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Finding Pole-zero pattern of transfer fcn and Stability of LTI system

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data

    The transfer function of an LTI system H(s) = (s^2 + 2)/(s^3+2s^2+2s+1)
    Find the followings

    i) pole-zero pattern of H(s)
    ii) Stability of the system
    iii) Impulse response h(t)

    2. Relevant equations

    Zero for which H(s) = 0 & Pole is for which H(s) = ∞

    3. The attempt at a solution

    Finding Zeros:
    Here H(S) =0, if s^2 + 2 =0, so how to find out the solution for s from the equation, I tried for different combination for imaginary values of i.

    Finding Poles:
    Here H(S) =∞, if s^3+2s^2+2s+1=0

    so, s^3+2s^2+2s+1 =(s+1) (s^2+s+1) . Can anyone find me the solution here to find out the poles?

    Please help me to find out the zeros & poles so that I can find out the stability of the system.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 18, 2014 #2

    donpacino

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    Gold Member

    there are 2 zeros,
    hint (-2)*(-2)=(2)*(2)

    also the zeros will be purely imaginary.


    there are 3 poles.
    one of them is -1

    you should be able to find the other 2
     
  4. Mar 19, 2014 #3
    Well, I found the zeros = √2*i & -√2*i
    poles = -1, (-1+√3*i)/2 & -(1+√3*i)/2

    Now, I am confused with the stability here? What type of stability is this LTI system? Anyone?
     
  5. Mar 19, 2014 #4

    donpacino

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    Gold Member

    take the inverse laplace transform of the transfer function, and evaluate it as t approaches inf. If the expression approaches infinity, then the system is unstable. If the system is purely sinusoidal then it is marginally stable. If not the system is stable.

    Now the being said there is a shortcut. The poles of a system determine stability. If any pole is in the right half plane, the system is unstable. If any pole is on the y axis the system is marginally stable. Else, the system is stable
     
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