Question

A causal discrete-time system is described by the following difference equation: Use Matlab to write a script to complete the following tasks. Turn in the output created by the Matlab "publish" utility. (a) Compute and plot the impulse response h[n], 0くn 〈 50. Use the function h=imp2(b, a , N ) to find the impulse response, and use the stem ) function to create the plot. (b) Let x[n] be defined by (n - 15)2 0n K 30 x[n] elsewhere Compute and plot the output yn for0 n K 75 using the function y-filter (b,a, x); c) Repeat 2b, this time using the function y conv (h,x) (d) Use your script to verify that the two results are the same.

Answer 1

Answer:-

Given differential equation is (In the question there is some ambiguity in the differential equation there is a sign missing between y(n), x(n) terms)

Given differential equation is (In the question there is some ambiguity in the differential equation there is a sign missing between y(n), x(n) terms) y(n) = 1.3(n - 1) - 0.72y(n - 2) - 0.081y(n - 3) + 0.3645y(n - 4) + 2.0x(n) + 2.8x(n - 1) + 1.6x(n - 2) - 0.4x(n - 3) - 1.2x(n - 4) In order to use the impz(b, a) we need to compute the transfer function and hence b(The coefficients int he denominator), a (coefficients in the numerator) . a) To find the transfer function, Applying z transform on either side, y(z) = y(z) (!.3z^-1 - 0.72z^-2 - 0.081z^-3 + 0.3645z^-4) + x(z) (2.0 + 2.8Z^-1 + 1.6z^-2 - 0.4z^-3 - 1.2z^-4) y(Z) (1 - 1.3z^-1 + 0.72z^-2 + 0.081z^-3 - 0.3645z^-4) = x(z)(2.0 + 2.8z^-1 + 1.6z^-2 - 0.4z^-3 - 1.2z^-4) H(z) = y(z)/x(z) = 2.0 + 2.8z^-1 + 1.6z^-2 - 0.4z^-3 - 1.2z^-4/1 - 1.3z^-1 + 0.72z^-2 + 0.081z^-3 - 0.3645z^-4) so b = [1 -1.3 0.72 0.081 0.3645]: a = [2 2.8 1.6 -0.4 -1.2]: percent Matlab code for part a. percent defining the numerator \r\n

In order to use the impz(b,a) we need to compute the transfer function and hence b(The coefficients int he denominator), a (coefficients in the numerator) .

a) To find the transfer function, Applying z transform on either side,

So

b = [1 -1.3 0.72 0.081 0.3645];
a = [2 2.8 1.6 -0.4 -1.2];

%Matlab code for part a.

% defining the numerator and denominator

b = [1 -1.3 0.72 0.081 0.3645];
a = [2 2.8 1.6 -0.4 -1.2];

%Computing impulse respnse and plotting
h = impz(b,a);

figure;stem(h);xlabel('n, samples');

ylabel('h(n)');title('Impulse response');

Output of this code:

B) In the following code the input signal is defined and the output of the filter is computed using filter function

% Defining the signal
n = 0:75;
for i = 0: 75
    if(i<=30)
        x(i+1) = (i)*(30-i);
    else
        x(i+1) =0;
    end
end
stem(n,x);xlabel('n, samples');ylabel('x(n)');title('Input signal');
% Finding the output using Fitler
yfilt = filter(b,a,x);
stem(n,yfilt);xlabel('n, samples');ylabel('y(n)');title('Output signal using filter function');

Output for the above code:

C) The Matlab code for part C is given below

ycon = conv(h,x);
n = 0:length(x)+length(h)-2;
figure,stem(n,ycon);xlabel('n, samples');ylabel('y(n)');title('Output signal using convolution');

The output of the code is

The output of the filter function and the output of convolution function are not the same. Filter function computes the output which will have same length of x while the conv function computes the full linear convolution which is of length =length(x)+length(h)-1.