Convolution not commutative for this case?

Discussion in 'Homework Help' started by jp1390, Oct 6, 2011.

  1. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Hi all, just needing some clarification on this question, or just another pair of eyes to check if I am doing this correctly. I am wanting to compute the convolution of v(t) and g(t). In doing so I wanted to try to test commutativity, but found that it wasn't working out.

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    When I tried the first convolution, I yielded the correct result over the desired regions of overlap, but for the second, I did not.

    Method 1
    y(t) = g(t)\ast v(t) = \int_{-\infty}^{\infty}g(\tau)v(\tau-t)d\tau

    Method 2
    y(t) = v(t)\ast g(t) = \int_{-\infty}^{\infty}v(\tau)g(\tau-t)d\tau (did not receive correct result)

    Region 1

    t <= 0;
    y(t_{1}) = 0

    Region 2

    t >= 0 and (t-2) <= 0
    0 <= t <= 2;

    y(t_{2}) = \int_{0}^{t}(exp{-\tau})(2exp{2(\tau -t)})d\tau

    = 2exp{-2t}\int_{0}^{t}exp{\tau}d\tau

    =2exp{-2t}(exp{t} - 1)

    The answer for this region should be:

    y(t_{2}) = 2exp{-t}(1 - exp{-t})

    I got the correct result when I flipped and shifted v(t) as in Method 1, which is weird because they should be commutative.

    Can anyone see where I went wrong?

    Thanks,
    JP
     
  2. Zazoo

    Member

    Jul 27, 2011
    114
    43
    These two results are the same, they are just factored differently. i.e if you distribute the outside term both are equal to:
    y(t) = 2exp{-t} - 2exp{-2t}
     
  3. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Whoa, I must have had a major brain fart. Thanks haha
     
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