How to compute the convolution of following fn

Y[n]=(-1)^n * 2^n u[2n+1]

Please solve it and share it.




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If you are posting class assignments and getting it done by others, it is unfair practice. However, if you are unable to understand some concepts, it is fine to ask here.

I am preparing for test so if you know how to solve it...
Do it...
I just tick the option by checking the answer from back of the chapter...

Convolution needs two functions: we convolute one function by another. See https://mathworld.wolfram.com/Convolution.html

You have given one function only.

There are two functions * is convolution sign between them look at it again

What is the function u[2n+1]?

if we call:
x = (-1)^2
h = 2^n*u[2n+2], then

y = x * h

making a graph, x will be an infinte alternating [... -1 1 -1 1 ...] with 1 for n=0
h will be, instead, zero up to n = -2, then will be [0.5 1 2 4 8 16 32 ....] with 0.5 for n=-1

To perform the convolution we have to flip and shift in time one of the two functions. Here it esasier to flip x because it will remain exactly the same (it's an even function). For n=0 we will have no shift of x the we will obtain the sum:

0.5 - 1 + 2 - 4 ...... --> infinite

shifting by one:

-0.5 + 1 - 2 + 4 ...... --> infinite

then it will diverge. I think there should be a typo in the book. If we substitute 2^n with 2^(-n) the previous sums will converge to:

1/0.5 - 1/1 + 1/2 - 1/4 --> 4/3
-1/0.5 + 1/1 -1/2 +1/4 --> -4/3

in this case the solution is 4/3*(-1)^n
in order to have the factor 8/3 we'll need the step function starting from n=-2 instead of n=-1 that is u[2n+4]