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Syntax:: Familiarization of Infinite Impulse Response (IIR) and Pole-Zero Response Filters

This document describes an experiment involving the design of various IIR filters using Butterworth, Chebyshev Type 1 and Type 2 filter designs. It involves designing low pass filters with different cutoff frequencies and specifications, analyzing the pole-zero plots and magnitude/phase responses. A test signal consisting of multiple sinusoids is generated and its frequency spectrum is plotted. Design steps for a multiple bandpass filter are also provided.

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Nico Loma
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347 views13 pages

Syntax:: Familiarization of Infinite Impulse Response (IIR) and Pole-Zero Response Filters

This document describes an experiment involving the design of various IIR filters using Butterworth, Chebyshev Type 1 and Type 2 filter designs. It involves designing low pass filters with different cutoff frequencies and specifications, analyzing the pole-zero plots and magnitude/phase responses. A test signal consisting of multiple sinusoids is generated and its frequency spectrum is plotted. Design steps for a multiple bandpass filter are also provided.

Uploaded by

Nico Loma
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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LOMA, Keith Nicole S.

ECE107L / B1

Experiment 4 – Familiarization of Infinite Impulse Response (IIR) and Pole-Zero Response Filters

1.) Using Butterworth filter, design a low pass filter with a sampling rate of 30,000 Hz and a cutoff
frequency of 5000 Hz and stop band frequency of 10,000 Hz. Use Rp = 0.1, Rs = 60. Determine values
for the following:
Syntax:
fs = 30000;
fc = 5000;
fsb = 10000;
wp = 2*fc/fs;
ws = 2*fsb/fs;
rp = 0.1;
rs = 60;
[N,wn] = buttord(wp,ws,rp,rs)
[b,a] = butter(N,wn)
Filter Order: 8
Filter Coefficients:
B = 0.0023 0.0186 0.0652 0.1305 0.1631 0.1305 0.0652 0.0186 0.0023
A = 1.0000 -1.5650 2.0512 -1.4970 0.8488 -0.3101 0.0797 -0.0119 0.0008

2.) Plot the poles and zeroes on the unit circle. Is the plot different from FIR filters? Explain.
Pole Zero Plot
Syntax:
zplane(b,a)
1

0.8

0.6

0.4

0.2
Imaginary Part

-0.2

-0.4

-0.6

-0.8

-1

-1 -0.5 0 0.5 1
Real Part

Magnitude & Phase Response


Syntax:
freqz(b,a)

-50

-100
Magnitude (dB)

-150

-200

-250

-300

-350
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

-100

-200
Phase (degrees)

-300

-400

-500

-600

-700

-800
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

Explain:
Yes, the produced plots differ from FIR. This is because the average sum of the previous sample
was used as the points of the FIR.

3.) With the same order and cutoff frequency, design a low pass Chebychev type 1 filter and
superimpose its magnitude response to the magnitude response of the Butterworth filter. What do you
observe?
Syntax:
[b1,a1] = cheby1(N,rp,wn)
b1 =

Columns 1 through 5

0.0006 0.0047 0.0164 0.0329 0.0411

Columns 6 through 9

0.0329 0.0164 0.0047 0.0006

a1 =

Columns 1 through 5

1.0000 -3.3465 6.5391 -8.5208 7.9732

Columns 6 through 9

-5.3937 2.5682 -0.7890 0.1215


Magnitude & Phase Response
Syntax:
freqz(b,a)
0

-50

-100
Magnitude (dB)

-150

-200

-250

-300

-350
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

-100

-200
Phase (degrees)

-300

-400

-500

-600

-700

-800
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

Syntax:
hold on
freqz(b1,a1)

0.1

0
Magnitude (dB)

-0.1

-0.2

-0.3

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45


Normalized Frequency ( rad/sample)

-100

-200
Phase (degrees)

-300

-400

-500

-600

-700

-800
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

What do you observe:


It can be observed in the Chebychev plot a visible ripple as compared to the smooth plot of the the
Butterworth.
4.) Using a Chebychev type 2, design a low pass filter with cutoff frequency of 5000 Hz, sampling
frequency of 30,000 Hz and an attenuation of greater than or equal 30 dB at 7500 Hz. Generate its
frequency response and pole-zero plot.

Syntax:
fs = 30000;
fc = 5000;
fsb = 7500;
rs = 60;
rp = 0.1;
wp = fc/(fs/2);
ws = fsb/(fs/2);
[N,wn] = cheb2ord(wp,ws,rp,rs);
[f,e] = cheby2(N,rs,wn);

Pole Zero Plot


Syntax:
zplane(f,e)

0.8

0.6

0.4

0.2
Imaginary Part

-0.2

-0.4

-0.6

-0.8

-1

-1 -0.5 0 0.5 1
Real Part

Magnitude & Phase Response


Syntax:
freqz(f,e)

-20
Magnitude (dB)

-40

-60

-80

-100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

0
Phase (degrees)

-200

-400

-600
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

5.) Create a test signal with 2000 sample points. Add five sinusoidal waves with amplitude of 2V and
frequencies of 2kHz, 3.7kHz, 6kHz, 7kHz, and 9kHz. Using sampling frequency of 20kHz, plot its
frequency spectrum.

Syntax:
t = [0:1999]/(20000/2);
y1 = 2*sin(2*pi*2000*t);
y2 = 2*sin(2*pi*3700*t);
y3 = 2*sin(2*pi*6000*t);
y4 = 2*sin(2*pi*7000*t);
y5 = 2*sin(2*pi*9000*t);
ytest = randn(1,2000)+y1+y2+y3+y4+y5;
plot(t,ytest)
10

-2

-4

-6

-8

-10
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

plot(w,abs([z(1:256)]))
z = fft(ytest,512);
w = (0:255)/256*(20000/2);
plot(w,abs([z(1:256)]))

500

450

400

350

300

250

200

150

100

50

0
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
6.) Design a multiple bandpass filter with sampling frequency of 20,000 Hz, pass band frequency of
3000, 6500, and 8500 Hz and stop band frequency of 0, 4800, 7200, and 10,000 Hz.

Syntax:
Fs = 20000;
pole1 = 3000*(2*pi/Fs);
pole2 = 6500*(2*pi/Fs);
pole3 = 8500*(2*pi/Fs);
[x,y] = pol2cart(pole1,-1.25);
p1 = x+y*i;
[x,y] = pol2cart(pole2,-0.11);
p2 = x+y*i;
[x,y] = pol2cart(pole3,0.21);

p3 = x+y*i;
P = ([p1 conj(p1) p2 conj(p2) p3 conj(p3)])

B = poly(P)
zero1 = 0;

zero2 = 4800*(2*pi/20000);

zero3 = 7200*(2*pi/20000);

zero4 = 10000*(2*pi/20000);

[x,y] = pol2cart(zero1,0.91);

z1 = x+y*i;

[x,y] = pol2cart(zero2,0.98);

z2 = x+y*i;

[x,y] = pol2cart(zero3,0.928);

z3 = x+y*i;

[x,y] = pol2cart(zero4,0.776);

z4 = x+y*i;
Z = ([z1 z2 conj(z2) z3 conj(z3) z4 conj(z4)])

A=poly(Z)

Determine the following values:

Roots for Poles:


Columns 1 through 2

-0.7347 - 1.0113i -0.7347 + 1.0113i

Columns 3 through 4

0.0499 - 0.0980i 0.0499 + 0.0980i

Columns 5 through 6

-0.1871 + 0.0953i -0.1871 - 0.0953i

Roots for Zeroes:


Columns 1 through 2

0.9100 + 0.0000i 0.0615 + 0.9781i

Columns 3 through 4

0.0615 - 0.9781i -0.5915 + 0.7150i

Columns 5 through 6
-0.5915 - 0.7150i -0.7760 + 0.0000i

Column 7

-0.7760 - 0.0000i

Multiple Bandpass Filter Coefficients:


Columns 1 through 5

1.0000 1.7438 1.9845 0.4564 0.0301

Columns 6 through 7

0.0010 0.0008

A=

Columns 1 through 5

1.0000 1.7020 1.5464 0.6995 -0.4502

Columns 6 through 8

-1.2221 -1.2346 -0.4532

7.) Generate its frequency response and pole-zero plot.

Pole Zero Plot


Syntax:
zplane(A,B)

0.8

0.6

0.4

0.2
Imaginary Part

2
0

-0.2

-0.4

-0.6

-0.8

-1

-1 -0.5 0 0.5 1
Real Part

Magnitude & Phase Response


Syntax:
freqz(B,A)

25

20

15
Magnitude (dB)

10

-5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

100

0
Phase (degrees)

-100

-200

-300

-400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)

8.) Filter the test signal using the multiple band pass filter.

Syntax:
signalfilter = filter(B,A,ytest);

yfilter = fft(signalfilter,512);

w = (0:255)/256*(Fs/2);

plot(w,abs([yfilter(1:256)]))

9.) Generate the frequency spectrum of the filtered signal. Plot output waveform below. What do you
observe?

Frequency Spectrum:
900

800

700

600

500

400

300

200

100

0
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

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