Collections of my colleague's lectures with respect to all

1. Assist. Prof. Dr. Khalaf S. Gaeid
Electrical Engineering Department
Tikrit University
gaeidkhalaf@gmail.com
+9647703057076
Lead , Lag Compensator in
Frequency Domain

2. 1.Introduction
2.Phase Lead or Lag Compensator
3.Digital Controller Design
4.Calculations of Frequency Response in w-domain
5.Design Approach
6.Design specific to Phase lag and lead Compensator
7.Examples
Contents

3. Introduction

8. A discrete time phase lead/lag compensator is designed by
using the bilinear transformation that maps the inside of the
unit circle in z-plane to the entire LHP of w-plane.
The phase lead/lag compensator is designed in w-domain
utilizing the techniques available for continuous time
compensator design, then map it back to the z -domain by the
bilinear transformation. The first order w-domain
compensator is given by analogy to s -domain as
2.Phase Lead or Lag Compensator

9. A pole is at w=−ωwp and a zero is at w=−ωw0

11. • Phase lead compensator: ωwp > ωw0
• Phase lag compensator: ω w0 > ω wp

13. 3.Digital Controller Design
Frequency Response Method
Design a phase lead or phase lag compensator
D(z) = Kd(z − z0)/(z − zp)
such that the (open loop) system satisfies a given phase
margin of
φm ≈(100 ξ ) so that the (closed loop) system has damping
ratio of ξ.

14. Since the frequency response method based on phase
margin and gain margin established in the Laplace domain (s-
domain) is used here, the frequency response of D(z)
combined with a plant transfer function G(z) with ZOH must
be transformed into w-domain by using the bilinear
transformationBecause, the stable region of the z-plane that is the inside of
a unit circle, is mapped into the entire LHP (Left Half Plane) of
the w-plane.
The parameters Kd, z0 and zp of the first order phase lead/lag
compensator are determined in terms of a0, ωw0 , ωwp defined
in the w-domain

15. 4.Calculations of Frequency Response in w-domain
In order to obtain frequency response (Bode plot) in w-
domain, do the following steps.
1. Find the transfer function in the z-transform,
G(z) = (1 − z−1)Z[G(s)/s]
2. Calculate the frequency response G(ejωT ) for 0 ≤ wT ≤π
with a proper step size. Record the actual frequency scale of
0 ≤ w ≤ωs = π/T
3. Calculate corresponding frequency ωw to the frequencies
used to calculate G(ejωT ).
ω w =2/TtanωT/2

16. 4. Make a frequency response table to help drawing the Bode
diagram.
5. Draw the Bode Diagram, both for Magnitude of G(ejωT )and
<G(ejωT ) using the frequency scale of ωw, not ω. This process
is often called as frequency pre-warping.

17. 5.Design Approach
•Digital controller design to determine D(z) by the frequency
response method uses phase margin φm as a key parameter.
•Design is to make the phase margin of the open loop
transfer function D(w)G(w) have a specified phase margin φm
at a frequency ωw1 .

18. •Finding the new cross-over frequency ωw1 requires several
conditions to be met, and also requires “try and error” until a
satisfactory value is found.
• We are designing (determining)

19. 6.Design specific to Phase lag Compensator
The first order phase lag compensator can be designed by
the formula similarly as in the case of phase lead
compensator. However, there is a basic difference in the
principle of compensating for a required phase margin.
• Phase lead compensators add a positive phase to increase
the phase margin without drastically changing the 0 dB
crossover frequency.
• A phase lag compensators is used to lower the overall gain
to shift the 0 dB crossover frequency to a much smaller
value so that a larger phase margin is obtained.

20. Since the 0 dB crossover frequency ωw1 is shifted to a lower
frequency where the phase response of the Bode diagram is
not affected by the compensator, selecting ω w1 can be done
on the Bodediagram of G(w) alone without being affected by the
compensator.
First, find a new crossover frequency ωw1 where the
condition,
< G(jωw1 ) = −1800 + φm + 50
is satisfied. 5 accounts for a slight phase angle decrease by
the phase lag compensator. The rule of thumb to set ωw0 in
the compensator’s transfer function,

22. Ex1:The servo motor is to control the horizontal (azimuth)
angle for pointing a radar antenna. The transfer function is
given by

26. Ex2.The servo motor is to control the horizontal (azimuth)
angle for pointing a radar antenna. The transfer function is
given by

31. Ex3. consider that the following system is subjected to a
sampled data control system with sampling time T = 0.2 sec.
Thus
The bi - linear transformation

32. will transfer Gz(z) into w -plane, as
We need first design a phase lead compensator so that PM of
the compensated system is at least 50° with Kv = 2 . The
compensator in w -plane is

33. Design steps are as follows.
• K has to be found out from the Kv requirement.
• Make ωmax = ωgnew.
• Compute the gain crossover frequency ωg and phase
margin of the uncompensated system after introducing K in
the system.
• At ωg check the additional/required phase lead, add safety
margin, find out φmax. Calculate α from the required φmax
• Since the lead part of the compensator provides a gain of
20 log101√𝜶, find out the frequency where the logarithmic
magnitude is - 20 log101√𝜶. This will be the new gain
crossover frequency where the maximum phase lead should
occur.
• Calculate τ from the relation

34. Using MATLAB command ''margin'', phase margin of the
system with K = 2 is computed as 31.6° with ωg = 1.26
rad/sec,

36. Thus the required phase lead is 50° - 31.6° = 18.4° . After
adding a safety margin of 11.6° , φ max becomes 30° . Hence
From the frequency response of the system it can be found
out that at ω = 1.75 rad/sec, the magnitude of the system is
Thus ωmax = ωgnew = 1.75 rad/sec. This gives
Thus the controller in w-plane is

37. The Bode plot of the compensated system is shown below.

38. Re-transforming the above controller into z -plane using the
relation we get the controller in z -plane, as
Try to design a phase lag compensator so that PM of the
compensated system is at least 50° and steady state error to
a unit step input is 0.1. The compensator in w -plane is
where,

39. Required PM is 50°. Let us put a safety margin of 5°. Thus
the PM of the system modified with K should be 55°.
By solving the above, ωg = 2.44 rad/sec. Thus the magnitude
at ωg should be 1

40. Putting the value of ωg in the last equation, we get K = 4.13
Thus,
If we place 1/ τ one decade below the gain crossover
frequency, then
Thus the controller in w -plane is
Re-transforming the above controller into z -plane using the
relation

41. we get
The Bode plot of the compensated system is shown below.

43. Operational amplifiers used to as a compensators

44. Thanks