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Homework Help: Control system stability

  1. Apr 12, 2012 #1
    Let's say I have the following transfer function:

    [tex]G(s)=\frac { s-1 }{ { s }^{ 4 }+2{ s }^{ 3 }+{ 3 }s^{ 2 }+{ 4s }+5 }[/tex]

    Which is run through MATLAB to obtain the pole-zero format:

    [tex]G(s)=\frac { s-1 }{ ({ s }^{ 2 }+2.576s+2.394)({ s }^{ 2 }-0.5756s+2.088) } [/tex]

    Using a quadratic solver such as this one, both poles are found to be complex.

    I still can't tell the difference between a pole and a zero in terms of system stability. From my understanding of poles and zeroes, roots that are located on the left hand side make the system stable while those on the right hand side make it unstable. Therefore, am I correct to assume that since the zero is 1, the system is unstable? And since the poles are complex, the system oscillates forever without reaching a steady state?
  2. jcsd
  3. Apr 12, 2012 #2
    Stability is determined by poles, not zeros. Poles on the complex axis make the system marginally stable, which means the system is unstable since a bounded input at the frequency on the axis will result in an unbounded output.
  4. Apr 12, 2012 #3

    D H

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    Staff Emeritus
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    Your transfer function has two poles that lie to the right of the imaginary axis, so it's unstable per the Nyquist criterion.
  5. Apr 12, 2012 #4
    I see, so the system is unstable because the poles are to the right of the y axis, making the response an increasing sinusoidal? Won't it be cancelled to a degree by the other two poles which work to reduce the response?

    If the zero has no effect on stability, what does it affect then?
    Last edited: Apr 12, 2012
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