breakaway point of root locus

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

  1. goldfinger

    Thread Starter New Member

    Jun 26, 2010
    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.