Root Locus: 4 Open Loop Poles at s=-2

[bill nye scienceguy!]

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

 

[rbj]

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 going to make me do the math to show you?

 

[bill nye scienceguy!]

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?

 

[rbj]

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.)

$$G_c(s) = \frac{K G(s)}{1 \ + \ K G(s)} = \frac{K}{1/G(s) \ + \ K}$$

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