Compute convolution y[n]=x[n]*h[n]: x[n]={2,0,1,-1,3}; h[n]={1,2,0,1}

My approach:

x=[2 0 1 -1 3];
h=[1 2 0 1];

% therefore 
y=conv(x,h)

y =

2 4 1 3 1 7 -1 3

[Constantino Carlos Reyes-Aldasoro]

What exactly is your question?

[James]

what if h[n]and x[n] given some function? like x[n]=u[n] and h[n]= 0.9^n*u[n-1]

[Walter Roberson]

@James

It would depend on whether that u[n-1] is the unit step function or not.

If it is then x[n] would be 0 for n < 0, and 1 for n >= 0 -- an infinite stream of 1's. And h[n] would be 0 for n <= 1, and 0.9^n for n > 1 -- an infinite stream of non-negative numbers. You cannot express that as a finite convolution sequence.

Let us see what it would turn out like for continuous functions:

sympref('heavisideatorigin', 1)

ans = 

1

syms n integer
x(n) = heaviside(n)

x(n) = 

h(n) = (sym(9)/sym(10))^n * heaviside(n-1)

h(n) = 

syms t tau
C(t) = int(x(tau) * h(t-tau), tau, 0, t)

C(t) = 

[C(-1), C(0), C(1)]

ans = 

assume(sign(t-1) == 1)
simplify(C)

ans(t) = 

[Paul]

The follow-up problem posed by James can be solved for a closed form solution using symsum and the shift property of the convolution sum. Seem like a HW problem so won't show it now.

Answer 1

[Image Analyst]

@Li Hui Chew, yes your approach is correct. Is that all you wanted - confirmation of your approach?