Chapter 3(b): Digital SignalProcessing

3.1 Introduction to Digital Signal Processing3.2 Analogue to Digital Conversion Process3.3 Quantisation and Encoding3.4 Sampling of Analogue Signals3.5 Aliasing3.6 Digital to Analogue Conversion3.7 Introduction to digital filters

3.7.1 Non-Recursive Digital filters3.7.2 Recursive Digital filters

3.8 Digital filter Realisation3.8.1 Parallel Realisation3.8.2 Cascade Realisation

3.9 Magnitude and Phase Responses3.10 Minimum/Maximum/Mixed Phase Systems3.11 All-pass Filters3.12 Second Order Resonant Filter3.13 Stability of a Second Order Filter3.14 Digital Oscillators3.15 Notch FiltersProf

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Question 2A digital communication link caries binary-coded words representing samples of an inputsignalx(t) = 3 cos(600t) + 2 cos (1800t) . The link is operated at 10,000 bits/s and each inputsample is quantised into 1024 different voltage levels.

(a) What is the sampling frequency?

(b) What are the frequencies in the resulting discrete-time signal x(n)?

[2 marks]

[4 marks]

kHzf s 11000

10000,10

condsamples/seebits/sampl

bits/sec

levels1024210sampleperbitsofNo 10

HzfHzfAnswer

Hzxkfff

ff

HzfHzf

sk

s

100;300:

100100019002

900;300

21

0

2

21

aliasingbyaffectedisitThereforeis

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Question 3Let x(t) have a spectrum X( f) as shown in figure 1.

The spectrum X() of the sampled signal x(t) is given by

If fs = 4 kHz, Sketch the digital spectrum X().

X(f)

f (kHz)2 4 6 8 10-8-10 -6 -4 -2

)(2;)2(1)( 1sf

k

fTT

kXT

X

Figure 1

[5 marks]

324)3(10

44)3(82

24)2(10

04)2(8

324310

44382

24210

0428

0

0

0

0

0

0

0

0

0

kxf

xfk

xf

xf

kxf

xfk

xf

xf

kfff sk

X()

2 4-4 -2

f (kHz)- 2 Prof

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3.7 Introduction to Digital Filters

There are two types of digital filters:

Recursive (there is at least one feedback path inthe filter)

Non-recursive (no feedback paths)

A linear time invariant discrete (LTD)system described by the following equationis commonly called a digital filter:

(3.9)10

L

kk

M

kk knybknxany

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where x[n] is the input signal, y[n] is the

output signal. a0, a1, a2, ...., aM; b1, b2,

b3, ..., bL are constants (filter coefficients).These coefficients determine thecharacteristics of the system.

when bk = 0 the filter is said to benon-recursive type

when bk 0 recursive type.

(3.9)10

L

kk

M

kk knybknxany

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3.7.1 Non Recursive Digital Filters (FIR)

If bk = 0, then the calculation of y[n] does notrequire the use of previously calculated samples ofthe output (see equation (3.9)).

This is recognised as a convolution sum.

knxanxanxaknxany M

M

kk

1100

M

k

knxkhny0Prof

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Therefore the impulse response is identical tothe coefficients, that is,

Any filter that has an impulse response offinite duration is called Finite ImpulseResponse (FIR) filter.

otherwise

Mnanh n

00

knxanxanxaknxany M

M

kk

1100

h(0) h(1) h(M)

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Example :

2

21

10

210

)()(

)(

21

zazaazXzY

zH

nxanxanxany

This is a (non-recursive) second order FIR filter

Property: A property of the FIR filter is that itwill always be stable.

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(a) Stability requires that there should be nopoles outside the unit circle. This condition isautomatically satisfied since there are no poles at alloutside the origin. (In fact, all poles are located atthe origin.)

(b) Another property of non-recursive filter is that wemake filters with exactly linear phase characteristics [4]

Note: The ability to have an exactly linear phaseresponse is one of the most important properties ofa LTD system (filter). When a signal passes througha filter, it is modified in amplitude and/or phase.The nature and extent of the modification of thesignal is dependent on the amplitude and phasecharacteristics of the filter.Prof

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The phase delay or group delay of thefilter provides a useful measure of how thefilter modified the phase characteristic of thesignal. If we consider a signal that consists ofseveral frequency components (eg. speechwaveform) the phase delay of the filter is theamount of time delay each frequencycomponent of the signal suffers in goingthrough the filter.

(3.10))(

)(_

pTdelayphase

[the negative of the phase angle divided byfrequency]Prof

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The group delay on the other hand isthe average time delay the compositesignal suffers at each frequency as itpasses from the input to the output ofthe filter.

(3.11))(

)(_d

Tdelaygroup g

[the negative of the derivative of thephase with respect to frequency]Prof

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A constant group delay means that signalcomponents at different frequencies receivethe same delay in the filter.

A linear phase filter gives same time delay to allfrequency components of the input signal. A filterwith a nonlinear phase characteristic will cause aphase distortion in the signal that passes through it.

()

() = -

-

Figure 3.12. Phase response of a linear phasefilter

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This is because the frequency components inthe signal will each be delayed by an amountnot proportional to frequency, therebyaltering their harmonic relationship. Such adistortion is undesirable in many applications,for example music, video etc.

A filter is said to have a linear phase responseif its phase response satisfies one of the followingrelationships:

(3.12)

ab

a

where ‘a’ and ‘b’ are constants.Profes

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Example

Two filter structures are shown below. Showthat both filters have linear phase.

z-1 z-1

+1 +1 +1

+

x[n]z-1 z-1

+1 -1

+

x[n]

y[n]y[n]

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z-1 z-1

+1 -1

+

x[n]

y[n]

cos21

1

1

1

21

2

21

j

jjj

jj

e

eee

eeH

zzzH

nxnxnxny

z-1 z-1

+1 +1 +1

+

x[n]

y[n]

sin2

sin2

22

1

1

2

2

2

2

2

j

jj

jjj

j

e

ee

jee

je

eH

zzH

nxnxny

phase: () = /2 - ,linear phase

phase: () = -,linear phaseProf

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3.7.2 Recursive Digital filter (IIR) Every recursive digital filter must contain at least one

closed loop. Each closed loop contains at least onedelay element.

For Recursive digital filters bk 0. Let a0 = a0 ,ak = 0 for k > 0, b1 = b1 & bk = 0 for k > 1

L

kk

M

kk knybknxany

10

11

0

10

1

1

zbazH

nybnxany

IIR filterProf

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A recursive filter is an infinite impulseresponse filter (IIR).

Examples :2

21

10)( zazaazH

22

111

1)(

zbzbzH

22

11

22

11

11

)(

zbzbzaza

zH

Zeros only

Poles and Zeros

(2nd order FIR filter)

2nd order IIR filter (all pole filter)

IIR filter

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Example :

The difference equation is: y[n] = x[n] + ay[n-1].

The DC gain of H(z) can be obtained by substituting = 0.

If dc gain is undesirable, introduce a constant gain

factor of (1-a), so that H(z) becomes

Note: Poles and zeros can be real or imaginary

101

1)( 1

a

azzH

aaeH j

11

11

|)( )0(0

111

)(

az

azH

dc gain = 1]1[][)1(][ naynxanyProf

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Example:

Consider a lowpass filter

(i) Determine ‘b’ so that |H(0)| = 1

(ii) Determine the 3dB bandwidth (here) for thenormalised filter in part (i)

101 abx[n],]ay[n-y[n]

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(i) Y(z) = aY(z) z-1 + b X(z) 11)(

azb

zH

jaeb

zH

1)(

11

)0(

a

bH

cos21

1

)sin()cos1(

1)(

sin)cos1(1

11

)(

222 aa

a

aa

aH

jaaa

aea

H j

we have |H(0)| =1

b = 1 - a

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Second Method:

2

2*2

cos21)1(

11

11

)()(|)(|aa

aae

aae

aHHH jj

222 |)0(|21|)(|1|)0(| H

cHH

(half-power point)

2

1

1

|H()|0 dB

3 dB

c

2cos21

1|)(|

aa

aH

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21

1

|H()|2

c

aaa

aaa

c

c

214

cos

cos21)1(

21

21

2

2

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Example: Consider a filter described by

2

2

)(azccza

zH where a & c are constants.

Show that the magnitude response |H()| isunity for all .

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1][][

)()(|)(|

2222

2222

2

2

2

2*2

jj

jj

j

j

j

j

eeaccaeeacac

ceaaec

ceaaec

HHH

|H()|

-

This is an all-pass filter.

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3.8 Digital filter Realisation [11]

1

1

1)()(

)(

)()()(

1

1

0

1

0

10

10

structure

poles

L

k

kk

zeros

M

k

kkL

k

kk

M

k

kk

L

k

kk

M

k

kk

L

kk

M

kk

zbza

zb

za

zXzY

zH

zzYbzzXazY

knybkxany

2

0

1

1

1

structure

zeros

M

k

kk

poles

L

k

kk

zazb

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-b1

-b2

-bL

X(z)+

z-1

z-1

z-1

a1

a2

aM

a0 Y(z)

z-1

z-1

z-1

Figure 3.13. Structure 1 or Direct Form 1

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a1

a2

aM

-b1

-b2

-bL

a0X(z) Y(z)

+ +

z-1

z-1

z-1

z-1

z-1

z-1

Figure 3.14. Structure 2 or Direct Form IIProfes

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In the case when L = M, we have Canonic form realisation.

A discrete-time filter is said to be canonic if it contains theminimum numbers of delay elements necessary to realise theassociated frequency response.

a0

a1

a2

aM

-b1

-b2

-bL

X(z)+

z-1

z-1

z-1

Y(z)+

Figure 3.15. Canonic form

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3.8.1 Parallel Realisation [11]

structureparallel

k

k

ii

LL

MM

zHzHzHzHzH

zbzbzbzazazaa

zH

_

3211

22

11

22

110

)(...)()()()(

1)(

(use partial fraction to obtain Hi(z))

Y(z)

…

H1(z)

H2(z)

Hk(z)

X(z)+

Figure 3.16. Parallel structureProf

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3.8.2 Cascade Realisation [11]

structurecascade

k

k

ii

LL

MM

zHzHzHzHzHzH

zbzbzbzazazaazH

_

3211

22

11

22

110

)(̂)...(̂)(̂)(̂)(̂)(

1)(

(Product of lower order transfer function ie. 1stor 2nd order sections)The cascade structure is the most popular form

Y(z)X(z)

Fig 3.17. Cascade structure

)(̂1 zH )(2̂ zH )(ˆ zH k

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Example:

A parallel realisation of a third order system

H(z) is given by

21

1

1

211

321

3252

)325)(2(19364023

)(

zzDzC

zB

A

zzzzzz

zH

H0(z) H1(z) H2(z)Profes

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21

1

1

21

1

1

21

1

1

6.04.01

)151

1523

(

5.01)5.2(

319

)6.04.01(5)23(

31

5.011

25

319

)(

325)23(

31

25

319

)(

zz

z

z

zzz

zzH

zzz

zzH

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-0.5

-0.4

-0.6

x[n]

+

z-1

z-1

+

-2.5

z-1

y[n]+ +

319

1523

151

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Example:

A cascade realisation of a third-order system is given by

21

21

1

1

21

21

1

1

21

21

1

1

21

21

1

1

321

321

6.04.018.34.36.4

5.01)5.05.0(

6.04.018.34.36.4

5.01)5.05.0(

)6.04.01(5191723

)5.01(2)1(

325191723

)2()1(

3891019364023

)(

zzzz

zz

zzzz

zz

zzzz

zz

zzzz

zz

zzzzzz

zH

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x[n] y[n]

0.5-0.5

0.5

z-1

+ +

3.4-0.4

4.6

z-1

+

3.8-0.6 z-1

+

Cascade

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Example:

Implement the following system in the cascade,direct form II and parallel structures. All coefficientsare real.

)1)(1(1

)().( 11

bzazzHa

cascade structure

x[n] y[n]

-az-1

+

b

z-1

+

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21)(11

)(

abzzbazH

y[n]x[n]+

-(a-b)z-1

+ab z-1

Direct form II

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1111 1111)(

bzba

b

azba

a

bzB

azA

zH

Parallel structure

-a

b

x[n]

z-1

z-1

y[n]baa

bab

+

+

+

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31)1(1

)()(

azzHb

a

x[n] y[n]

az-1

+

az-1

+

z-1

+

cascade structureS

No parallel structure exists because partialfraction expansion cannot be performed.Prof

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33221 3311

)(

zazaazzH

Direct Form II

a3

x[n] y[n]

3az-1

+

-3a2z-1

z-1

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cascadezbbzzaaz

zbazba

parallelbzza

zHc

)21)(21()()(21

)1(1

)1(1

)().(

221221

1221

2121

)1)(1(1

)1)(1(1

)( 1111

bzbzzazazH

y[n]x[n]a

z-1+

az-1

+

parallel structure

+

bz-1

+

bz-1

+

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-2(a+b)

a2+b2

y[n]+

2bz-1

-b2 z-1

+

2a z-1

-a2

x[n]

z-1

+

221221

2221

211

)21()()(21)(

zbbzzaazzbazbazH

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3.9 Magnitude and Phase Response [11]

We can show that the magnitude response isan even function of frequency

The phase response is an odd function offrequency

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Example:

Calculate the magnitude and phase response of the

3-sample averager given by

otherwise

nnh

0

1131

n n

nn zzzznhznhzH 1011

1 31

31

31

)()()(

11 131

)( zzzH Profes

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31

1131

)()(

jjjj

ezeeeezHH j

cos2131

)( H

Precautions must be taken when determining thephase response of a filter having a real-valuedtransfer function, because negative real valuesproduce an additional phase of radians.

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For example, let us consider the following linear-phaseform of the transfer function

H() = e-jkB()

real-valued function of that can take positive andnegative values.

Let phase angle is :

kjBkBH

kjBkBHsincossincos

)tan()cos()()sin()(

tan

kkBkB

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The phase function () includes linear phaseterm and also accommodates for the signchanges in B(). Since -1 can be expressed as ,phase jumps of will occur at frequencieswhere B() changes sign.

If B() > 0, then () = -k.

If B() < 0, then () = -k..

tan= tan(-k) = -k or () = -k

phase angle

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Let us get back to our example

32

32

32

32

and0)(0)(

0)(0)(

]cos21[31|)(|]cos21[

31)(

H

H

HH

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The appropriate sign of must be chosen to

make () an odd function of frequency.

()

-2 --2/3 2/3 2

|H()|

-2 --2/3 2/3 2

-

Odd function

Even function

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Example:

Find the magnitude and phase response of the following:

0)(,21

)2(,1)1(,21

)0( nhhhh

)(

]cos1[)(

21

121

21

21)(

21

21

21

121

)(

)(

2

21210

B

j

jjj

jj

eH

eee

eeH

zzzzzzH

The amplitude function is nevernegative (therefore there is no phasejumps of )Prof

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-

Even function2

|H()|

--

Odd function

()

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Example :

10)(

01)(case

otherwisennn

()|H()|

1

- -

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|H()|

1

-

-

()

-

k

(n-k)

n

h[n] = [n-k]H(z) = 1z-k

H() = 1e-jk

B() = 1 () = -k

Note: When phase exceeds range a jump of 2is needed tobring the phase back into range.

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Phase Jumps: From the previous examples, wenote that there are two occasions for which thephase function experiences discontinuities or jumps.

(1) A jump of 2occurs to maintain the phase

function within the principal value range of [-and ]

(2) A jump of occurs when B() undergoes achange of sign

The sign of the phase jump is chosen such that theresulting phase function is odd and, after the jump, lies in

the range [-and ].Profes

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Example: Magnitude and phase response ofcausal 3-sample average.

32

32

0)(

32

320)()(

]cos21[31

|)(||)(|;)(

]cos21[31

)(

131

31

31

31)(

31

31

31)(

otherwise020for

)(

221

31

B

B

BH

eH

eee

eeHzzzH

nnh

B

j

jjj

jj

-

()

-

1|H()|

- -/2 /2

2/3

-2/3

Phase is undefined at points |H()| = 0.Prof

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Example:

Determine and sketch the magnitude andphase response of the following filters:

4)(

8)(

121

)(

nxnyiiinxnxnyii

nxnxnyi

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2sin

2sin

222

1

21]1[

21)(

]1[21

)()]()([21

)(

22

222

222

222

11

j

jjj

jjj

jjjj

e

ejee

jjee

e

eeeeH

zzHzXzzXzY

|H()|

-

/2

-/2

()

-

(i)

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42

)(

4sin2

4sin222

1)(

1)()(

)()()(

)(

42

2444

48

88

B

j

jjjj

jj

e

eejjee

eeH

zzXzY

zXzzXzY

()

|H()|

/4 /2 3/2

-/2

/2

/4 /2 3/2

(ii)

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1

4

4

44

H

eH

zzHzXzzY

nxny

j

|H()|1

-

-

() = -4

/4

(iii)

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Example: Determine and sketch the magnitude and phase

response of 1st order recursive filter (IIR filter)

1 naynxny

phaseaa

aa

aaa

aaa

aaja

aaa

aeae

aeH

aeH

azzH

zXzY

j

j

j

j

cos1sintan)(

cos1sin

cos21cos1

cos21sin

tan

cos21sin

cos21cos1

11

11)(

11)(

11)(

)()(

1

2

2

22

1

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jj aeae

1

11

1

2cos211

|)(|aa

H

Even Symmetry|H()|

-

a= 0.5

()Odd Symmetry

-

Non-linear phase

Magnitude: |H()|2 = H()H*() [H*() is the complex conjugate]

Assuming 0 < a < 1

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Examples :

The gain k0 can be selected as 1 – a, so that the

filter has unity gain at = 0.

In this case, (for unity gain at = 0).

FilterPass-Low1

11

)().( 110

1

az

aazk

zHa

1

1

02 11

)().(

azz

kzHb

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The addition of a zero at z = -1 further attenuatesthe response of the filter at high frequencies

|H2()|

|H()|

-

Low-Pass Filter|H1()|

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(c) We can obtain simple high-pass filters by reflecting(folding) the pole-zero locations of the low-passfilters about the imaginary axis in the z-plane.

1

1

3 11

21)(

azzazH

-

1

|H3()|High pass filter

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1

1 14

nxnxnyzzH

1

1 15

nxnxny

zzH

(d)

(e)

High-Pass Filter

Low-Pass Filter

-

2|H5()|2

|H5()|2 = 2(1-cos)

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|H6()|2 = 4(1-cos)2

-

4

|H6()|2

(f) 21

6 1 zzH

317 1 zzH

(g)|H7()|2 = 8(1-cos)3

-

8

|H7()|2

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3.10 Minimum-phase, Maximum-phaseand Mixed phase systems [11]

Let us consider two FIR filters:

12

11

21

)(

211)(

zzH

zzH

21 ρ

|z|=1

= -0.5

|z|=1Profes

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H2(z) is the reverse of the system H1(z). This is dueto the reciprocal relationship between the zeros of

H1(z) & H2(z).

The magnitude characteristics for the two filters are

identical because the roots of H1(z) & H2(z) arereciprocal.

cos45

|)(||)(|

21

)(&21

1)(

21

21

HH

eHeH jj

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Phase:

θθ

θφ

θ

θθφ

cos2sin

tan)(

cos21

sintan)(

11

12

-

2 ()

-

-

1()

-

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Note: If we reflect a zero z = that is inside

the unit circle into a zero outside the unit

circle the magnitude characteristic of the system is

unaltered, but the phase response changes.

We observe that the phase characters 1() begins

at zero phase at frequency = 0 and terminates at

zero phase at the frequency = . Hence the netphase change.

Minimum phase filter

ρz

1

0011 Profes

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On the other hand, the phase characteristic for thefilter with the zero outside the unit circle undergoes anet phase change

As a consequence of these different phasecharacteristics, we call the first filter a minimum-phase system and the second system is called amaximum-phase system.

If a filter with M zeros has some of its zeros insidethe unit circle and the remaining outside the unitcircle, it is called a mixed-phase system.

radians012

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A minimum-phase property of FIR filtercarries over to IIR filter.

Let us consider

is called minimum phase if all its poles andzeros are inside the unit circle.

)()(

)(zAzB

zH

|z|=1

Re(z)Minimum phase

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If all the zeros lie outside the unit circle, thesystem is called maximum phase.

|z|=1

Re(z)Maximum phase

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If zeros lie both inside and outside the unitcircle, the system is called mixed-phase.

|z|=1

Re(z) Mixed phase

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Note: For a given magnitude response,the minimum-phase system is thecausal system that has the smallestmagnitude phase at every frequency

(). That is, in the set of causal andstable filters having the samemagnitude response, the minimum-phase response exhibits the smallestdeviation from zero phase.Prof

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Example:

Consider a fourth-order all-zero filter containing adouble complex conjugate set of zeros located at

. The minimum-phase, mixed phase and maximum

phase system pole-zero patterns having identicalmagnitude response are shown below.

47.0

jez

|z|=1

4

mixed-phase

=0.72

24

|z|=1

Minimum-phase

|z|=1

4

maximum-phase

2

21/Profes

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The magnitude response and the phase response ofthe three systems are shown below: The minimum-phase system seems to have the phase with thesmallest deviation from zero at each frequency.

|H()|

()

minimum phase

mixed-phase (In the case linear phase)

maximum phase

-

-2

-3

-4Profes

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Example: A third order FIR filter has a transfer function

G(z) given by

From G(z), determine the transfer function of anFIR filter whose magnitude response is identical

to that of G(z) and has a minimum phaseresponse.

)25)(21216()( 1 zzzzG

)25

1)(34

1)(23

1(12)( 111 zzzzG

>1

)52)(43)(32()( 111 zzzzG

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23

32

34

25

)52

1)(43

1)(32

1()(filterphaseMinimumThe 111 zzzkzP

lm(z)

Re(z)

|z|=1

-

- 43

52

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3.11 All-Pass Filters [11]

An all-pass filter is one whose magnitude response isconstant for all frequencies, but whose phaseresponse is not identically zero.

[The simplest example of an all-pass filter is a pure

delay system with system function H(z) = z-k]

A more interesting all-pass filter is one that isdescribed by

where a0 = 1 and all coefficients are real.

LL

LLLL

zaza

zazazaazH

1

1

01

11

1

1)(

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If we define the polynomial A(z) as

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

1)(

121

00

jezL

L

k

kk

zHzHHzA

zAzzH

azazA

i.e. all pass filter,

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Furthermore, if z0 is a pole of H(z), then is a

zero of H(z) {ie. the poles and zeros are reciprocalsof one another}. The figure shown below illustratestypical pole-zero patterns for a single-pole, single-zero filter and a two-pole, two-zero filter.

0

1z

a1

|z|=1

0

All-pass filter

a

|z|=1

r

0

All pass filter

0

(1/r, 0)

(1/r, -0)

(r, -0)

1

1

1

11

)(

az

zazH |a| < 1 for stabilityProf

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We can easily show that the magnitude response isconstant.

Phase response:

22

2

1*2

cos21

1cos

21

1

11

1

11

|)()()()(|)(

aaθaa

θa

ae

ea

ae

ea

zHzHθHθHθH

θj

θj

θj

θj

ez θj

cos)(2sin)(

tan)(

cos21sin)(cos)(2

11

1

11

)(

1

11

2

11

aaaa

aaaajaa

aeae

ae

eaH j

j

j

j

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()

-

a = 0.5

a = -0.5 a= -0.8

When 0 < a < 1, the zero lies on the positive realaxis. The phase over 0 is positive, at = 0 itis equal to and decreases until = , where it iszero.

When -1< a < 0, the zero lies on the negative realaxis. The phase over 0 is negative, startingat 0 for = 0 and decreases to -at = .Prof

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3.12 A second Order Resonant Filter

002

001

sincos

sincos0

0

jrrrep

jrrrepj

j

(A)1

1)(

212

2

22

11 bzbz

zzbzb

zH

z-1

z-1

+y[n]x[n]

-b1

-b2

0r p1

p2

All pole system has poles only (without counting thezeros at the origin)

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(B)cos2)(

)(

))(())(()(

20

2

2

22

2

2

21

2

21

12

2

00

00

rzrzz

rzeerzz

zH

rezrezz

pzpzz

bzbzzzH

jj

jj

2201 ,cos2 rbrb

2

10 2 b

bCos

sff0

02

0 = resonant frequency

Comparing (A) and (B), we obtain

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3.13 Stability of a second-order filter

Consider a two-pole resonant filter given by

b1 & b2 are coefficients

This system has two zeros at the origin and poles at

22

1121

2

2

11)(

zbzbbzbz

zzH

24

2, 211

21

bbbpp

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The filter is stable if the poles lies inside the unitcircle i.e. |p1| < 1 & |p2| < 1

For stability b2 < 1. If b2 = 1 then the system is anoscillator (Marginally stable)

Assume that the poles are complex

i.e. b12 – 4b2 < 0 b1

2 < 4b2 and

If b12 – 4b2 0 then we get real roots.

0,2 221 bbb

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The stability conditions define a region in the

coefficient plane (b1, b2) which is in the form of atriangle (see below)

The system is only stable if and only if the point

(b1, b2) lie inside the stability triangle.

b1

b2

-2 -1 0 1 2

1

-1

Real Poles

Complex Conjugate Poles

112 bb112 bb

4

21

2b

bparabola

12 b

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Stability Triangle

If the two poles are real then they must havea value between -1 and 1 for the system tobe stable.

01and01)2(4and4)2(

24and42

242

12

41

2121

212

212

21

21

12212

211

122

11

22

11

bbbbbbbbbb

bbbbbb

bbbb

bbb

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The region below the parabola (b12 > 4b2)

corresponds to real and distinct poles.

The points on the parabola (b12 = 4b2) result in real

and equal (double) poles.

The points above the parabola correspond tocomplex-conjugate poles.

b1

b2

-2 -1 0 1 2

1

-1

Real Poles

Complex Conjugate Poles

112 bb112 bb

4

21

2

bbparabola

12 b

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3.14 Digital Oscillators

A digital oscillator can be made using a second orderdiscrete-time system, by using appropriatecoefficients. A difference equation for an oscillatingsystem is given by

From the table of z-transforms we know that the

z-transform of p[n] above is

nAnp cos

21

1

cos21

cos1)(

zzθ

zθzP

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Let21

1

cos21

cos1)()(

)(

zzθ

zθzXzY

zP

Taking inverse z-transform on both sides, we obtain

1cos21cos2 nxnxnynyny

No Input term for an oscillatorx[n] = 0, x[n-1] = 0

So the equation of the digital oscillator becomes

21cos2 nynyny Profes

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So the equation of the digital oscillator becomes

21cos2 nynyny

and its structure is shown below.

b2 = -1

b1 = 2cos

y[n-2]y[n-1]+ z-1 z-1

y[n] = A cos(n)

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To obtain y[n] =Acos(n), use the following initialconditions:

y[0] = A cos(0.) = Ay[-1] = A cos(-1.) = A cos

The frequency can be tuned by changing the coefficient

b1 (b2 is a constant). The resonant frequencyof theoscillator is,

22cos 1

2

1 bbb

(For an oscillator b2 = 1)

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Example: A digital sinusoidal oscillator is shown below.

(a) Assuming 0 is the resonant frequency of thedigital oscillator, find the values of b1 and b2 forsustaining the oscillation.

x[n]

-b1

+z-1

z-1

y[n] = A sin(n+1)

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20

2

22

22

11

cos2

)(

))((1)(

00

00

rzrzK

rzeerzK

rezrezK

zbzbK

zH

jj

jj

b1 = -2 r cos0 ; b2 = r2

For oscillation b2 = 1 r = 1 b1 = -2 cos0Profes

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(b). Write the difference equation for the above figure.Assuming

x[n] = (Asin0)[n], and y(-1) = y(-2) = 0.

Show, by analysing the difference equation, that theapplication of an impulse at n = 0 serves the purpose ofbeginning the sinusoidal oscillation, and prove that theoscillation is self-sustaining thereafter.

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nAnynyny

nxnynybny

00

1

sin21cos2

21

y[0] = A sin0

y[1] = 2cos0 , Asin0 = A sin20

0 0 1

0 0

n = 0y[0] = 2cos0 y[-1] – y[-2] + A sin0 [0]

y[1] = 2cos0 y[0] –y[-1] + A sin0 [1]

n = 1

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y[2] = 2 cos0 y[1] – y[0] + A sin0[2]= 2 cos0 Asin20 – A sin0= 2A cos0 [2 sin0cos0] - sin0= A sin0 [4 cos20 –1 ] = A[3sin0 – 4 sin30]

n = 2

where sin30 = 3sin0 – 4 sin30

y[2] = A sin30 and so forth.

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By setting the input to zero and under certain initialconditions, sinusoidal oscillation can be obtainedusing the structure shown above. Find these initialconditions.

(x[n] = 0 for an oscillator)

n = 0 y[0] = 2cos0 y[-1] – y[-2]for oscillation, y[-1] = 0 no cosine terms

y[0] = -y[-2] y[-2] = -Asin0 (sine term is equired)

y[0] = 0-(-A sin0) = A sin0

Initial conditions:

nxnynyny 21cos2

y[-1] =0; y[-2] = -Asin0Profes

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Sine and cosine oscillators [1] Sinusoidal oscillators can be used to deliver the

carrier in modulators. In modulation schemes, bothsines and cosines oscillators and needed. A structurethat delivers sines and cosines simultaneously isshown below:

y[n]= cos(n)

x[n]= sin(n)

-sin

sin

cos

cos

z-1

+

z-1

+

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Proof:Trigonmetric equation for cos(n+1)is:cos(n+1)= cos(n)cos- sin(n) sinLet y[n] = cos(n) and x[n] = sin(n)

y[n+1] = cosy[n] – sinx[n]Replace n by n-1

y[n] = cosy[n-1] – sinx[n-1] (A)

Similarlysin(n+1)= sincos(n) + sin(n) cos

x[n+1] = siny[n] + x[n] cosReplace n n-1

x[n] = siny[n-1] + x[n-1] cos (B)

Using equations A & B above, the structure shown above canbe obtained.

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Exercise: An oscillator is given by the following coupled

difference equations expressed in matrixform.

Draw the structure for the realisation of thisoscillator, where 0 is the oscillationfrequency. If the initial conditionsyc[-1] = Acos0 and ys[-1] = -Asin0, obtainthe outputs yc[n] & ys[n] using the abovedifference equations.

11

cossinsincos

00

00

nyny

nyny

s

c

s

c

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1cos1sin

1sin1cos

00

00

nynyny

nynyny

scs

scc

yc[n]

ys[n]

-sin0

sin0

cos0

cos0

z-1

z-1

+

+

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n=0 ys[0] = sin0 (A cos0) + cos0 (-Asin0) = 0

n=0 yc[0] = cos0 (Acos0) - sin0(-Asin0) = A

n=1 yc[1] = cos0.A - sin0 .0 = Acos0

n=1 ys[1] = A sin0 + 0 = A sin0

n=2 yc[2] = cos0 yc[1] - sin0 ys[1]= cos0 A cos0 - sin0 A sin0 = A cos20

n = n yc[n] = A cos (n0)

similarly ys[n] = A sin(n0)Profes

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Exercise: For the structure shown below, write down the

appropriate difference equations and hence state thefunction of this structure.

θsin21

z-1

y1[n]

2cos

z-1

-1

-1y2[n]

+

+

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3.15 Notch filters [4] When a zero is placed at a given point on the

z-plane, the frequency response will be zeroat the corresponding point. A pole on theother hand produces a peak at thecorresponding frequency point.

Poles that are close to the unit circle give riselarge peaks, where as zeros close to or onthe unit circle produces troughs or minima.Thus, by strategically placing poles and zeroson the z-plane, we can obtain sample lowpass or other frequency selective filters(notch filters).Prof

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Example:

Obtain, by the pole-zero placement method,the transfer function of a sample digital notchfilter (see figure below) that meets thefollowing specifications: [4]

Notch Frequency: 50Hz 3db width of the Notch: ±5Hz Sampling frequency: 500 Hz

sffr 1

|H(f)|

50 250 f (Hz)0

The radius , r of the poles is determined by :

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To reject the component at 50Hz , place a pair ofcomplex zeros at points on the unit circle corresponds to50Hz. i.e. at angles of

To achieve a sharp notch filter and improved amplituderesponse on either side of the notch frequency , a pair ofcomplex conjugate zeros are placed at a radius r < 1.

937.050010

11

ππ

ff

rs

360

360

0.937

|z| =1

2.03650050

360 00

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21

21

2

2

2.02.02

2.02.02

2.02.0

2.02.0

878.05161.116180.11

)2.0cos(937.02878.0)2.0cos(21

)(937.0878.0)(1

937.0937.0)(

zzzz

zz

zeezeez

ezezezez

zH

jj

jj

jj

jj

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Question 4A digital filter structure is shown below. Calculate the dc gain of the filter in terms of the filtercooefficients.

Z-1

Z-1

X(z) a0

a2

+

b1

b2

Y(z)

+

+

[3 marks]

M(z)

Q(z)

P(z)

21

200

22

11

120

11

22

220

10

11

21

2

12

12

22

1)(

1)(

)()(

)()()()(

)()()(

)()()()(

)()()(

)()()(

bbaa

H

zbzbzaa

zHzXzY

zzYbzzYbzzXazXa

zzMzXazY

zYbzzYbzzXazM

zzYbzzXazQ

zYbzXazP

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Question 5The transfer function of a second order digital oscillator with an oscillation frequency of 0 isgiven by

(a) Show that the impulse response corresponding to the system function H(z) is given by

(c) The frequency of oscillation of the system H(z) can be determined using b1. Show that

where fs=sampling frequency and f0 =desired frequency of oscillation and f0 is thefrequency resolution

011

2cos2;

1)(

b

zbzz

zH

)sin()()sin(

)(0

0

nun

nh

)2sin(4 0

10

s

s

ffbf

f

[3 marks]

[4 marks]

)sin()()sin(

)(

)(1)cos(2)sin(

)()sin(1)cos(2

)sin()()sin(

0

0

02

0

0

02

00

nunnh

zHzz

znunzz

znun

Z:tabletransform-zFrom

)2sin(4)sin(4

2)sin(2)sin(2

)cos(2

0

1

0

10

00001

01

s

ss

s

ffbfbf

f

ff

b

b

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Question 6A difference equation for a particular filter is given by

y(n) = 0.12 x(n) – 0.1 x(n-2) + 0.82 x(n-3) – 0.1 x(n-4) + 0.12 x(n-6)

(a) Find the impulse response of the above filter

(b) Using minimum number of multiplications, draw an implementation for the above filter.

[2 marks]

[4 marks]

70)(

12.0)6(0)5(1.0)4(82.0)3(1.0)2(0)1(12.0)0(

nnh

hhhhhhh

for

)4()2(1.0)3(82.0)6()(12.0)( nxnxnxnxnxny

T T TT T T

+

+

+

y(n)

x(n)

0.82

-0.1

0.12x(n-3)

Implementation with 3 multiplications only

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Question 7Determine and sketch the approximate magnitude response for each of the following filters:

(a) y(n) = x(n) + x(n-1)

(b) y(n) = x(n) – x(n-1)

[4 marks]

[4 marks]

)cos(22

))sin())cos(1()(

1)(

1)(

22

1

H

eH

zzHj

)cos(22

))sin())cos(1()(

1)(

1)(

22

1

H

eH

zzHj

)(H

-

2

)(H

-

2

Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia

Question 8The zero locations of four FIR transfer functions of order 6 are sketched below. Does any oneof the FIR filters have a linear-phase response? If so, which one?

Im(z)

Re(z)1-0.5-1-2 0.5

0.5

Figure 1

2 zeros Im(z)

Re(z)1-0.5-1-2 0.5

0.5

Figure 2

Reflected zero

Im(z)

Re(z)1-0.5-1 0.5

0.5

Figure 3

Reflected zero2 zeros

Im(z)

Re(z)1-0.5-1-2 0.5

0.5

Figure 4

2 zeros

2 zeros

2 zeros

[4 marks]

Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia

Question 9(a) Give two reasons why you think IIR filters are superior to FIR filters

(b) State an application where a constant group delay filter is used.

(c) Draw a parallel and a cascade realisation of the filter)

41

1)(21

1(

2)(

11

zzzH

[6 marks]

[1 mark]

[2 marks]

filters.IIRequivalentintotransferedreadilybecabfiltersAnalogue

tionimplementaFIRtocomparedfiltersamethefortscoefficienlessrequirestionimplementaIIR

signalsvideoorsignalsmusicFiltering

x(n)2

0.5

z-1

+

0.25

z -1

+ y(n)

cascade

x(n)

0.5

z-1

+

y(n)

parallel

0.25

z-1

+-2

4

+

1111

41

1

2

21

1

4

41

121

1

2

zzzz Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia

Question 10(a) Sketch an approximate magnitude response from the pole-zero map given below:

(b) Determine and sketch the phase response of the following filter:y(n) = 0.5 x(n) + 0.5 x(n-2)

Im(z)

Re(z)1

-0.5

-1

0.5 x

x

[2 marks]

[3 marks]

)(H

- 2

)();cos()(

)21

)1(21

)(

)1(21)(

2

2

H

eeeeH

zzH

jjjj

)(

- 2

2

2

2

Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia

Question 11(a) A third order FIR filter has a transfer function G(z) given by

From G(z), determine the transfer function of an FIR filter whose magnituderesponse is identical to that of G(z) and has a minimum phase response.

(b) Determine the magnitude response of the following filter and show that it has an all-pas characteristic.

)52)(43)(32()( 111 zzzzG

1)1(

)( 1

1

azaza

zH

[5 marks]

[3 marks]

)52

1)(43

1)(32

1(12)( 111 zzzzP

Re{z}-5/2 -4/3 -3/4 -2/5 2/3 3/2

Im{z}

reflecting

filterPass-All

11

1)()().(

1)(

1)()1(

21)(

2

22*

*2

jj

jj

j

j

j

j

aeaeaaeaea

HHH

aeeaH

aeeaHzzH )(H

-

1

)25

1)(34

1)(23

1(12)( 111 zzzzG

Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia

Question 12(a) State two important properties of lattice filters

(b) Determine the lattice filter coefficients corresponding to thefollowing

FIR filter:

(b) Draw a lattice structure implementation for the above filter H(z)

21

31

21)( zzzH )()(

)()(

11

)()(

11

1

zAofpolynomialreversetheiszB

tcoefficienreflectionk

zBzzA

kk

zBzA

EquationsFilterLattice

mm

m

m

m

m

m

m

m

[2 marks]

[4 marks]

[3 marks]

• Low sensitivity

• Modularity

;Answer31

23

23

)1(

23

1

31

1

31

32

91

21

21

)(1)()(

)(2

1231

)(31

21)(31

)2(31

2111

12

2121

22

2221

212

21222

kkkk

zzzzz

kzBkzA

zAm

zzzBzzzAkk

z-1

+

+

x(n)

z-1

+

+

y2(n)

w2 (n)

y1(n)

w1(n)

y0(n)

w0 (n)

k1

k1

k2

k 2Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia

Question 13

Determine the stability region (triangle) for following the causal system

Your answer should include , sketch and equations of the boundaries of the region. Also,indicate the

region where complex conjugate poles exist.

22

111

1)(

zbzbzH [4 marks]

b1

b2

-2 -1 0 1 2

1

0

-1

b2 = 1

b2 =b1 - 1b2 = - b1 - 1

Real Poles

Complex ConjugatePoles

parabola b2 = b12

4

Profes

sor E

. Ambik

airaja

h

UNSW, A

ustra

lia