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Math Help - Discrete Time Convolution Integral

  1. #1
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    Discrete Time Convolution Integral

    Hello Everyone!

    I've been stuck on this DT convolution sum for quite sum time, it's because I have not much experience in calculating DT sums. Well here goes:

    x[n] = (1/2)^{n}.u[n+2]
    h[n] = (-3)^{n}.u[-n-1]

    Now, x[n] \ast h[n] = \displaystyle \sum ^{\infty} _{k = -\infty} (1/2)^k .u[k+2] . (-3)^{n-k}.u[-n-1+k]

    Now here, I saw that whenevr we have step functions we can remove them and change the limits of the sum, so I modified the sum to be from n+1 to infinity as:
    \displaystyle \sum ^{\infty} _{k=n+1}(1/2)^k . (-3)^{n-k}.
    But I think I am very wrong.

    Any help would be greatly appreciated!
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  2. #2
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    Nov 2009
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    If you mirror and shift h[n], the limits become:

    u[-n-1] is 1 for -n-1 >= 0 => n<=-1. If you mirror and shift that you end up with k-n<=-1 => k <= n-1, hence the upper limit.
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