Chapter 1 Linear System Design

1.4 LAG-LEAD COMPENSATOR 63

Acompensator having the characteristics of lag-lead network is called lag-lead

compensator. In a lag-lead network when sinusoidal signal is applied, both phaselag and
phase leadl occurs inthe output, but in different frequency regions. Phase lag occursin the
Iow frequency region and phase lead occurs in the high frequency region (i.e) the phase
ongle varies from lag to lead as the frequency is increased from zero to infinity.
Aleadcompensator basically increases bandwidth and speeds up the response and
decreases the maximumovershoot in the step response. Lag compensation increases the
low frequency gain and thus improves the steady state accuracy of thesystem, but reduces
the speed of responses due to reduced bandwidth.
Ifimprovements in both transient and steady state response are desired, then both a
lead compensator and lag compensator may be used simultaneously, rather than introducing
both a lead and lag compensator as separate elements. However it is economical to use a
single lag-lead compensator.
Alag-lead compensation combines the advantages of lag and lead compensations.
Lag-lead compensator possess two poles and two zeros and so such a compensation
increases the order of the system by two, unless cancellation of poles and zeros occurs in
the compensated system. JO
S-PLANE REPRESENTATION OF LAG-LEAD COMPENSATOR

The s-plane representation of lag-lead compensator is

shown in fig 1.15. The lag section has one real pole and one real T, T, T PT,
zero with pole to the right of zero. The lead section also has one Fig 1.15 :Pole-zero
real pole and one real zero but the zero is to the right of thepole. plot of lag-lead
compensator
Transfer function of
G,9)=
(s+1/T) (s+1/T,) ..1.36)
lag - lead compensator (s+1/$T) (s+1/aT,)
lag section lead section
where ß>1and 0 <a<1
REALISATION OF LAG-LEAD COMPENSATOR USING ELECTRICAL NETWORK
The lag-lead compensator can be realised by the R-C network shown in fig 1.16.
Let E(s) =Input voltage
and E,($) =Output voltage.
In the network shown in fig 1.16, the input voltage is applied to the series combination
ofR, C, R,and C, The output voltage, is obtained across series combination ofR, and
C, By voltage division rule,
 Advanced Control Theory
Chapter 1 Linear System Design
equations (1.37) and (1.38) we get,
R
c o m p a r i n g
65
64 On
1
R, +
sC, T, - R, C, .....(1.39)
Eo(s) = E,(s) 1/sC, R
E,(s) T, = R C,
(R.I+R*
sC, sC. T,T,
.....(1.40)
1/sC, R,R, C,C, = aß .....(1.4l)
1 1
1
sR,C, +1l Fig 1.16 : +
R,Cz R,C, BT, aT, ....(142)
sR,C, +1 Electrical
lag-lead compensato, RC1
E(s) sC
sR,C, +1 aB= T,T,
E,(s) RÊ (1.41) we get, .....(1.43)
R, sR,C, +1 sC Erom equ RR,C,C,
sCI sR,C, +1
1 sC,
R : (1.39), (1.40) and (1.43), wve can say that, aß = 1 -..(1.44)
sC
equations
From
sR,C, +1 (1.44)implies that a single lag-lead network doesnot allow an independent
(sR,C, + 1) (sR,C, + 1) The equ
o f aand
B. (But separate lag and lead netvwork will allowindependent choice of a
sRCa +(sR,C, +1) SR,C; +1) sR,C, +(sR,C, +1) (sR,C3 +1) choice
Hencein the transfer function ofelectrical lag-lead compensator replace a by I/B
(sR,C, + 1) sC, and B).
as shown below.
1 1
(s+1/T,) (s+1/T,)
RC,R,C;(s Rcs+ RC 1 Eo(5) G() (s+1/BT;) (s +B/T,)
.....(1.45)
1 E;(s)
SR,C, + R,C,R,Cz (st pC)(s+ R¿C,
Where B>1, T, = RC T,=&C, and g.c
On dividing the numerator and denonminator by R,C, RC, we get, COMPENSATOR
EREQUENCY RESPONSE OF LAG-LEAD
1
(s + Consider the transfer function of
lag-lead compensator.
E,(s) RC,
1
R,c,
E,(s) S
+(s+ Gs* R,C, (s+ 1/T) (s+1/ T,) (1+sT,) (1+sT, ) .....(1.46)
R,C R,C, aß
G.(s)
(s+ 1/ BT;) (s+1/aT, ) (1+ sß+,) (1+ saT,)
(s +
R,C, (st Rc, function of lag-lead compensator
is obtained by letting
1 1 *....(1.37) The sinusoidal transfer
s +s| + +
s=jo in equ (1.46).
RC, R,C
(1+ joT) (1+ jaoT,) .....(147)
The transfer function of lag-lead compensator is given by ..G(jo) =aß (1+ joBT;) (1+ joaT,)
equation(1.47) we can say
aß = 1. Hence from
G(s)= (s+1/T) (s+1 /T,) For a singlelag-lead
compensator,
(s+ 1/BT,) (s + 1/aT,) provides a dc gain of unity.
that the lag-lead compensator
(s +1/T, ) (s + 1 /T,) oF TECHNO (148)
(1+ joT ) (1 + joT,)
s² +s ((1 /BT,) + (1/aT, )) +1 / aßT, T, ..(1.38) ..G.(jo)= (1+ joBT,) (1+ joaT, y 63242
iCR, No
Pric Rs... .
 Thecn
has four corner
frequen
Contro/
Adi anced
shown in
cqu(l.48)
Fromthe hode plot, Chapter 1 Linear System
The
sinusoidal
transfer
function
o,, 0,<9,: Step-5: determine the new
frequency correspondinggto aphase gain crossover
Design
66
o.,. where
andthe, are ,.
,.. o,, and margin of frequency. which is the 67
and o = al; Lel, gcn New gain crossover
Here , and gen Phase of G((o) at o, frequency
=
gcn
the bode plot of 180°+ cn (or) c
similar to that of lag
compensator
lag-l ,-180°
B an anal sis shown in fig
1.17.
In the phase plot of uncompensated
is sketched as
compensator
phase of c is the new gain crossoversystem, the frequency corresponding to a
frequency o 2cn'
20-log(1B) Choose the gain crossover frequency
+20 db/dec of the lag
-b
-20db dec
greaterthan (i.e., choose O,, such that
o compensator, g sormewhat
in db
Step-6: Calculate ßof lag compensator.
Let,A|Gjo)|indb at o =ogCA
0°
Fig 1.17 :Bode plot of From the bode plot find Agc
lag-lead compensator Now, A,=20 log B (or) B=10
in deg
log o
Cion-7: Determine the transter function oflag section
iBT, 1T, 1/aT,
The zero of the lag compensator is
COMPENSATOR USING BODE PI 0 placed at a frequency one-tenth of o_
PROCEDURE FOR DESIGN OF LAG-LEAD
.Zero of lag compensator, z., =1/T, =0,/10
error constant and a l
The lag-lead compensator is employed only when a large
and design a
bandwidth are required. First design a lag section and then take a= 1/ß Now, T, = 10/0,c
section The step bystep procedure for the design oflag-lead compensator is given below Pole of lagcompensator, P= 1/PT,
Step-l:Determine the open loop gain K of the uncompensated system to satisfy t
Transfer function
specified error requirement. G() = (s+1/T) =B (1+sT)
of lag section (s+1/BT) (l+sßT,)
Step-2: Draw the bode plot of uncompensated system.
Step-3: From the bode plot determine the gain margin of the uncompensated system. Step-8: Determine the transfer function of lead section
Let ¢ =Phase of G(ja) at gain crossover frequency. Take, a = 1/ß
and y =Phase margin of uncompensated system.
Now, y =180°+ ¢ From the bode plot find o, which is the frequency at which the db gain is
If the gain margin is not satisfactory then compensation is required. -20log( 1/ a)
Step-4:Choose a new phase margin 1
Now I,
Let, y, =Desired phase margin m
va
Now, new phase margin, y,=Y, te
Choose an initial value of e = 50. Transfer function (l+sT,)
of lead section
G,(9) = (s+1/T)
(s+1/aT,) (l+ saT,)
 Advanced Control Theory
compensator. Chapter 1 Linear System Design
transfer function
of lag-lead <the poles and zeros of open loop
Determine the Slep-2: Mark transfer function and theedominant pole on 69
Step-9: G, (s)
68 compensator, G, (S) = G,(s) x the s-plane. Let the dominant pole be point P.
Transfer function of lag- lead
Step-3: Find the angle to be contributed by lead network to make the
X
point Pas apoint on
(1+sPT,) (1+ so root locus,

Let. = Angle to be contributed by lead network to make point, Pas a point on

Since a =
root locus.

(l+sT) (1+tsT,) Draw vectors from all open loop poles and zeros to point P. Measure the angle
G,(s) = contributed by the vectors. [For the procedure to find angle contribution by vectors
(1+sßT,)(1+ saT,) refer root locus in Appendix II].
of compensated system.
Step-10: Determine the open loop transfer function sum of angles sum of angles

compensator is connected in series with

G(s) as shown in foi Now, = contributed by poles contributed by zeros tnl 80°
The lag-lead
of uncompensated system)of uncompensated system,
(1+sT) (1+sT,)
Ge) where n is an odd integer, so that n180° is nearest to the difference between
(1+ sßT) (1+saT,)
angles contributed by poles and zeros.
section.
Step-4: Determine the pole and zero of the lead
Fig 1.18 : Block diagram of lag-lead compens ated system. Draw
Let point O be the origin of s-plane and point P be the dominant pole.
fig.1.19.
Open loop transfer function (1+sT) (1+tsT,) -x G(s) straight lines OP and AP such that AP is parallel to x-axis as shown in
APO [ZAPO] where the point C is on
G,(9) = Draw a line PCso as to bisect the angle
of compensated system (1+sßT,) (1 +saT,) the real axis. With line PC as reference, draw angles BPC and CPD such that
located on the real axis.
each equal to /2. Here the points B and D are
Step-11: Drawthe bode plot of compensated system and verify whether the specificati
are satisfied or not. If the specifications are not satisfied then choose anot Now the point Bis the location of thepole A
choice of asuch that, a< l/ß and repeat the steps 8 to 11. of thecompensator (-1/aT,) and the point LAOP
D is the location of the zero of the
PROCEDURE FOR DESIGN OF LAG-LEAD COMPENSATOR USING ROOT LOCUS a
compensator (-1/T,). Compute T, and
The lag-lead compensation is employed to improve both the transient and stea from the values of point D and B. LAOP
section.
state responses of asystem. First design alead section to realize the required , and Step-5:Determine the transfer function of lead 2
the dominant closed loop poles. Then determine the error constant of lead compensa
system. If it is satisfactory then only lead compensation will meet the requirement. I
eror constant has to be increased then design a lag section. The
step-by-step proceu" Transfer function)
sT B D
for the
design of lag-lead compensator is given below. G,(9) =
of lead section Fig 1.19.
Step-l : Determine the dominant pole, cT, aT;
s,
Step-6 :Determine the open loop gain, K.
The value of gain, K is
is the value of gain at s = s,.
where, = Damping ratio The open loop gain K system and by using the
determined from pole-zero
plot of lead compensated
),= Natural frequency of
oscillation, rad/sec. magnitude condition given
below.
 Advanced Control Theory

poles to s = SA Chapter 1 Linear System Design

from all
of vector lengths s=s the transfer function of
Product all zeros to Determine
70
K= of vector
lengths from Slep-10:
lag-lead compensator and compensated system. 71
Product
measredto
scale. For details of
(Note:The
length of vectors
should be
Appendix l).
magritud Transfer function of
root locus in
condition refer
lag-lead compensator G,(9)=G,(9)xG,(9) =. 1
function G() |S+
Open loop transfer Goz ($) =G; ($) x
system
of lead compensated
system. (s+1/T)(s+1/T,)
velocity error constant of leadcompensated G(s)
Step-7: Determine the (s+1/PT) (s+1/aT,)

Velocity error constant

of
Kyz = Lt s Gon(s)
lead compensated system Fig 1.20: Block diagram of lag-lead
compensated system
only lead compensation is sufficient| but
IfK, satisfies the requirement then
provide lag compensation.
K, is lessthan the desired value then The lag-lead compensator is connected in series with G(s) as shown in fig 1.20.
section.
Step-8 : Determine the parameter, ßoflag
Open loop transfer function of)
Let, K= Desired velocity error constant.
lag-lead compensated system Go(s) = 1
x G(s)
A=The factor by which Kis increased. S+
S+
Now, A=KK
Step-11: Check the velocity error constant of compensated system, if it is satisfactory
select B, such that B> A. [i.e., B=(1.1to 1.2) xA] then the design is accepted, otherwise repeat the design by modifying the
Siep-9: Detemine the transfer function of lag section. Choose the zero of lag section locations of poles and zeros of the compensator.
10% of thesecond pole of uncompensated system.
EXAMPLE 1.9
.Zero of lag section, z., =0.1 xsecond pole of G(s)
Consider the unity feedback system whose open loop transfer function is
Also, 2, =
G(s)=K/s(s+3)(s +6). Design a lag-lead compensator to meet the following
specifications. (i) Velocity error constant, K-80. (ii) Phase margin, y>35°.
SOLUTION
Pole of lag section, Pcl -1 Step-1 : Determine K
For unity feedback system,
Transfer function of lag section,
G (s) = Velocity error constant, K, = S’0
Lt sG(s)
1
S+
Given that, K=80.
 Chapter 1 Linear System Design
Advanced Control Theory
73

80 80
K = 80 At o W1, A= 20 1log = 20 log 44 db
0.5
72
= L s s(s +3) (s+ 6)
Lt
sG(s) $40 80 80
.. 1440
At o = 0cl A =20 log =20 log=28.5 db 28db
K=80 x 3 x6 =
= 80 (or) 1440
3x6
1440
+s/3)x
6(1+s/6) At o = 0c2 A =slope from o to o xlogc +A at (o =01)
.:.G(s) = s(s+ 3)(s+ 6) Sx 3(1
80
s(1+0.33s) (1 +0.167s)
6
-40 x log+28 = 16 db
3

uncompensated
system. At o =@h A=slope from oc to o, xlog+A at (o = 02)
Step-2: Bode plot
of
20
In G(s). put s =jo
=-60x log +16= -15 db
80 6

..G(jo) = (1+ j0.330) (1+ j0.1670)

jo
Let the points a, b, c and d be the points corresponding to frequencies o, o., , and
Magnitude plot
o, respectively on the magnitude plot. In asemilog graph sheet choose appropriate
The corner frequencies are
o. and . scales and fix the points a, b,c and d. Join the points by straight lines and mark the
rad/sec.
1/0.33=3 rad/sec and
o.,= 1/0.167 =6 slope on the respective region. The magnitude plot is shown in fig 1.9.2.
Here =
shows the sh
listed in table-1. Also the table
The various terms of G(jo)are
Phase Plot
corner frequency.
contributed by each term and the change
in slope at the
The phase angle of Gjo) as a function of o is given by
TABLE-1 = 2G(jo) =-90°-tan-'0.330- tan-'0.167o.
Term Corner frequency Slope Change in slope The phase angle of G(jo) are calculated for various values ofo and listed in table-2.
rad/sec db/dec db/dec
TABLE-2

80 6 10 20
-20 0.5 1.0 3.0
rad/sec
ZG(jo)
033 3 -20 -20 -20=40 deg -104 -118| -161-198 -222 -244
1+ j0.33
N-160

1+ j0.167o =6 --20 -40 -20 =-60 On the same semilog sheet take another y-axis, choose appropriate scale and
0.167
draw phase plot as shown in fig 1.9.2.
Choose a low frequency o, such that o, <o,, and choose a high Step-3 : Find phase margin of uncompensated system.
such that , > a frequen Let Phase of G(jo) at gain crossover frequency
Let w, = 0.5 rad/sec and o, = 20 and Phase margin of uncompensated system.
rad/sec.
Let A=|Gjo) in db From the bode plot of uncompensated system we get, =-226°.
Now, Y = 180o + ’ 180° -226° = -46°
Adanced Control Theony
Chapter 1 Linear System Design
phase margin From the bode plot of
Sep-4:
Choose a new
margin. y, = 35°
to a db pair of -12 db uncompensated system
is found to be 17 rad/sec.the frequency o,corresponding
75
phasc y, =Y, t¬
The desired compensatedsystem,
phase margin of .". Om=17 rad /sec
The
initial choice of e =5°
Let .:.T, =
), , + E
35° + 5° =
40°
O,va 17/007-0.22
gain croSsover
frequency
- Phase of G(jo) at oBCn Transfer function)
Step-$: Determine
newgain crossOver frequency
and
of lead section
(l+sT,) = 007(+022s)
Let om
New
_ . = y - 1 8 0 ° =
40°-180°=-140° (1+ seT,) (1+0.0154s)
Now. y, = 180° + corresponding to a phase of Sten-10: Determine the transfer function of lag-lead
found that the frequency compensator.
From the bode plot we
-140° is 1.8rad/sec.
Transfer function

Let o -Gain crossover frequency

of lagcompensator. of lag - lead compensator G,(9) =G,(5) xG,(9)
Choose o, such that, gen'
=14+2.5s)007+0225)
Let = 4 rad/sec. (1+35s) (1+0.0154s)
(1+2.5s) (1+0.22s)
Step-6 : Calculate ßoflag compensator
magnitude at o, is 23 db. (1+35s) (1+0.0154s)
From the bode plot we found that the db
=23 db. Step-11: Determine open loop transfer function of
.G(jo)|in db at (@ =@)=Ae compensated system.
B=10^1020 =14 The lag-lead compensator is connected in series with
Also, Age =20log ß ; G(s) as shown in fig 1.9.1.
(1+2.5s) (1+0.22s)
Step-7: Determine the transfer function oflag section.
80
(1+35s) (1+0.0154s) <l+033s) (1 +0.167s)
The zero of the lag compensator is placed at a frequency one-tenth ofo
Fig 1.9.1: Block diagram of lag-lead
:.Zero of lag compensator, Ze =
-1_gel compensated system.
10 Open loop transfer function)
80(1 +2.5s) (1+0.22s)
10 10 of compensated system G,() =s(1+35s) (1+0.0154s) (1+0.33s) (1 +0.167s)
Now,T, = -=*=2.5
4
Step-12: Bode plot of compensated system.
Put s =jo in G,(s)
Pole of lag compensator, Pei =
BT 14 x2.5 35 :.Go(jo)= 80(1+ j2 So)(1+ j0.220)
jo (1+ j3So) (1+ j0.01540)(1 +j0.330) (1+j0.167e)
Transfer function of
lag section
G,9-as)-4l"259) Magnitude plot
(l+ s$T,) (1+35s)
Step-8: Delermine the transfer function of lead section. There are six corner frequencies, which aregiven below.
Let a= 1/B; .a= 1/|4 = 0.07 = 0.03 rad /sec;
35 @ =0.4 rad/sec ; 0 = =3 rad/sec ;
2.5
The db gain (magnitude) 1
0.33
= 45 rad/ sec ; 0s=
correspond1ng to m =-20log =-20log Wo.07 =-115 dbs-12 db 0.22 0.167
=6 rad lsec ; ocó
00154
=65 rad/sec.
 567891
showshe s
-axisunt10db Y-axistunE
Advanced Control Theory table-3.
Alsothe table Chapter 1 unear System Design
listedin
(jo)are changeinslope atttthe corner frequency. 77
termsofG, the
The various temand
76 contributed
bycach 4

Change in slope 3

TABLE-J
Corner frequency Slope db/dec
db/dec 2

Term

80
rad/sec

-20
567891 Sate
40

jo -20 -20 =-40 067a

- 0.03 -20 4

3
35
l j350

0.4 +20
-40 +20=-20 2
usncyompsentaed
l+ j2.S0 25 Magni
plot
of tude

--3 -20 -20 -20=40 67891

0.33
1+ 0330

+20 -40 +20 =-20 5

= 4.5 4
I+ j0.220 0.22

GOad
3

1+j0.J67o 0s0167
1
=6 -20 -20-20 =-40
ec 4b/den of

1+ j0.0154o
1
0.0154
=65 -20 -40 -20 =-60 67891 20 Phase pBot of
uncompensaled sstém
plot
Bode
Choose alow frequency a, such that o, <@,, and choose a high frequencya
5 l.9.1:

2A23Idb3FH
4

such that o, >

Let o =0.01 rad/sec and o, = 80rad/sec
3
pMloagntioude E
Hr
2

Let A, =|G(jo)| in db.

80
5467891
Al o=o, A=20 log 0.01 = 78 db

80
At o : A20 log =685db 68 db
0.03 3

At o=os A,=-40 xlog 0.4

+68= 23 db
0.03
3
Pháse plot of
At o =0ds A =-20 conpensated ssten
xlog +23 =5 db 001
40
Ato= , A=40 xlog 0

+5=-2 db
 Advanced Control heony
6
+(-2) =-4
db Chapter 1 Linear System Design
log EXAMPLE 1.10
A, -20x
78 Ato @g. db 79
+(-4)= -45 Design a lag-lead
65
A,
40xlog 6
G(s) = K/s
compensator for a system with open loop transfer function
Ato -@. (s + 0.5) to satisfy the following
80
+(-45) =-$0
db dominant closed-loop poles, =0.5. (i)
specifications. (i) Damping ratio ot
-60x log 65 closed loop Undamped natural frequency of dominant
Ato -@,. A, poles, o, =5 rad/sec. (iii) Velocity
e magnitude plot of error constant, K, =80 sec.
of A, at various
Usingthe values shown in fig 1.9.2,.
frequencies the
compensa SOLUTION
SVstem is
drawn as Step-l: Determine dominant pole, s,.
Phase piot
ofo is given by
Dominant pole, s, =-Go, t jo,J1-?.
G,(j0) as afunction
The phase angle of - 90°- tan-135ao Given that, =0.5 and o, =5 rad/sec.
tan'2.So +tan'0.220
n
6.=LG,(o)= 0.33o- tan-' 0.167o.
-tan0.0154) - tan-
valuessofw and listed in
.S =-0.5 x5tjSxW1-0.5² =-2.5 +i43
calculated for various table
The phase angle of G,((o) are
Step-2: Draw thepole-zero plot
TABLE4
The pole-zero plot of open loop transfer functioD is shown fig
0.01 0.03 0.1 0.4 1 4 10 65 80 1.10.1. Poles are
represented by the symbol "x".The pole at point Pis the dominant
rad/sec pole, s,
Step-3:To find the angle to be contributed by lead network.
LG,G0)
-108-132 -152-138 -126 -144| -168| -221-228 Let ¢ = Angle to be contributed by lead network to
deg |-220 make point, P as a point on
root locus.
Using the values of ¢, listed in table-4, the phase plot of compensated system: sum of angles sum of angles
sketched as shown in fig 1.9.2. Now, = contributed by poles contributed by zeros tnl80°
Le .= Phase of G,(jo) at the gain crossover frequency of compensated'systen of uncompensated system of uncompensated system
and Y, = Phase margin of compensated system. From fig 1.10.1, we get,
From the bode plot of compensated system, we get, =-144°
Sum of angles contributed
Now.y, = 180° +.=180° 144° = 36°
by poles of uncompensated system =0, +0, = 120°+115°= 235°
CONCLUSION Since there is no finite zero in uncompensated system,
there is no angle
The phase margin of the compensated system is satisfactory. Hence the designi contribution by zeros.
acceptable.
RESULT
..= 235° t n180°
Let n= 1, .=235° - 180° = $5°
Transfer furnction of lag- lead (1+2.5s) (1+0.22s)
compensator, G,(s) = Step-4 :To find the pole and zero of the lead section
(1+35s) (1+0.0154s)
Open loop transfer function Draw a line AP parallel to x-axis as shown in fig 1.10.1. The bisector PC is
of compensated system 80 (1+2.5s) (1 +0.22s) drawn to bisect the angle APO. The angles CPD and BPC are constructed as
s(1+35s) (1+0.0154s) (1 +0.33s) (1+0.167S) shown in fig 1.10.1. Here ZCPD = ZBPC= /2 = 55°/2 =27.5° 27°.
 AdvancedC o n t r o l Theory
Chapter 1 Linear System Design
81
Erom fig 1.10.1., Pole of the lead section,p., =-10
80
Zero of the lead section, z,, =-2.65.
p t n e

Scate1em-0.5units We know that, z,, =-l/T, T,= 1/2.65 =0.377

We know that, P,, -l/aT, .T,= 1/10 (or) a= 1/(T,x 10) =0.265.
Step-5:
Transfer function of lead compensator
Transfer function of leadsection, G,(s) = (s+1/T,) (s+2.65)
(s+1/aT,) (s+10)
Slep-6: To find
gain, K

Open loop transfer function (s+ 2.65) K

Goz(s) =G($) xG(s) = X

of lead compensated system (s+10) s(s+05)

K(s+2.65)
s (s+0.5) (s+ 10)
Here the value of Kis given by the value of gain at the dominant pole s, on the
root locus. From magnitude condition Kis given by.

K=
Product of vector lengths from all poles to s= s
B Product of vector lengths from all zeros tos=SA
E From fig 1.10.1, we get,
A
K=*4*45x4.75x7.75-=42.8
4.3

180-tah20 42.8 (s+ 2.65)

s (s+0.5) (s+ 10)

Step-7: To find velocity error constant of leadcompensated system.

Let, K, =Velocity error constant of lead compensated system.
950,545 42.8 x 2.65
42.8 (s+2.65) --22,684
e8.6x543 Ky = Lt s. Gp(s)= Lt s s (s+0.5) (s +10) 05x 10
15.55775
5T75 Step-8 : To find the parameter, B
K 428
4.3 Let K,= Desired velocity error constant
A = The factor by which Kis increased
Now, A= Kvà/Ky =80/22.684=3.5267
Select B, such that B> A. Let B=4.
 Advanced Control Theory
82 Step-9: To findthe transfer function of lag section.
Let zero of lag section, z,, =0.1 xsecond pole of G(s) =0.1 «(-0.5)=

-1 1
.:.T = = 20
Also, Z¢l 0.05

=
-1 -1 -1 =-0.0125
Pole of lag section, Pci 80
BT 4x 20
(s+1/T) (s+ 0.05)
Transfer function of lag section, G(s) =
(s+1/BT) (s+0.0125)
Step-10 : Transfer function of compensated system
Transfer function of (s+0.05) (s+2.65)
G(s) = G,(s) x G,(9) =
lag -lead compensator (s+ 0.0125) (s+10)
The lag-lead compensator is connected in series with G(s) as shown in fig 1.1
(s+0.05) (s+ 2.65) 42.8
(s+0.0125) (s+10) s(s+0.5)

Fig 1.10.2 : Block diagram of lag-leadcompensated system

Open loop transfer function) 42.8 (s + 0.05) (s+ 2..65)
Go(s) s (s+0.0125) (s +0.5) (s +10)
of compensated system

Step-11 : Check velocity error constant of compensated system.

Velocity error constant 42.8 (s +0.05) (s +2.65)
Lt s
Kye = Lt s Go(s) = S’0
of compensated system) s(s+0.0125) (s +0.5) (s+10)
42.8x 0.05 x2.65 =90.7
0.0125x 0.5x 10
CONCLUSION
The velocity error constant of the compensated system satisfies t
requirement. Hence the design is accepted.
RESULT

Transfer function of lag-lead compensator, G:(s) = (s+0.05) (s+2.65)

(s+0.0125) (s+ 10)
Open loop transfer function of 42.8 (s+ 0.05) (s +2.65)
Go(s) =
lag -lead compensatedsystem s (s+0.0125) (s+0.5) (s+ 10)