Consider the closed loop transfer function:

R(s) + -

E(s) K

1 s(s+a)

C(s)

How do the poles of the closed-loop system change as a function of the gain K? The closed-loop transfer function is:

The characteristic equation:

Closed-loop poles:

Root Locus

When the gain is 0, the closed loop poles are the openloop poles Roots are real and distinct and for a positive a, in the left half of the complex plane.

Two coincident poles (Critically damped response)

Roots are complex conjugate with a real part of –a/2.

Im

K=0

X

K=0

X

Re 0

Root Locus

The locus of the roots of the closed loop system as a function of a system parameter, as it is varied from 0 to infinity results in the moniker, Root-Locus. The root-locus permits determination of the closed-loop poles given the open-loop poles and zeros of the system. For the system:

R(s) + -

E(s) G(s)

C(s)

H(s)

the closed loop transfer function is:

The characteristic equation:

Root Locus

when solved results in the poles of the closed-loop system. The characteristic equation can be written as:

Since G(s)H(s) is a complex quantity, the equation can be rewritten as: Angle criterion: Magnitude criterion:

The values of s which satisfy the magnitude and angle criterion lie on the root locus. Solving the angle criterion alone results in the root-locus. The magnitude criterion locates the closed loop poles on the locus.

Root Locus

Often the open-loop transfer function G(s)H(s) involved a gain parameter K, resulting in the characteristic equation:

Then the root-loci are the loci of the closed loop poles as K is varied from 0 to infinity. To sketch the root-loci, we require the poles and zeros of the open-loop system. Now, the angle and magnitude criterion can be schematically represented as, for the system:

Root Locus

Im

s A2

X

Im

θ2

X

θ2 -p2 s A4 θ1 -p1

X

A1 A4 θ4 -p4

X

-p2 A2 A1 θ1

B1 φ1 -z1

A3 θ4 -z1

Re

-p4

X

φ1

B1

-p1 A3

X

Re

θ3

X

θ3

X

-p3

-p3

Root Locus (Example)

Consider the system:

R(s) + -

E(s)

C(s)

The open loop poles and zeros are:

Determine root locus on the real axis: Select a test point on the positive real axis,

which indicates that the angle criterion is not satisfied. Consider a test point between 0 and -1, then

The angle criterion: indicates that the angle criterion is satisfied.

Root Locus (Example)

Consider a test point between -1 and -2, then

The angle criterion: indicates that the angle criterion is not satisfied. Consider a test point between -2 and -infinity, then

The angle criterion: indicates that the angle criterion is satisfied.

Thus, a point on the real axis lies on a locus if number of open-loop poles

plus zeros on the real axis to the right of this point are odd.

Root Locus (Example)

Asymptotes of the root-loci as s tends to infinity can be determined as follows: Select a test point far from the origin:

and the angle criterion is:

or, the angle of the asymptotes are:

For this example, they are 60o, -60o, -180o. To draw the asymptotes, we must find the point where they intersect the real-axis. Since, the transfer function for a test point s far from the origin permits us to approximate the transfer function as:

Root Locus (Example)

The angle criterion for the approximated system is:

Substituting

, we have

Taking the tangent of both sides, we have:

which can be written as:

Root Locus (Example)

The three equations result is three straight lines shown below, which are the asymptotes, which meet at the point s=-1. Thus, the abscissa of the intersection of the asymptotes and the real-axis is determined by setting the denominator of the approximated transfer function to zero and solving for s.

Im

-1

Re

Root Locus (Example)

The root loci starting from 0 and -1, break away from the real axis as K is increased. The breakaway...