Convolution

**Toonzaka** · February 19th 2008, 06:18 PM

These are my equations
x[n] = δ[n] + 2δ[n-1] - δ[n-3]
h[n] = 2δ[n+1] + 2δ[n-1]

They want to know what
y[n] = x[n] * h[n] (where * is the convolution)

I am getting so lost on this problem. I do not understand how to combine these two to get y[n]. I understand the definition of convolution, just I do not understand how to go about finding y[n]. Can someone please help me out on this problem.

Can you please show me the steps to get the answer.
I looked in the back and the answer is this...
y[n] = 2δ[n+1] + 4δ[n] + 2δ[n-1] + 2δ[n-2] - 2δ[n-4]
No idea how they get that... Can someone please show me the steps to get that answer?

**Peritus** · February 19th 2008, 11:04 PM

It's actually quite simple, first let us recall the shifting property of the delta function:

$\delta [n - k]*f[n] = f[n - k]$

$y[n] = x[n]*h[n] = \left( {\delta \text{[n] + 2}\delta \text{[n - 1] - }\delta \text{[n - 3]}} \right)*\left( {\text{ 2}\delta \text{[n + 1] + 2}\delta \text{[n - 1]}} \right) =$

$ = 2\delta \text{[n]*}\delta \text{[n + 1] + 2}\delta \text{[n]*}\delta \text{[n - 1]} $ $ \text{ + 4}\delta \text{[n - 1]*}\delta \text{[n + 1] + 4}\delta \text{[n - 1]*}\delta \text{[n - 1] - 2}\delta \text{[n - 3]*}\delta \text{[n + 1] - 2}\delta \text{[n - 3]*}\delta \text{[n - 1] = } $

$ \text{ = 2}\delta \text{[n + 1] + 2}\delta \text{[n - 1] + 4}\delta \text{[n] + 4}\delta \text{[n - 2] - 2}\delta \text{[n - 2] - 2}\delta \text{[n - 4] = } $

$ \text{ = 2}\delta \text{[n + 1] + 2}\delta \text{[n - 1] + 4}\delta \text{[n] + 2}\delta \text{[n - 2] - 2}\delta \text{[n - 4]} $

**Toonzaka** · February 20th 2008, 08:24 AM

For the second line
$ = 2\delta \text{[n]*}\delta \text{[n + 1] + 2}\delta \text{[n]*}\delta \text{[n - 1]} $ $ \text{ + 4}\delta \text{[n - 1]*}\delta \text{[n + 1] + 4}\delta \text{[n - 1]*}\delta \text{[n - 1] - 2}\delta \text{[n - 3]*}\delta \text{[n + 1] - 2}\delta \text{[n - 3]*}\delta \text{[n - 1] = } $

How do you get this is it that you take the function and then multiply it out?

**Peritus** · February 20th 2008, 09:21 AM

I'm not sure I understand your question, but maybe you're refering to the distributive property of the convolution:

$f[n]*\left( {g[n] + w[n]} \right) = f[n]*g[n] + f[n]*w[n]$

**Toonzaka** · February 20th 2008, 12:10 PM

If we were given a piece wise function such as...

x(t) =
t+1, 0 <= t <= 1

2-t, 1 < t <= 2

0, elsewhere

And give that
$h(t) = \delta \text{(t+2) + 2}\delta \text{(t+1)}$

And asked to find the convolution of this, how would we go about getting the answer?

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