Convolving 0.2(1.3)^n with sin(n*pi/2) in MATLAB

[user101]

Homework Statement

A DT system has the unit response h[n] = (0.2(1.3)^n)u[n]. You then input x[n] = sin(n*pi/2)u[n]. Find the convolution.

Homework Equations

None.

The Attempt at a Solution

>> n = 0:30;
>> h = 0.2*(1.3).^n;
>> x = sin(n*pi/2);
>> s = conv(h,x);

I left out u[n] because we know that u[n] = 1 whenever n >= 0 and 0 when n < 0. So, if I am correct, it is just multiplying the input by 1, if n >=0.

This should be right, just wanted to verify.

 

[wildman]

Looks good to me. Just remember that the far end of the convolution sum will not be valid because you are truncating it at 30. What the algorthm does is insert zeros at the front and back of the signals. This is all right on the low end because the signals are zero anyway below zero, but it will cause invalid results at the high end. This can be fixed by simply making the signals longer than you need.

 

[user101]

oh it didnt specify how far n should go, so i made it randoml from 0 to 30.