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Math Help - Convolution

  1. #1
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    Convolution

    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?
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  2. #2
    Senior Member Peritus's Avatar
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    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) =


    <br />
 = 2\delta \text{[n]*}\delta \text{[n + 1] + 2}\delta \text{[n]*}\delta \text{[n - 1]}<br />
<br />
\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] = }<br />

    <br />
\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] = }<br />

    <br />
\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]}<br />
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  3. #3
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    For the second line
    <br />
= 2\delta \text{[n]*}\delta \text{[n + 1] + 2}\delta \text{[n]*}\delta \text{[n - 1]}<br />
<br />
\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] = }<br />

    How do you get this is it that you take the function and then multiply it out?
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  4. #4
    Senior Member Peritus's Avatar
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    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]
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  5. #5
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    One more question!

    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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