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

• Yalanhar
In summary, the problem involves finding the value of y[n] for a given range of n based on the signals x[n] and h[n]. The signals are plotted and it is observed that they overlap for n >= 4. The formula used to calculate y[n] is incorrect as it does not take into account the overlap of the two signals. The correct formula involves flipping and shifting the signals and then finding the overlap, which will give a non-zero value for y[n]. It is important to focus on the overlap of the signals rather than the values of x[k] and h[m-k] separately. Drawing the signals in different stages can help in understanding the problem better.
Yalanhar
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

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##

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?

I recommend drawing this out in different stages. You almost have it!

"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.

## What is convolution and why is it important?

Convolution is a mathematical operation that combines two signals to produce a third signal. It is important because it allows us to understand how signals interact and how they can be manipulated to achieve desired outcomes.

## How do you perform convolution of two signals?

To perform convolution of two signals, you first flip one of the signals and then slide it over the other signal, multiplying the overlapping values at each step. The sum of these products gives the resulting signal.

## What are the limits of convolution?

The limits of convolution depend on the length of the two signals being convolved. The resulting signal will have a length equal to the sum of the lengths of the two original signals minus one.

## What is the significance of the sum of y[n] in convolution?

The sum of y[n] in convolution represents the total contribution of each point in the first signal to the resulting signal. It is important in understanding the overall effect of the convolution operation.

## How is convolution used in practical applications?

Convolution is used in a variety of practical applications, such as image and audio processing, pattern recognition, and signal filtering. It is also a fundamental concept in the field of digital signal processing.

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