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Signals and Systems: Response of LTI systems to Complex Exponentials

  1. Jul 9, 2007 #1
    How will we evaluate the integral for calculation of Amplitude Factor for an LTI system for which the input and output are related by a time shift of 3, i.e.,

    y(t) = x(t - 3)


    The answer is: H(s) = e^(-3s)

    I want to understand the Mathematics behind the evaluation of the integral.

    Thanks.
     
  2. jcsd
  3. Jul 9, 2007 #2

    rbj

    User Avatar

    i'm not sure how the title of your post is related to the content.

    by definition, on the left side you have,

    [tex] Y(s) = \mathcal{L} \left\{y(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} y(t) \,dt [/tex]

    and, also by definition,

    [tex] X(s) = \mathcal{L} \left\{x(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt [/tex]

    and, it turns out that for LTI systems, after you show that the convolution operator and some impulse response is what relates the input x(t) to the output y(t), then

    [tex] Y(s) = H(s) X(s) [/tex]

    where

    [tex] H(s) = \mathcal{L} \left\{h(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} h(t) \,dt [/tex]

    and h(t) is the impulse response of the LTI system and completely characterizes the input/output relationship of the LTI system.

    now, in your specific case,

    [tex] Y(s) = \mathcal{L} \left\{y(t)\right\} = \mathcal{L} \left\{x(t-3)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t-3) \,dt [/tex]

    which, after you do a trivial substitution of variable of integration is

    [tex] Y(s) = \int_{-\infty}^{+\infty} e^{-s(t+3)} x(t) \,dt = e^{-s \cdot 3} \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt = e^{-s \cdot 3} \cdot X(s) [/tex]

    which means that

    [tex] H(s) = e^{-s \cdot 3} [/tex].
     
    Last edited: Jul 9, 2007
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