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Systems Modeling - Sinusoidal Inputs

  1. Aug 23, 2015 #1
    1. The problem statement, all variables and given/known data
    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)?

    2. Relevant equations


    3. The attempt at a solution
    I have no attempt to the solution as I just know that the transfer function, G(s), is y(s)/x(s).
    I don't see how to get yss(t) from G(s).
     
  2. jcsd
  3. Aug 23, 2015 #2
    Consider a sinusoidal input [itex]x(t) = x_m \cos(\omega t) u(t)[/itex] that produces a steady state output [itex]y_{ss}(t)[/itex] and a network that has the transfer function [itex]H(s)[/itex]. Then the Laplace transform of the output is: [tex]Y(s) = H(s) L \left ( x_m \cos(\omega t) u(t) \right ) = H(s) L \left ( \left ( \frac{x_m}{2} e^{j \omega t} + \frac{x_m}{2} e^{-j \omega t}\right )u(t) \right )[/tex]
    So
    [tex]Y(s) = \frac{x_m}{2} H(s) \frac{1}{s - j \omega} + \frac{x_m}{2} H(s) \frac{1}{s + j \omega}.[/tex]

    I was taught this in the context of electrical systems. My professor then went on to say that the poles of [itex]H(s)[/itex] influence the transient solution, while the poles of [itex]\frac{1}{s \pm j \omega}[/itex] make up the steady state solution. This feels kind of hand-wavy, but it makes sense when you consider that the transient solution is influenced by the particular system, but the steady state solution (long term) is obviously influenced by the driving force (for instance, consider pushing a spring with a sinusoidally varying force). In this case, if the driving force is a sinusoid, then the laplace transform of the driving force will look like the terms above. So we have [tex]Y(s) = \frac{k_1}{s - j \omega} + \frac{k_2}{s + j \omega} + (terms \ involving \ poles \ of \ H(s))[/tex] which implies that [itex]k_1 = Y(s)(s - j \omega)[/itex] evaluated at [itex]s = j \omega[/itex], and by extension, [itex]k_2 = k_1^*[/itex], or the complex conjugate of [itex]k_1[/itex]. So that means [tex]k_1 = \frac{x_m}{2} H(j \omega), k_2 = k_{1}^*.[/tex]

    Finally, you have to remember this little shortcut when taking inverse Laplace transforms: if [itex]Y(s) = \frac{k_1}{s + a - i b} + \frac{k_2}{s + a + i b}[/itex], then [itex]y(t) = 2 |k_1| e^{-a t} \cos(b t + \theta)[/itex]. In this case, [itex]a = 0[/itex], and [itex]\theta[/itex] is the angle of the complex number [itex]k_1[/itex].

    Putting those things together, when we take the inverse Laplace transform of the equation above, we get [tex]y(t) = x_m |H(j \omega)| \cos(\omega t + \theta) + (transient \ terms)[/tex]
    Or, ignoring the transient terms, [tex]y_{ss}(t) = x_m |H(j \omega)| \cos(\omega t + \theta)[/tex] where [itex]\theta[/itex] is the angle of [itex]H(j \omega)[/itex]
     
  4. Aug 24, 2015 #3
    Thanks @axmls , still seems like a pretty long and laborious task. But basically, the magnitude of the transfer function enables us to get the amplitude of yss(t) and the angle of the transfer function is the phase difference between the input and the output?
     
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