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Lead Compensators Design Using Frequency Response Techniques
Ahmed Abu‐Hajar, Ph.D. Frequency Response of Feedback System
Transfer Function of Feedback System: Let ω180 be the frequency at which the phase of L(jω) is 1800
Important Notes: 1‐ Positive Feedback, system is not stable (oscillate at ω180)
Frequency Response of Feedback System
2‐ Infinite (or very high gain @ ω180
Stability Considerations
bili
id
i
Let assume that there is a frequency ωx such that
3‐ Stable and Tolerable Gain at ω180
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Phase Margin & Gain Margin
Phase Margin (PM)
Let ω0 be the frequency at which the Open Loop Gain is 0dB (unity gain).
PM must be positive PM > 30
PM > 30o‐ 60o.
0dB
Gain Margin (GM)
GM
@ω180, |L(jω180)| << 1 (0dB). ω0
ω180
GM must be positive GM > 6dB
GM > 6dB.
PM
-180o
ω0
ω180
Lead Compensator Design using Frequency response The Transfer Function: Km
z
p
ϕm
The Maximum Phase ϕm
z
ωm p
The gain is increased because of Kc and because of the zeros which gives+20dB/Dec
ωm: The frequency @ which the phase is ϕm
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Designing Lead Compensators
Design Objective 1‐ Increase the phase at the frequency 0dB crossing, hence increases PM
2‐ Improves the velocity error constant Kv by selecting the proper Kc.
Steps: 1‐ Find the Loop Gain L(s) = G(s)H(s).
2‐ Determine K= αKc that would give the desired Kv.
3‐ Draw Bode Plot of L1(s) = KG(s)H(s). ‐Must compensate for K changes the PM requirement. ‐ Evaluate the PM. 4‐ Add 5o‐12o because of the increase in the gain due to the zero
5‐ Determine α from the required PM. ‐The new zero crossing occurs at frequency is where ωm is located. It i h
th
i f L1(s) is given as: ( )i i
‐It is where the gain of L
6‐ Determine the location of the zero and the pole as: 7‐ Choose the value of Kc from K = αKC.
8‐ Check the GM and PM requirements. If not met, reiterate the process.
(Usually, you need to go to step 4 and add more phase).
Lead Compensator Example
R(s)
G(s)
C(s)
Design a lead compensator using Bode Plot Method such that
1‐ The velocity error constant Kv = 20 sec‐1
2‐ Phase Margin PM = 50o
3‐ Gain Margin GM > 10.
R(s)
Gc(s)
G(s)
C(s)
Solution
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Lead Compensator Example
Draw Bode Plot of L1(s)
Using approximated bode plot PM is found to be 17o.
Using Matlab, exact PM was found to be 17.9o.
We need to evaluate ϕm of the compensator to get 50o + (5o ‐12o) The maximum phase of the compensator
Lead Compensator Example
Solve for α
The gain (Km) caused by the early zero The new zero crossing (ωm) is where the L1(jω) = ‐ 6.2 dB
From the Graph, that is located at ωm = 9 rad/sec
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Lead Compensator Example
The zero and the pole and KC of the compensators are found as follows
The compensator is given as
Lead Compensator Example
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Lead Compensator Example
Lead Compensator Example
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Lag Compensators Design Using Frequency Response Techniques
Ahmed Abu‐Hajar, Ph.D. Lag Compensator
The Transfer Function
Km
p
The Objective:
p
z
z
To provide attenuation (reduction in magnitude) after z.
By reducing the magnitude before the 0dB crossing (ω0) , the phase margin improves.
The location of z and p must be before ω0.
The attenuation is determined by the ratio β
Design Objectives: Improving the steady state error while improving the PM. 7
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Lag Compensator Example
R(s)
C(s)
G(s)
Design a Lag Compensator using Frequency Response method such that: Design a Lag Compensator using Frequency Response method such that:
1‐ The velocity error constant Kv = 5 sec‐1
2‐ Phase Margin PM = 40o
R(s)
Gc(s)
G(s)
C(s)
Solution
The velocity error constant is give as: Lag Compensator Example
K changes the 0dB crossing frequency ω0.
Investigate the effect of K on ω0.
Draw Bode Plot of L1(s)
Uncompensated the PM is 30o
Due to K =5, the PM is ‐12o
Not stable system.
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Lag Compensator Example
Now, we need to locate the frequency at which we get a phase margin
From the graph, @ ω = 0.7, PM = 40o. Add 12o correction factor, (acquired by trial and error)
@ ω0 = 0.5= PM = 52. Choose the 0dB crossing @ω0 = 0.5
Choose z = 0.01 before ω0  p = 0.01
The compensated PM = 40.7o
Lag Compensator Example
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Lag Compensator Example
Lag Compensator Example
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