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Pole zero plot

  1. Jan 7, 2013 #1
    I am attempting the question shown in the attachment.

    It can be seen that the poles are located at -2 ± 3j which expressed in terms of s is (s + 2)2 + 32.

    This is the denominator, but how is the numerator of the transfer function found?


    Looking at the Laplace look-up table I would want the numerator to be 3 giving:

    3/((s + 2)2 + 32) so that i could use e-atcosωt to perform the reverse Laplace transform in part b.

    Is this correct?

    Attached Files:

    Last edited: Jan 7, 2013
  2. jcsd
  3. Jan 8, 2013 #2

    rude man

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    There are no zeros in your plot, ergo there is no numerator other than a constant. The constant cannot be determined from the plot (unless it's contained in those funny numbers within the white part of the plot. I have never seen a plot like that before.) You can assume it's 3 but any other real constant is OK also. That should be obvious since L-1{cF(s)} → cf(t), c a constant.

    My table says L-1{1/[(s+a)2 + b2]} → (1/b)e-atsin(bt).
  4. Jan 8, 2013 #3
    Yes you are correct, I have misread the table and copied the entry from the line above.
  5. Jan 8, 2013 #4

    rude man

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    BTW you could also have done the inversion by partial fraction expansion, if you're comfy with manipulating complex numbers just a reminder probably ...
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