Second order pole positions and rise time

  • #1
Dear All

I'm stuck on something that seems to be contradictory. I was under the impression that the further the 2 closed loop poles are in a 2nd order system to the left of the root locus, the higher the damping. Surely high damping means longer rise time? But other sources say that the further to the left the poles are, the shorter the predominant time constant and the shorter the rise time?

Please can someone explain which is right as they seem to contradict?

Many thanks

Regards

Paul Harris
 

Answers and Replies

  • #2
272
0
Dampening is only a function of the cosine of the angle between real axis and the pole. Thus, [tex]\zeta=0 [/tex] at the imaginary axis and [tex]\zeta=1 [/tex] at the real axis. What you read on the websites is correct because the further away you are from the imaginary axis, the larger negative your eigenvalue, e.g. compare [tex] e^{-t}[/tex] to [tex] e^{-100t}[/tex] which one dies out faster?

Also high dampening is proportional to high rise time but not the only factor. In a second order system rise time can be approximated as [tex]t_r = \frac{1+1.1\zeta + 1.4 \zeta^2}{\omega_n} [/tex].
 
Last edited:
  • #3
272
0
Also I forgot to mention. Lines of constant dampening lie along a given angle, not along the real axis. So the further out you place a pole, the smaller the angle becomes, the less dampening and the shorter the rise time. So either way you look at it, rise time has to decrease the farther it is from the imaginary axis.
 
  • #4
Thanks for the reply viscousflow :) I see what your saying for the most part. But surely for a fixed imaginary part (+/- jwd) moving the poles left would decrease the angle and cause dampening to INCREASE since it's cosine ?

Thanks

Paul
 
  • #5
272
0
Yes, sorry. It was pretty late when I wrote that, but dampening does increase!
 

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