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