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Laplace Transform of Noncausal systems

  1. Jul 27, 2010 #1
    This is some homework from my Analog Signal Processing class, it's a cool class and I've not had much trouble, but this is giving me a bit of a problem and I was hoping someone could help me see what I'm doing... Here's the problem:

    Sketch, Get the Laplace Transform of, and Sketch the complex plane w/ region of convergence of the following functions:

    f(t) = u(t) - u(t+8)

    g(t) = rect((t+8)/3)

    Of course there is the Laplace transform integral of integral(e^(-st)f(t) dt,t,0,infinity), which is what I've used. Also, g(t) also is equal to u(t+19/2)-u(t+13/2), which makes it a bit easier to connect to known transforms.

    Doing the integration, I get 0 for both signals, which is because they are not causal (they have a value before t=0).
    Looking through the text, I see that a non-causal signal has the same transform as a(t)*u(t) <-multiplication, not convolution. Applying this, I would still get zero. However, someone was asking about this after class and the professor implied that this is not correct. Now, I'm not 100% sure if he was talking about the answer of zero, or if he was arguing about the Region of Convergence of zero. Am I right that the transform comes to zero, or am I missing something? If I am correct, I was thinking that with a transform of zero, the ROC would be all values, but I'm not sure if that is true, or if it is no value. I was thinking all values, because it wouldn't matter what you're putting in, nothing is getting through, so all values are in play. Am I lost, or am I just not seeing something? Thanks for the help.
  2. jcsd
  3. Jul 27, 2010 #2
    you're doing a unilaterial laplace on anticauasal signals which is sort of useless. are you supposed to do this (maybe to "show what happens"?) Or are you supposed to do a bilateral laplace on these signals? Yes, the answers will be zero with the unilateral laplace, because the signals are not even in the limits of integration. You're integrating 0*e^whatever from 0 to infinity, which is zero.
  4. Jul 27, 2010 #3
    We're just kind of getting used to doing transforms. There are others that we are doing that aren't difficult because they are easily integrable or they are in our transform list.
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