control systems engineering · 2016-11-21 · control systems engineering (chapter 8.root locus...

Control Systems Engineering(Chapter 8. Root Locus Techniques)

Prof. Kwang-Chun [email protected]: 02-760-4253 Fax:02-760-4435

Dept. Information and Communication Eng.2

Introduction

In this lesson, you will learn the following : The definition of a root locusHow to sketch root locusHow to use the root locus to find the poles of a closed

loop systemHow to use root locus to describe the changes in

transient responseHow to use root locus to design a parameter value to

meet a desired transient response


Introduction

What is Root Locus?Root Locus is graphical presentation of the closed-loop

poles as the parameter is variedA powerful method of analysis and design for stability

and transient responseAbility to provide solution for system of order higher

than twoA graphic representation of a system’s stability

Why do we need to use Root Locus?We use Root Locus to analyze the transient

qualitatively


Introduction

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 feedback control system qualitatively

The control system problem: The poles of the closed-loop transfer function are more

difficult to find The closed-loop poles change with changes in system

gainThe poles of T(s) are not immediately known without factoring

the denominator, and they are a function of K


Introduction

Since the system transient response and stability are dependent upon the poles of T(s), we have no knowledge of the system performance unless we factor the denominator for specific values of K

The Root Locus will be used to give us a vivid picture of the poles of T(s) as K varies


Introduction

Complex Numbers and Their Representation as Vectors:Any complex number, ,described in Cartesian

coordinates can be graphically represented by a vector, as shown in the following figure (a)

The complex numbers also can be described in the polar form with magnitude , and angle , as

If the complex number is substituting into a complex function, F(s), another complex number will result

For example, if F(s)=(s+a), then substituting the complex number yields which is another complex number, as shown in figure (b)

j

M M

s j ( ) ( )F s a j


Introduction

Since F(s) has a zero at –a, we have alternate representation 5 2( 7) s js


Introduction

Example:Evaluation of a complex function via vectorsMagnitude and angle of F(s) at any point ( )Solution:

3 4s j

( 3 4 1)( 3 4)( 3 4)( 3 4 2)

20 116.6(5 126.9 )( 17 104.0 )

20 (116.6 126.9 104.0 )5 170.217 114.3

jF jj j

M

3 4s j


Defining the Root Locus

Consider security camera that can automatically follow a subject (auto tracking)

2 10s s K Coming from


Defining the Root Locus

(Pole plot from Table 8.1) (Representation of the paths of theclosed-loop poles as the gain is varied )

Call Root Locus!


Properties of the Root Locus

The root locus technique can be used to analyze and design the effect of a loop gain upon the system’s transient response and stabilityThe closed loop transfer function for a system with a gain, K, is

From this equation, a pole, s, exists when the characteristic polynomial in the denominator becomes zero, or

(For control system of slide 5)



Here, -1 represented in polar form asAlternatively, a value of s is a closed loop pole if

and

Let’s demonstrate the relationship:For control system of slide 9,

From Table 8.1, the closed-loop poles exist at -9.47 and -0.53 when the gain is 5

Substituting the pole at -9.47 for s and 5 for K yields

1 (2 1)180k

( ) ( )(s 10)

KKG s H ss

( ) ( ) 1KG s H s (Thus, it is a point on the root locus!)



If these calculation is repeated for the point , the angle equals an odd multiple of

That is, is a point on the root locus for some value of gain

Now we proceed to evaluate the value of gain

For a point , the angle does not equal

Thus, it is not a point on the root locus, or is not a closed-loop pole for any gain

5 5j 180

( ) ( ) 180(s 10) ( 5 5)(5 5) 50( 50 135 )( 50 45 )

K K K KKG s H ss j j

5 5j

( ) ( ) 150KKG s H s 50K

(Can confirm from Table 8.1!)2 2j

( ) ( ) 149

(s 10) ( 2 2)(8 2) ( 8 135 )( 68 14 ) 8 68K K K KKG s H s

s j j


Sketching the Root Locus

The following five basic rules allow us to sketch the root locus using minimal calculationsRule 1: Number of the branches

Each closed loop pole moves as the gain is varied If we define a branch as the path that one pole traverses, then

there will be one branch for each closed loop poleOur first rule defines the number of branches of the root locus:

The number of the branches of the root locus are equal to the number of closed loop poles

Rule 2: Symmetry Physically realizable system can not have the complex

coefficients K in their transfer functionsThus, we conclude:

The root locus is symmetrical about the real axis



Rule 3: Real-axis segment Using some mathematical properties of root locus plot, we get

On the real axis, for K>0 the root locus exist to the left of an odd number of real axis, finite open loop poles and/or finite open loop zeros

symmetrical



Rule 4: Starting and ending pointsConsider the closed loop transfer function

As K approaches to zero, the closed loop poles approaches the poles of G(s)H(s)

Similarly, as K approaches to infinity, the closed loop poles approaches the zeros of G(s)H(s)

We conclude that The root locus begins at the finite and infinite poles of G(s)H(s)

and ends at the finite and infinite zeros of G(s)H(s)



Rule 5: Behavior at infinityA function can have infinite poles and zerosEvery function of s has an equal number of poles and zerosFor example, consider the function

This function has two poles at -2 and -3 and one zeros at -1 Because of the number of poles and zeros must be equal, this

function has a zero at infinity! (Ref. G(s)=1/s: , a pole at s=0) This rule tell us the following conclusion:

In this system, the root locus begins at the poles at -1 and -2 and ends at zeros at -3 and -4

(Complete Root Locus)

( ) 0G



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, , and angle, , as follows:

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

Example: Sketch the root locus for the system shown in figure

a a

finite poles finite zeros (2 1),#finite poles #finite zeros #finite poles #finite zerosa a

k



Solution: Begin by calculating the asymptotes:

If the values for k continued to increase, the angels would begin repeat

The number of lines obtained equals the difference between the number of finite poles and the number of finite zeros

Rule 4 states that the root locus begins at the open loop poles and ends at the open loop zeros



For this example, there are more open loop poles than open loop zeros

Thus, there must be zeros at infinity The asymptotes tell us how we get to these zeros at infinity! This figure shows the complete root locus as well as the asymptotes

that were just calculated

Root locus with the 3 asymptotes



Notice that we have made use of all the rules learned so far The real axis segment lie to the left of an odd number of poles and/or

zeros The locus starts at the open-loop poles and ends at the open-loop

zeros For this example, there is only one open loop finite zero and three

infinite zeros Rule 5, then, tell us that the three zeros at infinity are at the ends of

the asymptotes

Now, consider four additional rules for Refining the SketchOnce 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


Refining the Sketch

Rule 6: Real Axis Breakaway and Break-in PointsNumerous root loci appear to break away from the real axis as

the system poles move from the real axis to the complex planeAt other times the loci appear to return to the real axis as a pair

of complex poles becomes realThis is illustrated in the figure

The breakaway point is found at the maximum point, and the break-in point is found at the minimum point

(Variation of gain along the real axis )


Refining the Sketch

The figure shows a root locus leaving the real axis between -1 and -2 and returning to the real axis between +3 and +5

The point where the locus leaves the real axis, , is called the breakaway point, and the point where the locus returns to the real axis, , is called break-in point

At the breakaway or break-in point, the branches of the root locus form an angle of 180/n with the real axis, where n is the number of closed loop poles arriving or departing from the single breakaway or break-in point on the real axis

Thus for the poles shown in the figure the branches at the breakaway point form 90° angles with the real axis

We now show how to find the breakaway and the break-in points There is a simple method for finding the points at which the root

locus breaks away from and breaks into real axis The first method is to maximize and minimize the gain, K, using

differential calculus

1

2


Refining the Sketch

For all points on the root locus, because of the KG(s)H(s)= -1 as we pointed out before, K = -1 / G(s)H(s)

For points along the real axis segment of the root locus where breakaway and break-in points could exist,

Hence along the real axis,

Hence if we differentiate this equation with respect to and set the derivative equal to zero, we can find the points of maximum and minimum gain and hence breakaway and break-in points

s

1 1( ) ( ) ( ) ( )s

KG s H s G H

0, ?dKd


Refining the Sketch

Example: Find the breakaway and break-in points for the root locus of the

figure using differential calculus Solution:

Using the open loop poles and zeros, we present the open loop system whose root locus is shown in figure as follows

But for all points along the root locus, KG(s)H(s)=-1, andalong the real axis, . Hence

Solving for K, we find

s


Refining the Sketch

Differentiating K with respect to and setting the derivative equal to zero yields

Solving for , we find =-1.45 (break-away point) and 3.82 (break-in point), which are the breakaway and break-in points


Refining the Sketch

Rule 7: -Axis CrossingsThe axis crossing is a point on the root locus that

separates the stable operation of the system from the unstable operation

The value of at the axis crossing yields the frequency of oscillation

The gain at the -axis crossing yields the maximum positive gain for system stability

To find -axis crossing, we can use the Routh-Hurwitz criterion as follows: Forcing a row of zeros in the Routh table will yield the gain; going

back one row to the even polynomial equation and solving for the roots yields the frequency at the imaginary axis crossing

jj

j

j


Refining the Sketch

Example: For the system of the following figure, find the frequency and gain,

K, for which the root locus crosses the imaginary axis For what range of K is the system stable?

Solution: The closed loop transfer function for the system is

Simplifying some of the entries by multiplying any row by a constant, we obtain the Routh array shown in table


Refining the Sketch

A complete row of zeros yields the possibility for imaginary axis roots For positive value of gain only the s1 row can yield a row of zeros for

K>0 Thus, , and K = 9.65 Forming the even polynomial by using the s2 row with K=9.65 We obtain , and s is found to

be ±j1.59 Thus the root locus crosses the -axis at ±j1.59 at a gain of 9.65 We conclude that the system is stable for

j0 9.65K


Refining the Sketch

Rule 8: Angles of Departure and Arrival In order to sketch the root locus more accurately, we want to

calculate the root locus departure angle from the complex poles and the arrival angle to the complex zeros

If we assume a point on the root locus close to the point, we assume all angles drawn from all other poles and zeros are drawn directly to the pole that is near the point

Thus, the only unknown angle in the sum is the angle drawn from the pole that is close

We can solve for this unknown angle, which is also the angle of departure from this complex pole

Hence from the figure(a)

(Angles of Departure)


If we assume a point on the root locus close to a complex zero, the sum of angles drawn from all finite poles and zeros to this point is an odd multiple of 180°

Except for the zero that is close to the point, we can assume all angles drawn from all other poles and zeros are drawn directly to the zero that is near the point

Refining the Sketch


Thus, the only unknown angle in the sum is the angle drawn from the zeros that is close

We can solve for this unknown angle, which is also the angle of arrival to this complex zero

Hence from the figure(b)

Refining the Sketch

(Angles of Arrival)


Refining the Sketch

Example:Given the unity feedback system of following figure, find the

angle of departure from the complex poles and sketch the root locus

Solution:Using the poles and zeros of G(s)= (s+2)/[(s+3)(s2+2s+2)] as

plotted in the figure, we calculate the sum of angles drawn to a point close to the complex pole, -1+j1, in the second quadrant

Thus,


Refining the Sketch


Refining the Sketch

Rule 9: Plotting and calibration the Root locusWe might want to know the exact coordinates of the root locus

as it crosses the radial line representing an overshoot valueOvershoot value can be represented by a radial line on the s-

planeWe learnt in Chapter 3,

the value of zeta on the s-plane is

Given percent overshoot value we can calculate the zeta value

Radial line representing overshoot value on the s-plane

cos

2 2

ln(%OS /100)ln (%OS /100)


Refining the 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

Next step is to find the angle of the radial line

2 2 2 2

ln % /100 ln(20 /100) 0.456ln % /100 ln 20 /100

OS

OS

1

cos 0.456 coscos 0.456 62.87

j


Refining the 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

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 ̊,

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

Odd multiple of 180 ̊ :　(2k+1)180 ̊, k = 1, 2, 3, 4, ….180 ̊ , 540 ̊, 900 ̊, 1260 ̊, …


0.45

Then, we must then calculate the value of gainRefer to the previous root locus that we have calculated, we will

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

Refining the 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

At a point s,

Refining the Sketch

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


Refining the Sketch

Example:Sketch the root locus for the system shown in figure

and find the breakaway point on the real axis and the range of K within which the system is stable

Solution: The root locus of the system is shown in following figure The real axis segment is found to be between -2 and -4, using

Rule 3 The root locus starts at the open loop poles and ends at the open

loop zeros, using Rule 4 The root locus crosses the -axis at ±j3.9, using Rule 7j


Refining the Sketch

The exact point and gain where the locus crosses the 0.45 damping ratio line: K=0.47 at , using Rule 9

To find breakaway point use the root locus algorithm to search the real axis between -2 and -4 for the point that yields maximum gain, using Rule 6

Naturally all points will have the sum of their angles equal to an odd multiple of 180°

A maximum gain of 0.0248 is found to at the point -2.88

Thus, the breakaway point is between the open loop poles on the real axis at -2.88

Finally, the system is stable for K between 0 and 15

3.4 116.7


Homework Assignment #8

control systems engineering · 2016-11-21 · control systems engineering (chapter 8.root locus...

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