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The transfer function of a delayed system x(t-T)

  1. Nov 12, 2008 #1
    Hi, im not looking for the answer, i just want to know how i go about solving it

    Ive got to find the transfer function of this equation:
    [​IMG]

    I know that the transfer function G(s) is found by:

    y(s)=G(s)x(s)

    As hard as i try i cannot seem to get x(s) through my workings.

    From Laplace i have found the LHS = y(s^2+6s+45) and the RHS to be: e^-sT (not sure about the RHS)

    so rearranging i can get y(s) = 1/(s^2 +6s +45) * e^-sT

    but i dont know quite what do do here. Ive tried inverse laplace to turn it into a heaviside function, but it didn't seem to help. Any help would be much appreciated, thanks.
     
  2. jcsd
  3. Nov 12, 2008 #2
    That should be LHS [itex]= Y(s)(s^2+6s+45)[/itex] and RHS [itex]= X(s)e^{-sT} [/itex].
     
  4. Nov 12, 2008 #3
    Ok thx, so it would be [itex] Y(s) = e^{-sT}/(s^2 + 6s +45)X(s) [/itex]
    In the form Y(s) = G(s)X(s)

    So would the transfer function be:
    [itex] G(s) = e^{-sT}/(s^2 + 6s +45) [/itex] ?
     
  5. Nov 13, 2008 #4
    can anyone tell me if this is the right answer? thank you

    [itex]G(s) = e^{-sT}/(s^2 + 6s +45)[/itex]
     
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