Time and Frequency-Domain Analysis and Design of Phase-Lead and
Phase-Lag controllers (Compensators)
In the previous lectures we looked at PID controllers. PID controllers are the simplest
form of controllers that make use of the derivative and integration operations in control
systems compensation.
Root locations and root locus are typically used in the design of controllers, it would
make a perfect sense to describe a compensator in terms of the pole and zero locations.
A simple passive realizable controller is of the form:
Gc ( s) 
sz
s p
This controller is easily realized by passive elements (resistors and a capacitor)
The controller if of the high-pass filter type (phase –lead) if p > z and is of the lowpass filter type (phase –lag) if p<z Think about this – make sure it makes sense !
Phase-Lead Controller
The network realization of a phase-lead controller is shown below
The compensator’s transfer function is:
Eout ( s )
R2
 Gc ( s ) 
Ein ( s )
R1  R2
1  R1Cs
R R
1  1 2 Cs
R1  R2
Let
a
R1  R2
R2
a 1
And
T
R1R2
C
R`  R2
Then, the transfer function of the compensator becomes:
Gc ( s ) 
1  1  aTs 


a  1  Ts 
Gc ( s ) 
s 1
aT
s 1
T
a 1
(1)
a 1
(2)
Note:
The transfer function of the phase – lead controller network has a real zero at
s  1
aT
s-plane is:
and a real pole at
s  1
T
and the representation in the
S-Plane
pole
zero
0
1
T
1
aT
The zero is always to the right of the pole. This is why the phase –lead
controller can improve the relative and absolute stability of a closed loop
control system.
Phase – Lead:
Stabilize the system
Increase the PM – thus reduce the overshoot
Increase the gain crossover frequency – reduce the rise time.
Note: we will work with equation (1) – ignore the 1/a term – it is simply
gain and that is taken care of when calculating K for the system to be
compensated.
- For the compensator in (1):
o The gain at high frequencies is approximately = 20log(a)
o The gain at low frequencies is = 0dB
o The lower cut-off frequency = 1/aT
o The upper cut-off frequency = 1/T
o The maximum phase shift of the phase lead is:
a  1

 a  1
 max  sin 1
The corresponding frequency is:
max 
1
T a
The gain at this frequency is:
1
Gain  adB  10 log a
2
The above realities are used in the design.
o The Bode plot of a typical Phase –lead controller is:
Let the phase lead controller be:
Gc ( s) 
1  5s
1 s
Bode Diagram
15
Magnitude (dB)
(discuss the result)
10
5
Phase (deg)
0
60
30
0
-2
10
10
-1
0
10
Frequency (rad/sec)
10
1
10
2
Let us calculate some of the values discussed above:
 5 1
o
  42
 5  1
 max  sin 1 
max 
1
 0.45 rad / sec
1 5
Design Steps:
1 – Choose the gain, K, to satisfy the steady state error requirement
2- Draw the Bode diagram of KG(s)
3- Determine the new crossover frequency (wg) – the frequency at
which the system has the desired phase margin.
4- Construct a table of phase, a, and Gain (for the compensator).
In some cases, you may find out that one lead compensator may
not be enough.
5- calculate T from the chosen a and the new crossover frequency.
6- implement the compensator design and test the final result.
7- if o.k – done, otherwise, repeat the process.
Example:
For the sun seeker control system . The seeker is typically mounted on
space vehicles. The purpose of the controller is that the seeker tracks the
sum with high accuracy.
The input is the reference angle of the solar ray  r and the output is the
angle of the vehicle’s axis o
After some block diagram manipulations, the simplified block diagram is
shown below:
The design specifications are:
- the steady state value of e(t) due to a unit ramp input for  r (t) should be
less than or equal to 0.01 rad /(rad/sec.) of the final steady state output
velocity. [ The steady state error due to a unit ramp input should be less
than or equal to 1% ess  0.01 due to a unit ramp input]
- The peak overshoot should be less than 10%
Solution:
o
G

r 1  G
E= R – Y =
E
RR
G
R
1
1

 2
1  G 1  G s 1  2500 K
s ( s  25)
s ( s  25)
s 2 s ( s  25)  2500 K 
Applying the final value theorem:
lim ess (t )  lim sE ( s) 
t 
s 0
25
0.01

2500 K
K
Therefore, for a steady state error to be less than or equal to 1%, K
must be greater than or equal to 1.
For the worst case, when K =1,
The characteristic polynomial is:
s 2  25s  2500  0
Which yields the following information:
n  2500,

  50 rad / sec.
25
 0.25
100
Thus the predicted % overshoot is:

Max overshoot =
e
1 2
 100
Let us simulate using Matlab – to verify:
= 44.43 %
Step Response
1.5
System: Gcl
Peak amplitude: 1.44
Overshoot (%): 44.4
At time (sec): 0.0642
Amplitude
1
System: Gcl
Rise Time (sec): 0.0254
0.5
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.3
0.35
0.4
0.45
Time (sec)
Step Response
1.5
Amplitude
1
0.5
0
0
0.05
0.1
0.15
0.2
0.25
Time (sec)
Let us plot the %overshoot vs. zeta.
clear all;clc
Z=[];
OS=[];
for z=0:.01:0.9999
os=100*exp((-z*pi)/(sqrt(1-z^2)));
Z=[Z z];
OS=[OS os];
end
plot(Z,OS)
grid
xlabel('zeta')
ylabel('% overshoot')
title('')
100
90
80
% overshoot
70
60
50
40
30
20
10
0
P.M =
0
0.1
0.2
0.3
0.4
0.5
zeta
0.6
1 

2



1

 
1
tan 2 
1
 

4
2
2
  4  1  2  




0.7
0.8
0.9
1
Let us plot this relationship of Phase margin vs. damping ratio:
clear all
Z=[];
PM=[];
for z=0:.01:1
pm=atand(2*z*(1/((4*z^4+1)^0.5-2*z^2))^0.5);
Z=[Z z];
PM=[PM pm];
end
plot(Z,PM)
grid
xlabel('zeta')
ylabel('PM - degrees')
title('Phase lead design')
Phase lead design
80
70
PM - degrees
60
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
zeta
0.6
0.7
Now. Let us plot the phase margin vs. % overshoot
clear all;clc
Z=[];
OS=[];
PM=[];
for z=0:.01:0.9999
os=100*exp((-z*pi)/(sqrt(1-z^2)));
pm=atand(2*z*(1/((4*z^4+1)^0.5-2*z^2))^0.5);
Z=[Z z];
0.8
0.9
1
OS=[OS os];
PM=[PM pm];
end
plot(OS,PM)
grid
xlabel('%overshoot')
ylabel('% Phase Margin')
title('')
80
70
% Phase Margin
60
50
40
30
20
10
0
0
10
20
30
40
60
50
%overshoot
70
80
90
100
From the plot, a 10% overshoot corresponds to a phase margin of 58o
Let us draw the Bode diagram of the uncompensated system.
clear all;close;clc
K=1;
num=K*[2500];
den=[1 25 0];
G=tf(num,den);
bode(G)
Bode Diagram
60
Magnitude (dB)
40
20
0
-20
-40
Phase (deg)
-60
-90
System: Guncomp
Phase Margin (deg): 28
Delay Margin (sec): 0.0104
At frequency (rad/s): 47
Closed loop stable? Yes
-135
-180
0
10
1
2
10
10
3
10
Frequency (rad/s)
As can be seen, the phase margin for the open loop system is 28 degrees at
the wp = 47 rad/sec.. (Check the plot of % OS vs. phase margin – it agrees
with our predictions. )
We want the phase margin to be at 60 degrees (according to %os vs. PM).
(Add a phase lead) Thus the phase lead controller should add :
60-28 = 32 degrees of phase in the vicinity of the gain –crossover frequency.
But, when adding the phase lead controller, the magnitude response will be
affected in such a way that the gain crossover frequency will shift higher.
Thus, when designing the phase lead, allow for some marginal error to
account for the inevitable phase drop off.
a  1

 a  1
 max  sin 1
a
1  sin 32
 3.25
1  sin 32
Since 20log(a) = 10.25dB (the phase lead will cause a 10.25 shift in the
magnitude) Thus the new gain crossover frequency will shift.
The problem is now; we want the amount of phase shift correction to occur
at the new gain crossover frequency.
Choose the geometric mean to locate the new crossover frequency.
At the attenuation of -10.25/2 = -5.125dB, the corresponding frequency is:
Bode Diagram
60
40
Magnitude (dB)
20
System: Guncomp
Frequency (rad/s): 64.6
Magnitude (dB): -5.05
0
-20
-40
-60
-90
-120
Phase (deg)
System: Guncomp
Phase Margin (deg): 28
Delay Margin (sec): 0.0104
At frequency (rad/s): 47
Closed loop stable? Yes
-150
-180
0
1
10
2
10
10
Frequency (rad/s)
65 rad/sec.
Now, we can calculate T
T= 1
 0.0085
w
a
m

Substituting, we get:
Gc ( s) 
1 1  aTs
a 1  Ts
The open loop transfer function of the compensated system is:
G( s) 
1 2500(1  aTs )
a s( s  25)(1  Ts )
=
1
2500(1  0.0277 s)
3.25 s( s  25)(1  0.0085s)
3
10
The Frequency response of the compensated system is:
Bode Diagram
40
Magnitude (dB)
20
0
-20
-40
-60
-80
Phase (deg)
-100
-90
-120
-150
-180
0
10
System: Gcom
Phase Margin (deg): 67.3
Delay Margin (sec): 0.0457
At frequency (rad/sec): 25.7
Closed Loop Stable? Yes
10
1
2
10
Frequency (rad/sec)
10
3
10
4
Plots of uncompensated and compensated are:
Bode Diagram
Magnitude (dB)
50
0
-50
Phase (deg)
-100
-90
-120
-150
-180
0
10
System: Gcom
Phase Margin (deg): 67.3
Delay Margin (sec): 0.0457
At frequency (rad/sec): 25.7
Closed Loop Stable? Yes
10
1
System: Gunc
Phase Margin (deg): 28
Delay Margin (sec): 0.0104
At frequency (rad/sec): 47
Closed Loop Stable? Yes
2
10
Frequency (rad/sec)
10
3
10
4
a=3.25;
T=0.0085;
K=1;
den1=[1 0];
den2=[1 25];
den3=[T 1];
num1=[K*2500];
num2=(1/a)*[T*a 1];
NUM=conv(num1,num2);
DEN=conv(den1,conv(den2,den3));
Gcom=tf(NUM,DEN)
bode(Gcom)
% GCL=feedback(Gcom,1);
% step(GCL)
hold
DENunc=conv(den1,den2);
Gunc=tf(num1,DENunc)
bode(Gunc)
Let us look at the transient response:
Step Response
1.5
System: GCLunc
Peak amplitude: 1.44
Overshoot (%): 44.4
At time (sec): 0.0642
System: GCL
Peak amplitude: 1.05
Overshoot (%): 4.81
At time (sec): 0.122
System: 1GCLunc
Rise Time (sec): 0.0254
Amplitude
System: GCL
Rise Time (sec): 0.0561
0.5
0
0
0.05
0.1
0.15
0.2
0.25
Time (sec)
clear all;clf;close;
a=3.25;
T=0.0085;
0.3
0.35
0.4
0.45
K=1;
den1=[1 0];
den2=[1 25];
den3=[T 1];
num1=[K*2500];
num2=(1/a)*[T*a 1];
NUM=conv(num1,num2);
DEN=conv(den1,conv(den2,den3));
Gcom=tf(NUM,DEN)
% bode(Gcom)
GCL=feedback(Gcom,1);
step(GCL)
hold
DENunc=conv(den1,den2);
Gunc=tf(num1,DENunc)
% bode(Gunc)
GCLunc=feedback(Gunc,1);
step(GCLunc)
Phase-Lag Controller
The network realization of a phase-lag controller is shown below
R1
R2
Ein
Eout
C1
The compensator’s transfer function is:
Eout ( s)
1  R 2 C1 s
 Gc ( s) 
Ein ( s)
1  ( R1  R2 )C1 s
Let
a
R1  R2
R2
a 1
And
T  R2 C1
Then, the compensator’s transfer function is:
𝐺𝑐 (𝑠) =
1 + 𝑇𝑠
1 + 𝑎𝑇𝑠
Note, aT > T
Observations:
1- The complex plane will have a zero at s = -1/T and a pole at s = -1/aT
2- Thus, the pole is to the right of the zero (opposite to the phase lead
compensator.)
3- For the phase lag compensator, the output lags the input in phase
angle for all positive frequencies.
4- Also, the magnitude of the output decreases from unity at near 0
rad/sec. to 1/a as   ∞
Demonstrate the phase lag network’s frequency response for the case where
a = 5, T = 2.
Bode Diagram
0
Magnitude (dB)
-2
-4
-6
-8
-10
-12
-14
0
-5
Phase (deg)
-10
-15
-20
-25
-30
-35
-40
-45
-3
10
10
-2
-1
10
Frequency (rad/sec)
10
0
Notice the phase lag and magnitude attenuation.
The lower cutoff frequency is: 1/aT = 0.1 rad/sec.
The upper cutoff frequency is 1/T =0.5 rad/sec.
Is it obvious why the magnitude plot levels off after the upper cutoff
frequency?
10
1
The frequency at which the maximum phase lag occurs, ωm, is found by:
𝜔𝑚 = 1⁄
𝑇 √𝑎
For this example, ωm = 0.2236 rad/sec.
The maximum phase lag is given by:
1−𝑎
)
1+𝑎
For our example, the maximum phase lag that may be obtained is: 41.8
degrees ( -41.8)
𝜃𝑚𝑎𝑥 = 𝑠𝑖𝑛−1 (
Design steps with a phase lag controller:
These steps are demonstrated using an example.
The objective of the phase lag controller in the frequency domain is to move
the gain crossover frequency where the desired phase margin is realized,
while keeping the phase curve relatively unchanged at the new gain
crossover frequency.
Let us refer to the sun-seeker example used in the phase lead design.
The requirements are as follows:
1- the steady state error due to a unit ramp input should be less that 0.01
2- the maximum overshoot of the step response should be less than 5%
3- thr rise time should be less than or equal to 0.5 sec.
4- the settling time should be less than or equal to 0.5 sec.
5- due to noise, the bandwidth of the system should be less than 50
rad/sec.
notice, the rise time and settling time requirements have been relaxed.
The open loop compensated system is:
𝐺𝑓𝑤𝑑 = 𝐺𝑐 𝐺 =
2500𝐾(1 + 𝑇𝑠)
𝑠(𝑠 + 25)(1 + 𝑎𝑇𝑠)
Solution:
1- as was done earlier with phase lead, let K =1 (for worst case)
2- Then the open loop uncompensated system will have a %overshoot of
44.4% and a damping ratio of 0.25.
3- For effective phase –lag control, the pole and zero of the controller
function must be placed close together and both are relatively close to
the origin.
4- Select a using the equation:
𝒂=
𝐾 𝑡𝑜 𝑟𝑒𝑎𝑙𝑖𝑧𝑒 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑝𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒
𝐾 𝑡𝑜 𝑟𝑒𝑎𝑙𝑖𝑧𝑒 𝑡ℎ𝑒 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑑𝑎𝑚𝑝𝑖𝑛𝑔
to do this:
- draw the root locus of the uncompensated system and determine K to
realize the desired damping. Note, from the plot of % overshoot vs.
zeta, we obtain a damping ratio of 0.7071 to correspond to
% overshoot of 4.32%
Root Locus
15
System: Gol
Gain: 0.125
Pole: -12.5 + 12.5i
Damping: 0.707
Overshoot (%): 4.32
Frequency (rad/sec): 17.7
10
Imaginary Axis
5
0
-5
-10
-15
-30
-25
-20
-15
-10
-5
0
Real Axis
Therefore,
a = 1/0.125 = 8.
5- if T is sufficiently large, then the pole and zero will be sufficiently
close to the origin
6- Let us demonstrate two case of choices of T
a. T = 100
b. T = 10
The corresponding Step response of each is shown below:
5
a) T =100
compensated system- Phase lag: T =100, a =8
1.4
System: GCL
Peak amplitude: 1.04
Overshoot (%): 4.39
At time (sec): 0.254
1.2
System: GCL
Settling Time (sec): 0.34
Amplitude
1
System: GCL
Rise Time (sec): 0.122
0.8
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25
Time (sec)
0.3
0.35
0.4
0.45
0.5
Red= uncompensated, Blue = Compensated
40
20
0
-20
-40
-60
-80
0
10
1
10
2
10
and the complete frequency response od the compensated and
uncompensated system is:
3
10
Red= uncompensated, Blue = Compensated
40
magnitude dB
20
0
-20
-40
-60
-80
0
10
1
2
10
10
3
10
Frequency rad/sec.
Red= uncompensated, Blue = Compensated
-80
phae - degrees
-100
-120
-140
-160
-180
0
10
1
2
10
10
Frequency rad/sec.
pole(GCL)
ans =
-12.4956 +12.4956i
-12.4956 -12.4956i
-0.0100
The compensated system has a phase margin of 65 at 11.4 rad/sec.
3
10
b- T = 10;
Step Response
1.4
System: GCL
Peak amplitude: 1.05
Overshoot (%): 5.04
At time (sec): 0.255
1.2
1
System: GCL
Settling Time (sec): 0.362
System: GCL
Rise Time (sec): 0.121
Amplitude
0.8
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Time (sec)
Red= uncompensated, Blue = Compensated
40
magnitude dB
20
0
-20
-40
-60
-80
0
10
1
2
10
10
3
10
Frequency rad/sec.
Red= uncompensated, Blue = Compensated
-80
phae - degrees
-100
-120
-140
-160
-180
0
10
1
2
10
10
Frequency rad/sec.
pole(GCL)
3
10
ans =
-12.4559 +12.4561i
-12.4559 -12.4561i
-0.1007
The compensated system has a phase margin of 65.1 at 11.4 rad/sec.
notice – barely any change
Notice: The response is basically insensitive to T, therefore, pick a large T
and vary a to obtain desired response. No matter what a is (as long as a>1),
the steady state error is NOT violated.
Fun
Let us try a phase –lead –lag controller
Using results above, the following phase –lead- lag network is obtained.
𝑠 + 1⁄𝑎 𝑇
𝑠 + 1/𝑇2 1
1 1
𝐺𝑐 (𝑠) = (
)(
)
𝑠 + 1/𝑇1
𝑠 + 1⁄𝑎 𝑇 𝑎
2 2
discuss with class
let a1=70, T1 = 0.0004, a2= 5, T2 = 20
a=3.25;
T=0.0085;
The resulting system’s response is:
Step Response
1.4
System: GCL
Peak amplitude: 1.05
Overshoot (%): 5.07
At time (sec): 0.156
1.2
System: GCL
Time (sec): 0.216
Amplitude: 1.02
Amplitude
1
System: GCL
Time (sec): 0.0876
Amplitude: 0.91
0.8
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
Time (sec)
controller parameters are:
0.25
0.3
0.35
T1 = 0.00085, a1= 20
T2 = 100, a2 = 2;
a2=5;
numc1=[1 1/(a1*T1)];
denc1=[1 1/T1];
Gc1=tf(numc1,denc1);
numc2=[1 1/T2];
denc2=[1 1/(a2*T2)]*1/a2;
Gc2=tf(numc2,denc2);
Gc = Gc1*Gc2