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Convolution Problem

  1. Jul 29, 2013 #1
    1. The problem statement, all variables and given/known data

    The unit impulse response of an LTIC system is h(t) = [itex]e^{-t}u(t)[/itex]. Find the system's (zero-state) response y(t) if the input f(t) is [itex]e^{-2t}u(t-3)[/itex].

    2. Relevant equations

    y(t) = f(t) * h(t) = [itex]∫^{∞}_{-∞}f(t)h(t-\tau)d\tau[/itex]
    [itex]f_{1}(t) * f_{2}(t ) = c(t)[/itex]
    [itex]f_{1}(t) * f_{2}(t - T) = c(t - T)[/itex]

    3. The attempt at a solution

    I'm not sure how to apply the shifting property because here in f(t) I have the unit step function only which is shifted and not the exponential. Is it possible to apply the shifting property above for this problem? I don't see how I can apply it for the reason mentioned above.

    Thanks for any help.
     
    Last edited: Jul 29, 2013
  2. jcsd
  3. Jul 29, 2013 #2

    vela

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    Hint: Use ##t = (t-3)+3## in the exponential in f(t).
     
  4. Jul 30, 2013 #3
    Thanks. I was able to do it. I'm however now stuck on a some what more challenging problem.

    [itex]-\delta(t) * e^{t}u(-t) \stackrel{Δ}{=} ∫^{\tau = ∞}_{\tau = -∞}e^{t}u(-\tau)(-\delta(t-\tau))d\tau[/itex]

    I know that

    [itex]f(t) * \delta(t-T) = f(t - T)[/itex]

    But I'm unsure how to apply this here. I guess in my case [itex]T = 0[/itex] but how do I tell with the negative sign of the dirac delta function? Is my solution

    [itex]-e^{t}u(-t)[/itex]?

    Thanks for help.
     
  5. Jul 30, 2013 #4

    vela

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    You can pull the negative sign right out of the integral, right? So you should be able to see that
    $$[-\delta(t)]*e^t u(-t) = -[\delta(t)*e^t u(-t)]$$
     
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