CONTROL SYSTEMS THEORY

A Graphical Tool

CHAPTER 7STB 35103

Objectives To learn the definition of a root locus.

To sketch a root locus.

To refine your sketch of a root locus.

To use the root locus to find the poles of a closed-loop system.

To use the root locus to describe qualitatively the changes in transient response and stability of a system as a system parameter varied.

Introduction What is root locus?

Root locus is graphical presentation of the closed-loop poles as the parameter is varied.

Why do we need to use root locus? We use root locus to analyze the transient

qualitatively. E.g. We can use root locus to analyze qualitatively the effect of varying gain upon percent overshoot, settling time and peak time.

We can also use root locus to check the stability of our system qualitatively.

Introduction What kind of system that uses root

locus? Usually we use root locus to analyze the

feedback control system qualitatively.

K in the feedback system is called a gain. Gain is used to vary the system in order to get a different output response.

Introduction What does a

root locus look like? The line with arrow

is the movement path for closed loop poles as the gain is varied on the s-plane.

Figure 1 – An example of root locus.

1 Pole 1 zero

Drawing the root locus First step in drawing the root locus is to draw

the open-loop poles and zeros in the s-plane.

The most important thing that we must understand in the movements of the poles is, it is the closed-loop poles that move not the open-loop poles.

Next slide shows an example of poles movements when gain is varied.

Drawing the root locus Given a feedback system for a video

camera that can automatically follow a subject.

(a)Feedback system(b)Closed loop transfer function of

the system, T(s)

Drawing the root locus Based on the video camera feedback

system, the open-loop transfer function is

21( ) 1

10

10

KG s K

s s

K

s s

The modified unity feedback system for an open-loop transfer function can be displayed as

1 2K K K

Drawing the root locus We are going to start drawing the root

locus for the video camera. The first step is to draw the poles and zeros of the system’s open-loop transient response.

*Fill up a table with the corresponding K and pole values

Drawing the root locus Next step is to plot the poles values on the

s-plane by varying the gain, K, value.

Drawing the root locus Join the poles with solid lines and you will

the shape of the locus (path)

Drawing the root locus The process of drawing a root locus is time

consuming. If the system is complex, the calculation will be much harder especially if you use hand calculation.

An alternative approach is to sketch the root locus instead of drawing the root locus.

Sketching the root locus In order to sketch the root locus we must

follow these five rules.1. Number of branches. We know that the closed loop

poles move when the gain, K, is varied. A branch is a path where the a pole moves. The first rule is the number of branches of the root locus equals to the number of closed-loop poles.

Sketching the root locus2. Symmetry. The root locus is symmetrical

about the real axis.

real axis

imaginary axis

Sketching the root locus3. Real-axis segment. On the real-axis, for K

> 0 the root locus exists to the left of an odd number of real-axis, finite open-loop poles and/or finite open-loop zeros. Below is an example of real-axis segment

Blue lines s-plane above are parts of the real-axis where the root locus exists.

Sketching the root locus4. Starting and ending points. Where does

the root locus begin (zero gain) and end (infinite gain)?

The root locus begins at the open-loop poles and ends at open-loop zeros.

1 Pole 1 zero

Sketching the root locus5. Behavior at infinity. A function can have finite

and infinite poles and zeros.

A function must have an equal number of poles and zeros, finite and infinite.

If a function approaches infinity when s approaches infinity – pole at infinity

If a function approaches zero when s approaches infinity – zero at infinity

Sketching the root locus5. Behavior at infinity. The root locus

approaches straight lines as asymptotes as the locus approaches infinity. Further, the equation of the asymptotes is given by the real-axis intercept, σa and angle θa as follows.

finite poles finite zeros

#finite poles #finite zeros

2 1

#finite poles #finite zeros

a

a

k

Where k = 0, ±1, ±2, ±3 and the angle is given in radians with respect to the positive extension of the real axis

Sketching the root locus We will sketch a root locus in Example 8.2

Problem: Sketch the root locus for the system shown in figure below.

Sketching the root locus It will greatly help our locus design if we know

the asymptotes for our locus. Based on the open-loop transient response for

this system, the finite poles are at 0, -1, -2, and -4 and the finite zero is at -3. So, there are 4 poles but only 1 zero. Hence there must be 3 infinite zeros, 3 asymptotes.

#Asymptotes = #finite poles – #finite zeros Using the equation to calculate the asymptotes

crossing with the real axis,

1 2 4 3 4

4 1 3a

Sketching the root locus The angles of the lines that intersect at -4/3

given the equation for θa

2 1

#finite poles #finite zeros

/ 3 for 0

for 1

5 / 3 for k 2

7 / 3 for k 3

a

k

k

k

As k continues to increase, the angle would begin to repeat. The number of asymptotes equals the difference between the number of finite poles and the number of finite zeros.

There are 3 asymptotes so 3 values of k=0,1,2 each representing the angle of each asymptote

Sketching the root locus Root locus with the 3 asymptotes

Plotting and calibrating the root locus sketch Once we sketch the root locus using the

five rules discussed in the previous slides, we may want to accurately locate points on the root locus as well as find their associated gain.

We might want to know the exact coordinates of the root locus as it crosses the radial line representing an overshoot value.

Plotting and calibrating the root locus sketch Overshoot value can be represented by a

radial line on the s-plane.

Radial line representing overshoot value on the s-plane

Plotting and calibrating the root locus sketch We learnt in Chapter 3, the value of zeta

on the s-plane is

Given percent overshoot value we can calculate the zeta value

cos

100/%ln

100/%ln22 OS

OS

Plotting and calibrating the root locus sketch Example

Draw a radial line on an S-plane that represents 20% overshoot.

Solution Overshoot is represented by zeta (damping ratio) on

the S-plane. So, the first step is to find the value of zeta.

2 2

2 2

ln % /100

ln % /100

ln(20 /100)

ln 20 /100

0.456

OS

OS

Plotting and calibrating the root locus sketch

Next step is to find the angle of the radial line.

1

cos

0.456 cos

cos 0.456

62.87

j

Plotting and calibrating the root locus sketch The point where our root locus intersect with the

n percent overshoot radial line is the point when the gain value produces a transient response with n percent overshoot.

Our root locus intersect with the radial line. Meaning the gain at the intersection produces transient response with zeta = 0.45

Plotting and calibrating the root locus sketch We know that when our root locus

intersect with the %OS radial line, the gain during that intersection will produce transient response with the same %OS.

Since we just sketch the root locus, we do not know the exact coordinate of the intersection between radial line and the root locus.

Plotting and calibrating the root locus sketch A point on the radial line is on the root locus if

the angular sum (zero angle –pole angles) in reference to the point on the radial line add up to an odd multiple of 180 ̊, Odd multiple of 180 ̊ (2k+1)180 ̊, k = 1,2,3, 4, ….

180 ̊ , 540 ̊, 900 ̊, 1260 ̊, …

We must then calculate the value of gain. The calculation for the gain by multiplying the pole lengths drawn to that point and dividing by the product of the zero length drawn to that point.

Plotting and calibrating the root locus sketch Refer to the previous root locus that we have calculated, we

will find the exact coordinate as it crosses the radial line representing 20% overshoot

0.45

Plotting and calibrating the root locus sketch We can find the point on the radial line that crosses the root

locus by selecting a point with radius value then add the angles of the zeros and subtract the angles of the poles ( =zero - pole =2 – [1 + 3 + 4 + 5]

Theory : Odd multiple of 180 ̊ 180 ̊ , 540 ̊, 900 ̊, 1260 ̊…point on the radial line is on the locus

Plotting and calibrating the root locus sketch We will calculate the angle and magnitude

using the sine, cosine and tangent rules

2 2 2 2 cosb a c ac

Plotting and calibrating the root locus sketch Let us take r = 0.747 and check if this

point on the radial line intersects with the root locus

Plotting and calibrating the root locus sketch

We will calculate the value of A, B, C, D, and E and also θ1, θ2, θ3, θ4, and θ5

E length is equal to the radius r because of the poles is at origin.

5

0.747

180 63.256

116.744

E

ED

1

θ5θ4

2 2 2

2

1 2 (1)cos(63.256)

0.747 1 2(0.747)cos(63.25)

0.886

0.886

0.941

D E E

D

4

4

sin sin( )

0.941sin(63.256)

0.941

45.141

E D

Plotting and calibrating the root locus sketch

We will then calculate the length of C and θ3 You can either take the triangle E C or the

triangle D and C but the easiest is triangle E and C.

EC

2

θ5θ3

2 2 2

2 2 2

2 2 (2)cos

0.747 2 2(0.747)(2)cos(63.256)

1.793

C E E

C

C

3

3

sin sin( )

0.747sin(63.256)

1.793

21.842

E C

Plotting and calibrating the root locus sketch Calculate the length of B and θ2

EB

3

θ5θ2

2 2 2

2 2 2

3 2 (3)cos

0.747 3 2(0.747)(3)cos(63.256)

2.746

B E E

B

B

2

2

sin sin( )

0.747sin(63.256)

2.746

14.06

E B

Plotting and calibrating the root locus sketch Calculate the length of A and θ1

2 2 2

2 2 2

4 2 (4)cos

0.747 4 2(0.747)(4)cos(63.256)

3.725

A E E

A

A

EA

4

θ5θ1

1

1

sin sin( )

0.747sin(63.256)

3.725

10.316

E A

Plotting and calibrating the root locus sketch We can calculate the gain, K, value where

root locus intersect the radial line.

poles length

zeros length

3.725 1.793 0.941 0.747

2.7461.71

K

A C D EK

B

hasil darab

Plotting and calibrating the root locus sketch For root locus that do not have zeros, the

equation to find the gain is

poles lengthK

hasil darab

Transient Response Design via Gain Adjustment In our previous example, the root locus

crossed the 0.45 damping ratio line with a gain of 1.71.

Does this mean that the system will respond with 20% overshoot, the equivalent to a damping ratio of 0.45?

It must be emphasized that the formulas describing percent overshoot, settling time, and peak time were derived only for a system with two closed-loop poles and no closed-loop zeros.

Transient Response Design via Gain Adjustment If our system has additional poles and zeros,

we can still assume the system with only two poles and no zero if it fulfills the requirement below:

Higher order poles are much farther into the left half of the s-plane measured from the jω-axis. (we assume a factor of five times farther from jω-axis than the dominant second-order pair. )

Closed-loop zeros near the closed-loop second order pole pair are nearly canceled by the close proximity of higher-order closed-loop poles.

Transient Response Design via Gain Adjustment

Closed-loop zeros not canceled by the close proximity of higher-order closed-loop poles are far removed from the closed-loop second order pole pair.

The best

Transient Response Design via Gain Adjustment Skill-Assessment Exercise 8.6

a) Sketch the root locusb) Using a second order approximation, design the

value of K to yield 10% overshoot for a unit step input (Hint r = 3.431).

c) Estimate the settling time, peak time, and steady-state error for the value of K designed in (b)

d) Determine the validity of your second-order approximation

( )( 2)( 4)( 6)

KG s

S S S

Transient Response Design via Gain AdjustmentIf we look back at the equation for peak time, Tp and settling time Ts.

21

n

pT

Imaginary

nsT

4

Real

Transient Response Design via Gain Adjustment Solution

K=45.55 Ts=1.97

Tp=1.13

Kp=0.949, Comparing this value to the real part of the

dominant pole, -2.028, we find that it is not five times further. The second-order approximation is not valid

( ) 0.51stepe