Feedback Control System THE ROOT-LOCUS DESIGN METHOD Dr.-Ing. Erwin Sitompul Chapter 5 http://zitompul.wordpress.com
Feb 25, 2016
Feedback Control System
THE ROOT-LOCUS DESIGN METHOD
Dr.-Ing. Erwin Sitompul
Chapter 5
http://zitompul.wordpress.com
7/2Erwin Sitompul Feedback Control System
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
7/3Erwin Sitompul Feedback Control System
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
7/4Erwin Sitompul Feedback Control System
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
7/5Erwin Sitompul Feedback Control System
Example 3: Plotting a Root Locus
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
7/6Erwin Sitompul Feedback Control System
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
7/7Erwin Sitompul Feedback Control System
Example 3: Plotting a Root Locus
Realaxis
Imagaxis
1 20–1–2–3–4
1
2
–1
–2
3
–3
RULE 5
00, 0K
7/8Erwin Sitompul Feedback Control System
Example 3: Plotting a Root Locus1 ( ) 0KL s
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
7/9Erwin Sitompul Feedback Control System
Example 3: Plotting a Root Locus
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
point
Not on the root locus
0
7/10Erwin Sitompul Feedback Control System
Example 3: Plotting a Root Locus
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
7/11Erwin Sitompul Feedback Control System
Example 3: Plotting a Root Locus
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
7/12Erwin Sitompul Feedback Control System
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
7/13Erwin Sitompul Feedback Control System
Example 4: Plotting a Root Locus
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
7/14Erwin Sitompul Feedback Control System
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
7/15Erwin Sitompul Feedback Control System
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
7/16Erwin Sitompul Feedback Control System
Example 4: Plotting a Root Locus1 ( ) 0KL s
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
7/17Erwin Sitompul Feedback Control System
Realaxis
Imagaxis
1 20–1–2–3–4
1
2
–1
–2
3
–3
–9
Example 4: Plotting a Root Locus
00, 0K
RULE 5
7/18Erwin Sitompul Feedback Control System
Example 4: Plotting a Root Locus1 ( ) 0KL s
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
7/19Erwin Sitompul Feedback Control System
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 point0
RULE 6–3
7/20Erwin Sitompul Feedback Control System
Realaxis
Imagaxis
1 20–1–2–3–4
1
2
–1
–2
3
–3
–9
Example 4: Plotting a Root Locus
0
After examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot.
90–3
7/21Erwin Sitompul Feedback Control System
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
7/22Erwin Sitompul Feedback Control System
Conclusion: Example 3 and Example 4
2
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.
7/23Erwin Sitompul Feedback Control System
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
7/24Erwin Sitompul Feedback Control System
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).
7/25Erwin Sitompul Feedback Control System
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.
7/26Erwin Sitompul Feedback Control System
( ) 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 Compensation
7/27Erwin Sitompul Feedback Control System
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
7/28Erwin Sitompul Feedback Control System
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 Compensation
7/29Erwin Sitompul Feedback Control System
First Design Using Lead Compensation 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
7/30Erwin Sitompul Feedback Control SystemReal 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
First Design Using Lead Compensation 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 s
4.396 0.4629.517
Design requirement fulfilled with:
22( ) 100
10sD ss
7/31Erwin Sitompul Feedback Control System
Second Design Using Lead Compensation The closed-loop system of G(s) = 1/[s(s+1)] is now required
to have a pole at (corresponds to ωn = 7 and ζ = 0.5).
0 3.5 3.5 3r j
Now using , the value of K and z must be determined.
3( )20
s zD s Ks
• 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
7/32Erwin Sitompul Feedback Control System
Second Design Using Lead Compensation
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
7/33Erwin Sitompul Feedback Control System
1 72.585
Second Design Using Lead Compensation0 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
7/34Erwin Sitompul Feedback Control System
Second Design Using Lead Compensation1.902 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
7/35Erwin Sitompul Feedback Control System
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.
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
7/36Erwin Sitompul Feedback Control System
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
7/37Erwin Sitompul Feedback Control System
Design Using Lag Compensation 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)
7/38Erwin Sitompul Feedback Control System
Design Using Lag Compensation 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
7/39Erwin Sitompul Feedback Control System
Example: Compensation Design
Let 1( )( 2)( 3)
G ss s
( ) s aD s Ks b
and
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
7/40Erwin Sitompul Feedback Control System
Example: Compensation Design
1( ) ( )( 2)( 3)
L s G ss s
Before compensation
Realaxis
Imagaxis
1 20–1–2–3–4
1
2
–1
–2
3
–3
desired closed-loop poles
7/41Erwin Sitompul Feedback Control System
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
7/42Erwin Sitompul Feedback Control System
Example: Compensation Design
Realaxis
Imagaxis
1 20–1–2–3–4
1
2
–1
–2
3
–3
: roots of the compensation
1( )( 2)
L s Ks s
7/43Erwin Sitompul Feedback Control System
Example: Compensation Design
Realaxis
Imagaxis
1 20–1–2–3–4
1
2
–1
–2
3
–3
K?
1( )( 2)
L s Ks s
7/44Erwin Sitompul Feedback Control System
1 ( ) 0L s
Solving for K using the characteristic equation,
Example: Compensation Design
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
7/45Erwin Sitompul Feedback Control System
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