Find the convolution of x(n) and h(n). where where the convolution is note the symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for , where n=0,1,2,3.... Now, Plugging in, Simplifying, The argument passed into the unit step must be or else the value of the unit step function is 0. So if I need to change my limits of summation so that the two unit step functions resolve to 1. That is, if then . if then . So the limits on k would be, if if I am bamboozled by the two lower limits on k. How can I do a summation with two lower limits? Would the summation diverge?
Signals and systems eh. Chapter 2 proplem . Funny thing is i got stuck on the same thing yesterday. The summation limits change with each n chosen which makes calculating the summation to a closed form a bit more tedious. Most likely it needs to be solved by looking at different scenarios when you plug in different "ranges" of n. K needs to be larger or equal to 4 so that u[k-4] be nonzero. From this we could say that we could start summing from k = 4. however let's say we have y[7]. Which means that u[2-n+k] won't be nonzero till k = 5. So with increasing n we can't start summing till k = n-2. So for n between 0 and 5 you can start summing from k = 4. While for n >= 6 you have to begin the summation at K = n-2.
Hi, thanks for the reply. I was able to solve it shortly after posting. It's like you said, the limits depend on values of n. So the ultimate solution is piecewise.