Root Locus question

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?

 

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 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.