# breakaway point of root locus

Discussion in 'Homework Help' started by goldfinger, Jul 2, 2010.

1. ### goldfinger Thread Starter New Member

Jun 26, 2010
3
0
Hi!!
I am confused regarding the calculation of breakaway point of root locus.

I know that:

1.breakaway point lies on root locus.
2.It is determined from the roots of dk/ds= 0.
3.substituting breakaway point s=a in the original characteristic eqn should yield +ve value of K.

In the eg. below

G(s)H(s)= K(s+1)/ s2(s+3.6) Poles are at s=0,0,3.6
Zeros are at s= -1

The segment b/w s= -1 & s= -3.6 is the part of the root locus.
Now, the characteristic eqn is : s2(s+3.6)+ K(s+1)=0.

so, dk/ds=0 implies s=0, -1.65±j0.936

It is mentioned that s=0 is the breakaway point.

But, s=0 doesnt lies on root locus. So how come s=0 is the breakaway point??

In another eg:

G(s)H(s)= K/s( s2+6s +12)
Poles at s=0, -3± j1.73

The entire -ve real axis is the part of root locus.

Now,K= -( S3+6s2 +12) so dk/ds =0 implies s= -2

s= -2 lies on root locus but sub s= -2 inabove eqn yields -ve value of K. So why s= -2 is the breakaway point.