10 Nov'08Comp30291 DMP Section 51 University of Manchester School of Computer Science Comp 30291 : Digital Media Processing Section 5 z-transforms & IIR-type.

10 Nov'08 Comp30291 DMP Section 5 1

University of ManchesterSchool of Computer Science

Comp 30291 : Digital Media Processing

Section 5

z-transforms & IIR-type digital filters


• Order is maximum of N and M. • Recursive if any b j is non-zero.

• A 2nd order recursive filter has the difference-equation:

y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2]

N

i

M

jji jnybinxany

0 1

][ ][ ][

Introduction

• General causal digital filter has difference equation:


y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2]

Signal-flow-graph for 2nd order recursive diff equn

z-1

z-1

z-1

z-1

+

+

+

+a0

a1

a2

x[n]

-b2

-b1

y[n]


Derivation of ‘z-transform’

transform)-(z ][ ][)( where

)(

][

][

][][

n

n

m

m

n

m

mn

m

mn

m

mn

znhzmhzH

zHz

zmhz

zzmh

zmhny

If {x[n]} with x[n] = zn is applied to an LTI system with impulse-response {h[n]}, output would be, by convolution :


‘z-transform’ of {h[n]}

n

nznhzH ][)(

• If input is {zn}, output is {H(z).zn}

• z may be any real or complex number for which series converges.

• For causal & stable IIR, the series converges when |z| 1.

• Replacing z by ej gives frequency-response:

n

njj enheH ][)(


Visualising H(z) On Argand diagram (‘z-plane’), z = ej lies on unit circle.

Imaginary part of z

Real part of z1

z = ej


Example 5.1

Find H(z) for the non-recursive difference equation:

y[n] = x[n] + x[n-1]

Solution:

{h[n]} = { ... , 0, 1, 1, 0, ... },

therefore

H(z) = 1 + z - 1


Example 5.2 • Find H(z) for the recursive difference equation:

y[n] = a 0 x[n] + a 1 x[n-1] - b 1 y[n-1]

Solution: If x[n] = z n then y[n] = H(z) z n ,

y[n-1] = H(z) z n - 1

• Substitute to obtain:

H(z) z n = a 0 z n + a 1 z n - 1 - b 1 H(z) z n - 1

H(z) = a 0 + a 1 z - 1 - b 1 H(z) z - 1

(1 + b1z-1) H(z) = a 0 + a 1 z - 1

-zen except wh 1

)( 111

110 bzb

zaazH

• When z = -b1, H(z) =


Find H(z) for the 2nd order difference-equation y[n] = a 0 x[n] + a1x[n-1] + a2x[n-2] - b1 y[n-1] - b2 y[n-2] The same method gives: H(z) z n = a0zn + a1zn - 1 +a2zn - 2 - b1H(z) zn - 1 - b2H(z)zn-2

a 0 + a 1 z - 1 + a 2 z - 2

H(z) = b 0 + b 1 z - 1 + b 2 z - 2 with b0 = 1.


Consider difference-equation for general digital filter

N M y[n] = a i x[n i] b j y[n j] i=0 j=1

• The same method gives:

1 with ...

...)( 02

21

10

22

110

bzbzbzbb

zazazaazH

MM

NN

• This is the ‘system function’ of the digital filter.

• Also referred to as ‘transfer function’


• Ratio of 2 polynomials in z-1

• Equal to z-transform of impulse-response when this converges.

• Easily derived from difference-equation & signal-flow graph.

• Replacing z by ej gives frequency-response

System function


Example 5.3 Give a signal-flow graph for the system function:

22

11

22

110

1)(

zbzb

zazaazH

Solution: The difference-equation is: y[n] = a0 x[n] + a1 x[n-1] + a2 x[n-2] - b1 y[n-1] - b2 y[n-2]Represented by the signal-flow graph below:

2nd order or ‘bi-quadratic’ IIR section in ‘direct form 1’.

z-1

z-1

z-1

z-1

+

+

+

+a0

a1

a2

x[n]

-b2

-b1

y[n]


Label ‘z-1’ box inputs & outputs as shown:

To implement ‘direct form 1’ biquad as a program

z-1

z-1

z-1

z-1

+

+

+

+a0

a1

a2

X

-b2

-b1

Y

X1

X2

Y1

Y2


X1=0; X2=0; Y1=0; Y2=0;

while 1

X = input(‘X=’);

Y = a0*X + a1*X1 + a2*X2 - b1*Y1 - b2*Y2;

disp(sprintf(‘Y = %f’ , Y) ) ; % output Y

X2 = X1; X1 = X ;

Y2 = Y1; Y1 = Y ;

end;

In MATLAB using floating point arithmetic


y = filter([a0 a1 a2], [b0 b1 b2], x );

In MATLAB using Signal Processing toolbox:


K=1024;A0=round(a0*K); A1=round(a1*K); A2=round(a2*K); B1=round(b1*K); B2=round(b2*K);X1=0; X2=0; Y1=0; Y2=0;while 1 X = input(‘X=’) ; Y = A0*X + A1*X1 + A2*X2 - A1*Y1 - A2*Y2 ; Y = round(Y/K); % Divide by arith right shift disp(sprint(‘Y=%f’,Y)); % Output Y X2 = X1; X1 = X ; % Prepare for next time Y2 = Y1; Y1 = Y ;end;

In MATLAB using fixed point arith & shifting


Look again at ‘Direct Form 1’ signal-flow-graph

• It may be thought of as two signal-flow-graphs:

z-1

z-1

z-1

z-1

+

+

+

+a0

a1

a2

x[n]

-b2

-b1

y[n]

Non-recursive part

Recursive part

x[n] y[n]


Re-ordering LTI systems

• It may be shown than if we have two LTI systems as shown:

L1 L2x[n] y[n]

L2 L1x[n] y[n]

then re-ordering L1 & L2 does not change the behaviour of the overall system.

• Only guaranteed to work for LTI systems


Alternative signal-flow-graph Look again at ‘Direct Form 1’:

Re-order the two ‘halves’ & then simplify to ‘direct form 2’:

x[n]

z-1

z-1

z-1

z-1

+

+

+

+a0

a1

a2 -b2

-b1

y[n]

a2

x[n]

z-1

z-1

+

+a0

a1

-b2

z-1

z-1

+

+

-b1

y[n]

a2

z-1

z-1

+

+a0

a1

y[n]x[n]

-b2

+

+

-b1


‘Direct Form II’ signal-flow-graph

Its difference equation is: y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2]i.e. exactly the same as Direct Form 1

22

11

22

110

1)(

zbzb

zazaazH

It is a 2nd order (bi-quad) section whose system function is:a2

z-1

z-1

+

+a0

a1

y[n]x[n]

-b2

+

+

-b1

W

W1

W2

• Direct form II economises on ‘delay boxes’.

• Notice labels: W,W1 & W2


W1 = 0; W2 = 0; %For delay boxes while 1 X = input(‘X=’) ; % Input to X W =X - b1*W1 - b2*W2; % Recursive part Y = W*a0 + W1*a1 + W2*a2; % Non-rec. part W2 = W1; W1 = W; % For next time disp(sprintf(‘Y=%f’,Y)); % Output Y using disp end; % Back for next sample

Program to implement Direct Form II using normal arithmetic


K=1024;A0=round(a0*K); A1=round(a1*K); A2=round(a2*K); B1=round(b1*K); B2=round(b2*K);W1 = 0; W2 = 0; %For delay boxes while 1 X = input(‘X=’) ; % Assign X to input W =K*X - B1*W1 - B2*W2; % Recursive part W =round( W / K); % By arith right-shift Y = W*A0+W1*A1+W2*A2; % Non-rec. part W2 = W1; W1 = W; %For next time Y = round(Y/K); %By arith right-shift disp(sprintf( ‘Y=%f’, Y)); %Output Y using dispend; % Back for next sample

Direct Form II in fixed point arithmetic with shifting


• Re-express as:

1 with ...

...)( 02

21

10

22

110

bzbzbzbb

zazazaazH

MM

NN

Poles & zeros of H(z)• For a discrete time filter:

MMMM

NNNN

NM

bzbzbz

azazazazzH

...

...)(

22

11

22

110

• Now factorise numerator & denominator:

))...()((

))...()(()(

21

210

M

NNM

pzpzpz

zzzzzzzazH


Poles & zeros of H(z) continued

• z 1 , z 2 ,..., z N , are ‘zeros’. p 1 ,p 2 ,..., p N , are ‘poles’.

• H(z) generally infinite with z equal to a pole.

• H(z) generally zero with z equal to a zero.

• For a causal stable system all poles must satisfy p i < 1.

i.e. on Argand diagram: poles must lie inside unit circle.

• No restriction on the positions of zeros.


Design of IIR ‘notch’ filter

• Design a 4th order 'notch' filter to eliminate an unwanted sinusoid at 800 Hz without severely affecting rest of signal. The sampling rate is FS = 10 kHz.• One way is to use the MATLAB function ‘butter’ as follows:

FL = 800 – 25 ; FU = 800+25; [a b] = butter(2, [FL FU]/(FS/2),’stop’);

a = [0.98 -3.43 4.96 -3.43 0.98]b= [ 1 -3.47 4.96 -3.39 0.96]

freqz(a, b); freqz(a, b, 512, FS); % Better graph axis([0 FS/2 -50 5]); % Scales axes

• Notch has -3 dB frequency band: 25 + 25 = 50 Hz.

MATLAB response


Gain/phase response of notch filter

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000-400

-300

-200

-100

0

Frequency (Hz)

Ph

as

e (

de

gre

es

)

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

-40

-20

0

Frequency (Hz)

Ma

gn

itu

de

(d

B)


Gain/phase responses of notch filter

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000-400

-300

-200

-100

0

Frequency (Hz)

Pha

se (

degr

ees)

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

-40

-20

0

Frequency (Hz)

Mag

nitu

de (

dB)


• How sharp is the notch? • Can answer this question by specifying notch’s -3 dB bandwidth.• Have just designed what Barry calls a 4th order band-stop filter.• MATLAB calls it a 2nd order band-stop filter. • For a sharper notch, decrease -3 dB bandwidth• But this will decrease its ‘depth’,

i.e. the attenuation at the ‘notch’ frequency.• If necessary, increase the order to 6 (3) or 8 (4).

Details


Sketch the same gain-response • The -3dB frequencies are at ( 800 + ) and ( 800 - ) Hz. • Given (= 25 Hz say) can sketch gain-response:

1

FS/2

F

Gain

0.5

0

800 800 + 800 -

0 dB

-3 dB


Implement the 4th order ‘notch’ filter

MATLAB function gave us:

a = [0.98 -3.43 4.96 -3.43 0.98]

b= [ 1 -3.47 4.96 -3.39 0.96]

Transfer (System) Function is:

4321

4321

96.039.396.447.31

98.043.396.443.398.0

zzzz

zzzzzH


z-1

z-1

z-1

z-1

+

+

+

+

+

+

+

+

0.98

3.47

-4.96

3.39

-0.96

0.0.98

x[n] y[n]

-3.43

4.96

-3.43

A direct Form 2 implementation of the 4th order IIR notch filter


Problems with ‘direct form’ IIR implementations

•Implementation on previous slide works fine in MATLAB.

•But ‘direct form’ IIR implementations of order >2 are rarely used.

•Sensitivity to round-off error in coeff values will be high.

•Also range of ‘intermediate’ signals in z-1 boxes is high.

•High wordlength floating point arithmetic hides this problem

•But in fixed point arithmetic, great difficulty occurs.

•Instead we use ‘cascaded biquad sections’


Using biquad (2nd order) IIR sections

H(z)x[n] y[n]

Given 4th order H(z), instead of:

we prefer to arrange two biquad sectns as follows:

H1(z) H2(z)x[n] y[n]G


Converting to the new implementation

• Get a & b for 4th order H(z) as before:

[a b] = butter(2, [FL FU]/(FS/2),’stop’);• Then execute:

[SOS G] = tf2sos(a,b) • MATLAB responds with:

SOS = 1 -1.753 1 1 -1.722 0.9776

1 -1.753 1 1 -1.744 0.9785

G = 0.978

Transfer function to 2nd order sectns

First sectn

2nd sectn


H(z) may now be realised as:

x[n] 0.978

-1.753

1.722

-0.978

y[n]

-1.7531.744

-0.979

Fourth order IIR notch filter realised as two biquad (SOS) sections


Example

A digital filter with a sampling rate of 200 Hz is required to eliminate an unwanted 50 Hz sinusoidal component of an input signal without affecting the magnitudes of other components too severely. Design a 4th order "notch" filter for this purpose whose 3dB bandwidth is not greater than 3.2 Hz.

Solution method:FS=200; FL=50-1.6; FU=50+1.6;[a b]=butter(2,[FL,FU]/(FS/2), ‘stop’);[SOS G] = tf2sos(a,b)


• Many design techniques for IIR digital filters have adopted ideas of analogue filters.

• Can transform analogue ‘prototype’ transfer function Ha(s) into H(z) for an IIR digital filter. • Analogue filters have infinite impulse-responses.

• Many gain-response approximations exist which are realisable by analogue filters

• e.g. Butterworth low-pass approximation which can be transformed to high-pass, band-pass & band-stop.

IIR digital filter design by bilinear transformation


Butterworth low-pass gain approximation of order n

nC

aG2)/(1

1)(

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Freq (radians/s)

Ga

in(a

bs

olu

te)

Butterworth LP gain responses (Cut-off = 1)

n=2

n=4

n=7

At C , Gain 0.71 i.e -3 dB


• Can transform Ha(s), with gain-response Ga(), to H(z) for an IIR digital filter with similar gain-response G().

• Many ways exist.

• Most famous is ‘bilinear transformation’.

• Replaces s by 2(z-1)/(z+1) to transform Ha(s) to H(z).

• Fortunately MATLAB does all this for us.

Transformations from Ha(s) to H(z) for IIR digital filter


Properties of bilinear transformation

(i) Order of H(z) = order of Ha(s)

(ii) If Ha(s) is causal & stable, so is H(z).

(iii) G() = Ga() where = 2 tan(/2)

So gain of analog filter at radians/second becomes gain of digital filter at radians/sample where = 2tan( /2).


Frequency warping

By (iii), from - to mapped to in range - to .

-3.14

-2.355

-1.57

-0.785

0

0.785

1.57

2.355

3.14

-12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12

Radians/secondRa

dia

ns/

sam

ple


• Shape of G() will change under the transformation.

• If Ga() is Butterworth, G() will not have exactly the same shape, but we still call it Butterworth.

• Mapping approx linear for in the range -2 to 2.

• As increases above 2 , a given increase in produces smaller and smaller increases in .

Frequency warping (cont)


• G() becomes more and more compressed as .

• Illustrate for an analog gain-response with ripples:

(a): Analogue gain response (b): Effect of bilinear transformation

Ga() G()

Comparing Ga() with G()


‘Prototype’ analogue transfer function

• Although the shape changes, we would like G() at its cut off C to the same as Ga() at its cut-off frequency.

• If Ga() is Butterworth, it is -3dB at its cut-off freq

• So we would like G() to be -3 dB at its cut-off C.

• Achieved if analogue prototype is designed to have its cut-off frequency at C = 2 tan(C/2).

C is the ‘pre-warped’ cut-off frequency.

• Designing analog prototype with cut-off freq 2 tan(C/2) guarantees that the digital filter will have its cut-off at C.


• Let required cut-off frequency C = /4 radians/sample.• Need prototype transfer fn Ha(s) for 2nd order Butt low-pass filter with 3 dB cut-off at 2tan(C/2) = 2 tan(/8) radians/second.• C = 2 tan(/8) = 0.828• I happen to remember that the transfer fn for a 2nd order Butt low-pass filter with cut-off C is:

2)/()/(21

1)(

CC

ass

sH

• If you don’t believe me, check that replacing s by j and taking the modulus gives G() = 1/[1+(/C)2n] with n=2.

• Set C = 0.828 in this formula, then replace s by 2(z-1)/(z+1).

• Gives us H(z) for an IIR digital filter. That’s it!

Design 2nd order IIR lowpass digital filter by bilinear transfm


Resulting IIR digital filter

21

21

2

2

33.094.01

210.098

4.36.92.10

12)(

zz

zz

zz

zzzH

x[n] y[n]0.098

2 0.94

--0.33


Design of 2nd order IIR low-pass digital filter by bilinear transform using MATLAB

• If required cut-off freq is /4 radians/sample, type: [a b] = butter(2, 0.25)• MATLAB gives us:

a = [0.098 0.196 0.098]b = [1 -0.94 0.33]

The required expression for H(z) is therefore:

21

21

33.094.01

098.0196.0098.0)(

zz

zzzH

21

21

33.094.01

2 1 098.0 )(

zz

zzzH

To save multipliers it is a good idea to re-express this as:


Realise by ‘direct form 2’ signal-flow graph

21

21

33.094.01

21098.0

zz

zzzH

x[n] y[n] 0.098

2 0.94

-0.33


Higher order IIR digital filters

Example: Design 4th order Butterwth-type IIR low-pass digital filter with 3 dB c/o at fS / 16. .

Solution: Relative cut-off frequency is /8. Typing:[a b] = butter(4, 0.125)

gives the response:a = 0.0009 0.0037 0.0056 0.0037 0.0009b = 1 -2.9768 3.4223 -1.7861 0.3556

4321

4321

3556.0786.1422.3977.21

00093.00037.0056.0037.000093.0

zzzz

zzzzzH

4321

4321

3556.0786.1422.3977.21

464100093.0

zzzz

zzzz


z-1

z-1

z-1

z-1

+

+

+

+

+

+

+

+

0.00093

2.977

-3.422

1.79

-0.356

0.00093

x[n] y[n]

0.0037

0.0056

0.0037

A direct Form 2 implementation of 4th order IIR filter


0.00093x[n]

z-1

z-1

z-1

z-1

+

+

+

+

+

+

+

+2.977

-3.422

1.79

-0.356

y[n]

4

6

4

Better ‘direct Form 2’ implementation of 4th order IIR filter


• Higher order IIR digital filters not normally implemented in Direct Form 1 or 2.• Instead, implement as cascaded biquad (sos) sections by typing: [a b] = butter(4, 0.125);

[sos G] = tf2sos(a,b) • MATLAB responds with:sos = 1 2 1 1 -1.365 0.478 1 2 1 1 -1.612 0.745G = 0.00093

48.0365.1

12.

74.06.1

1200093.0

2

2

2

2

zz

zz

zz

zzzH

Cascaded bi-quad sections

‘transfer function’ to ‘second order sections’


H(z) may be realised as:

x[n] 0.00093

2 1.6

-0.74

y[n]

2 1.36

-0.48

Fourth order IIR Butterworth filter with cut -off fs/16


Better realisation of H(z)

•At =0, gain of 1st section is (1+2+1)/(1-1.612+0.0745) =30.12

•At =0, gain of 2nd section is (1+2+1)/(1-1.365+0.488) = 35.56

•Make gain of each section one at =0 by scaling as follows:

Fourth order IIR Butterworth filter with cut-off Fs/16

x[n] 0.033

2 1.6

-0.74

0.028 y[n]

2 1.36

-0.48

1/30.12 1/35.56


Analogue gain resp.

-0.1

0.1

0.3

0.5

0.7

0.9

1.1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

Radians/second

Ga

in

Gain response of 4th order IIR filter

-0.1

0.1

0.3

0.5

0.7

0.9

1.1

0 0.785 1.57 2.355 3.14

Radians/sample

Ga

in

Gain-responses for 4th order analog & IIR digital filter


Compare gain-response of 4th order Butt low-pass transfer function used as a prototype, with that of derived digital filter.

• Both are 1 (0 dB) at zero frequency.• Both are 0.707 (-3 dB) at the cut-off frequency.

• Analogue gain approaches 0 as whereas digital filter gain becomes exactly zero at = .

• Shape of Butt gain response is "warped" by bilinear transfn.

• For digital filter, cut-off rate becomes sharper as because of the compression as .


High-pass band-pass and band-stop IIR filters

Example: 2nd (4th)order bandpass filter with L = /4 , u = /2.Solution: [a b] = butter(2,[0.25 0.5])a = 0.098 0 -0.195 0 0.098b = 1 -1.219 1.333 -0.667 0.33freqz(a,b);[sos G] = tf2sos(a,b)sos = 1 2 1 1 -0.1665 0.5348 1 -2 1 1 -1.0524 0.6232G = 0.098

Many people call this 4th order - MATLAB calls it 2nd order.


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

-100

0

100

200

Normalized Frequency ( rad/sample)

Ph

as

e (

de

gre

es

)

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

-40

-30

-20

-10


Ma

gn

itu

de

(d

B)

2nd (4th) order IIR bandpass


Arranged as 2 biquad sections

21

21

11

21

62.005.11

21

54.017.01

211.0)(

zz

zz

zz

zzzH

x[n] 0.1

-2 .17

-0.54

1 y[n]

2 1.05

-0.62


Higher order band-stop IIR filterExample: 4th (8th)order bandpass filter with L = /4 , u = /2.Solution: [a b] = butter(4,[0.25 0.5],’stop’)a = 0.35 -1.15 2.815 -4.24 5.1 -4.24 2.815 -1.15 0.35b = 1 -2.472 4.309 -4.886 4.477 -2.914 1.519 -0.5 0.12freqz(a,b);[sos G] = tf2sos(a,b)sos =1 -.828 1 1 -0.351 0.428 1 -.828 1 1 -0.832 0.49 1 -.828 1 1 -0.046 0.724 1 -0.828 1 1 -1.244 0.793G = 0.347


Gain & phase resp of IIR band-stop filter

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-800

-600

-400

-200

0


Ph

as

e (

de

gre

es

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-40

-30

-20

-10

0


Ma

gn

itu

de

(d

B)


Band-pass as arrangement of 4 ‘sos’ biquads

H1(z) H2(z) H3(z) H4(z)

• Careful with scaling for band-pass & high-pass sections

• Can make gain =1 in pass-band for each section.

• Not at =0 this time !!!


•. Must use

[sos G] = tf2sos ([a4 a3 a2 a1 a0], [b4 … b0 ])

• ‘help tf2sos’ to find out abt this function

Transfer-function to second order sections: ‘tf2sos’


•Option of cascading high pass & low-pass digital filters to give band-pass or band-stop filters must be used with care.

•It is much simpler & avoids the factorisation problem.

•Then make sure that analogue prototype is wide-band

• High-pass IIR filters are designed by:

[a b] = butter(4,0.125,’high’);

Wide-band band-pass & band-stop filters


Comparison of IIR and FIR digital filters

Advantage of IIR type digital filters:

Economical in use of delays, multipliers and adders.

Disadvantages:

(1) Sensitive to coefficient round-off inaccuracies & effects of overflow in fixed point arith. These effects can lead to instability or serious distortion.

(2) An IIR filter cannot be exactly linear phase.


Advantages of FIR filters:

(1) may be realised by non-recursive structures which are simpler and more convenient for programming especially on devices specifically designed for DSP. (2) FIR structures are always stable.(3) Because there is no recursion, round-off and overflow errors are easily controlled. (4) An FIR filter can be exactly linear phase.

Disadvantage of FIR filters:

Large orders can be required to perform fairly simple filtering tasks.


Problems 1 Find H(z) for the following difference equations

(a) y[n] = 2x[n] - 3x[n-1] + 6x[n-4] (b) y[n] = x[n-1] - y[n-1] - 0.5y[n-2]

2 Show that passing {x[n]} thro’ H(z) = z - 1 produces {x[n-1]}.3. Give difference equation & give a signal-flow graph for: 1 + 3 z - 1 + 2z - 2

H(z) = 1 + 0.9z - 1

Plot poles & zeros & roughly sketch its gain response.4 Calculate the impulse response of the digital filter with

1 H(z) = 1 - 2 z - 1

Draw its signal flow graph, plot its poles and zeros and comment on them.5. Plot poles & zeros for the following difference equation: y[n] = x[n] - 0.9x[n-1] + 0.81x[n-2] + 0.95y[n-1] - 0.9025y[n-2] on z-plane and estimate gain and phase responses


6 Write high level language program, or give flow diagram to implement difference equation given in Prob. 2.6.7. Modify your program for prob 7 so that it uses only integers.8. If LTI systems L1 & L2, with imp-responses {h1[n]} & {h2[n]} are arranged as below, calculate overall impulse- response. Show that this is affected by interchanging L1 & L2 .

L 1 L 2 x[n] y[n]

{ h 1 [n] } { h 2 [n] }

9. A 3rd order low-pass IIR filter is required with 3 dB cut-off at f S /4. Apply bilinear transfn to design it & give its signal flow graph as 2nd & 1st order sections in cascade. 10. Give program to implement 3rd order IIR filter above in floating point arithmetic. Then do it in fixed point arithmetic.11. Low-pass IIR digital filter required with cut-off at f s/4 & stop-band attenuation at least 20 dB for all frequencies above 3f s /8 & below f s /2. Design by bilinear transfn using MATLAB.


12. Design a 4th order band-pass IIR digital filter with lower & upper cut-off frequencies at 300 Hz & 3400 Hz when fS = 8 kHz.13. Design a 4th order band-pass IIR digital filter with lower & upper cut-off frequencies at 2000 Hz & 3000 Hz when fS = 8 kHz.14. What limits how good a notch filter we can implement on a fixed point DSP processor? In theory we can make notch sharper & sharper by moving poles closer & closer to zeros. What limits us in practice. I wonder how sharp a notch we could get in 16-bit fixed pt arithmetic?15. What order of FIR filter would be required to implement a /4 notch approximately as good as a 2nd order IIR /4 notch with 3 dB bandwidth 0.2 radians/sample?16. What order of FIR low-pass filter would be required to be approx as good as the 2nd order IIR low-pass filter (/4 cut-off) designed in these notes?

10 Nov'08Comp30291 DMP Section 51 University of Manchester School of Computer Science Comp 30291 : Digital Media Processing Section 5 z-transforms & IIR-type.

Documents

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nov08comp30291 dmp section