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Convolution problem

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Convolve,

    [tex]2e^{-t}u(t) \ast e^{t}u(-t) \quad \text{Where u(t) is the step function}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]2e^{t}\int_{?}^{?}e^{-\tau}u(\tau) e^{-\tau}u(-t+\tau)d \tau[/tex]

    I'm not sure what the limits on my integral should be, the u(-t) is confusing me. Shouldn't the second function only be nonzero from -infinity to 0 and the other from to +infinity?

    Thanks again!
     
    Last edited: Oct 8, 2012
  2. jcsd
  3. Oct 8, 2012 #2

    LCKurtz

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    The limits are from ##-\infty## to ##\infty##, but of course the step functions can cut the interval down by being zero in places. But perhaps an easier approach would be to remember that the Fourier transform ##\mathcal F## satisfies$$
    \mathcal F(f * g) =\hat f (\omega)\cdot \hat g(\omega)$$So transform the functions, multiply the transforms, and invert the result. That would give ##f*g## and you can likely just use transform tables and be done with it.
     
  4. Oct 14, 2012 #3
    We haven't learnt that technique yet nor do we have access to tables so is there a way we can mechanically do the integration?

    How would it be done?
     
  5. Oct 14, 2012 #4

    jbunniii

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    For what values of [itex]\tau[/itex] is [itex]u(\tau)[/itex] nonzero?

    For what values of [itex]\tau[/itex] is [itex]u(-t + \tau)[/itex] nonzero?
     
  6. Oct 14, 2012 #5

    LCKurtz

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    Well, start with this:$$
    2e^{t}\int_{-\infty}^{\infty}e^{-\tau}u(\tau) e^{-\tau}u(-t+\tau)d \tau$$Then, since ##u(\tau)## is zero for ##\tau <0## and 1 for ##\tau > 0## you can write it as$$
    2e^{t}\int_{0}^{\infty}e^{-\tau}\cdot 1 \cdot e^{-\tau}u(-t+\tau)d \tau$$Next you know ##u(\tau - t)## is only going to be 1 if ##\tau > t##, so that needs to be taken into account in the limits. When you work on it you will see it matters whether ##t > 0## or not. See if you can take it from there.
     
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