MATLAB Matlab Code for Convolution is this right?

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The discussion revolves around calculating the convolution of a discrete-time system with a given unit response h[n] = (0.2(1.3)^n)u[n] and an input signal x[n] = sin(n*pi/2)u[n]. The user implements the convolution in MATLAB, defining n from 0 to 30, and correctly notes that the unit step function u[n] is equal to 1 for n >= 0, thus simplifying the computation. However, there is a caution regarding the truncation of the convolution result at n = 30, as it may lead to invalid results due to the insertion of zeros at the ends of the signals. The user acknowledges this limitation and suggests that extending the signals beyond the required range could mitigate the issue, emphasizing the importance of signal length in convolution calculations.
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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.
 
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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.
 
oh it didnt specify how far n should go, so i made it randoml from 0 to 30.
 

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