example 3: plotting a root locus

Lecture 7Feedback Control Systems

President University Erwin Sitompul FCS 7/1

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com2 0 1 2


Example 3: Plotting a Root LocusNow, sketch the root locus for:

1 1,2 31, 0, 4z p p

3, 1n m • There are 3 branches of locus.• Two starting from s = 0 and

one from s = –4.• There will be two zeros at

infinity.

RULE 1

2

11 0( 4)sK

s s

2

1( )( 4)sL s

s s

Chapter 5 The Root-Locus Design Method


Example 3: Plotting a Root Locus

Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 2

1 1,2 31, 0, 4z p p



Example 3: Plotting a Root Locus180 360 ( 1)

ll

n m

180 360 ( 1)3 1

l

i ip zn m

90 180 ( 1)l 90 , 270

4 ( 1)3 1

1.5 Center of Asymptotes

Angles of Asymptotes

1 1,2 31, 0, 4z p p




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 3

90°

270°

90 ,270l 1.5

Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros.

RULE 4



Example 3: Plotting a Root Locus1 ( ) 0KL s

2

11 0( 4)sK

s s

3 24 0s s Ks K

Replacing s with jω0,3 2

0 0 0( ) 4( ) ( ) 0j j K j K 3 20 0 04 0j jK K

2 30 0 04 ( ) 0K j K

≡ 0 ≡ 0

204K 3

0 020

KK

Points of Cross-over00, 0K




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 5

00, 0K




2 3 2

( ) 1 1( )( ) ( 4) 4b s s sL sa s s s s s

( ) ( )( ) ( ) 0da s db sb s a sds ds

The root locus must have a break-away point, which can be found by solving:

2 3 2( 1) (3 8 ) ( 4 ) 1 0s s s s s 3 2 3 23 11 8 0 4 0s s s s s

1 2,30, 1.75 0.9682s s j

3 22 7 8 0s s s




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 6

1 20, 1.75 0.9682s s j

On the root locusThe break-away

pointNot on the root locus

0




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

0

After examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot.

90




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result



Example 4: Plotting a Root LocusNow, the characteristic equation is changed a little bit:

1 1,2 31, 0, 9z p p

3, 1n m • There are 3 branches of locus.• Two starting from s = 0 and

one from s = –9.• There will be two zeros at

infinity.

RULE 1

2

11 0( 9)sK

s s

2

1( )( 9)sL s

s s




RULE 2

1 1,2 31, 0, 9z p p

Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

–9



Example 4: Plotting a Root Locus180 360 ( 1)

ll

n m

180 360 ( 1)3 1

l

i ip zn m

90 180 ( 1)l

90 , 270

9 ( 1)3 1

4 Center of Asymptotes

Angles of Asymptotes

1 1,2 31, 0, 9z p p



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

–9

Example 4: Plotting a Root Locus 90 , 270l

4

90°

270°

RULE 3

Not applicable.RULE 4




2

11 0( 9)sK

s s

3 29 0s s Ks K

Replacing s with jω0,3 2

0 0 0( ) 9( ) ( ) 0j j K j K 3 20 0 09 0j jK K

2 30 0 09 ( ) 0K j K

≡ 0 ≡ 0

209K 3

0 020

KK

Points of Cross-over00, 0K



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

–9


00, 0K

RULE 5




2 3 2

( ) 1 1( )( ) ( 9) 9b s s sL sa s s s s s

( ) ( )( ) ( ) 0da s db sb s a sds ds

The root locus must have a break-away point, which can be found by solving:

2 3 2( 1) (3 18 ) ( 9 ) 1 0s s s s s 3 2 3 23 21 18 0 9 0s s s s s

1 2,30, 3s s

3 22 12 18 0s s s



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

–9

Example 4: Plotting a Root Locus1 2,30, 3s s

On the root locusThe break-away

point

On the root locusAt the same time, the break-in

and the break-away point

0 RULE 6–3



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

–9


0

After examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot.

90–3



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

–9

Example 4: Plotting a Root LocusThe final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as:


Final Result


Conclusion: Example 3 and Example 42

11 0( 4)sK

s s

2

11 0( 9)sK

s s

The characteristic equations can be so similar, yet the resulting root locus plots are very different.

It is very important to examine each rule carefully.



Real Axis

Imag

Axi

s

-10 -5 0

-5

0

5

The Effect of Adding Poles to a System

Real Axis

Imag

Axi

s

-3 -2 -1 0-1.5

-1

-0.5

0

0.5

1

1.5

Real Axis

Imag

Axi

s

-3 -2 -1 0

-0.1

-0.05

0

0.05

0.1

1( )( 1)

L ss

1( )( 1)( 3)

L ss s

1( )( 1)( 3)( 5)

L ss s s

If a pole is added to a system: The root locus is pulled to the right. The stability tends to decrease. The settling time tends to increase

(for the same value of ζ, the value of ωd decreases).

α = –2 α = –3



The Effect of Adding Zeros to a System

Real Axis

Imag

Axi

s

-10 -5 0

-5

0

5

1( )( 1)( 3)( 5)

L ss s s

Real Axis

Imag

Axi

s

-6 -5 -4 -3 -2 -1 0

-10

-8

-6

-4

-2

0

2

4

6

8

10

Real Axis

Imag

Axi

s

-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0

-6

-4

-2

0

2

4

6

Real Axis

Imag

Axi

s

-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0

-8

-6

-4

-2

0

2

4

6

8

( 6)( )( 1)( 3)( 5)

sL ss s s

( 4)( )

( 1)( 3)( 5)sL s

s s s

( 2)( )( 1)( 3)( 5)

sL ss s s

α = –1.5 α = –2.5 α = –3.5

If a zero is added to a system: The root locus is pulled to the left. The stability tends to increase. The settling time tends to

decrease (for the same value of ζ, the value of ωd increases).



Design Using Dynamic Compensation If the process dynamics are of such a nature that a

satisfactory design cannot be obtained by adjustment of K alone, then some modification or compensation of the process dynamics must be done.

Two compensation schemes have been found to be particularly simple and effective:Lead compensation, approximates the function of PD

control and acts mainly to speed up a response by lowering the rise time and decreasing the transient overshot.

Lag compensation, approximates the function of PI control and is usually used to improve the steady-state accuracy of the system.

The techniques to select the parameters of each compensation schemes will be discussed now.



( ) s zD s Ks p

A compensation scheme is written generally in the form of a transfer function:

If z < p, it is called lead compensation. If z > p, it is called lag compensation.

The characteristic equation of the system is:1 ( ) ( ) 0D s G s

1 ( ) 0KL s

Design Using Dynamic CompensationChapter 5 The Root-Locus Design Method


Design Using Lead Compensation

1( )( 1)

G ss s

Consider a second-order position control system with normalized transfer function:

The root locus of the system will be compared for:1( )D s K

22( )

10sD s Ks

32( )20

sD s Ks



Real Axis

Imag

Axi

s

-1 -0.8 -0.6 -0.4 -0.2 0-0.5

0

0.5

Real Axis

Imag

Axi

s

-10 -8 -6 -4 -2 0

-15

-10

-5

0

5

10

15

Real Axis

Imag

Axi

s

-20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0

-5

-4

-3

-2

-1

0

1

2

3

4

5

1( )D s K 22( )

10sD s Ks

32( )20

sD s Ks

α = –0.5 α = –4.5 α = –9.5

Selecting the value of z and p is usually done by trial and error, which can be minimized with experience.

In general, the zero is placed near the closed-loop ωn, as determined by time domain specification, the pole is located at a distance 5 to 20 times the value of the zero location.

If the pole is too far to the right, the root locus moves back too far toward its uncompensated shape, while if the pole is too far the left, sensor noise will be amplified too much.

Design Using Lead CompensationChapter 5 The Root-Locus Design Method


First Design Using Lead Compensation Chapter 5 The Root-Locus Design Method

Find a compensation for G(s) = 1/[s(s+1)] that will provide overshoot of no more than 20% and rise time of no more than 0.25 sec.

r

d

t

22

2 2

(ln )(ln )

p

p

MM

2

2 2

(ln 0.2)(ln 0.2)

0.208

2 2.044

d

rt

2.0440.25d 8.176,

21d

n

2

8.176 9.1871 0.456

0.208 0.456

1 1sin sin 0.456 0.473


First Design Using Lead CompensationChapter 5 The Root-Locus Design Method

Real Axis

Imag

Axi

s

-10 -8 -6 -4 -2 0-10

-5

0

5

10

108642

108642

0.94

0.80.64 0.5 0.38 0.28 0.170.08

0.94

0.80.64 0.5 0.38 0.28 0.170.08 9.187n

0.456

Giving grid to the root locus plot of using D2(s)G(s), and using the calculation results:

Area of Eligible Roots

For 100,K

2,3 4.396 8.4413s j 2( 4.396) (8.4413)n

9.517 rad s4.396 0.4629.517

Design requirement fulfilled with:

22( ) 100

10sD ss


Second Design Using Lead CompensationChapter 5 The Root-Locus Design Method

The closed-loop system of G(s) = 1/[s(s+1)] is now required to have a pole at (corresponds to ωn = 7 rad/s and ζ = 0.5). 0 3.5 3.5 3r j

Now using , the value of K and z must be determined. 3( ) ( ) ( 20)D s K s z s

• There is exactly one location for the zero where the angle ψ1 will fulfill the phase condition of r0.

• If the zero is placed on that location, r0 will be on the root locus.s = –z

1



1 2 3

11 tan (3.5 3 16.5)

20.174

13 tan (3.5 3 3.5)

120

12 tan (3.5 3 2.5)

112.411

1 1 2 3 180 360 ( 1)i i l 1 20.174 112.411 120 180 360 ( 1)l

1 252.585 180 360 ( 1)l 1 72.585

1



1 72.585

0 3.5 3.5 3r j

1

The position of the zero can now be calculated:

s = –z

tan 72.585 3.5 3 ( 3.5 ( ))z 3.1883.5 3 3.5 3.188

3.188z

5.402

31 ( ) ( ) 0D s G s Solving for K using the characteristic equation,

3.5 3.5 3

5.402 11 020 ( 1) s j

sKs s s

1.902 3.5 3 11 016.5 3.5 3 ( 3.5 3.5 3)( 2.5 3.5 3)

jKj j j



1.902 3.5 3 11 016.5 3.5 3 3.5 3.5 3 ( 2.5 3.5 3)

jK

j j j

6.3536 11 017.5784 7 6.5574

K

126.996 127K

Thus, the compensation that will make to be on the root locus is: 0 3.5 3.5 3r j

35.402( ) 127

20sD ss

0r

0r


Design Using Lag Compensation Once satisfactory dynamic response has been obtained,

perhaps by using one or more lead compensations, we may discover that the low-frequency gain (the value of the relevant steady-state error constant, such as Kv, Ka) is still too low.

In order to increase this constant, it is necessary to do so in a way that does not upset the already satisfactory dynamic response.



Design Using Lag Compensation The new compensation D(s) should yield a significant gain at

s = 0 to raise the steady-state error constant but is nearly unity at the higher frequency ωn.

The result is:

The value of z and p are small compared with ωn, yet D(0) = z / p can be adjusted to be big enough to adjust the steady-state gain.

( ) s zD s Ks p

, z p



To study the effects of lag compensation, we use again the result of the “Second Design Using Lead Compensation”.

The uncompensated closed-loop system is:G(s) = 1/[s(s+1)]

It is required to have a pole at (corresponds to ωn = 7 and ζ = 0.5).

The obtained lead compensation is D(s) = 127(s+5.402)/(s+20).

At the operating point, the velocity constant is given by:

Design Using Lag Compensation

0 3.5 3.5 3r j

0

5.402 1lim 12720 ( 1)s

sss s s

0limv s

K s L s

34.30• Suppose not big enough• Required Kv = 100



Design Using Lag CompensationChapter 5 The Root-Locus Design Method

To obtain Kv = 100, an additional lag compensation is designed, with: z/p = 3 to increase the velocity constant by 3 at s = 0.a pole at p = –0.01 to keep the values of both z and p

very small so that the lag compensation would have little effect around ωn = 7, the dominant dynamics already obtained previously by the lead compensation.

The overall open-loop transfer function with lead-lag compensation is now given by:

lag lead( ) ( ) ( ) ( )L s D s D s G s 0.03 5.402 11270.01 20 ( 1)

s ss s s s

• Remark: The design using lag compensation is performed after adjusting the gain K and performing the design using lead compensation(s)


Design Using Lag CompensationChapter 5 The Root-Locus Design Method

The root locus of the new L(s) is plotted below.The root locus near the

lag compensation

The transient response corresponding to the lag-compensation zero will be very slowly decaying, with small magnitude, and might seriously influence the settling time.

The lag pole-zero combination must be placed at the highest frequency possible without shifting the dominant roots.

Dominant root, hardly affected


Example: Compensation DesignChapter 5 The Root-Locus Design Method

Let and1( )( 2)( 3)

G ss s

( ) s aD s Ks b

Using root-locus techniques, find values for the parameters a, b, and K of the compensation D(s) that will produce closed-loop poles at s =–1 ± j for the system shown below.

Unity Feedback System

Problem 5.24 FPE


Example: Compensation Design1( ) ( )

( 2)( 3)L s G s

s s

Before compensation

Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

Desired closed-loop poles



Example: Compensation Design

Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

Desired closed-loop poles

: roots of the compensation

The zero of D(s) cancels the pole of G(s)

The pole of D(s) is placed in such a way that the desired closed loop poles are on the future root locus3( ) sD s K

s

1( ) ( ) ( )

( 2)L s D s G s K

s s




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

: roots of the compensation

1( )( 2)

L s Ks s




Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

K?

1( )( 2)

L s Ks s



1 ( ) 0L s Solving for K using the characteristic equation,


11 0( 2)

Ks s

( 2)K s s

1( 2) s js sK ( 1 )(1 )j j 2

The compensation D(s) can now completely be written as:3( ) 2 sD ss



Homework 7 No.1, FPE (5th Ed.), 5.23. No.2, FPE (5th Ed.), 5.30. No.3

Consider the unity feedback system, with:

(a) Show that the system cannot operate with a 2%-settling-time of 2/3 second and a percent overshoot of 1.5% with a simple gain adjustment.

(b) Design a lead compensator so that the system meets the transient response characteristics of part (a). Specify the compensator’s pole, zero, and the required gain.

( )( 3)( 5)

KG ss s


Deadline: 30.10.2012, at 07:30.

example 3: plotting a root locus

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