Lecture 10 – Root Locus

All possible root locations of a system

→ For higher order systems the ``degree'' of polynomial in the characteristic equation is high!

Finding root becomes difficult

By changing `k' roots of the characteristic equation change from `poles' to `zero' of the `open loop' T.F.

Vector representation in complex plane

origin of complex plane to s' .

F(s)=s+b for any complex s , we'll end up to as a complex number

at any s , the function value is a complex number (in general) which has a ``magnitude'' and a ``phase''

What is the function value at s=-3+j4

4 We need some criteria to check if the trial points ( s ) on the complex plane can be accepted as root of the characteristic equation

Any point like ` s ' which satisfies both conditions would be our point of interest (root of characteristic eq)

The locus of points satisfying the phase condition constitutes the root locus plot of the system.

Steps of Drawing

1- Find the OL zeros and poles and show on the s plane

2- Check the validity of point on `Re' axis

So if the number of zeros/ poles on the right hand side of the zone is `odd', then it passes the phase condition

Complex poles/zeroes not effect the acceptable zone at the real axis

if the number of poles and zeros are not the same

Then there will be some asymptotes for the root locus

poles : n ⇒ n-m : number of asymptotes

Number of Branches is equal to ` n '

Centroide of asymptotes

Determine the breaking points

Branches leave (break out) or enter (break in) the real axis at 90^°

The characteristic eq

repeated root occurs at point of break in/out ⇒ dkds=0

All break points are solution of this polynomial. But not all solutions are break in/out points

Only root of the given expression are acceptable that give a positive gain!

The other approach is that the break in/out point must be on the valid zones of the real axis.