Conv produces different answers for different steps

Dear all,

I have this piece of code I wrote to plot two signals - x(t) and h(t) - alongside their time range then find the convolution of the two signals.

It works fine with a step size of 1. Howerver, when I try to change the step size - to have better graphs, I get an error in the vectors dimensions. Hence, I can't plot the graphs.

fs = 1;                                             % Sampling rate - step size
x = [2*((-2:1)+2).^2 18*1.^(2:3) -9*((4:5)-5)];     % x(t) function
tx1 = -2;   tx2 = 5;    tx = tx1:fs:tx2;            % time for function x
h = [3*(0:2) -6*((3:4)-3)];                         % h(t) function
th1 = 0;    th2 = 4;    th = th1:fs:th2;            % time for function h

figure(1)
subplot(2,1,1); plot(tx,x); title('x(t)');
subplot(2,1,2); plot(th,h); title('h(t)');

t = ((tx1+th1):fs:(tx2+th2));                      % time for function yc (conv)
yc = conv(x,h);                                    % Continuous convolution

figure(2)
subplot(2,1,1); stem(yc);   xlabel('--->n');  ylabel('y(n)'); title('Discrete Convolution');
subplot(2,1,2); plot(t,yc); xlabel('t');      ylabel('y(t)'); title('Continuous Convolution');

When I change fs to 0.1, I get this error. Is there a way to solve this issue?

Error using plot
Vectors must be the same length.

Thanks in advance for your answers.

Answer 1

[Ameer Hamza]

It happens because when you decrease 'fs', the length of tx increase, but the value of vector 'x' is hardcoded

x = [2*((-2:1)+2).^2 18*1.^(2:3) -9*((4:5)-5)]; 

Its length remains the same, so the plot function gives an error. You need to increase its length of 'x' to be of the same size as the time vector.

As a side note, remember that conv() is a function for discrete-time convolution. If you try to change the step size, then you also need to multiply it with the step-size to get the correct result

yc = conv(x,h)*fs;

[Aqeel Mohamed]

Its length remains the same, so the plot function gives an error. You need to increase its length of 'x' to be of the same size as the time vector.

Thanks for your answer.

How could I do this? I tried this but it still giving the same error.

x = [2*((-2:fs:1)+2).^2 18*1.^(2:fs:3) -9*((4:fs:5)-5)];     % x(t) function

Thanks for the side note also.

[Ameer Hamza]

Can you show the equstion of x(t)?

[Aqeel Mohamed]

[Ameer Hamza]

Check the following code to see how these equations can be correctly implemented

fs = 0.1;                                             % Sampling rate - step size
tx1 = -2;   tx2 = 5;    tx = tx1:fs:tx2;            % time for function x
x = zeros(size(tx));                                % x(t) function
x(-2<=tx & tx<1) = 2*(tx(-2<=tx & tx<1)+2).^2;
x(1<=tx & tx<3) = 18;
x(3<=tx & tx<5) = -9*(tx(3<=tx & tx<5)-5);

th1 = 0;    th2 = 4;    th = th1:fs:th2;            % time for function h
h = zeros(size(th));                               % h(t) function
h(0<=th & th<2) = 3*(th(0<=th & th<2)+2);
h(2<=th & th<4) = -6*th(2<=th & th<4)+3;


figure(1)
subplot(2,1,1); plot(tx,x); title('x(t)');
subplot(2,1,2); plot(th,h); title('h(t)');

t = ((tx1+th1):fs:(tx2+th2));                      % time for function yc (conv)
yc = conv(x,h)*fs;                                 % Continuous convolution

figure(2)
subplot(2,1,1); stem(yc);   xlabel('--->n');  ylabel('y(n)'); title('Discrete Convolution');
subplot(2,1,2); plot(t,yc); xlabel('t');      ylabel('y(t)'); title('Continuous Convolution');

[Aqeel Mohamed]

Works great!

Thank you very much Ameer :)

[Ameer Hamza]

I am glad to be of help.