Root Locus drawing

Thread Starter

Halim.Akiki

Joined Dec 29, 2014
44
Hello. I'm having trouble finding the angle of departure of a function: G(s) = K/[s(s^2 + 4s +5)].
This function has:
no zeros n = 0
poles m = 3 s = 0; s = -2+j; s = -2-j
the root loci i on the negative x axis
the angles off departure are 60 and -60
the point of intersection is -4/3
a breakaway point at 0
k = -(s^3 + 4s^2 + 5s)
dk/ds = 0 yields s = 2 and s = 1.85
 

shteii01

Joined Feb 19, 2010
4,644
Hello. I'm having trouble finding the angle of departure of a function: G(s) = K/[s(s^2 + 4s +5)].
This function has:
no zeros n = 0
poles m = 3 s = 0; s = -2+j; s = -2-j
the root loci i on the negative x axis
the angles off departure are 60 and -60
the point of intersection is -4/3
a breakaway point at 0
k = -(s^3 + 4s^2 + 5s)
dk/ds = 0 yields s = 2 and s = 1.85
I'm having trouble finding the angle of departure

the angles off departure are 60 and -60
 

Thread Starter

Halim.Akiki

Joined Dec 29, 2014
44
I'm having trouble finding the angle of departure

the angles off departure are 60 and -60
Hello. I'm having trouble finding the angle of departure of a function: G(s) = K/[s(s^2 + 4s +5)].
This function has:
no zeros n = 0
poles m = 3 s = 0; s = -2+j; s = -2-j
the root loci i on the negative x axis
the angles off departure are 60 and -60
the point of intersection is -4/3
a breakaway point at 0
k = -(s^3 + 4s^2 + 5s)
dk/ds = 0 yields s = 2 and s = 1.85
My bad! 60 are the angles of asymptotes not the angles of departure.
 

t_n_k

Joined Mar 6, 2009
5,455
There being no zeros, the angle of locus departure at any pole Pj would be

\(180^o-\sum (\text {angles \ subtented \ by \ all \ other \ poles \ to \ P_j}) \)

So for the pole at the origin this would be

180-(-26.565+26.565)=180 deg
 

Thread Starter

Halim.Akiki

Joined Dec 29, 2014
44
There being no zeros, the angle of locus departure at any pole Pj would be

\(180^o-\sum (\text {angles \ subtented \ by \ all \ other \ poles \ to \ P_j}) \)

So for the pole at the origin this would be

180-(-26.565+26.565)=180 deg
Thank you!
 
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