DIGITAL SIGNAL PROCESSING(MATLAB) ECE LAB REPORT, Assignments of Digital Signal Processing

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Submitted By:-

Arunabh Bora

Roll No:- 180104004

th

Sem

Electronics and Communication Engineering

Aim:- To find the z transform of a finite sequence

Theory :- In mathematics and Signal Processing the z transform converts a

discrete time signal which is a response of real or complex number into a

complex frequency domain representation

Mathematically 𝑍𝑍

∞

𝑛𝑛=

−𝑛𝑛

Matlab Code:-

%Z transform of finite sequences

clc;

close all;

clear all;

syms 'z';

disp('If you input a finite duration sequence x(n),we will

give you its z-transform');

nf=input('Please input the initial value of n = ');

nl=input('Please input the final value of n = ');

x= input('Please input the sequence x(n)= ');

syms 'm';

syms 'y';

f(y,m)=(y*(z^(-m)));

disp('Z-transform of the input sequence is displayed below');

k=1;

for n=nf:1:nl

answer(k)=(f((x(k)),n));

k=k+1;

end

disp(sum(answer));

Aim :- To plot the poles and zeros of

1−𝑧𝑧

−

−2𝑧𝑧

−

1−1.75𝑧𝑧

−

+1.25𝑧𝑧

−

−0.375𝑧𝑧

−

Theory :-

The roots of solution of variable S in characteristic equation are called poles of

the system of transfer function. The roots of solution of variable as in the

numerator polynomials are called zeros of the system of closed loop transfer

function.

Matlab Code :-

%to plot poles and zeros

clc;

clear all;

close all;

H=[1 1 2];

Q=[1 -1.75 1.25 -0.375];

zplane(H,Q);

grid on;

Result :-

Conclusion :-

From the above experiment we get our result and it’s successfully

executed.

Aim :- (𝑖𝑖)Find Residue Term and convert back to 𝑥𝑥(𝑛𝑛)

𝑧𝑧

3𝑧𝑧

2

−4𝑧𝑧+

(𝑖𝑖𝑖𝑖) Find 𝑧𝑧

−

Theory :- In complex analysis a discipline within mathematics the Residue

theorem sometime called ‘Cauchy’s Residue theorem’ is a powerful tool to

evaluate line integrals of analytic function over closed curve can often be used

to compute real integrals and infinite series as well.

Matlab Code :-

%Residue term and connect back to x(n)

clc;

clear all;

close all;

b = [0 10];

a= [3 -4];

[r,p,c]= residuez(b,a);

%Residue term and connect back to x(n)and find z^-

clc;

clear all;

close all;

r=[0.5000,-0.5000];

p=[1.0000,-0.3333];

c=[?];

[b,a]=residuez(r,p,c);?

Aim :- Determine inverse z transform of

1+0.4√2 𝑧𝑧

−

1−0.8√2𝑧𝑧

−

+0.64𝑧𝑧

−

Theory :- In mathematics and Signal Processing the inverse z transform converts

frequency domain signal to discrete time domain signal.

Matlab Code :-

%Determine Inverse Z transform

clc;

clear all;

close all;

b= [1+0.4*sqrt(2) 0];

a= [1-0.8*sqrt(2) 0.64];

[r,p,c]=residuez(b,a);

M=abs(p);

Q=angle(p);

Result :-

Conclusion :-

From the above experiment we get our result and it’s successfully

executed.

Aim:- Given difference equation

(𝑖𝑖) Find 𝐻𝐻(𝑧𝑧)

(𝑖𝑖𝑖𝑖) Sketch the poles and zeros

(𝑖𝑖𝑖𝑖𝑖𝑖) Plot �𝐻𝐻(𝑒𝑒

𝑗𝑗𝑗𝑗

𝑗𝑗𝑗𝑗

(𝑖𝑖𝑖𝑖) Find ℎ(𝑛𝑛)

𝐴𝐴𝑛𝑛𝐴𝐴: − Given that 𝑦𝑦

−

𝑌𝑌(𝑧𝑧)

𝑋𝑋(𝑧𝑧)

1

1−0.9𝑧𝑧

−

Matlab Code:-

%poles and zeros

clc;

clear all;

close all;

b=[1,0];

a=[1,-0.9];

zplane(b,a);

grid on;

(𝑖𝑖𝑖𝑖𝑖𝑖)

%residuez,angle,absolute

clc;

clear all;

close all;

b=[1,0];

a=[1,-0.9];

[r,p,c]=residuez(b,a);

[H,W]=freqz(b,a,100);

M= abs(H);

Q= angle(H);

Aim:- To generate DFS and IDFS function.

Theory:- The term discrete function series(DFS) is intended for use instead of

DFS where the original function is periodic defined over an infinite integral.

−𝑗𝑗

2𝜋𝜋𝑛𝑛𝜋𝜋

𝑁𝑁

𝑁𝑁−

𝑛𝑛=

𝑗𝑗

2𝜋𝜋𝑛𝑛𝜋𝜋

𝑁𝑁

𝑁𝑁−

𝑛𝑛=

Matlab Code :-

clc;

clear all;

close all;

xn = [0,1,2,3];

N = 4;

Xk = dfs(xn,N);

xn = idfs(Xk,N);

function [Xk] = dfs(xn,N)

n = [0:1:N-1]; % row vector for n

k = [0:1:N-1]; % row vector for k

WN = exp(-j2pi/N);

nk = n'*k;

WNnk = WN .^ nk; %DFS matrix

Xk = xn * WNnk;

end

function [xn] = idfs(Xk,N)

n = [0:1:N-1]; % row vector for n

k = [0:1:N-1]; % row vector for k

WN = exp(-j2pi/N);

nk = k' * n;

WNnk = WN .^(- nk); %IDFS matrix

xn = (Xk*WNnk)/N;

end

Result :-

Conclusion :-

From the above experiment we get our result and it’s successfully

executed.

%N=

N = 20;

k = 0:1:N-1; % sample index

wk = 2pik/N;

zk = exp(j*wk); % samples of z

Xk = (zk)./(zk-0.7); % DFS as samples of X(z)

xn = real(idfs(Xk,N)); % IDFS

xtilde = xn'*ones(1,2);

xtilde = (xtilde(:))'; % Periodic sequence

subplot(2,2,3); stem(0:39,xtilde);axis([0,40,-0.1,1.5])

xlabel('n'); ylabel('xtilde(n)'); title('N=20')

%N=

N = 40;

k = 0:1:N-1; % sample index

wk = 2pik/N;

zk = exp(j*wk); % samples of z

Xk = (zk)./(zk-0.7); % DFS as samples of X(z)

xn = real(idfs(Xk,N)); % IDFS

xtilde = xn'*ones(1,1);

xtilde = (xtilde(:))'; % Periodic sequence

subplot(2,2,4); stem(0:39,xtilde);axis([0,40,-0.1,1.5])

xlabel('n'); ylabel('xtilde(n)'); title('N=40')

function [xn] = idfs(Xk,N)

% Computes Inverse Discrete Fourier Series

% ----------------------------------------

% [xn] = idfs(Xk,N)

% xn = One period of periodic signal over 0 <= n <= N-

% Xk = DFS coeff. array over 0 <= k <= N-

% N = Fundamental period of Xk

n = [0:1:N-1]; % row vector for n

k = [0:1:N-1]; % row vecor for k

WN = exp(-j2pi/N); % Wn factor

nk = n'*k; % creates a N by N matrix

of nk values

WNnk = WN .^ (-nk); % IDFS matrix

xn = (Xk * WNnk)/N; % row vector for IDFS

values

end

Result:-

Conclusion :-

From the above experiment we get our result and it’s successfully

executed.

n1=n+1;

if(rem(n,2)~=0)

n1=n;

n=n-1;

end

y=boxcar(n1);

%LPF

b=fir1(n,wp,y);

[h,o]=freqz(b,1,256);

m=20*log10(abs(h));

subplot(221);plot(o/pi,m);

ylabel('Gain in dB-->');xlabel('(a) Normalised frequency --

');

%HPF

b=fir1(n,wp,'high',y);

[h,o]=freqz(b,1,256);

m=20*log10(abs(h));

subplot(222);plot(o/pi,m);

ylabel('Gain in dB-->');xlabel('(b) Normalised frequency --

');

%BAND PASS FILTER

wn=[wp ws];

b=fir1(n,wn,y);

[h,o]=freqz(b,1,256);

m=20*log10(abs(h));

subplot(223);plot(o/pi,m);

ylabel('Gain in dB-->');xlabel('(c) Normalised frequency --

');

%BAND STOP FILTER

b=fir1(n,wn, 'stop', y);

[h,o]=freqz(b,1,256);

m=20*log10(abs(h));

subplot(224);plot(o/pi,m);

ylabel('Gain in dB-->');xlabel('(d) Normalised frequency --

');

Result:-

Conclusion :-

From the above experiment we get our result and it’s successfully

executed.

[b,a]=butter(n,wn,'high');

w=0:.01:pi;

[h,om]=freqz(b,a,w);

m=20*log10(abs(h));

an=angle(h);

subplot(423);plot(om/pi,m);

ylabel('Gain in dB-->');xlabel('(a) Normalised frequency --

');title('High Pass Filter');

subplot(424);plot(om/pi,an);

ylabel('Phase in radians-->');xlabel('(b) Normalised frequency

-->');title('High Pass Filter');

%BAND PASS FILTER

w1=2*wp/fs;

w2=2*ws/fs;

%[n]=buttord(w1,w2,rp,rs);

wn=[w1 w2];

[b,a]=butter(n,wn,'bandpass');

w=0:.01:pi;

[h,om]=freqz(b,a,w);

m=20*log10(abs(h));

an=angle(h);

subplot(425);plot(om/pi,m);

ylabel('Gain in dB-->');xlabel('(a) Normalised frequency --

');title('Band Pass Filter');

subplot(426);plot(om/pi,an);

ylabel('Phase in radians-->');xlabel('(b) Normalised frequency

-->');title('Band Pass Filter');

%BAND STOP FILTER

w1=2*wp/fs;

w2=2*ws/fs;

wn=[w1 w2];

[b,a]=butter(n,wn,'stop');

w=0:.01:pi;

[h,om]=freqz(b,a,w);

m=20*log10(abs(h));

an=angle(h);

subplot(427);plot(om/pi,m);

ylabel('Gain in dB-->');xlabel('(a) Normalised frequency --

');title('Band Stop Filter');

subplot(428);plot(om/pi,an);

ylabel('Phase in radians-->');xlabel('(b) Normalised frequency

-->');title('Band Stop Filter');

Result:-

Conclusion :-

From the above experiment we get our result and it’s successfully

executed.