Quadratic Zero/Pole Question

Discussion in 'Homework Help' started by jp1390, Aug 22, 2011.

  1. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Hi all, this is a quick question regarding bode plots.

    Say you have a quadratic term in the numerator or denominator like the following:

    H(s) = \frac{1}{s^2 + 5s + 6}

    H(s) = \frac{1}{6((\frac{s}{sqrt{6}})^2 + \frac{5}{6}s + 1)}

    Now looking at that most of us would probably say, oh factor the bottom quadratic into two simple poles:

    H(s) = \frac{1}{(s + 2)(s + 3)}

    H(s) = \frac{1}{6(\frac{s}{2} + 1)(\frac{s}{3} + 1)}

    but say, the quadratic pole/zero was not easily seen as factorable and was left in it's standard form.

    Instead of having two poles at s = -2 and s = -3, there would be a single quadratic pole at s = -sqrt{6}.

    This is a completely different bode plot compared to having two simple poles. How are they related?

    Thanks
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    If the second order polynomial is not factorable by inspection then I would resort to the quadratic equation to obtain the factors.

    hgmjr
     
  3. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Thanks for the reply! Okay, maybe I have figured this out quicker than I thought. If you have complex poles, you leave it in the quadratic form and have a single quadratic poles, otherwise you will have two distinct simple poles?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    That is basically true. Always remember that the quadratic equation will allow you to find the poles of a second order polynomial whether they are complex or simple.

    hgmjr
     
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