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Simple Zero State Response Signals Question

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Let's say I've got simple y''(t) + 2y' + 1 = x(t), with maybe y(0) = 3 and y'(0)=1. the input x(t) is 12cos(20t). Using a laplace transform, determine zero state response of the output y(t).

    I know eventually I end up with the Y(s) and X(s) form, and then I have to use the partial fractions crap using A, B, C etc to get y(t). Butt my real question is, since the input is sinusoidal, am I going to get an angular response for this too instead of just the magnitude?

    2. Relevant equations

    I know the transformation (using a Laplace Table) for this x(t) to X(s) is 12s/(s^2 + 400), but do I have to use the transfer function to find an angular response too?

    3. The attempt at a solution
    N/A
     
  2. jcsd
  3. Dec 1, 2012 #2

    gneill

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    Staff: Mentor

    If your result ends up with both sine and cosine terms, you might combine them into a single cosine term which will have a phase constant. Is that what you mean?
     
  4. Dec 1, 2012 #3

    rude man

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    Laplace-transform each term in your equation!

    P.S. I don't know what "zero state response" means.
     
  5. Dec 1, 2012 #4

    gneill

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    A zero state response is the response of the D.E. when the initial conditions y(0), y'(0), etc., are ignored (set to zero) but the driving function is still in place.

    A zero input response leaves the initial states alone, but sets the driving function to zero.
     
  6. Dec 2, 2012 #5

    rude man

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    Ya live and ya learn! :smile:
     
  7. Dec 3, 2012 #6
    Your initial conditions are not zero so this is not a zero state response.

    Yes you are going to get the full response. If you go the long root and do an inverse laplace on all the terms (including any you may have introduced due to non-zero initial conditions as above), you will find transient terms that will die off (assuming the system is stable) and that are the characteristic modes of the system and then you will find a term that is a steady state sinusoid for this case with a sinusoid input. This term will invert to a sinusoid of the same frequency as the input but modified by phase and magnitude, just like LTI systems are supposed to respond to a sinusoid.

    The quicker way to find this steady state response is to find the transfer function, set s=jw and compute the magnitude and phase response of the system. If you watch closely, this is exactly what you do going the long root to find the coefficient of the laplace output term containing the steady sinusoid. Then your input sinusoid will have its amplitude modified by this magnitude and its phase modified by the phase of the transfer function. But again, this is the steady state term only, which is what you are usually interested in.
     
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