Vlsies Lab Assignment i Sem Dsp_cdac-2

7/27/2019 Vlsies Lab Assignment i Sem Dsp_cdac-2

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Jay Kothari Page 1 of 67

Date: 23-09-2013

M.E. Electronics & Communication Engineering

(VLSI &Embedded Systems Design)

Semester: III

Subject Name: Digital Signal Processing

Subject Code: 735206

Lab-Assignment:

Problem 1

Design and Implement a Digital Butterworth Low Pass System with the following

specifications:

o Passband Frequency: 1000Hz

o Passband Attenuation: 1db

o Stopband Frequency: 12000Hz

o Stopband Attenuation: 80db

o Sampling Rate: 16000Hz

Obtain and Plot the impulse response of the given digital low pass filter.

Obtain and Plot the Magnitude and Phase responses of the given digital low pass filter.

Verify the cut-off frequency of the filter from its Magnitude Response.

Implement the Difference Equation by passing the samples of a sinusoidal signal of

amplitude 0.9, frequency 1000Hz and number of cycles equal to 1000.

Verify the output signal amplitude with the theoretical and expected amplitude.

MATLAB Code:-

clc; clear all; close all;

fs=16000;

ts=1/fs;

fp=1000;

fstop=12000;

rp=1;

rs=80;

ap=10^(-rp/20);

as=10^(-rs/20);

ap_lin=(1/(ap*ap))-1;

as_lin=(1/(as*as))-1;

Wp=2*pi*fp;

Ws=2*pi*fstop;


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N=0.5*(log(ap_lin)-log(as_lin))/(log(Wp)-log(Ws));

N=ceil(N);

disp(N);

Wc=Wp/((ap_lin)^(1/(2*N)));

k=0:1:N-1;

sk=Wc*exp(j*(((pi*(2*k+1)/(2*N))) +(pi/2)));

num=Wc^N;

den=poly(sk);

%%Frequency Transformation

[bm,ak]=bilinear(num,den,fs);

figure

impz(bm,ak);

figure

freqz(bm,ak);

title('Frequency Response')

A=0.9;

f0=1000;

ts=1/fs;

cycle=1000;

D=cycle/f0;

t=0:ts:D-ts

%%Sinusoidal Signal Passing through Filter

x=A*sin(2*pi*f0*t);

m=990*(fs/f0);

y=filter(bm,ak,x);

figure('Name','Sin() signal response of Butterworth Filter','NumberTitle','off')

subplot(2,1,1)

plot(t(m:end),x(m:end));

title('Sine Signal')

ylabel('Magnitude');

xlabel('Time');

subplot(2,1,2)

plot(t(m:end),y(m:end))

ylabel('Magnitude'); xlabel('Time');

title('Sine Signal Response')


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OUTPUT:-


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Problem 2

Design and Implement a Digital Chebyshev Low Pass System with the following

specifications:

o Passband Frequency: 1000Hz

o Passband Attenuation: 1db

o Stopband Frequency: 12000Hz

o Stopband Attenuation: 80db

o Sampling Rate: 16000Hz







MATLAB Code:-

clc

clear all

close all

fpass = 1000;

fstop = 12000;fs = 16000;

nf = fs/2;

ts = 1/fs;

Wpass = 2*pi*fpass;

Wstop = 2*pi*fstop;

rp = 2;

rs = 20;

ap = 10^(-rp/20);

as = 10^(-rs/20);

epsilon = sqrt((1/(ap*ap))-1);

alp = sqrt((1/(as*as))-1);

N = acosh(alp/epsilon)/acosh(Wstop/Wpass);

N = ceil(N);

k = 0:1:2*N-1;


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uk = ((2*k+1)*pi)/(2*N);

v = asinh(1/epsilon)/N;

sk = Wpass*(sin(uk)*sinh(v)+j*cos(uk)*cosh(v));

sk = sk(N+1:end);

disp(sk);

%% Frequency Transformation

zk = exp(sk*ts);

ak = poly(zk);

if(mod(N,2)==0)

kd = sum(ak)/sqrt(1+(epsilon^2));

else

kd = sum(ak);

end

bm= kd;

figure

impz(bm, ak);

figure

freqz(bm, ak);


%%Sinusoidal Signal

A=0.9;

f0=1000;

ts=1/fs;

cycle=1000;

D=cycle/f0;

t=0:ts:D-ts;

x=A*sin(2*pi*f0*t);

m=990*(fs/f0);

y=filter(bm,ak,x);

figure

subplot(2,1,1)


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plot(t(m:end),x(m:end));



xlabel('Time');

subplot(2,1,2)

plot(t(m:end),y(m:end));



xlabel('Time');

OUTPUT:-

Command Window Output:-

1.0e+003 *

-2.5253 - 5.1104i -2.5253 + 5.1104i


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Problem 3

Design and Implement an FIR system of order 127 and using Hamming Window samples

and cut-off frequency 2000Hz and sampling rate fs = 32000.







MATLAB Code:-

clc;

clear all;

close all;

cf=2000;

order = 127;

win = window(@hamming,order);

fs=32000;

nf=fs/2;

ts=1/fs;

ncf=cf/nf;

bm=fir1(order-1,ncf,'low',win);

ak=1;

figure

impz(bm,ak);

figure

freqz(bm,ak);

title('Frequency Response');

A=0.9;

f0=1000;

cycles=1000;

D=cycles/f0;

t=0:ts:D-ts;

x=A*sin(2*pi*f0*t);

y=filter(bm , ak, x);


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m=fix(990*(fs/f0));

figure

subplot(2,1,1);

plot(t(m+1:end),x(m+1:end));



xlabel('Time');

subplot(2,1,2);

plot(t(m+1:end),y(m+1:end));



xlabel('Time');


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OUTPUT:-


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Problem 4

Calculate and plot Cn = (2*A*tau/T)*sinc(2*n*f0*tau) with A = 10, T = 1 milli sec and tau

= 0.1 milli sec on n*f0 axis to depict 5 zero crossing points on both the negative and positive

frequency axes.

Repeat the above with T = 10000 milli sec and compare the following obtained in both the

cases:

o First Zero Crossing Frequency

o Total Average Power, Partial and Percentage average powers contributed by the

complex exponentials distributed between positive and negative first zero

crossing frequencies.

o Inter Sample Distance in the frequency domain.

MATLAB Code:-

clcclear all

close all

msec=10^(-3);

A = 10;

%% For T=0.1 msec

T=1*msec;

f0 = 1/T;

tau = 0.1*T;

fz1 = 1/(2*tau);

disp(['First zero crossing frequency : ',num2str(fz1)])

N1 = fz1/f0;

N5 = 5*N1;

n = -N5: 1: N5;

cn = (2*A*tau/T)*sinc(2*n*f0*tau);

figure

subplot(2,1,1)

stem(n*f0,cn);

title('Sinc Function(T=0.1*msec)')


xlabel('n*fs');

TAP = 2*A*A*tau/T;

N5 = 5*N1;


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disp(['Ineter Sample Distance:',num2str(ISD)]);

OUTPUT:-

Command Window:-

First zero crossing frequency : 5000

Total Average Power:20

Partial Average Power:18.0576

Percentage Average Power:90.2878

Ineter Sample Distance:1000


Total Average Power:0.002Partial Average Power:0.0018056


Ineter Sample Distance:0.1


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Problem 5

Implement DFT and IDFT functions using the related mathematical relations as the basis.

Find the DFT samples of X(k) of a rectangular pulse of amplitude 10 and having its pulse

width equal to 2 msecs.

Plot the magnitude and phase parts of X(k) along the frequency axis f.

Find the average power of the above signal using time domain and frequency domain

signal samples

Find the DFT samples of X(k) of an exponentially decaying pulse given by alpha*exp(-

alpha*t)*u(t).



signal samples

Find the DFT samples of X(k) of a triangular pulse of maximum amplitude 100 and pulse

width 4 msec and centered around origin.



signal samples

MATLAB Code:-clc

clear all

close all

A = 10;

ms = 10^-3;

tau = 1*ms;

fs = 32000;

ts = 1/fs;

nq = fs/2;

% Rectangular PulseL = tau/ts;

M = 2*L + 1

N = input('Enter the value of N>M : ');

x = [ones(1,1) ones(1,L) zeros(1,N-M) ones(1,L)];

n = 0:1:N-1;

figure('Name','Rectangular pulse','NumberTitle','off')


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plot(n,x);

title('Rectangular Pulse')

X = dft(x,N);

k = 0:1:N-1;

f = k*fs/N;

figure('Name','DFT version of rectangular pulse','NumberTitle','off')

subplot(2,1,1);

stem(f, abs(X));


subplot(2,1,2);

stem(f, angle(X));

title('Phase Response');

%Exponentially Decaying Function

alp = 10000;

fc = alp/(2*pi);

fm = 100*fc;

fs = 2*fm;

ts = 1/fs;

T = 100/alp;

f0 = 1/T;

N = T/ts;

n = 0:1:N-1;

x = ts*alp*exp(-alp*n*ts);

figure

plot(n,x);

title('Exponential decaying signal')

X = dft(x,N);

k = 0:1:N-1;

f = k*fs/N;

figure('Name','DFT version of exponential decaying signal','NumberTitle','off')

subplot(2,1,1);


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stem(f, abs(X));

title('Frequncy Response');

subplot(2,1,2);

stem(f, angle(X));


disp(['fc = ',num2str(fc)]);

TAP = (alp * (1 - exp(-2*alp*T)))/(2*T);

N1 = fix(fc/f0);

n = -N1:1:N1;

cn = (alp*f0)./(alp+j*2*pi*n*f0);

PAP = sum(abs(cn).*abs(cn));

PCAP = PAP * 100 / TAP;

disp(['Partial Average Power of the exponentially decaying curve :',num2str(PAP)]);

disp(['Total Average Power of the exponentially decaying curve : ',num2str(TAP)]);

disp(['Percentage Average Power of the exponentially decaying curve : ',num2str(PCAP)]);

% Triangular Pulse

tau = 2*ms;

fs = 32000;

ts = 1/fs;

nq = fs/2;

t = -tau:ts:-ts;

x1 = 50*t + 100;

t = 0:ts:tau;

x2 = -50*t + 100;

T = 4*tau;

f0 = 1/T;

L = tau/ts;

M = 2*L + 1

N = input('Enter the value of N>M : ');

x = [x2 zeros(1,N-M) x1];

X = dft(x,N);


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k = 0:1:N-1;

f = k*fs/N;

figure

subplot(2,1,1);

stem(f, abs(X));

subplot(2,1,2);

stem(f, angle(X));

TAP = 1/T*sum(x.^2)

N1 = T/(2*tau);

fz1 = 1/(2*tau);

k1 = fz1*N/fs;

X1 = [X(1:k1+1) X(N-k1:end)];

PAP = 1/(2*k1)*sum(abs(X1).^2)

PCAP = 100* PAP/TAP

OUTPUT:-

Command Window:-

M =

65

Enter the value of N>M : 128fc = 1591.5494

Partial Average Power of the exponentially decaying curve :245817.2134

Total Average Power of the exponentially decaying curve : 500000

Percentage Average Power of the exponentially decaying curve : 49.1634

M =

129

Enter the value of N>M : 256

TAP =

1.6109e+008

PAP =

7.6577e+007

PCAP = 47.5376


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Problem 6

Calculate and plot Cn = (*f0)/(+j*2**n*f0) with = 10000, f0 = 100Hz on n*f0 axis to

depict 100 frequency components on both the negative and positive frequency axes.

Identify the first Half Power frequency.

Calculate and display the Total Average Power of a of the signal xp(t). Calculate the partial amount of average power contributed by the complex exponentials

between the Half-Power Frequency points on the frequency axis.

Find the percentage of average power contributed by the same complex exponentials.

o Repeat the above exercise with = 10000.

MATLAB Code:-

clc

clear all

close all

f0 = 100;

%% For alp=10000

alp = 10000;

T = 1/f0;

N = 100;

n = -N:1:N;


figure

stem(n*f0, abs(cn));


xlabel('Frequency');


figure

stem(n*f0, angle(cn));




TAP = (alp/(2*T))*(1-exp(-2*alp*T));

fc = alp/(2*pi);

disp(['Half Power Frequency:',num2str(fc)]);


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N1 = fix(fc/f0);

n = -N1:1:N1;


PAP = sum(abs(cn).*abs(cn));

PCAP = PAP*100/TAP;

disp(['Partial Average Power :',num2str(PAP)]);

disp(['Total Average Power : ',num2str(TAP)]);

disp(['Percentage Average Power : ',num2str(PCAP)]);

OUTPUT:-

Command Window:-

Half Power Frequency:1591.5494

Partial Average Power :245817.2134

Total Average Power : 500000

Percentage Average Power : 49.1634


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Problem 7

Find and plot the impulse response of the discrete time system equivalent to RC system.

Apply DFT/FFT for the obtained impulse response to get H(k) and plot the magnitude and

Phase responses of the system across frequency values.

Identify the half-power frequencies from the obtained plot.

Implement the difference equation of the system by passing a sinusoidal signal of frequency

f0 and amplitude 10.

MATLAB Code:-

clc

clear all

close all

kohms = 10^3;

mfarads = 10^(-6);R = 10*kohms;

C = 0.01*mfarads;

alp = 1/(R*C);

fc = alp/(2*pi);

fm = 100*fc;

fs = 2*fm;

ts = 1/fs;

a = exp(-alp*ts);

T = 100/alp;

N = T/ts;

n = 0:1:N-1;

h = alp*ts*a.^n;

impz(h,1);

title('Impulse Response in Time Domain');

H = dft(h, N);

k = 0:1:N-1;

f = k*fs/N;

figure

subplot(2,1,1);

plot(f, abs(H));


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ylabel('Magnitude');subplot(2,1,2);

plot(f, angle(H));




bm = alp*ts;

ak = [1, -a];

figure

impz(bm, ak, 512);

title('Impulse Response ')

figure

freqz(bm, ak)


A = 10;

f0 = 1000;

cycles = 1000;

D = cycles/f0;

t = 0:ts:D-ts;

x = A*sin(2*pi*f0*t);

y = filter(bm, ak, x);

m = fix(990*fs/f0);

figure

subplot(2,1,1)

plot(t(m+1:end), x(m+1:end));

title('Sine Signal ')


xlabel('Time');

subplot(2,1,2)

plot(t(m+1:end), y(m+1:end));



xlabel('Time');


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OUTPUT:-


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Problem 8

Implement Goertzel Function on MATLAB and using the function implement a DTMF

Signal Detection System on MATLAB.

Matlab Code:

close all;clear allfigure(1)

% DTMF tone generator

fs=8000;

t=[0:1:204]/fs;

x=zeros(1,length(t));

x(1)=1;

y852=filter([0 sin(2*pi*852/fs) ],[1 -2*cos(2*pi*852/fs) 1],x);

y1209=filter([0 sin(2*pi*1209/fs) ],[1 -2*cos(2*pi*1209/fs) 1],x);

y7=y852+y1209;

subplot(2,1,1);plot(t,y7);grid

ylabel('y(n) DTMF: number 7');

xlabel('time (second)')Ak=2*abs(fft(y7))/length(y7);Ak(1)=Ak(1)/2;

f=[0:1:(length(y7)-1)/2]*fs/length(y7);

subplot(2,1,2);plot(f,Ak(1:(length(y7)+1)/2));grid

ylabel('Spectrum for y7(n)');

xlabel('frequency (Hz)');

figure(2)

% DTMF detector (use Goertzel algorithm)

b697=[1];

a697=[1 -2*cos(2*pi*18/205) 1];

b770=[1];

a770=[1 -2*cos(2*pi*20/205) 1];b852=[1];

a852=[1 -2*cos(2*pi*22/205) 1];

b941=[1];

a941=[1 -2*cos(2*pi*24/205) 1];

b1209=[1];

a1209=[1 -2*cos(2*pi*31/205) 1];

b1336=[1];

a1336=[1 -2*cos(2*pi*34/205) 1];

b1477=[1];

a1477=[1 -2*cos(2*pi*38/205) 1]

[w1, f]=freqz([1 -exp(-2*pi*18/205)],a697,512,8000);

[w2, f]=freqz([1 -exp(-2*pi*20/205)],a770,512,8000);

[w3, f]=freqz([1 -exp(-2*pi*22/205)],a852,512,8000);

[w4, f]=freqz([1 -exp(-2*pi*24/205)],a941,512,8000);

[w5, f]=freqz([1 -exp(-2*pi*31/205)],a1209,512,8000);

[w6, f]=freqz([1 -exp(-2*pi*34/205)],a1336,512,8000);

[w7, f]=freqz([1 -exp(-2*pi*38/205)],a1477,512,8000);

subplot(2,1,1);plot(f,abs(w1),f,abs(w2),f,abs(w3), ...

f,abs(w4),f,abs(w5),f,abs(w6),f,abs(w7));grid


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xlabel('Frequency (Hz)');

ylabel('BPF frequency responses');

yDTMF=[y7 0];

y697=filter(1,a697,yDTMF);


y852=filter(1,a852,yDTMF);y941=filter(1,a941,yDTMF);




m(1)=sqrt(y697(206) 2+y697(205)^2- ...

2*cos(2*pi*18/205)*y697(206)*y697(205));

m(2)=sqrt(y770(206) 2+y770(205)^2- ...

2*cos(2*pi*20/205)*y770(206)*y770(205));

m(3)=sqrt(y852(206) 2+y852(205)^2- ...

2*cos(2*pi*22/205)*y852(206)*y852(205));

m(4)=sqrt(y941(206) 2+y941(205)^2- ...2*cos(2*pi*24/205)*y941(206)*y941(205));

m(5)=sqrt(y1209(206)^2+y1209(205)^2- ...

2*cos(2*pi*31/205)*y1209(206)*y1209(205));

m(6)=sqrt(y1336(206)^2+y1336(205)^2- ...

2*cos(2*pi*34/205)*y1336(206)*y1336(205));

m(7)=sqrt(y1477(206)^2+y1477(205)^2- ...

2*cos(2*pi*38/205)*y1477(206)*y1477(205));

m=2*m/205;

th=sum(m)/4; %based on empirical measurement

f=[ 697 770 852 941 1209 1336 1477];

f1=[0 4000];

th=[ th th];x

subplot(2,1,2);stem(f,m);grid

hold; plot(f1,th);


ylabel('Absolute output values');

Output:


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Problem 9

Assuming an ideal and symmetrical frequency response of a low pass filter with the following

specifications given by

o H(w) = 1; for 0 w /2 and

o

H(w) = 0; for

/2 < w

Design a FIR system with the order given by 128 (Even number) using Frequency Sampling

Method.

MATLAB Code:

clc;

clear all;

close all;

f = [0 0.25 0.25 1]; m = [1 1 0 0];

b = fir2(128,f,m);

[h,w] = freqz(b,1,128);

plot(f,m,w/pi,abs(h))legend('Ideal','fir2 Designed')

title('Comparison of Frequency Response Magnitudes'

Output:


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Problem 10

Calculate and plot Cn = (2*A*tau/T)*sinc(2*n*f0*tau) with A = 10, f0 = 1000Hz and tau =

0.1* T on n*f0 axis to depict 5 zero crossing points on both the negative and positive

frequency axes.

Identify the first zero crossing frequency.

Calculate and display the Total Average Power of a Rectangular Pulse Train.

Calculate the partial amount of average power contributed by the complex exponentials

between the first zero crossing points on the frequency axis.

Find the percentage of average power contributed by the same complex exponentials.

Repeat the above exercise with tau = 0.05*T.

MATLAB Code:

clc

clearall

close all

%Case A

A = 10;

mili = 10^-3;

T = 1*mili;

f0 = 1/T;

tau = 0.1*mili;

fz1 = 1/2/tau;

disp(['First zero crossing frequency for T = 1ms: ',num2str(fz1)]);f5 = 5*fz1;

n = -f5/f0:1:f5/f0;


figure('Name','plot of Cn upto 5zero crossing frequency','NumberTitle','off')

subplot(2,1,1);

stem(n*f0,cn);

title('n*f0 vs Cn plot for 5 zero crossing frequency for T=1ms');

grid on;

% Total Average Power

TAP = 2*A^2 * tau/T;

N1 = T/(2*tau);n = -N1:1:N1;

% Partial Average Power

PAP = sum(cn.*cn);

% Partial Average Power Contributed by exponential

PACP = 100*PAP/TAP;

disp(['Total Average Power for Tau = 0.1ms: ',num2str(TAP)]);

disp(['Partial Average Power for Tau = 0.1ms: ',num2str(PAP)]);

disp(['Percentage Average Power Contributed by Exponential for Tau = 0.1ms: ',num2str(PACP)]);


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disp(['Intersample Distance In Frequency Domain for Tau = 0.1ms: ',num2str(f0)]);

%Case B

A = 10;

mili = 10^-3;

T = 1*mili;

f0 = 1/T;tau = 0.05*mili;

fz1 = 1/2/tau;

disp(['First zero crossing frequency for Tau = 0.05ms: ',num2str(fz1)]);

f5 = 5*fz1;

n = -f5/f0:1:f5/f0;


subplot(2,1,2);

stem(n*f0,cn);

title('n*f0 vs Cn plot for 5 zero crossing frequency for T=0.05ms');

grid on;

% Total Average PowerTAP = 2*A^2 * tau/T;

N1 = T/(2*tau);

n = -N1:1:N1;

% Partial Average Power

PAP = sum(cn.*cn);

% Partial Average Power Contributed by exponential

PACP = 100*PAP/TAP;

disp(['Total Average Power for Tau = 0.05ms: ',num2str(TAP)]);

disp(['Partial Average Power for Tau = 0.05ms: ',num2str(PAP)]);

disp(['Percentage Average Power Contributed by Exponentioal for Tau = 0.05ms:

',num2str(PACP)]);

disp(['Intersample Distance In Frequency Domain for Tau = 0.05ms: ',num2str(f0)]);


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OUTPUT:-

Command Window:-










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Problem 11

Given three sinusoids with the following amplitudes and phases

5.018002cos525.012002cos5

5002cos5

3

2

1

ttx

ttx

ttx

Create a MATLAB program to sample each sinusoid and generate a sum of three

sinusoids, that is x(n) = x1(n)+x2(n)+x3(n) using a sampling rate of 8000Hz and plot the sum

x(n) over a range of time that will exhibit approximately 0.1 second.

Use the user defined dft() function and MATLAB built-in function fft() to compute the

DFT coefficients and plot and examine the examine the spectrum of the signal.

MATLAB Code:-

clc;

clear all;

close all;

fs=8000;

ts=1/fs;

n=0:(0.1*fs)-1;

x1=5*cos(2*pi*500*ts*n);

x2=5*cos(2*pi*1200*ts*n+0.25*pi);

x3=5*cos(2*pi*1800*ts*n+0.5*pi);

x=x1+x2+x3;

stem(n,x)

title('Sine signals Summation (x signal)')

xlabel('n(samples)')

ylabel('Magnitude')

M=length(x);

%DFT

X=dft(x,M);

k=0:M-1;


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f=k*fs/M;

figure

subplot(2,1,1)

plot(f,abs(X));

title('Mangnitude Vs Frequency(DFT)');

xlabel('Freq.')

ylabel('Magnitude')

subplot(2,1,2)

plot(f,angle(X))

title('Phase Vs Frequency(DFT)');

xlabel('Freq.')

ylabel('Magnitude')

%FFT

X1=fft(x,M);

figure

subplot(2,1,1)

plot(f,abs(X1));

title('Magnitude Vs Frequency(FFT)');

xlabel('Freq.')

ylabel('Magnitude')

subplot(2,1,2)

plot(f,angle(X1))

title('Phase Vs Frequency(FFT)');

xlabel('Freq.')

ylabel('Magnitude')


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OUTPUT:-_


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Problem 12

Using the user defined dft() function obtain the spectral coefficients of

o Rectangular Window

o Hamming Window

o Hanning Window

o Blackman Window

Assume window length of 255 and plot the magnitude and phase spectra of all the windows

individually and compare them

MATLAB Code:-

clc;

close all;

clear all;

win=window(@rectwin,255);

figure

subplot(2,4,1)

plot(win);

bm=dft(win,255)

ak=1;

subplot(2,4,5);

freqz(bm, ak);

title('Magnitude & Phase Spectra of Rectangular Window');

win=window(@hann,255);

figure

subplot(2,4,2);

plot(win);

bm=dft(win,255);

ak=1;

subplot(2,4,6);

freqz(bm, ak);

title('Magnitude & Phase Spectra of Hanning Window');

win=window(@hamming,255);

figure


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subplot(2,4,3);

plot(win);

bm=dft(win,255);

ak=1;

subplot(2,4,7);

freqz(bm, ak);

title('Magnitude & Phase Spectra of Hamming Window');

win=window(@blackman,255);

figure

subplot(2,4,4);

plot(win);

bm=dft(win,255);

ak=1;

subplot(2,4,8);

freqz(bm, ak);

title('Magnitude & Phase Spectra of Blackman Window');


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OUTPUT:-


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Problem 13

Using MATLAB, design a fourth-order digital lowpass Chebyshev filter with cut-off

frequency of 1.5kHz and a 0.5dB ripple at a sampling rate of 8,000Hz.

Determine the transf er function and difference equation

Plot the Magnitude and Phase Response of the filter.

MATLAB Code:-

clc;

close all;

clear all;

N=4;

fpass = 1500;

fs = 8000;

nf = fs/2;

ts = 1/fs;

Wpass = 2*pi*fpass;

rp = 0.5;

ap = 10^(-rp/20);

epsilon = sqrt((1/(ap*ap))-1);

k = 0:1:2*N-1;

uk = ((2*k+1)*pi)/(2*N);

v = asinh(1/epsilon)/N;

sk = Wpass*(sin(uk)*sinh(v)+j*cos(uk)*cosh(v));

sk = sk(N+1:end);

disp(sk);

zk = exp(sk*ts);ak = poly(zk);

if(mod(N,2)==0)

kd = sum(ak)/sqrt(1+(epsilon^2));

else

kd = sum(ak);

end


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bm= kd ;

figure

freqz(bm,ak);

title('Magnitude and Phase Response');

TF=tf(bm,ak)

OUTPUT:-

Command Window:-

1.0e+003 *

-1.6527 - 9.5780i -3.9899 - 3.9673i -3.9899 + 3.9673i -1.6527 + 9.5780i

Transfer function:

(0.303-1.31e-017i)

---------------------------------------------------------s^4 - (1.662-7.772e-016i) s^3 + (1.665-1.332e-015i)

s^2 (0.9257-7.772e-016i)s

+(0.244-2.359e-016i)


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Problem 14

Design a second order digital bandstop Butterworth filter with a centre frequency of

1.8kHz and a bandwidth of 200Hz and a passband ripple of 3dB at a sampling rate of

8,000Hz.

Determine the transfer function and difference equation of the system.

Use MATLAB to plot the magnitude and phase frequency responses.

MATLAB Code:-

clc; clear all; close all;

N=2;

fs=8000;

nf=fs/2;

ts=1/fs;

fp=1902.77;

fstop=1702.77;

fmp=fp/nf;

fms=fstop/nf;

rp=3;

ap=10^(-rp/20);

ap_lin=(1/(ap*ap))-1;

Wp=2*pi*fp;

Ws=2*pi*fstop;

Wc=Wp/((ap_lin)^(1/(2*N)));

fc=Wc/(2*pi);

fmc=fc/nf;

k=0:1:N-1;

sk=Wc*exp(j*(((pi*(2*k+1)/(2*N))) +(pi/2)));

disp(sk);

num=Wc^N;

den=poly(sk);

[bm,ak]=bilinear(num,den,fs);

[num1,den1] = iirlp2bs(bm, ak, fmc, [fms, fmp]);

tf(num1,den1)

figure

freqz(num1,den1);


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OUTPUT:-

Command Window:-

1.0e+003 *

-8.4638 + 8.4638i -8.4638 - 8.4638i

Transfer function:

(0.8714+2.512e-016i) s^4 - (0.5394+1.55e-016i) s^3 + (1.826+5.229e-016i) s^2 - (0.5394

+1.533e-016i) s + (0.8714+2.458e-016i)

---------------------------------------------------------

s^4 - (0.5767-2.176e-018i) s^3 + (1.81-1.452e-017i) s^2 - (0.5022-3.828e-018i) s+ (0.7594-5.339e-

018i)


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Problem 15

Given a DSP system with a sampling rate set up to be 8,000Hz, develop a 800Hz single-tone

generator using a digital IIR filter by completing the following steps

o Determine the digital IIR filter transfer function

o Determine the DSP equation

o Plot the tone thus generated for a duration of 0.01 sec

MATLAB Code:-

clc;

close all;

clear all;

fs=8000;

ts=1/fs;

t=0:ts:0.01;

x=zeros(1,length(t));

x(1)=1;

y=filter([0 sin(2*pi*800/fs) ],[1 -2*cos(2*pi*800/fs) 1],x);

plot(t,y);grid

ylabel('y(n)');

xlabel('time (second)');

TF=tf([0 sin(2*pi*800/fs) ],[1 -2*cos(2*pi*800/fs) 1])


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OUTPUT:-

Command Window:-

Transfer function:

0.5878

-----------------

s^2 - 1.618 s + 1


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Problem 16

Given x(0) = 1, x(1) = 1, x(2) = 0, x(3) = -1, use the Goertzel algorithm to compute the

following DFT coefficients and their amplitude spectra

o X(0)

o X(1)

o |X(0)|2

o |X(1)|2

MATLAB Code:-

clc;

close all;

clear all;

x(1) = 1;

x(2) = 2;

x(3) = 3;

x(4) = 4;

N=4;

X1=goertzel(x,1);

subplot(4,1,1);

stem(abs(X1));

title('Amplitude Spectra of|X(0)|')

xlabel('N(Samples)')

ylabel('Amplitude')

X2=goertzel(x,2);

subplot(4,1,2);

stem(abs(X2));


ylabel('Amplitude')

title('Amplitude Spectra of|X(1)|')

x1=abs(X1)*abs(X1);

subplot(4,1,3);

stem(abs(x1));


ylabel('Amplitude')


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title('Amplitude Spectra of|X(0)|^(2)')

x2=abs(X2)*abs(X2);

subplot(4,1,4);

stem(abs(x2));


ylabel('Amplitude')

title('Amplitude Spectra of|X(1)|^(2)')

disp(['X(0):',num2str(X1)])

disp(['X(1):',num2str(X2)])

disp(['|X(0)|^(2):',num2str(x1)]);

disp(['|X(1)|^(2):',num2str(x2)]);

OUTPUT:-

Command Window:-

X(0):10

X(1):-2+2i

|X(0)|^(2):100

|X(1)|^(2):8


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Problem 17

Design a 31-tap lowpass & highpass FIR filters whose cut-off frequency is 2,500Hz using the

following window functions. Assume that the sampling rate is 8kHz.

o Hamming Window

o Hanning Window

o Blackman Window

MATLAB Code:-

n=31;

fs=8000;

nf=fs/2;

wn=2500;

Wn=wn/nf;

%Low Pass Filter

lphamm = fir1(n,Wn,hamming(n+1));

figure('Name','Lowpass Filter using Hamming Window','NumberTitle','off')

freqz(lphamm,1,fs);

title('Lowpass Filter using Hamming Window')

lphann = fir1(n,Wn,hanning(n+1));

figure('Name','Lowpass Filter using Hanning Window','NumberTitle','off')

freqz(lphann,1,fs);

title('Lowpass Filter using Hanning Window')

lpblac = fir1(n,Wn,blackman(n+1));

figure('Name','Lowpass Filter using Blackman Window','NumberTitle','off')

freqz(lpblac,1,fs);

title('')

%High Pass Filter

if mod(n,2)==1

n=n+1;

end

lhhamm = fir1(n,Wn,'high',hamming(n+1));

figure('Name','Highpass Filter using Hamming Window','NumberTitle','off')

freqz(lhhamm,1,fs);

title('Highpass Filter using Hamming Window')


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lhhann = fir1(n,Wn,'high',hanning(n+1));

figure('Name','Highpass Filter using Hanning Window','NumberTitle','off')

freqz(lhhann,1,fs);

title('Highpass Filter using Hanning Window')

lhblac = fir1(n,Wn,'high',blackman(n+1));

figure('Name','Highpass Filter using Blackman Window','NumberTitle','off')

freqz(lhblac,1,fs);

title('Highpass Filter using Blackman Window')

OUTPUT:-


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Problem 18

Design a 41-tap bandpass FIR filter with the lower and upper cot-off frequencies being

2,500Hz and 3,000Hz respectively, using the following window functions. Assume a sampling

rate of 8kHz.

o Hanning Window

o Hamming Window

o Blackman Window

List the FIR filter coefficients and plot the frequency responses for each design.

MATLAB Code:-

clc;

close all;

clearall;

N=41;fs=8000;

nf=fs/2;

wp=2500;

ws=3000;

wp1=wp/nf;

ws1=ws/nf;

wn=[wp1 ws1];

w = hamming(N+1);

bphamm = fir1(41,wn,w);

figure('Name','Bandpass Filter using Hamming Window','NumberTitle','off')

freqz(bphamm,1,fs)title('Bandpass Filter using Hamming Window')

w=hanning(N+1);

bphann = fir1(41,wn,w);

figure('Name','Bandpass Filter using Hanning Window','NumberTitle','off')

freqz(bphann,1,fs)

title('Bandpass Filter using Hanning Window')

w=blackman(N+1);

bpblac= fir1(41,wn,w);

figure('Name','Bandpass Filter using Blackman Window','NumberTitle','off')freqz(bpblac,1,fs)

title('Bandpass Filter using Blackman Window')


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OUTPUT:-_


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Problem 19

Given the difference equation with the input-output relationship of a certain initially relaxed

DSP system (all initial conditions are zero)

y(n)-0.4 y(n-1)+0.29 y(n-2) = x(n) + 0.5 x(n-1)

Find the impulse response h(n) and frequency response H(w) of the above filter (using

MATLAB)

Find the step response of the system (using MATLAB)

MATLAB Code:-

clc;

clear all;

close all;

num=[1 0.5];den=[1 -0.4 0.29];

sys=tf(num,den,-1);

z=tf('z',-1)

impz(num,den);

figure

freqz(num,den);


figure

step(sys)

title('Step Response of System')


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OUTPUT:-

Command Window:-

Transfer function:

z + 0.5

------------------

z^2 - 0.4 z + 0.29

Sampling time: unspecified


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Problem 20

Write a MATLAB program to read speech data from a PCM formatted speech signal and for

passing it through Pre-Emphasis and De-Emphasis systems.

Also write the output signal from the de-Emphasis System into an output PCM formatted

signal.

MATLAB Code:-

close all;

clearall

fs = 8000; % Sampling rate

%Pre-Emphasis

alpha =0.9; % Degree of pre-emphasis

figure(1);

freqz([1 (-alpha)],1,512,fs); % Calculate and display

frequency responses

t=0:(1/fs):1

x1=sin(2*pi*100*t)+sin(2*pi*1000*t)+sin(2*pi*10000*t);%test signal on behalf of speech signal

figure(2);

y = filter([1 -alpha],1,x1); % Filtering test signal

subplot(2,1,1),

plot(x1,'k')

ylabel('test signal')

title('test signal')

subplot(2,1,2),plot(y,'k');gridylabel('Filtered samples')

xlabel('Number of samples');

title('Pre-emphasized test signal')

figure(3);

N = length(x1); % Length of test signal

Axk =(abs(fft(x1.*hamming(N)')))/ N; % Two-sided spectrum of test signal

Ayk = (abs(fft(y.*hamming(N)')))/ N; % Two-sided spectrumof pre-emphasized test signal

f=[0:N/2]*fs/N;

Axk(2:N)=2*Axk(2:N); % Get one-sided spectrum of test signal

Ayk(2:N)= 2*Ayk(2:N); % Get one-sided spectrum of filtered test signalsubplot(2,1,1),

plot(f,Axk(1:N/2 + 1),'k');grid

ylabel('Amplitude spectrum Ak')

title('Original test signal');

subplot(2,1,2),plot(f,Ayk(1:N/2 + 1),'k');grid

ylabel('Amplitude spectrum Ak')


title('Preemphasized test signal');


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%De-Emphasis

y1=filter(1,[1 (-alpha)],y);

figure(4)

title('Original test signal');

subplot(2,1,2)

subplot(2,1,1)

plot(x1,'k')plot(y1,'k')


title('De-emphasized test signal')

OUTPUT:-_


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Documents

Vlsies Lab Assignment i Sem Dsp_cdac-2