Digital Filter Design - uCoz

Digital Filter Design

Chapters 9 and 10Chapters 9 and 10Appendix A and Appendix B

Dr. Iyad Jafar

Outline

I t d tiIntroduction

C Id l Filt TCommon Ideal Filter Types

Selecting FilterT peSelecting Filter Type

Digital IIR Filter DesignDigital IIR Filter Design

Digital FIR Filter DesignDigital FIR Filter Design

2

IntroductionWhat is filtering?

Remove unwanted frequency content in the signal

For LTI systems this can be done using convolution inFor LTI systems, this can be done using convolution inthe with the impulse response of the filter h[n] in thetime domain or multiplication in the frequency domaintime domain or multiplication in the frequency domainwith the frequency response

Filter h[n]

3 x[n] y[n]

IntroductionDigital filter design is the process of identifying theimpulse h[n] response or the transfer function H(z) of aimpulse h[n] response or the transfer function H(z) of asystem that is capable of passing certain frequencycontent of an input signal x[n] and blocking others withcontent of an input signal x[n] and blocking others withthe following considerations

H(z) is realizableH(z) is realizable H(z) is causal H( ) h f fH(z) meets the frequency response specifications H(z) meets the phase response specificationsComplexity of H(z)Is H(z) an IIR or FIR system? y

4

Common Ideal Filter Types The simplest way to design filters is to consider thefrequency response in the ideal casefrequency response in the ideal caseFrequency response of an ideal digital lowpass filter

H( jω)Passband

1

H(ejω)

Stopband

ωω ω π-π -ωc ωc

Cutoff frequency

5

Common Ideal Filter Types Frequency response of an ideal digital highpass filter

1

|H(ejω)|

ωωπ-π -ωc ωc

Frequency response of an ideal digital band-pass filter

1

|H(ejω)|

Frequency response of an ideal digital band pass filter

1

6ω

π-π ωc1 ωc2-ωc1 -ωc2

Common Ideal Filter Types Frequency response of an ideal digital band-reject filter

| ( jω)|

1

|H(ejω)|

ωωω π-π ωc1

These filter are easy to specify !

ωc2-ωc1-ωc2

These filter are easy to specify ! They have unity magnitude in the passband and zero in h b d d h h h ( l f i !)the stopband, and they have zero phase (real functions!)

7Problems?!!!

Common Ideal Filter Types Example 1 – the impulse response of an ideal lowpass filter

0 050.05

0.8

1

0

h[n]

0.4

0.6

H(e

j)

-200 -100 0 100 200-0.05

n-4 -2 0 2 40

0.2

The impulse response is Non causal not absolutely summable and doubly infinite??

n

Non causal, not absolutely summable, and doubly infinite?? Not realizable or stable!?

Solution8

SolutionMagnitude Approximation and ignore the phase!

Magnitude Approximation of Digital Filters

Magnitude approximation g ppincludes

The magnitude responseg pspecifications in the stop and pass bands are given with acceptable

ltolerance

The transition between the pass and stop bands spans a range of frequencies

9

frequencies

Magnitude Approximation of Digital Filters

10

Magnitude Approximation of Digital Filters

Alternatively, magnitude specifications may by given iny, g p y y gnormalized form where the maximum magnitude in thepassband is assumed to be unityp y

11

Magnitude Approximation of Digital Filters

12

Magnitude Approximation of Digital Filters

2Example 2

13

Selection of Filter TypeThe desired filter could be IIR or FIR!For IIR filters, the transfer function should be a real rational ,function of z-1

In this case, H(z) should be stable with lowest order N to reduceIn this case, H(z) should be stable with lowest order N to reduce computational complexityFor FIR filters, the transfer function is a polynomial in z-1 with , p yreal coefficients

For reduced complexity, N should be the smallest

14

p y

Selection of Filter TypeComparison between FIR and IIR

15

Digital IIR Filter DesignGiven the digital filter specifications (δp, δs, ωc, N, FT….), we want to find

that approximates the magnitude response, stable, with lowest order N.

The most common approach is to convert the digital filterf l l f d hspecifications into analog lowpass specifications to determine the

transfer function H(s) of the analog lowpass filter. Then, theobtained H(s) is converted into the digital filter transfer functionobtained H(s) is converted into the digital filter transfer functionH(z).Why this approach?Why this approach?

Analog approximation techniques are highly advanced They result in closed-form solution

16

yMany applications require the digital simulation of analog filters

Digital IIR Filter Design

Digital Filter Convert to Analog Filter

S ifi i i

Spectral Transformation into an Analog Lowpass g ta te

SpecificationsSpecifications Using

Bilinear Transformation

to a a og owpassFilter Prototype (If

needed)

Design the analog Convert H(s) into G(z)

U i Bili

Spectral Transformation to the Required Analog g g

Lowpass Filter H(s)Using Bilinear

Transformation

q gFilter

(If needed)

• We need to know how to • Design lowpass analog filtersDesign lowpass analog filters• Perform spectral transformation between different types of

analog filters

17

g• Perform bilinear transformation

Analog Filter SpecificationsTypical magnitude response of an analog lowpass filter

18

Analog Filter SpecificationsMagnitude specifications can be given in normalized form

19

Analog Lowpass Filter PrototypesButterworth Approximation

20

Analog Lowpass Filter PrototypesButterworth Approximation

21

Analog Lowpass Filter PrototypesButterworth Approximation

22

Analog Lowpass Filter PrototypesButterworth Approximation

E ample 3Example 3

23

Analog Lowpass Filter PrototypesButterworth ApproximationExample 3 – continuedExample 3 continued

In Matlab this can be obtained usingIn Matlab, this can be obtained using

[N,Wc] = buttord(1000,5000,1,40,'s')N = 4

24

N = 4Wc = 1581 Hz

Analog Lowpass Filter PrototypesButterworth Approximation

25

Analog Lowpass Filter PrototypesButterworth ApproximationExample 4 - Determine the transfer function H(s) of aExample 4 Determine the transfer function H(s) of a analog butterworth lowpass filter of order 2 and

2 1000 rad / secSolution

c 2 1000 rad / sec

1 2N 2 Two roots and ⇒j (2 2 1)/4

1 2 1000 e 1000 j1000

j (2 4 1)/42 2 1000 e 1000 j1000

2( 2 1000)( )a( 2 1000)H (s)

(s 1000 j1000)(s 1000 j1000)62 10

26 a 2 6

2 10H (s)s 2000s 2 10

Analog Lowpass Filter PrototypesButterworth ApproximationExample 4 – continuedExample 4 continuedIn Matlab (N = 2, )

% compute the numerator and denumerator of H (s)c 2 1000 rad / sec

% compute the numerator and denumerator of HLP(s)>> [num,den] = butter(2,sqrt(2)*1000,'s') ; % determine the transfer function% determine the transfer function>> tf(num,den)

2e006

% plot the magnitude of the frequency response

--------------------s^2 + 2000 s + 2e006

% plot the magnitude of the frequency response >> W = 0: 0.1 : 6000 ; >> H = freqs(num, den, W);

27

eqs( u , de , );>> plot(W,20*log10(abs(H));

Analog Lowpass Filter PrototypesButterworth ApproximationExample 4 – continuedExample 4 continuedIn Matlab

0

-5

0

15

-10

| (dB

)

-20

-15

|H(

)|

-25

280 1000 2000 3000 4000 5000 6000

-30

(rad/sec)

Analog Lowpass Filter PrototypesOther Approximations

29

Design of Other Analog FiltersHow about highpass, bandpass, band-stop filter? Do we need to specify magnitude approximations for these filters?Given the analog lowpass filter prototypes, can we design

h f l ?other filters? We consider frequency or spectral transformations! (Appendi B)(Appendix B)Basically,

The filter specifications are mapped into a lowpass filterThe filter specifications are mapped into a lowpass filter specifications. The lowpass filter is designed p gThe inverse spectral transformation is applied to obtain the desired filter

30

We consider here the design of analog highpass filter as an example. (refer to Appendix B for other filters).

Design of Other Analog Filters

31

Design of Analog Highpass Filter

32

Design of Analog Highpass FilterExample 5. Design an analog highpass filter such that

Stopband edge frequency of 2 Krad/sec with minimum attenuation of 5 dBPassband edge frequency 4 Krad/sec with maximum attenuation of 1 dBdB

Solution p pˆ

ˆ

p s sˆ* Let 1 rad/sec, then maps to 2 rad/sec for

lowpass filter

c* Now, we compute N and for the lowpass filter as we did in Example 3. This cgives N 2 and 1.6493 rad / sec* The transfer function is obtained as in Example 4 to be

2.72H ( )33

LP.7

s ^ 2 2.332 H (

s)

.s

2 72

Design of Analog Highpass FilterExample 5. Continued

LP D ˆ* Now, we transform H (s) into H (s) by substitutingˆ 4000p p 4000 s

ˆ ˆs s* This gives the desired transfer function of the analog This gives the desired transfer function of the analog highpass filter

02 12 25

D 2 6

2.97ˆ ˆs 1.108 10 sˆH (s 10ˆ3430 s 5.882 1

)s 0

34

Design of Analog Highpass FilterExample 5. ContinuedIn Matlab

% Direct approach [N,Wc] = buttord(4000,2000,1,5,'s')[numHP denHP] butter(N Wc 'high' 's');

% Step-by-step approach wp_HP = 4000 ; ws_HP = 2000 ; wp_LP = 1 ;

LP 1 * HP / HP [numHP,denHP] = butter(N,Wc,'high','s');tf(numHP,denHP)W = 0:0.1:6000 ; hHP = freqs(numHP denHP W);

ws_LP = 1 * wp_HP / ws_HP ;[N,Wc] = buttord(wp_LP,ws_LP,1,5,'s');[numLP,denLP] = butter(N,Wc,'s');[numHP denHP] = lp2hp(numLP denLP 4000); hHP = freqs(numHP,denHP,W);

plot(W,20*log10(abs(hHP)),'linewidth',2)axis([0 max(W) -20 0])xlabel('\Omega' 'fontweight' 'b' 'fontsize' 14)

[numHP,denHP] = lp2hp(numLP,denLP,4000);tf(numHP,denHP)W = 0:0.1:6000 ; H HP = freqs(numHP denHP W); xlabel( \Omega , fontweight , b , fontsize ,14)

ylabel('|H_D(\Omega)| (dB)','fontweight','b','fontsize',14)set(gca,'fontsize',14,'fontweight','b')

H_HP = freqs(numHP,denHP,W);plot(W,20*log10(abs(H_HP)),'linewidth',2)axis([0 max(W) -20 0])xlabel('\Omega','fontweight','b','fontsize',14) set(gca, fontsize ,14, fontweight , b )

grid onxlabel( \Omega , fontweight , b , fontsize ,14)ylabel('|H_D(\Omega)| (dB)','fontweight','b','fontsize',14)set(gca,'fontsize',14,'fontweight','b')

35

(g , , , g , )grid on

Design of Analog Highpass FilterExample 5. Continued

00

-5

B)

-10

HD

()|

(d

-15

|H

0 1000 2000 3000 4000 5000 6000-20

36

0 1000 2000 3000 4000 5000 6000 (rad/sec)

Bilinear Transformation

37

Bilinear Transformation

38

Bilinear Transformation

3.1416

0

39 -10 -5 0 5 10-3.1416

Bilinear Transformation

40

Bilinear TransformationExample 6. Design a digital lowpass IIR filter with thefollowing specificationsωp = 0.2π rad/sample with maximum passband ripple of 0.5dB

0 6 d/ l ith i i t b d tt ti fωs = 0.6π rad/sample with minimum stopband attenuation of15 dB

SolutionSolution

(1) Prewrap frequencies to obtain the analog lowpass filter specs

p p

(1) Prewrap frequencies to obtain the analog lowpass filter specstan( / 2) tan(0.2 / 2) 0.3249 rad/sec

tan( / 2) tan(0 6 / 2) 1 3764 rad/secs stan( / 2) tan(0.6 / 2) 1.3764 rad/sec

41

Bilinear TransformationExample 6. Continued

c

2

(2) Compute N and of the lowpass filter

1⎛ ⎞ 210 2

120log 0.5 0.12201

1

⎛ ⎞⇒⎜ ⎟

⎝ ⎠⎛ ⎞ 2

10

2 2

120log 15 A 31.6228A

l ((A 1) / ) 1/ k1

⎛ ⎞ ⇒⎜ ⎟⎝ ⎠

⎡ ⎤ ⎡ ⎤⎛ ⎞2 210 1

1010 s p

log ((A 1) / ) 1/ k1N = log 22 log ( / ) 1/ k⎡ ⎤ ⎡ ⎤⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥

2a s 2N 2

s c

1 1H ( j )1 ( / ) A

⇒ c 0.5804 rad/sec

42

Bilinear TransformationExample 6. Continued

(3) Find H (s)a

N 2c

a N

(3) Find H (s)

0.5804H (s)(s 0 4104 + 0 4104i)(s 0 4104 0 4104i)

kk 1

(s 0.4104 + 0.4104i)(s 0.4104 - 0.4104i)(s p )

0 3396a 2

0.3396H (s)s 0.8202s 0.3369

(4) Find H(z)

1

1

1 2

1 za 1 2s

(4) Find H(z)0.1561 0.3122 z 0.1561z X(z)H(z) H (s)

1 0 6147z 0 2393z Y(z)11 z 1 0.6147z 0.2393z Y(z)

(5) The system as a difference equationy[n] 0 6147y[n 1] 0 2393y[n 2]

43

y[n] 0.6147y[n 1] 0.2393y[n 2] 0.561x[n] 0.3122 x[n 1] 0.1561x[n 2]

Bilinear TransformationExample 6. Continued

(6) Test the stability of the system Examine the locations ( ) y yof the poles. Since all poles are inside the unit circle, then we can define a stable and causal system

1

0.5

Par

t

0 2

Imag

inar

y

-0.5

I

44 -1 -0.5 0 0.5 1

-1

Real Part

Design of Analog Highpass FilterExample 6. ContinuedIn Matlab

% Direct approach [[N,wc] = buttord(0.2,0.6,0.5,15,'z')[numD denD] butter(N wc 'z')

% Step-by-step approachWp = tan(0.2*pi/2);Ws = tan(0 6*pi/2); [numD,denD] = butter(N,wc,'z')

w = 0:0.001:pi; h = freqz(numD,denD,w);figure

Ws = tan(0.6 pi/2);[N,Wc] = buttord(Wp,Ws,0.5,15,'s')[numA,denA] = butter(N,Wc,'s')[numD denD] = bilinear(numA denA 0 5) figure

zplane(numD,denD)figureplot(w 20*log10(abs(h)) 'linewidth' 2)

[numD,denD] bilinear(numA,denA,0.5)w = 0:0.001:pi; h = freqz(numD,denD,w);figure plot(w,20 log10(abs(h)), linewidth ,2)

axis([0 max(w)+1 -20 .1])xlabel('\omega','fontweight','b','fontsize',14)ylabel('|H(z)| (dB)','fontweight','b','fontsize',14)

figureZplane(numD,denD)figureplot(w,20*log10(abs(h)),'linewidth',2) ylabel( |H(z)| (dB) , fontweight , b , fontsize ,14)

set(gca,'fontsize',14,'fontweight','b')grid on

p ( , g ( ( )), , )axis([0 max(w)+1 -20 .1])xlabel('\omega','fontweight','b','fontsize',14)ylabel('|H(z)| (dB)','fontweight','b','fontsize',14)

45

set(gca,'fontsize',14,'fontweight','b')grid on

Design of Analog Highpass FilterExample 6. Continued

00

-5

B)

-10

H(e

j)|

(d

-15

|H

0 0 2 0 4 0 6 0 8 1-20

46

0 0.2 0.4 0.6 0.8 1 (/ )

Digital FIR Filter DesignFor FIR filters, the FIR transfer function is apolynomial in z-1 with real coefficientspolynomial in z with real coefficients

For reduced complexity N should be the smallestFor reduced complexity, N should be the smallestThere are many techniques for the design ofdigital FIR filters

Windowed Fourier seriesFrequency samplingComputer-based

47

p

Digital FIR Filter Design

48

Digital FIR Filter Design

49

Digital FIR Filter Design

50

Digital FIR Filter Design

51

Digital FIR Filter Design

52

Digital FIR Filter Design

0.8

1

0.8

1

0.4

0.6

Hd(e

j)

0.4

0.6

h d[n]

-4 -2 0 2 40

0.2

-100 -50 0 50 100

0

0.2

100 50 0 50 100n

11

Truncation2M+1 samples

0.6

0.8

1

]0.6

0.8

1

n]

Delay by M samples

0.2

0.4

h t[n]

0.2

0.4

h t[n

53 -10 -5 0 5 10

0

n0 5 10 15 20

0

n

Digital FIR Filter DesignImpulse response of typical filters

0 8

1

0.6

0.8

(ej

)

0.4

Hd(

-3 14 -wc 0 wc 3 140

0.2

3.14 wc 0 wc 3.14

54

Digital FIR Filter DesignImpulse response of typical filters

0 8

1

0.6

0.8

d(ej

)

0 2

0.4

Hd

-3.14 -wc 0 wc 3.140

0.2

55

Digital FIR Filter DesignImpulse response of typical filters

0.8

1

0.6

Hd(e

j)

0.2

0.4

H

-3.14-wc2 -wc1 0 wc1 wc2 3.140

56

Digital FIR Filter DesignImpulse response of typical filters

0 8

1

0.6

0.8

(ej

)

0.4

Hd(

-3 14-wc2 -wc1 0 wc1 wc2 3 140

0.2

3.14 wc2 wc1 0 wc1 wc2 3.14

57

Digital FIR Filter DesignHow to perform truncation?

Truncation of the target impulse response ht[n] is representedTruncation of the target impulse response ht[n] is representedby multiplying the desired impulse response hd[n] with awindow function w[n]

The simplest window function is the rectangular window

0.8

1

0.4

0.6

wR

[n]

-M 0 M0

0.2

58 Any problems?

-M 0 Mn

Digital FIR Filter DesignRecall that multiplication in time is equivalent to convolution in frequencyq y

Thus, Ht(ejω) is obtained by periodic convolution between Hd(ejω) and ψ(ejω)

1

0.8

1

0.6

0.8

Hd(e

j)

0.4

0.6

0.8

WR

(ej

)

0.2

0.4

0 2

0

0.2

W

59

-4 -2 0 2 40

-50 0 50-0.2

Digital FIR Filter Design

0.8

0.6

0 2

0.4

Ht(e

j)

0

0.2

-15 -10 -5 0 5 10 15-0.2

60

Digital FIR Filter Design

61

Digital FIR Filter Design

62

Digital FIR Filter Design

63

Digital FIR Filter Design

64

Digital FIR Filter DesignEstimating the filter order

Kaiser’s Formula

10 p s20 log ( ) 13N

14 6( ) / 2

Bellanger’s Formula

s p14.6( ) / 2

g

10 p s2 log (10 )N 1p

s p

N 13( ) / 2

65

Digital FIR Filter DesignExample 7. Design a digital lowpass FIR filter withthe following specificationsthe following specifications

δp = 0.1 , δs = 0.1, Fp = 1 KHz , Fs = 4KHz F 10KHFT = 10KHz

Solutionωp = 2π * 1000 / 10000 = 0.2 πωs = 2π * 4000 / 10000 = 0.8 π

1020log ( 0.01) 13N 1.59 214 6(0 8 0 2 ) / 2

⎡ ⎤⎢ ⎥

Thus, the minimum length of the filter is L = N+1 = 3 Let ω = (ω + ω )/2 = 0 5 π

14.6(0.8 0.2 ) / 2

66

Let ωc = (ωs + ωp)/2 = 0.5 π

Digital FIR Filter DesignExample 7. ContinuedF L 3 2M+1 3 M 1 Th t t d filtFor L = 3, 2M+1 = 3, so M = 1. The truncated filter starts from n = -1. E l i h [ ] f 1 0 1Evaluating hLP[n] , for n = -1, 0, 1

gives hLP[n] = {0.3183 0.5000 0.3183}

Delaying by M = 1 h [n] = {0 3183 0 5000 0 3183}

67

hLP[n] = {0.3183 0.5000 0.3183}

Digital FIR Filter DesignExample 7. Continued 1.2

1.4

N=2

wc = 0 5*pi ; 0.6

0.8

1

|Ht(e

j)|

wc = 0.5 pi ;N = 2 ; n = -floor((N+1)/2):floor((N+1)/2);

i ( * ) / i /0.2

0.4

y = sin(wc*n) ./pi./n;y(find(isnan(y))) = wc/pi;[h,w] = freqz(y);

0 5

1

N=2

0 0.2 0.4 0.6 0.8 10

figureplot(w/pi,abs(h));figure -0.5

0

0.5

j)g

plot(w/pi,unwrap(angle(h)))

2

-1.5

-1

(e

68 0 0.2 0.4 0.6 0.8 1-2.5

-2

Digital FIR Filter DesignExample 7. Continued

Eff t f filt dEffect of filter order1.4

N=2

1

1.2

N=2N=8N=20N=50

0.8

1

(ej

)|

0.4

0.6|Ht(

0

0.2

690 0.2 0.4 0.6 0.8 1

0

Digital FIR Filter DesignExample 7. Continued

Eff t f t ti ith H i i dEffect of truncation with Hamming window1.4

N=2

1

1.2

N=2N=8N=20N=50

0.8

1

(ej

)|

0.4

0.6|Ht(

0

0.2

700 0.2 0.4 0.6 0.8 1

0

Digital FIR Filter DesignMatlab functions

k i dkaiserordfir1fir2hamming hannbalckmankaiserfdatoolfdatool

71