj

eX

eHeY

13


Phase response for rational system function:

N

k

jk

M

k

jk

o

oj

ed

eca

beH

1

1

1

1)(

Group delay for rational system function:

N

k

jk

M

k

jk

j

edd

d

ecd

deH

1

1

]1arg[

]1arg[)(grd


Frequency Response of A single Zero:Consider transfer function of a system as

11)( azzH

then with jrea jjeHjjj ereeeHeH

j 1)()( )(

then magnitude is

and magnitude in dB is

)cos(21)( 22

rreH j

)]cos(21[Log20)(Log20 21010 rreH j

14


then phase is

)cos(1

)sin(tan)( 1

r

reH j

group delay is derivative of the (unwrapped) phase function

d

eHdeHgrd

jj ))(()]([

Example: Consider two cases : r=0.9, =0 and r=0.9, =:


15


magnitudes of vectors give the magnitude response

31

3)( vv

v

e

reeeH

j

jjj

jezz

azzH

)(


phases of vectors give the phase response

31313

)(

vv

e

reeeH

j

jjj

16


Example: Consider a second order system with

11

1

11

1)(

zrezre

zezH

jj

j

21

3)(vv

veH j


jez

j zHeH

)()(

:

17


Example: Consider a third order system with

211

211

7957.04461.11683.01

0166.11105634.0)(

zzz

zzzzH


Relation Between Magnitude and Phase:In general, knowledge about magnitude does not provide information about phase, and vice versa.But, for rational system functions, with some additional information such as number of poles and zeros, magnitude and phase responses provide information about each other.Consider

jez

jjj

zHzH

eHeHeH

|)/1()(

)()()(

**

*2

with

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(

18


N

kk

M

kk

o

o

zd

zc

a

bzH

1

*

1

*

**

)1(

)1()/1(

Then

N

kkk

M

kkk

o

o

zdzd

zczc

a

bzHzHzC

1

*1

1

*12

**

)1)(1(

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

Notice that poles and zeros of C(z) occur in conjugate reciprocal pairs. If one pole/zero is inside the unit circle there is another outside.


If H(z) is causal and stable, then all poles must be inside the unit circle, with this we can identify the poles. But zeros of H(z) can not be uniquely identified from zeros of C(z) with this constraint alone.

Example: Consider two stable systems with

14/14/

11

1 8.018.01

5.0112)(

zeze

zzzH

jj

and

14/14/

11

2 8.018.01

211)(

zeze

zzzH

jj

19


Pole/zero plots are


zezezeze

zzzz

zHzHzC

jjjj 4/4/14/14/

11

**111

8.018.018.018.01

5.01125.0112

)/1()()(

zezezeze

zzzz

zHzHzC

jjjj 4/4/14/14/

11

**222

8.018.018.018.01

211211

)/1()()(

Observe that (with )

zzzz 21215.015.014 11

then

)()( 21 zCzC

1224 zz

20


Example: Consider pole/zero plot of C(z) for a system, determine H(z).

For a causal and stable system, poles of H(z) are: p1, p2,p3.For real ak, bk, zeros/poles occur in complex conjugate pairs.


All-Pass Systems:Consider following stable system function

1

*1

1)(

az

azzH ap

j

jj

j

jj

ap ae

aee

ae

aeeH

1

)1(

1)(

**

then constant )(but ,1)( jap

jap eHeH

This is called an all-pass system .A general all-pass system has the following form

cr M

k kk

kkM

k k

kap zeze

ezez

zd

dzAzH

11*1

1*1

11

1

)1)(1(

))((

1)(

21


Example: Consider pole/zero plot of a typical all-pass system


Example: First order all-pass system with a real pole: z=0.9 (z=-0.9)

22


Example: Second order all-pass system with poles:4/9.0 jez


Notice that group delay for causal all-pass systems are positive (unwrapped/continuous phase is nonpositive).All-pass systems may be used as phase compensators.They are also useful in transforming lowpass filters into other frequency-selective forms.

23


Block-Diagram Representation:LTI systems with difference equation represetation (rational system function) may be imlemented by converting to an algorithm or structure that can be realized in desired technology. These structures consists of basic operations of addition, multiplication by a constant and delay.In block diagram representation:


Example: Consider the second order system :][]2[]1[][ 21 nxbnyanyany o

with

22

111

)(

zaza

bzH o

Block diagram representation is

24


For a system with higher order difference equation

M

kk

N

kk knxbknyany

01

][][][

with system function

N

k

kk

M

k

kk

za

zbzH

1

0

1)(

Rewriting the equation as

][][

][][][

1

01

nvknya

knxbknyany

N

kk

M

kk

N

kk

where

M

kk knxbnv

0

][][


N+MDelayelement

Direct Form I:

25


Previous diagram is implementation of

M

k

kkN

k

kk

zbza

zHzHzH0

1

21

1

1)()()(

or

)()()()(0

1 zXzbzXzHzVM

k

kk

)(1

1)()()(

1

2 zVza

zVzHzY N

k

kk


We can rearrange the system function

)(1

1)()()(

1

2 zXza

zXzHzW N

k

kk

)()()()(0

1 zWzbzWzHzYM

k

kk

In the time domain

M

kk knwbny

0

][][

][][][1

nxknwanwN

kk

26


M = N


Direct Form II:

Max(N,M)Delay elements

27


Example:Consider the system with

21

1

9.05.11

21)(

zz

zzH


Flow Graph Representation:Similar to the block diagram representation:

28



)()()( 41 zXzWzW )()( 12 zWzW )()()( 23 zXzWzW

)()( 31

4 zWzzW

)()()( 42 zWzWzY


1

1

1)(

z

zzH

)(1

)(1

1

zXz

zzY

29


Structures for IIR Systems:Some considerations:Computaional complexity(no. of multiplication, delay )Finite precision, Ease of implementation, etc.

Direct Forms:We have already seen direct forms.

Cascade Form:We factor numerator and denominator polynomials of H(z)

21

21

1

1*1

1

1

1

1*1

1

1

)1)(1()1(

)1)(1()1()( N

kkk

N

kk

M

kkk

M

kk

zdzdzc

zgzgzfAzH


We have cascade of first or second order subsystems

s

ss

N

k kk

kk

o

N

k kk

kkokN

kk

zaza

zbzbb

zaza

zbzbbzHzH

12

21

1

22

~1

1

~

12

21

1

22

11

1

1

1

1)()(

30



11

11

21

21

25.015.01

11

125.075.01

21)(

zz

zz

zz

zzzH


Parallel Form:H(z) may be written as sum of subsystems

crf N

k kk

kkN

kk

k

kN

kk zdzd

zeB

zc

AzCzH

01*1

1

00

1

)1)(1(

)1(

1)(

N

k kk

kokN

kk zaza

zeezCzH

f

02

21

1

11

0

1

1)(

31




112121

25.01

25

5.01

188

125.075.01

21)(

zzzz

zzzH

32


Transposed Forms :Reverse the directions of all branches while keeping the values as they were and reverse roles of input and output. Transposed forms may be useful in finite precision implementation.

Example:Consider the second order system with


Structures for FIR Systems:Consider an FIR system with following input-output relation

][][0

knxbnyM

kk

Observe that the impulse response of this system is

otherwise,0

0,][

Mnbnh n

Direct form (or tapped delay line or transversal filter) :

33


Transposed direct form structure:


ss N

kkkok

N

kk zbzbbzHzH

1

22

11

1

)()()(

Cascade form:

34


Linear-Phase FIR Systems:Finite impulse response (FIR) systems with linear-phase have symmetry properties such as

MnnMhnh 0],[][

MnnMhnh 0],[][or

For M even and odd. Thus there are four types of linear phase FIR filters.


)2/sin(

)2/5sin()( 2

jj eeH

otherwise,0

40,1][

nnh

Example: Consider

35


Example: Consider

]2[]1[2][][ nnnnh

)].cos(1[2

2

21)(

1

111

21

j

jjj

jjj

e

eee

eeeH


Assume M is even

M

Mk

M

k

M

k

knxkh

MnxMhknxkh

knxkhny

12/

12/

0

0

][][

]2/[]2/[][][

][][][

With, in the last term, k=M-l, l=0M/2-1

12/

0

12/

0

][][

]2/[]2/[][][][

M

l

M

k

lMnxlMh

MnxMhknxkhny

36


][][ nMhnh If

]2/[]2/[

])[][]([][12/

0

MnxMh

kMnxknxkhnyM

k

If ][][ nMhnh

])[][]([][12/

0

kMnxknxkhnyM

k


Finite-Precision Numerical Effects:Input Quantization:We have seen earlier that continuous-time signals are sampled, quantized and coded first

37


Example:


In twos complement binary system leftmost bit is the sign bit. If we have (B+1)-bit twos complement fraction of the following form

Baaaa 210 Value is B

Baaaa 2222 22

11

00

Example: Binary Code Numeric value

110 4/3

101 2/1

011 4/1

38


Quantizer step size is12

2

B

mX

With][][

^^

nxXnx Bmwhere

)complement s(two'1][1^

nx B

Analysis of Quantization Errors:We observe that the quantized samples are in general different from the true values. The difference is the quantization error

].[][][^

nxnxne

and2/][2/ ne


A simplified model of quantizer is

Assumptions about e[n]:e[n] is stationarye[n] is uncorrelated with x[n]e[n] is white noisee[n] is uniformly distributed

39


Example:

3bit quantizer

Error for 3bit

Error for 8bit

Sinusoidal signal


Mean value of e[n] is zero

012/

2/

deee

Variance (or power) of e[n] is

12

1 22/

2/

22

deee

For (B+1)-bit quantizer with full-scale value Xm

12

2 222 mXB

e

40


Signal-to-Noise Ratio (SNR) is defined as the ratio of signal variance (power) to noise variance. Expressed in dB

x

m

m

xB

e

x

XB

X

10

2

22

102

2

10

log208.1002.6

212log10log10SNR

If

6dB-SNRSNR,2/

dB 6SNRSNR,1

xx

BB


Coefficient Quantization in IIR Systems:Consider

N

k

kk

M

k

kk

za

zbzH

1

0

1)(

If the coefficients are quantized, we get

N

k

kk

M

k

kk

za

zbzH

1

^

0

^

^

1)(

wherekkkkkk aaabbb

^^

,

41


Each pole or zero will be affected by all the errors in the coefficient quantization. If the poles (or zeros) are close to each other (clustered), then quantization of coefficients may cause large shifts of poles (or zeros).Direct form structures are more sensitive to coefficient quantization than the other forms (parallel, cascade,...)


Example: Consider a bandpass IIR elliptic filter of order 12 implemented in cascade form of 2nd order subsystems and direct form.

42


Passband cascade unquantized

Passband cascade 16-bit


Passband parallel 16-bit

Direct form 16-bit

43


Poles of Quantized 2nd order subsystems :Because of robustness, parallel and cascade forms are used more than direct forms.We can further improve the robustness, by improving implementation of the 2nd order subsystems. Consider the following implementation in direct form:


When coefficients are quantized, a finite number of pole possitions possible.

4bits 7bits

Circles correspond to r2, vertical lines to 2rcos

44


Consider the following coupled form


4bits 7bits

45


Coefficient Quantization in FIR Systems:Consider FIR system with transfer function

M

n

nznhzH0

][)(

If][][][

^

nhnhnh

)()(][)(0

^

zHzHznhzHM

n

n

then

where

M

n

nznhzH0

][)(and

)()()(^

jjj eHeHeH


Example: Consider FIR filter of order 27

46



Effects of Round-off Noise:Analysis of Direct Form IIR Structures:Consider Nth-order difference equation for Direct Form I

M

kk

N

kk knxbknyany

01

][][][

Assume that all signal values and coefficients are represented by (B+1)-bit fixed point binary numbers. Therefore, each multiplication is followed by a quantizer, then

M

kk

N

kk knxbQknyaQny

01

^^

][][][

47



An alternative representation is as following

48


Rounding or truncation of a product bx[n] is represented by a noise source of the form

][][][ nbxnbxQne

Assumptions about e[n]s:Each e[n] is stationaryEach e[n] is uncorrelated with x[n] and other e[n]sEach e[n] is uniformly distributed

For (B+1)-bit rounding

2/2][2/2 BB ne


For (B+1)-bit truncation

0][2 neB

Mean and variance for rounding are

0e 122 22

B

e

Mean and variance for truncation are

2

2 Be

12

2 22B

e

49


Now, lets try to determine effect of quantization noise on the output of the system. We can redraw the system as

][][][][][][ 43210 nenenenenene


Since all the noise sources are

12

255

22222222

043210

B

eeeeeee

For the general Direct Form I case

12

21

22

B

e NM

Now, we observe that

][][][1

neknfanfN

kk

For rounding mean of the output noise is zero.The variance for rounding or truncation is

22

2 ][12

21

n

ef

B

f nhNM

is impulse response for ][ nh ef )(/1)( zAzH ef

50


Example: Consider the following system

.1,1

)(1

aazb

zH

][][ nuanh nef

Then the noise variance(power) at the output is

2222

2

1

1

12

22

12

22

aa

Bn

n

B

f


Analysis of Direct Form FIR systems:Consider

M

k

knxkhny0

][][][

12

21,][][][

22

0

B

f

M

kk Mnenenf

51


End of Part 2

3

Documents

jeh2digital signal processing

jehdigital signal processing

nsdigital signal processing

zh i9digital signal

varying signal

zzzzhthe inverse system

zzzh ithe inverse system

phase response