Root Locus question

  • #1
If have a system with 4 open loop poles, all at say s=-2, how would the locus approach each of the four asymptotes?

thanks
 

Answers and Replies

  • #2
rbj
2,226
9
are you really the bill nye on the radio? just curious. after looking at your history, i somehow doubt it.

this thing sounds like what happens with a Moog 4-pole Low Pass Filter. once you surround it with a loop and gain, you just solve the new transfer function for the poles and you can see that they move outward from the original place like 4 corners of a square centered at the original location. are you gonna make me do the math to show you?
 
  • #3
unfortunately not, sorry!

anyway, do you mean that the open loop transfer function G(s)=1/(s+2)^4 would become something like Gc(s)=K/[(s+2)^4+K]?
I still can't really visualise what the locus would look like though. Does it just expand outwards from s=-2, following the four asymptotes?
 
  • #4
rbj
2,226
9
anyway, do you mean that the open loop transfer function G(s)=1/(s+2)^4 would become something like Gc(s)=K/[(s+2)^4+K]?
I still can't really visualise what the locus would look like though. Does it just expand outwards from s=-2, following the four asymptotes?
yes it does. but it's not really asymptotes. the loci of the roots really are the four corners of a square centered at s=-2. as K gets bigger, the square gets bigger (not proportionately). (i am presuming that K is the signed, linear gain in series with G(s) and that the feedback is negative feedback with unity gain.)

[tex] G_c(s) = \frac{K G(s)}{1 \ + \ K G(s)} = \frac{K}{1/G(s) \ + \ K} [/tex]

now, to find the poles of Gc(s), you set the denominator to zero and solve for s. if you are not already familiar with the nth roots of unity (involving complex numbers), you should get that down first.
 
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