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SIGNALS AND CONTROL SYSTEMS

Instructor : Dr. Raouf Fareh

Fall Semester 2014/2015

Week 11 Root Locus Technique

Introduction

In the preceding chapters we discussed how the performance of a feedback system can be described in terms of the location of the roots of the characteristic equation in the s-plane.

We know that the response of a closed-loop feedback system can be adjusted to achieve the desired performance by judicious selection of one or more system parameters. It is very useful to determine how the roots of the characteristic equation move around the s-plane as we change one parameter.

The locus of roots in the s-plane can be determined by a graphical method. A graph of the locus of roots as one system parameter vary is known as a root locus plot. The root locus is a powerful tool for designing and analyzing feedback control systems and is the main topic of this chapter.

We will show that it is possible to use root locus methods for design when two or three parameters varies.

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What is Root Locus ?

The characteristic equation of the closed-loop system is 1 + K G(s) = 0

The root locus is essentially the trajectories of roots of the characteristic equation as the parameter K is varied from 0 to infinity.

A camera control system:

How the dynamics of the camera changes as K is varied ?

A simple example: pole locations

(b) Root locus.(a) Pole plots from the table.

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The 7 Steps to the Root Locus

Step 1: Write the characteristic equation

1 01 0

Find the m zeros zi and n poles pj of P(s)

Locate the poles and zeros on the s-plane with selected symbols (o-zero,

X-pole)

The RL (root locus) starts at the n open-loop poles

The RL ends at the open loop zeros, m of which are finite, n-m of which

are at infinity

The 7 Steps to the Root LocusStep 1: Write the characteristic equation

The closed loop transfer function of the system is

The characteristic equation for this closed-loop system is obtained by setting the denominator polynomial to zero, i.e.

Example 1

Zeros= roots of numerator of P(s)

Poles= roots of denominator of P(s)1 0

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The 7 Steps to the Root LocusStep 1: Write the characteristic equation

Example 2

The characteristic equation for this system is

Notice that the adjustable gain K does not appear as a multiplying factor for KG(s)H(s), as in the example 1. By dividing both sides of the above Equation by the sum of the terms that do not contain K, we get

1

5 1

0

n 3

The 7 Steps to the Root LocusStep 2: Locate the segments of the RL on the real axis

They lie in sections of the real axis at the left of an odd number of poles and zeros Determine the number of separate branches (or loci). The number of branches is equal to the number of poles n The root locus is symmetrical with respect to the horizontal real axis (because roots are either real or complex conjugate) Root Locus starts from poles and ends to the zeros

Existence on the Real AxisThe root locus exists on the real axis to the left of an odd number of poles and zeros.

Example

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The 7 Steps to the Root LocusStep 3: Asymptotes

Branches of the root locus which diverge to (i.e. to open-loop zeros at ) are asymptotic to the lines with angle

Where is the number of open loop-poles, ! is the number of open- loop

zeros and

The asymptotes intersect the real axis at a point called the pivot or centroid given by

" $%&'

%() +,

&-,()

!

. 20 1 . 180

!, 0 = 0,1, , ( ! 1)

The 7 Steps to the Root LocusExample:Step1-3

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The 7 Steps to the Root LocusExample:Step1-3

The 7 Steps to the Root LocusStep 4: intersection with the imaginary axis

The actual point at which the root locus crosses the imaginary axis can be evaluated using the Routh-Hurwitz criterion

When the root locus crosses the imaginary axis, there is a zero in the first column of the Routh-Hurwitz table, and other elements of the row containing the zero are also zero.

A zero-entry appears in the first column, and all other entries in that row are also zero

Solution:

Return to the previous row and form the Auxiliary Polynomial, qa(s), The auxiliary polynomial is the polynomial immediately precedes the zero entry in Routh array.

The order of the auxiliary polynomial is always even and indicates the number of symmetrical roots pair.

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The 7 Steps to the Root LocusStep 4: intersection with the imaginary axis

Example

05 = 26 8

) 27

6 27

Intersection of RL with imaginary axis at +2j and -2j

The 7 Steps to the Root LocusStep 4: intersection with the imaginary axis

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The 7 Steps to the Root LocusStep 4: intersection with the imaginary axis

The 7 Steps to the Root LocusStep 5: Breakaway points

Breakaway points occur on the locus where two or more loci converge or diverge. They often occur on the real axis, but they may appear anywherein the s-plane.

As K increases, the poles starts moving towards each other until they meet at the breakawaypoint, from which the break away in opposite directions.

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The 7 Steps to the Root LocusStep 5: Breakaway points

. 8 $ 1

1

9:

b. Obtain ;

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The 7 Steps to the Root LocusStep 6: Angle of departure

Note that conjugate pole moves in a direction that preserves conjugate symmetry

The 7 Steps to the Root LocusStep 7: Sketch Root Locus

Join the segments that have been drawn

with a smooth curve

Curve should be as simple as possible

Curve must respect conjugate symmetry of poles and zeros of a system with real

inputs and real outputs

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Root Locus: Example 1

(45 and 315) (135 and 225)

? =20 1

!. 180@; 0 0,1, , ! 1

Root Locus: Example 1

We assume that a test point s1 is very close to the pole or zero on the RL so that the angles

from other poles and zeros to s1 are known. The only unknown angle is the angle of the tangent at s1.

At s1 the sum of angles from zeros minus the angles from poles must be equal to 180 deg.

Angles of departure

From z1=-3 to p1=-1+j

From p2=0 to p1=-1+j

From p3=-1-j to p1=-1+j

From p4=-5 to p1=-1+j

From p5=-6 to p1=-1+j

" $%&'

%() +,

&-,()

!

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Root Locus: Example 1

From z1=-3 to p1=-1+j

From p2=0 to p1=-1+j

From p3=-1-j to p1=-1+j

From p4=-5 to p1=-1+j

From p5=-6 to p1=-1+j

Root Locus: Example 1

For K=35 this row =0

Replace K=35 in this row to find A(s)

Intersection with jw-axis

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Root Locus: Example 1

. 8 $ 1

1

9:

b. Obtain ;

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Additional of poles/zeros to G(s)H(s)

Adding a pole to GH has the general effect of pushing the RL towards the RHP, thus making the system less stable (more oscillatory).

Adding a zero to GH has the general effect of pushing the RL towards the LHP, thus making the system more stable (less oscillatory).

Calculating of K on RL:

Once the RL is constructed, the values of K at any point s1 on the RL can be determined as follows:

Additional of poles/zeros to G(s)H(s)

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Root Locus: Example 2

Example:

Sketch the RL.

Root Locus: Example 2

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Root Locus: Example 3

Example:

Root Locus : Example 3

Example: