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Stability of a system with poles and zeros at infinity

  1. Jan 7, 2009 #1
    I have a system transfer function

    H(s) = 1/(e^s + 10)

    This system has both poles and zeros at infinity and -infinity.

    Can anybody tell me if this is a stable system. Thanks.
     
  2. jcsd
  3. Jan 7, 2009 #2

    berkeman

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    Staff: Mentor

    You must show your own work before we can help you. You know that.
     
  4. Jan 7, 2009 #3

    mheslep

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    Gold Member

    As shown H(s) has no zeros, and it has a pole at e^s=-10, not at +/-infinity
     
  5. Jan 7, 2009 #4
    e^s = -10 is the pole. You cannot solve this equation. ln(-10) does not exist. That is why i concluded the pole is at infinity. Is my conclusion wrong?
     
  6. Jan 7, 2009 #5
    Yes you can solve it. The solutions are complex. One solution is s = i*pi+Ln(10)
     
  7. Jan 7, 2009 #6
    So the pole is at 2.3 (ln 10) +j 3.14. SO this is an unstable system. Am I correct? Also fourier transform does not exist for this am I right? Since it not abolutely integrable within - infinity and infinity.
     
  8. Jan 7, 2009 #7

    mheslep

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    http://en.wikipedia.org/wiki/Control_theory
     
  9. Jan 8, 2009 #8

    CEL

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    As The Electrician said, this is ONE of the solutions. There is an infinity of them, in the form ln(10) + j(2k+1)pi.
     
  10. Jan 9, 2009 #9
    So stability depends om the value of k? Can somebody point me to a good link which teaches how to solve equations like e^x = -a;
     
  11. Jan 10, 2009 #10

    CEL

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    k is an integer variable that can take any value between minus infinity and plus infinity. Since for an infinity of values of k the poles lie on the RHP, the system is unstable.
    For an explanation on solving your equation see
    http://mathforum.org/library/drmath/view/61830.html
     
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