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
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