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DSP MatLab: Assessed exercise IIR and FIR filter design

IIRlow-pass filter Butterworth

signal sampled frequency = 40kHz

cut-off frequency = 8kHz.transition band = 4kHz.

gain in pass-band = 0dB.

attenuation in stop-band = 40dB.

This difference equation defines how the input signal is related to the output signal

where P is the forward filter order, bi are the forward filter coefficients, Q is the feedback

filter order, ai, are the feedback filter coefficients, x(n) is the input signal and y(n) is theoutput signal.

Filter order

n = 8

n Factors of Polynomial Bn(s)8 = (s2 + 0.3902s + 1)(s2 + 1.1111s + 1)(s2 + 1.6629s + 1)(s2 + 1.9616s + 1)

LP PreWarping

Wd = 2 pi 8000

2 Wd delta(t)Walfa = --------- tan ---------------

delta(t) 22 pi 8000 (1/40000)

Walfa = 2(40 000) tan -------------------------- = 58123.40 = 2 pi 58123

21

H(s) = -------------------------------------------------------------------------------------(s2 + 0.3902s + 1)(s2 + 1.1111s + 1)(s2 + 1.6629s + 1)(s2 + 1.9616s + 1)

s = s / 2 pi 58123

Bilinear transformation

2 (1 Z -)s = ----------------------------

delta (t) (1 + Z -)

b = 0.0001 0.0008 0.0028 0.0056 0.0070 0.0056 0.0028 0.0008 0.0001a = 1.0000 -1.5906 2.0838 -1.5326 0.8694 -0.3192 0.0821 -0.0122 0.0009

0.0001+0.0008Z-+0.0028Z-+0.0056Z-+0.0070Z-+0.0056Z-+0.0028Z-6+0.0008Z+0.0001Z-HLP(z) = --------------------------------------------------------------------------------------------------------------------

1-1.5906Z-+2.0838Z--1.5326Z-+0.8694Z--0.3192Z-+0.0821Z-6-0.0122Z-+0.0009Z-

Luis Alfredo Mateos Guzman

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IIRlow-pass filter Butterworth

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

-400

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-200

-100

0Butterworth Lowpass Filter, Magnitude,order=8

frequency

|H(f)|dB

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

-4

-2

0

2

4Butterworth Lowpass Filter, Phase,order=8

frequency

angleH(f)rad

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

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0.8

1

Real Part

ImaginaryPart

Pole-Zero, Butterworth Low Pass Filter

% Butterworth Filter[n,fn] =buttord(frequency_pass_band,frequency_stop_band,ripple_pass_band,ripple_stop_band)

[b,a] = butter(n,fn)

[H,w] = freqz(b,a,512,40000);

figure;% Magnitude Responseplot(w, 20*log10(abs(H)));title(['Butterworth Lowpass Filter,Magnitude,order=',num2str(n)]);

% Plotting the phase responseplot(w,angle(H));title(['Butterworth Lowpass Filter,

Phase,order=',num2str(n)]);grid on

% Pole-Zero Plotz = roots(b);p = roots(a);figure;zplane(z,p);title('Pole-Zero, Butterworth Low Pass Filter');

b = 0.0001 0.0008 0.0028 0.0056 0.0070 0.0056 0.0028 0.0008 0.0001a = 1.0000 -1.5906 2.0838 -1.5326 0.8694 -0.3192 0.0821 -0.0122 0.0009


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IIRhigh-pass filter Butterworth





Filter order

Is the same formula used in the low-pass filter, we use the value obtained in the low-passfilter

n = 8

n Factors of Polynomial Bn(s)

8 = (s2 + 0.3902s + 1)(s2 + 1.1111s + 1)(s2 + 1.6629s + 1)(s2 + 1.9616s + 1)

Low-Pass to High-Pass Filter

Since the High-Pass filter has a cut-off frequency of 8KHz, the filter transfer function may beobtained from the Low-Pass filter, HLP(Z) by substitution Z to -Z.

From the HLP(Z):

b = 0.0001 0.0008 0.0028 0.0056 0.0070 0.0056 0.0028 0.0008 0.0001

a = 1.0000 -1.5906 2.0838 -1.5326 0.8694 -0.3192 0.0821 -0.0122 0.0009

0.0001+0.0008Z-+0.0028Z-+0.0056Z-+0.0070Z-+0.0056Z-+0.0028Z-6+0.0008Z+0.0001Z-HLP(z) = --------------------------------------------------------------------------------------------------------------------

1-1.5906Z-+2.0838Z--1.5326Z-+0.8694Z--0.3192Z-+0.0821Z-6-0.0122Z-+0.0009Z-

This yields the High-Pass filter transfer function HHP(Z):

b_HP = 0.0001 -0.0008 0.0028 -0.0056 0.0070 -0.0056 0.0028 - 0.0008 0.0001a_HP = 1.0000 1.5906 2.0838 1.5326 0.8694 0.3192 0.0821 0.0122 0.0009

0.0001-0.0008Z-+0.0028Z--0.0056Z-+0.0070Z--0.0056Z-+0.0028Z-6-0.0008Z+0.0001Z-HLP(z) = -------------------------------------------------------------------------------------------------------------------

1+1.5906Z-+2.0838Z-+1.5326Z-+0.8694Z-+0.3192Z-+0.0821Z-6+0.0122Z-+0.0009Z-


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IIRhigh-pass filter Butterworth

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

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200Butter Highpass Filter, Magnitude ,order=8

f

|H(f)|

dB

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

-4

-2

0

2

4Butter Highpass Filter, Phase ,order=8

f

angleH(f)rad

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

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0.6

0.8

1

Real Part

ImaginaryPart

Pole-Zero Butter HighPass Filter

[n,fn] =buttord(frequency_pass_band,frequency_stop_band,ripple_pass_band,ripple_stop_band);

[z,p,k] = buttap(n);N = n;b = k*poly(z); % Numerator polynomiala = poly(p); % Demnominator[b1,a1] = lp2hp(b,a,2*pi*12000); % LowPass to High Passtransformation

[b2,a2] = bilinear(b1,a1,40000, 12000); %Bilinear Transformation , frequency_pass_band=Pre-warping freq.%Or trying the Z -> -Z%a2 = [1.0000 1.5906 2.0838 1.53260.8694 0.3192 0.0821 0.0122 0.0009];

%b2 = [0.0001 -0.0008 0.0028 -0.00560.0070 -0.0056 0.0028 - 0.0008 0.0001];

[H3,w3] = freqz(b2,a2,512,40000);figure;

% Pole-Zero Plotz = roots(b2);p = roots(a2);figure;zplane(z,p);title('Pole-Zero, Butterworth HighPass Filter');

b_HP = 0.0001 -0.0008 0.0028 -0.0056 0.0070 -0.0056 0.0028 - 0.0008 0.0001a_HP = 1.0000 1.5906 2.0838 1.5326 0.8694 0.3192 0.0821 0.0122 0.0009


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IIRband-pass filter Butterworth

signal sampled frequency = 40kHzcut-off frequency = 12kHz.

transition band = 8kHz.

gain in pass-band = 0dB.attenuation in stop-band = 40dB.

Low-Pass to Band-Pass Filter

For the Band-Pass filter, the prototype Low-Pass filter should have cut-off frequency of

WCLP = 2 pi (12 000 4 000)

WCLP = 2 pi 8000

Thus the Low-Pass filter HLP(Z) may also be used to design the Band-Pass filter. The otherparameter that is required is alfa, which may be found:

pi(2 pi 12000) + 2 pi 8000Cos --------------------------------

2 pi 40000alfa = -----------------------------------------------------

pi(2 pi 12000) - 2 pi 8000Cos --------------------------------

2 pi 40000

The Low-Pass to Band-Pass transformation then simplifies to Z- to -Z-

From the HLP(Z):

b = 0.0001 0.0008 0.0028 0.0056 0.0070 0.0056 0.0028 0.0008 0.0001a = 1.0000 -1.5906 2.0838 -1.5326 0.8694 -0.3192 0.0821 -0.0122 0.0009

0.0001+0.0008Z-+0.0028Z-+0.0056Z-+0.0070Z-+0.0056Z-+0.0028Z-6+0.0008Z+0.0001Z-HLP(z) = --------------------------------------------------------------------------------------------------------------------

1-1.5906Z-+2.0838Z--1.5326Z-+0.8694Z--0.3192Z-+0.0821Z-6-0.0122Z-+0.0009Z-

This yields the Band-Pass filter transfer function HBP(Z):

From matlab:

HBP(Z)

a = [1.0000 -0.8085 3.2043 -1.8508 3.7585 -1.4247 1.9031 -0.3663 0.3486];b = [0.0010 -0.0000 -0.0040 -0.0000 0.0060 -0.0000 -0.0040 -0.0000 0.0010];


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IIRband-pass filter Butterworth

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

-400

-200

0

200Butterworth BandPass Filter, Magnitude response,order=8

f

|H(f)|dB

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

-4

-2

0

2

4Butterworth BandPass Filter, Phase response,order=8

f

angleH(f)rad

frequency_sample = 40000;frequency_pass_band1 = 8000;

frequency_pass_band2= 12000;

wo =2*pi*sqrt(frequency_pass_band1*frequency_pass_band2);%Centre Frequency rad/s wo =2*pi*sqrt(Fpb1*Fpb2);

bw = (frequency_pass_band2-frequency_pass_band1)*2*pi; %Bandwidthrad/s bw = (Fpb2-Fpb1)*2*pi;N =4;[z,p,k] = buttap(N);b = k*poly(z); % Numerator polynomiala = poly(p); % Demnominator

[bt,at] = lp2bp(b,a,wo,bw); % Low Pass toBandpass Transformation[b2,a2] = bilinear(bt,at,frequency_sample,frequency_pass_band1); % BilinearTransformation matchinglower pass frequency

%Butterworth Filter[H,w] = freqz(b2,a2,512,40000);

% Pole-Zero Plotz = roots(b2);p = roots(a2);figure;

zplane(z,p);title('Pole-Zero, Butterworth BandPass Filte

a = 1.0000 -0.8085 3.2043 -1.8508 3.7585 -1.4247 1.9031 -0.3663 0.3486b = 0.0010 -0.0000 -0.0040 -0.0000 0.0060 -0.0000 -0.0040 -0.0000 0.0010


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FIRlow-pass filter Hanning

signal sampled = 40kHz




This difference equation defines how the input signal is related to the output signal

where P is the filter order, bi are the forward filter coefficients, x(n) is the input signaland y(n) is the output signal.

HD(W) = 0 {2 pi /5 < w < pi / delta (t)}

HD(W) = 1 {0

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FIRlow-pass filter Butterworth

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1500

-1000

-500

0

Normalized Frequency ( rad/sample)

Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50


Magnitude(dB)

0 50 100 150 200 250 300 350-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

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0.35

0.4FIR

N / 100

% 1 Design a low pass filter FIR Filter type% Filter Specs

% Hanning = 3.1 / delta t = 3.1 x 40000 /4000 = 31

%15 1 15n=[-15:1:15];

%Stop-band = 8000 Hz%Stop-band = 8 pi / 20 delta t = 2 pi / 5delta(t)

y = 2/5*sinc(2/5.*n);

% Wn = (1 + cos (31 pi / 15)w=0.5*(1+cos(n.*pi/15));

FIR=y.*wfigure;plot(FIR)grid on;title(['FIR']);xlabel('N / 100');ylabel('');% plot the magnitud and phasefreqz(FIR)

% help to plot allfvtool(FIR);

[a values]Columns 1 through 7 0 -0.0002 -0.0006 0.0015 0.0046 -0.0000 -0.0116

Columns 8 through 14 -0.0105 0.0148 0.0330 -0.0000 -0.0632 -0.0564 0.0895

Columns 15 through 21 0.2994 0.4000 0.2994 0.0895 -0.0564 -0.0632 -0.0000

Columns 22 through 28 0.0330 0.0148 -0.0105 -0.0116 -0.0000 0.0046 0.0015

Columns 29 through 31 -0.0006 -0.0002 0

FIRhigh-pass filter Hanning


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gain in pass-band = 0dB.attenuation in stop-band = 40dB.

This formula defines how the input signal is related to the output signal

where P is the filter order, bi are the forward filter coefficients, x(n) is the input signaland y(n) is the output signal.

HD(W) = 0 {0 < w < pi /5 }

HD(W) = 1 {pi /5

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1500

-1000

-500

0

500


Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50


Magnitude(dB)

0 50 100 150 200 250 300 350-0.4

-0.3

-0.2

-0.1

0

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0.4FIR

N / 100

% 1 Design a high pass filter FIR Filter type% Filter Specs% Hanning = 3.1 / delta t = 3.1 x 40000 /4000 = 31n=[-15:1:15]

%Stop-band = 12000 Hz%Stop-band = 12 pi / 20 delta t = 3 pi / 5delta t

const=2/5;part1=sinc(n/5)part2=cos(4*pi*n/5)y=const.*part1.*part2

%y = 2/5*sinc(2/5.*n);

figureplot (y)grid on;title(['FIR']);xlabel('N / 100');ylabel('');

% Wn = (1 + cos (31 pi / 15)

w=0.5*(1+cos(n.*pi/15));

[a values]Columns 1 through 7 0 -0.0002 0.0006 0.0015 -0.0046 -0.0000 0.0116

Columns 8 through 14 -0.0105 -0.0148 0.0330 0.0000 -0.0632 0.0564 0.0895

Columns 15 through 21 -0.2994 0.4000 -0.2994 0.0895 0.0564 -0.0632 0.0000

Columns 22 through 28 0.0330 -0.0148 -0.0105 0.0116 -0.0000 -0.0046 0.0015

Columns 29 through 31 0.0006 -0.0002 0

FIRband-pass filter Hanning


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signal sampled frequency = 40kHzcut-off frequency = 12kHz.

transition band = 8kHz.gain in pass-band = 0dB.


HD(w) = 0 {0 < w < pi /10,4 pi /10 < w

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1000

-500

0

500


Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0


Magnitude(dB)

0 50 100 150 200 250 300 350-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4FIR

N / 100

% 1 Design a band pass filter FIR Filter type% Filter Specs% Hanning = 3.1 / delta t = 3.1 x 40000 /4000 = 31n=[-15:1:15]

%Stop-band1 = 12000 Hz and%Stop-band2 = 4000Hz%Stop-band1 = 12 pi / 40 delta t = 3 pi / 10%Stop/band2 = 1 pi / 10 delta t

const=1/5;part1=sinc(n/10)part2=cos(pi*n/2)

y=const.*part1.*part2

figureplot (y)grid on;

title(['FIR']);xlabel('N / 100');ylabel('');

% Wn = (1 + cos (31 pi / 15)w=0.5*(1+cos(n.*pi/15));FIR=y.*w

[a values]Columns 1 through 7 0 0.0005 0.0000 -0.0030 0.0000 -0.0000 0.0000

Columns 8 through 14 0.0209 -0.0000 -0.0660 0.0000 0.1263 -0.0000 -0.1790

Columns 15 through 21 0.0000 0.2000 0.0000 -0.1790 -0.0000 0.1263 0.0000

Columns 22 through 28 -0.0660 -0.0000 0.0209 0.0000 -0.0000 0.0000 -0.0030

Columns 29 through 31 0.0000 0.0005 0

Bonus work.


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