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Transfer function=Laplace of the impulse response?

  1. Nov 22, 2014 #1
    I saw in some books, that:
    Y(s)/X(s) = H(s)
    where,
    Y(s) is the laplace of the output
    X(s) is the laplace of the input
    H(s) is the laplace of impulse response.

    How to prove it? In the book Benjamin Kuo, he only mentions it without proof, and did not find it in the book of Oppennheim.
     
  2. jcsd
  3. Nov 22, 2014 #2
    For continuous-time systems, you can represent the impulse input using the Dirac delta function:
    $$
    x(t) = \delta(t)
    $$
    Since:
    $$
    X(s) = \mathcal{L}\{x(t)\}(s) = \mathcal{L}\{\delta(t)\}(s) = 1
    $$
    It follows that the Laplace transform of its impulse response ##Y(s)## gives its transfer function:
    $$
    Y(s) = H(s)X(s) = H(s)
    $$
     
  4. Nov 22, 2014 #3
    Thanks, helped me.
     
  5. Nov 22, 2014 #4
    Another way:
    $$ y(t) = x(t) \ast h(t) $$
    $$ \mathcal{L}\{y(t)\} = \mathcal{L}\{x(t) \ast h(t)\} $$
    $$ \mathcal{L}\{y(t)\} = \mathcal{L}\{x(t)\}\mathcal{L}\{h(t)\} $$
    $$ Y(s) = H(s)X(s) $$
     
    Last edited: Nov 22, 2014
  6. Nov 23, 2014 #5
    Not sure I understand. Are you looking for a proof of the convolution theorem?
     
  7. Nov 23, 2014 #6
    First I tried to find a way to understand why H (s) is the Laplace of impulse response.
    The first thing I thought was that I should start by the convolution of x and h.
    Then you answered me in a way, and a few hours later I saw this on a website:
    $$ \mathcal {L} \{x (t) \ast h (t) \} = \mathcal {L} \{x (t) \} \mathcal {L} \{h (t) \} $$
     
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