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Convolution with Impulse Function

  1. Oct 13, 2012 #1
    Is this true:

    h(t)* Heaviside(t-t0) = h(t0)

    If this is true saves my work a lot. It appears not to be true I tried proving it...
     
  2. jcsd
  3. Oct 13, 2012 #2

    cepheid

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    I don't think that's true. It's true if you replace the Heaviside step function with a Dirac delta function.
     
  4. Oct 13, 2012 #3
    Sorry I meant:

    h[n]*IMPULSE[n-t0] = h[n-t0]

    I tried it with a question that wanted to do a convolution that would expand out as follows.

    x[n] = δ[n] + 2δ[n-1] - δ[n-3]
    h[n] = 2δ[n+1] + 2δ[n-1]

    x[n] * h[n] = 2δ[n]*δ[n+1] + 2δ[n]*δ[n-1] + 4δ[n-1]δ[n+1] + 4δ[n-1] - 2δ[n-3]δ[n+1] - 2δ[n-3]δ[n-1]

    x[n] * h[n] = 2δ[n+1] + 2δ[n-1] + 4δ[n+1] + 4δ[n-1] - 2δ[n+1] - 2δ[n-1]

    x[n] * h[n] = 4δ[n+1] + 4δ[n-1]

    But it turns out answer is wrong. So I crapped out.
     
  5. Oct 13, 2012 #4
    It is only true for h(n) multiplied by δ[n-t0]

    Not for Convolution.

    AM I CORRECT?
     
  6. Oct 13, 2012 #5

    cepheid

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    Yeah ok. It is true for multiplication: x[n]δ[n - a] = x[a]. The property I was thinking of for convolution was that the convolution of any function with the Dirac delta function (or unit impulse function in discrete time) is just equal to the function itself:

    x[n]*δ[n] = x[n]

    which follows if you just think of the definition of a convolution sum. A function in discrete time can be represented as a sequence of shifted and scaled unit impulses. What also follows fairly intuitively from that is just that:

    x[n]*δ[n + a] = x[n + a]
     
  7. Oct 14, 2012 #6
    I have a hard time finding that to be correct only because it fails to yield the right answer in the example above.
     
  8. Oct 14, 2012 #7

    cepheid

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    Oh okay. It must be wrong then. :rolleyes:

    How about if you correct the problem in red below:

    The term in red should be 4δ[n-1]*δ[n-1]. Also, how about if you show the steps in your latest attempt?
     
  9. Oct 14, 2012 #8
    These theorems are right in all sense..
    x(n)*delta(n)=x(n)
    x(n)*delta(n-n0)=x(n-n0)
    x(n).delta(n)=x(0)
    x(n).delta(n-n0)=x(n0)
    and u have done it in a wrong way and so u got it wrong. That just meant that u r wrong not the theorom...
     
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