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Systems Modelling Question - Sinusoidal inputs (Important)

  1. Aug 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi,
    When finding the steady-state response to a sinusoidal input, since "s" is a complex number, (a + jw), why do we substitute "s" with only the imaginary part (jw) in the transfer function, G(s) , to get G(jw), rather than substituting the whole complex number to get G(a + jw) ?
    Also, how does finding G(s) help us to get the steady-state part of the response anyway?
    Quite confused, please please please help!

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Aug 23, 2015 #2

    Hesch

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    Gold Member

    If you know the Laplace transform of a model, and you want to find its response to some sine-function, you should give the model an input, something like:

    Inp(s) = 1 / ( s2 + a2 ) , which is the Laplace transform of

    Inp(t) = sin(at) / a.

    I think that jω as input is used to determine amplification and phase ( Bode plot ) as a function of ω.
     
    Last edited: Aug 23, 2015
  4. Aug 23, 2015 #3
    @Hesch , okay, so after finding G(jω), knowing the amplitude and phase will help us get the steady-state part of the response?
    Also, I still don't get why, in G(s), we substituted "s" with just "jω" only while "s" actually equals "a+jω" and not just "jω".
     
  5. Aug 23, 2015 #4
    Finding the complete response (steady-state and transient) is a long and laborious task. My lecturer's notes read (since at our stage of the course, we're mostly interested in the steady-state part of the solution and not so much the transient) :

    "The simple method for finding the steady-state part of the response to a sinusoidal input is simply to use the imaginary part of "s", substituting "jω" in place of "s" in the transfer function".

    Then he goes on to show how to get the magnitude of the transfer function, G(s) / G(jω), and then on to get the phase.

    Please explain what the magnitude of the transfer function has to do with the steady-state part of the solution, yss(t)?
     
  6. Aug 23, 2015 #5

    Hesch

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    Gold Member

    I don't understand what you mean by "steady state". Is it the same as "final value"? then read this:

    https://en.wikipedia.org/wiki/Final_value_theorem

    There is a remark as to sine-functions too.
     
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