How to Correctly Convolve x[n] with a Unit Step Function?

[redundant6939]

Homework Statement

Find convolution of x[n] (graph in attachment) and h[n] where h[n] = u[n]

Homework Equations

The Attempt at a Solution

- flipped the h[n] to have h[-n]
- moved to the left once (h[-1-n]) to align
- multiplied h and x and it gives me all zeros

Is this correct or I'm missing something?

 

[rude man]

I don't know about your graphical algorithm but if you use the definition

y[n] = x[n]*h[n] = ∑ (k = -∞ to +∞) x[k] h[n-k]

then I get a non-zero sequence of numbers:

a+0.5, b+0.5, c+0.5, c+0.5, d+0.5, d+0.5, d+ 0.5, ...

wher a, b , c, and d are positive integers for you to find.

 

[rude man]

Here's a good demo of how the graphical approach works:
http://www-rohan.sdsu.edu/~jiracek/DAGSAW/4.3.html

 

[lazyaditya]

Just think x(n) as sum of shifted and scaled impulses . Since you know the convolution of a signal h(n) with shifted impulse will be shifted h(n) and then you add them .

 

[rezeew]

Your approach is on the right track, but there are a few things to consider. First, the unit step function u[n] is defined as 1 for n ≥ 0 and 0 for n < 0. So when you flip h[n] to h[-n], you actually get a shifted version of the original unit step function, with h[-n] = 1 for n ≤ 0 and h[-n] = 0 for n > 0. This means that when you multiply h and x, you will only get non-zero values for n ≤ 0, which will not give you the full convolution.

To fix this, you can either shift x[n] to the right instead of h[-n], or you can redefine h[n] as h[n] = u[-n]. Either way, you should end up with a convolution that is non-zero for all values of n, and you can then proceed with the rest of the convolution process.