Chapter 5
THE ROOT-LOCUS DESIGN METHOD
Feedback Control System
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Example 3: Plotting a Root Locus
Now, sketch the root locus for:
s 1
1 K 2
0
s ( s  4)
 L( s ) 
s 1
s 2 ( s  4)
n  3, m  1
z1  1, p1,2  0, p3  4
Erwin Sitompul
RULE 1
• There are 3 branches
of locus.
• Two starting from s = 0
and one from s = –4.
• There will be two zeros
at infinity.
Feedback Control System
7/2
Example 3: Plotting a Root Locus
z1  1, p1,2  0, p3  4
3
Imag
axis
2
1
RULE 2
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/3
Example 3: Plotting a Root Locus
180  360(l  1)
l 
nm
180  360(l  1)

3 1
 90  180(l  1)
 90, 270
p  z


i
nm
4  (1)

3 1
 1.5
Erwin Sitompul
i
Angles of Asymptotes
z1  1, p1,2  0, p3  4
Center of Asymptotes
Feedback Control System
7/4
Example 3: Plotting a Root Locus
90°
l  90, 270
  1.5
3
Imag
axis
2
1
–4
–3
–2
–1
0
–1
RULE 4
Not applicable. The angles
of departure or the angles of
arrival must be calculated
only if there are any complex
poles or zeros.
Erwin Sitompul
1
2
Real
axis
RULE 3
–2
270°
–3
Feedback Control System
7/5
Example 3: Plotting a Root Locus
1  KL(s)  0
s 1
1 K 2
0
s ( s  4)
s3  4s 2  Ks  K  0
Replacing s with jω0,
( j0 )3  4( j0 )2  K ( j0 )  K  0
 j03  402  jK0  K  0
K  4  j(K0   )  0
2
0
≡0
3
0
≡0
K  402
K0  03
K  02
 K  0, 0  0
Points of Cross-over
Erwin Sitompul
Feedback Control System
7/6
Example 3: Plotting a Root Locus
3
K  0, 0  0
Imag
axis
2
RULE 5
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/7
Example 3: Plotting a Root Locus
1  KL(s)  0
b( s )
s 1
s 1
L( s ) 
 2
 3
a( s) s ( s  4) s  4s 2
The root locus must have a break-away point, which
can be found by solving:
da ( s )
db( s )
b( s )
 a(s)
0
ds
ds
(s  1)  (3s2  8s)  (s3  4s2 ) 1  0
3s3  11s 2  8s  0  s3  4s 2  0
2s3  7s 2  8s  0
s1  0, s2,3  1.75  j0.9682
Erwin Sitompul
Feedback Control System
7/8
Example 3: Plotting a Root Locus
s1  0, s2  1.75  j 0.9682
On the root locus
The break-away
point
3
Not on the
root locus
2
1
–4
–3
Imag
axis
–2
–1
0
0
1
RULE 6
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/9
Example 3: Plotting a Root Locus
After examining RULE 1 up to
RULE 6, now there is enough
information to draw the root
locus plot.
3
2
90 1
–4
–3
Imag
axis
–2
–1
0
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/10
Example 3: Plotting a Root Locus
The final sketch, with direction of
root movements as K increases
from 0 to ∞ can be shown as:
3
Imag
axis
Final Result
2
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/11
Example 4: Plotting a Root Locus
Now, the characteristic equation is changed a little bit:
s 1
1 K 2
0
s ( s  9)
 L( s ) 
s 1
s 2 ( s  9)
n  3, m  1
z1  1, p1,2  0, p3  9
Erwin Sitompul
RULE 1
• There are 3 branches
of locus.
• Two starting from s = 0
and one from s = –9.
• There will be two zeros
at infinity.
Feedback Control System
7/12
Example 4: Plotting a Root Locus
z1  1, p1,2  0, p3  9
3
Imag
axis
2
1
RULE 2
–9
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/13
Example 4: Plotting a Root Locus
180  360(l  1)
l 
nm
180  360(l  1)

3 1
 90  180(l  1)
 90, 270
p  z


i
nm
9  (1)

3 1
 4
Erwin Sitompul
i
Angles of Asymptotes
z1  1, p1,2  0, p3  9
Center of Asymptotes
Feedback Control System
7/14
Example 4: Plotting a Root Locus
l  90, 270
  4
90°
3
Imag
axis
2
1
–9
–4
–3
–2
–1
0
–1
1
2
Real
axis
RULE 3
–2
RULE 4
270°
–3
Not applicable.
Erwin Sitompul
Feedback Control System
7/15
Example 4: Plotting a Root Locus
1  KL(s)  0
s 1
1 K 2
0
s ( s  9)
s3  9s 2  Ks  K  0
Replacing s with jω0,
( j0 )3  9( j0 )2  K ( j0 )  K  0
 j03  902  jK0  K  0
K  9  j(K0   )  0
2
0
≡0
3
0
≡0
K  902
K0  03
K  02
 K  0, 0  0
Points of Cross-over
Erwin Sitompul
Feedback Control System
7/16
Example 4: Plotting a Root Locus
3
K  0, 0  0
Imag
axis
2
RULE 5
1
–9
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/17
Example 4: Plotting a Root Locus
1  KL(s)  0
b( s )
s 1
s 1
L( s ) 
 2
 3
a( s) s ( s  9) s  9s 2
The root locus must have a break-away point, which
can be found by solving:
da ( s )
db( s )
b( s )
 a(s)
0
ds
ds
(s  1)  (3s 2  18s)  (s3  9s 2 ) 1  0
3s3  21s 2  18s  0  s3  9s 2  0
2s3  12s 2  18s  0
s1  0, s2,3  3
Erwin Sitompul
Feedback Control System
7/18
Example 4: Plotting a Root Locus
s1  0, s2,3  3
3
On the root locus
At the same time, the break-in
and the break-away point
On the root locus
–3
The break-away
point
–9
–4
–3
–2
Imag
axis
2
RULE 6
1
–1
0
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/19
Example 4: Plotting a Root Locus
After examining RULE 1 up to
RULE 6, now there is enough
information to draw the root
locus plot.
3
2
–3
90
–9
–4
–3
Imag
axis
–2
–1
1
0
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/20
Example 4: Plotting a Root Locus
The final sketch,
with direction of
root movements as
K increases from 0
to ∞ can be shown
as:
3
Imag
axis
Final Result
2
1
–9
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/21
Conclusion: Example 3 and Example 4
s 1
1 K 2
0
s ( s  4)
s 1
1 K 2
0
s ( s  9)
 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.
Erwin Sitompul
Feedback Control System
7/22
The Effect of Adding Poles to a System
1.5
0.1
0.05
5
0
Imag Axis
0.5
Imag Axis
Imag Axis
α = –2
1
0
-0.5
-0.05
-3
-2
Real Axis
L( s ) 
-1
1
( s  1)
0
0
-5
-1
-0.1
α = –3
-1.5
-3
-2
L( s ) 
Real Axis
-1
1
( s  1)( s  3)
0
-10
-5
Real Axis
0
1
L( s ) 
( s  1)( s  3)( s  5)
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).
Erwin Sitompul
Feedback Control System
7/23
The Effect of Adding Zeros to a System
L( s ) 
Imag Axis
5
1
( s  1)( s  3)( s  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).
0
-5
-10
-5
Real Axis
0
10
α = –1.5
8
6
8
α = –2.5
6
4
2
0
-2
2
Imag Axis
Imag Axis
4
Imag Axis
α = –3.5
6
4
0
-2
2
0
-2
-4
-4
-4
-6
-6
-8
-6
-8
-10
-6
-5
-4
-3
Real Axis
-2
-1
0
-5
-4.5
-4
-3.5
-3
-2.5
-2
Real Axis
-1.5
-1
-0.5
0
-5
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
( s  6)
( s  4)
( s  2)
L( s ) 
L( s ) 
L( s ) 
( s  1)( s  3)( s  5)
( s  1)( s  3)( s  5)
( s  1)( s  3)( s  5)
Erwin Sitompul
Feedback Control System
7/24
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.
Erwin Sitompul
Feedback Control System
7/25
Design Using Dynamic Compensation
 A compensation scheme is written generally in the form of a
transfer function:
sz
D( s )  K
s p
 If z < p, it is called lead compensation.
 If z > p, it is called lag compensation.
 The characteristic equation of the system is:
1  D(s)G(s)  0
1  KL(s)  0
Erwin Sitompul
Feedback Control System
7/26
Design Using Lead Compensation
 Consider a second-order position control system with
normalized transfer function:
1
G( s) 
s( s  1)
 The root locus of the system will be compared for:
D1 (s)  K
s2
D2 ( s )  K
s  10
s2
D3 ( s )  K
s  20
Erwin Sitompul
Feedback Control System
7/27
Design Using Lead Compensation
0.5
5
15
α = –0.5
10
α = –4.5
α = –9.5
4
3
0
Imag Axis
Imag Axis
Imag Axis
2
5
0
-5
0
-1
-2
-10
-3
-4
-15
-0.5
1
-5
-1
-0.8
-0.6
-0.4
Real Axis
-0.2
0
-10
-8
-6
-4
Real Axis
-2
0
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
s2
s2
D2 ( s )  K
D3 ( s )  K
D1 (s)  K
s  10
s  20
 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.
Erwin Sitompul
Real Axis
Feedback Control System
7/28
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.
1
1
  sin   sin 0.456
(ln M p ) 2
2
(ln
0.2)
2 2
 0.208
2 
2
2
  (ln M p )   (ln 0.2)
 0.473
  0.208  0.456
tr
 



d
    
d

2
 


tr
   2.044
d
8.176
2.044

 9.187
 8.176, n 
d 
2
2
0.25
1 
1  0.456
Erwin Sitompul
Feedback Control System
7/29
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
  0.456
10
0.64
0.8
Imag Axis
5
0.94
0.5
10
0.38 0.28 0.170.088
6
4
2
0
n  9.187
For K  100,
s2,3  4.396  j8.4413
n  (4.396) 2  (8.4413)
0.94
-5
0.8
-10
-10
0.64
-8
Erwin Sitompul
2
4
6
0.5 0.38 0.28 0.170.088
100
-6
-4
-2
Real Axis
 9.517 rad s
 
4.396
 0.462
9.517
Design requirement
fulfilled with:
D2 ( s)  100
s2
s  10
Feedback Control System
7/30
Second Design Using Lead Compensation
 The closed-loop system of G(s) = 1/[s(s+1)] is now required
to have a pole at r0  3.5  j3.5 3 (corresponds to ωn = 7
and ζ = 0.5).
sz
Now using D3 ( s )  K
, the value of K and z must be
s  20
determined.
1
s = –z
Erwin Sitompul
• 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.
Feedback Control System
7/31
Second Design Using Lead Compensation
1  tan1 (3.5 3 16.5)
 20.174
1
1
2 3   tan1 (3.5 3 2.5)
2
 112.411
3  tan1 (3.5 3 3.5)
 120
   
i
i
1
 1  2  3  180  360(l 1)
1  20.174 112.411  120  180  360(l  1)
1  252.585  180  360(l  1)  1  72.585
Erwin Sitompul
Feedback Control System
7/32
Second Design Using Lead Compensation
r0  3.5  j3.5 3
The position of the zero can now be calculated:
tan 72.585  3.5 3 (3.5  ( z))  3.188
3.5 3  3.5  3.188
 5.402
z
3.188
Solving for K using the characteristic equation,
1
1  D3 (s)G(s)  0
s = –z
1  72.585
s  5.402 1
1 K
s  20 s( s  1) s 3.5 j 3.5
0
3
1.902  j3.5 3
1
1 K
0
16.5  j3.5 3 (3.5  j3.5 3)(2.5  j3.5 3)
Erwin Sitompul
Feedback Control System
7/33
Second Design Using Lead Compensation
1 K
1.902  j3.5 3
1
16.5  j3.5 3 3.5  j3.5 3 (2.5  j3.5 3)
0
6.3536
1
1 K
0
17.5784 7  6.5574
K  126.996  127
r0
Thus, the compensation that
will make r0  3.5  j3.5 3
to be on the root locus is:
s  5.402
D3 ( s )  127
s  20
Erwin Sitompul
r0
Feedback Control System
7/34
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:
sz
,
z p
D( s )  K
s p
 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.
Erwin Sitompul
Feedback Control System
7/35
Design Using Lag Compensation
 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 r0  3.5  j3.5 3
(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:
s  5.402
1

K v  lim s  L  s   lim s 127 
s 0
s 0
s  20 s( s  1)
 34.30
• Suppose not big enough
• Required Kv = 100
Erwin Sitompul
Feedback Control System
7/36
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:
L(s)  Dlag (s)  Dlead (s)  G(s)
s  0.03
s  5.402
1

127 


s  0.01
s  20 s( s  1)
• Remark: The design using lag compensation is
performed after adjusting the gain K and performing the
design using lead compensation(s)
Erwin Sitompul
Feedback Control System
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Design Using Lag Compensation
 The root locus of the new L(s) is plotted below.
Dominant root,
hardly affected
The root locus near the
lag compensation
 The transient response corresponding to the lagcompensation 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.
Erwin Sitompul
Feedback Control System
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Example: Compensation Design
Problem 5.24 FPE
Let
G( s) 
sa
1
and D( s )  K
sb
( s  2)( s  3)
Using root-locus techniques, find values for the parameters a,
b, and K of the compensation D(s) that will produce closedloop poles at s =–1 ± j for the system shown below.
Unity Feedback System
Erwin Sitompul
Feedback Control System
7/39
Example: Compensation Design
Before compensation
1
L( s )  G ( s ) 
( s  2)( s  3)
3
Imag
axis
desired closedloop poles
2
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/40
Example: Compensation Design
3
Imag
axis
desired closedloop poles
2
The zero of D(s) cancels
the pole of G(s)
–4
1
–3
–2
–1
0
–2
Erwin Sitompul
2
–1
: roots of the
compensation
s3
D( s )  K
s
 L( s)  D( s)G( s)  K
1
Real
axis
–3
The pole of D(s) is
placed in such a way
that the desired closed
loop poles are on the
future root locus
1
s( s  2)
Feedback Control System
7/41
Example: Compensation Design
1
L( s )  K
s( s  2)
3
Imag
axis
2
1
–4
: roots of the
compensation
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/42
Example: Compensation Design
1
L( s )  K
s( s  2)
3
Imag
axis
2
K?
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
7/43
Example: Compensation Design
Solving for K using the characteristic equation,
1  L(s)  0
1
1 K
0
s( s  2)
K  s(s  2)
K  s(s  2) s1 j  (1  j)(1  j)  2
The compensation D(s) can now completely be written as:
s3
D( s )  2
s
Erwin Sitompul
Feedback Control System
7/44
Homework 7
 No.1, FPE (5th Ed.), 5.23.
 No.2, FPE (5th Ed.), 5.30.
 No.3
Consider the unity feedback system, with:
K
G( s) 
( s  3)( s  5)
(a) Show that the system cannot operate with a 2%-settlingtime 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.
Erwin Sitompul
Feedback Control System
7/45