Quadratic Zero/Pole Question

Thread Starter

jp1390

Joined Aug 22, 2011
45
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
 

hgmjr

Joined Jan 28, 2005
9,029
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
If the second order polynomial is not factorable by inspection then I would resort to the quadratic equation to obtain the factors.

hgmjr
 

Thread Starter

jp1390

Joined Aug 22, 2011
45
If the second order polynomial is not factorable by inspection then I would resort to the quadratic equation to obtain the factors.

hgmjr
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?
 

hgmjr

Joined Jan 28, 2005
9,029
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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