Quadratic Zero/Pole Question

[jp1390]

Hi all, this is a quick question regarding bode plots.

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

[latex]H(s) = \frac{1}{s^2 + 5s + 6}[/latex]

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

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

[latex]H(s) = \frac{1}{(s + 2)(s + 3)}[/latex]

[latex]H(s) = \frac{1}{6(\frac{s}{2} + 1)(\frac{s}{3} + 1)}[/latex]

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 [latex]s = -sqrt{6}[/latex].

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

Thanks

 

[Ratchit]

jp1390,

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

.

You can always find the roots by the quadratic formula, even if they are not integers. Factoring out 6 puts the equation into a nonstandard form, but it does not change the roots. I don't under your statement which I quoted above.

Ratch

 

[jp1390]

I figured it out actually. To be able to draw the Bode plot you must first try to factorize the quadratic pole/zero into simple real poles/zeros. If they are complex, then you have a quadratic root and it must be left in the unfactorized form.