Convolution of 2 Signals: Finding the Sum and Limits of y[n]

AI Thread Summary
The discussion focuses on the convolution of two signals, x[n] and h[n], where x[n] is defined for n >= 4 and h[n] for n <= 2. The participant is trying to calculate y[n] but is confused about the limits of summation and the conditions under which y[n] has values. It is clarified that the overlap of the two signals is crucial for determining y[n], and that the multiplication of x[k] and h[m-k] will yield non-zero results when they overlap, specifically for m >= 6. The advice includes visualizing the process in stages and simplifying the problem to better understand the convolution. Understanding the overlap is key to resolving the participant's confusion.
Yalanhar
Messages
39
Reaction score
2
Thread moved from the technical forums, so no Homework Template is shown.
##x[n] = (\frac{1}{2})^{-2} u[n-4]##
##h[n] = 4^{n} u[2-n]##

So I plotted x[k] and h[n-k] in picture

172248458_2995201717409947_6353737937217694665_n.jpg


but x[n] is 0 for n < 4, therefore ##y[n]## only has value for n >= 4. Therefore my sum is like that:

##y[n]=\sum_{k=4}^{\infty} 4^{n-k} (-\frac{1}{2})^k##

##y[n]=-4^{n} \sum_{k=4}^{\infty} (\frac{1}{8})^k##

however, the answer is
for n <= 6,
##y[n]=4^{n}( \sum_{k=0}^{\infty} (-\frac{1}{8})^k - \sum_{k=0}^{3} (-\frac{1}{8})^k) ##

for n > 6,
##y[n]=4^{n} (\sum_{k=0}^{\infty} (-\frac{1}{8})^k - \sum_{k=0}^{n-1} (-\frac{1}{8})^k )##

What is my mistake? I'm having difficulties with this subject and I am not sure when I use k as limits or n. Also, if both signals extend to ##+\inf##, does that means that y[n] always has a value?
 
Physics news on Phys.org
I recommend drawing this out in different stages. You almost have it!

convolution_example.jpg


"Flip and shift." You flipped the ##h[n]## no problem and you can see immediately that since ##x[k]## goes to infinity and ##h[m-k]## goes to infinity there is some overlap and the multiplication of the two will give you a non-zero answer :)

You shift ##h[m-k]## by adjusting the values of ##m##. When ##m=6## they both suddenly overlap on the right side. It doesn't matter that the left side is zero because the multiplication of the two is the overlap... it's non-zero for when ##m \geq 6##.

It didn't matter that that ##x[k]## where ##k < 4## it's zero. Your concern is with the overlap since you're multiplying ##x[k] \cdot h[m-k]##. The overlap is your concern, and so while it is indeed zero in that region your product (the multiplication of them) is not zero.

Forgive me if the variables or letters I used are confusing. I'm use to using ##t## and ##\tau##, but I was trying to align with what I thought you were using.

Does this help? Again: Try drawing in different stages and break this into a smaller and easier problems instead of one big one.
 
Back
Top