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Transfer functions - Complex poles

  1. Dec 7, 2011 #1
    Hi,

    i'm having a bit of trouble understanding how a system modelled by a transfer function in the laplace domain with no zero's and a complex conjugate pair for the poles causes an oscillation when the unit step function is applied. Any help would be appreciated

    Many thanks.
     
  2. jcsd
  3. Dec 7, 2011 #2
    The complex pair does not "cause" the oscillation.
    They indicate there will be one.

    Think of the complex plane this way.
    A pole located along the x axis is showing the speed of an exponential response to a disturbance input. The farther to the left, the larger the negative number, the faster the exponential decays. If you were in the right half of the plane, for positive values of x, the exponential would grow, not decay and you have instability. This is why the left hand plane is the stable region.

    Now poles off the x axis appear in pairs. If the pair is right on the y axis, that corresponds to a pure oscillation which never stops. No dissipation to drain the energy. It is a pure oscillator. The higher up the axis, the faster the oscillation.

    If the poles move off the axis in to the left half plane, say at (-1, 1) and (-1,-1), you now have a combination [by multiplication] of the oscillation and the exponential decay. It is a damped harmonic oscillator.

    So the following points correspond to exponentially decaying oscillations which are ....
    (-0.1, 0.1) and (-0.1,-0.1) very low freq, very slowly decaying
    (-0.1, 1000) and (-0.1,-1000) very high freq, very slowly decaying
    (-1000, 0.1) and (-1000,-0.1) very low freq, very fast decaying
    (-1000, 1000) and (-1000,-1000) very high freq, very fast decaying

    The X axis value indicates the exponential rate [negative is decay, positive is growth => instability]
    The Y axis value indicates the oscillatory rate

    Hope that helps
     
    Last edited: Dec 8, 2011
  4. Dec 8, 2011 #3
    Yeh that is helpful thanks, i was wondering if anyone could explain it in terms of Laplace domain equations and time responses etc as that is what im struggling with.

    many thanks
     
  5. Dec 8, 2011 #4
    Laplace domain equations ??
    Do you mean the Laplace Transform Integral ?
     
  6. Dec 9, 2011 #5
    Sorry i didnt explain myself very well, say you have a system whos transfer function is;

    H(s)= [itex]\frac{5}{s^2 + 4S + 5}[/itex]

    When the input is a step response, the output will be a decaying sinusoid before it reaches a steady state, I would like some help in understanding how you get to a time domain response h(t) which has a sinusoid in it.
     
  7. Dec 9, 2011 #6
    You mean the input is a "step function"
    The output resulting from the input step function is the "step response"

    To get the output Laplace transform, you multiply the Laplace Transforms of the input step function and the system's impulse response. The output time domain function is then given by the Inverse Laplace Transform of this multiplication of transforms.

    Note the Laplace Transform of a system's time domain Impulse Response is the system Laplace Transform

    Anytime the poles are off the X axis of the Complex Plane, there will be oscillatory behavior. In fact for the most part, only single pole systems have no oscillatory behavior. This is because all you need is 2 energy storage components in a system to get oscillation. Any two pieces of metal [in an electronic system] constitute a capacitor and thus a storage device.

    The optimum response is when the dominant poles [those closest to the origin] are at 45 degrees to the X [real] axis. This gives the fastest response to a step function input while also resulting in minimum overshoot.
     
  8. Dec 10, 2011 #7
    Sorry for my mix up with terminology,

    yeh i understand the process of doing it but i dont know how i would do it in this scenario given that the poles are complex and im looking for a sine or cos in the time domain response. Many thanks.
     
  9. Dec 10, 2011 #8
    OK
    You say you know the response will be a decaying [exponentially] sinusoid.
    That's right for most systems with more than one pole, which is practically all systems.
    So how does a sinusoid appear in the result ?

    Both Laplace and Inverse Laplace Transform formulas are Integrations.
    And both involve "e" to some power of either "s" [the complex frequency] or "t" [time]
    "e" you should know can be written as the sum of sinusoids.
    And from Calculus you recall that integrals of trig functions are usually also trig functions.
    So the Inverse Transform integral will in general be sinusoid when going to the time
    domain.
     
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