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Controller Types
• Lead compensator
when is very small, can be approximated as PD control
• Lag compensator
For large , can be approximated as PI control
• Lead-Lag compensator
Can be approximated as PID control
11
1( ) , 0<
wC w
wK
1( ) , 1
1ww
wC K
Lead Compensator Transfer function:
Frequency response:
If is given, pole and zero of is obtained as
The gain will be determined from the specification
11
1( ) , 0<
wC w
wK
max max, ( )C w
max
max
1 sin
1 sin
max
1
KssE
1 1 1
20log 20log(1/ )K
20log K
Lead Characteristic
Comparison:
Key Idea: The lead compensator will enlarge the gain crossover
frequency and increase the PM. However, the GM can not be designed from
the lead compensator
maxnew oldPM PM
max (new)gc
Design Example (1)
Design a lead compensator for the digital control system below
so that the PM is 50, the GM is at least 10 DB
and the static velocity error
Sol. Obtain the discrete-time plant (by Matlab or by hand)
C(z) ZOH
-
( )R s
+ 0.2T
( )Y s
( 1)
K
s s
2vK
1 1
2{ { }}
( 1( ) )
)(1G z
K
s sz
Z L
2
(0.01873 0.01752)
1.8187 0.8187
K z
z
Design Example (2)
By using the bilinear mapping, we obtain
The controller is in the form of
1 ( /2) 1 0.1
1 ( /2) 1 0.1
( ) ( ) |w w T wz
w T w
G w G z
2
2
( 0.000333 0.09633 0.9966)
0.9969
K w w
w w
11(
1) , 0<w
wC
ww
Design Example (3) Design technique:
Step 1 Compute the gain that satisfies the required
Step 2 Set , find Bode plot of
Matlab command: bode(2*Gw)
Step 3 Determine PM from the Bode diagram
Here, we get PM=30 (approximately)
Matlab command: [Gm,Pm,Wcg,Wcp] = margin(2*Gw)
vK
0lim ( ) ( ) 2wvw
wK wC w G w K
2K ( )wG w
2
2
2( 0.000333 0.09633 0.9966)( )
0.9969w
w wG w
w w
Design Example (5)
Step 4 Estimate required phase lead
Step 5 Compute the lead factor
Step 6 Find the new “gain cross over freq.” from
By reading the Bode diagram, we get
max new oldPM PM
30 8 25 80
(1 sin 28 ) / (1 sin 28 ) 0.361
| ( ) | 10log( ) dB 4.4251/ dBw gcG j
1.7 rad/sgc
Design Example (6)
Step 7 Get corner frequencies for zero and pole
The lead compensator is
Step 8 Verify margins from the Bode plot of
Step 9 If everything is OK, obtain the controller in z-plane
1 10.9790
1.7 0.361
0.3534
gc
0.9790 1(
1
1)
0.3534 1w
wC w
w
w
w
( ) ( )w wC w G w
2 1
1
2.3798 1.9387( ) ( ) |
0.5589w z
wT z
zC z C w
z
Discussion (Lead Comp.)
Advantage
• Improve phase margin
• Improve high-frequency performance
• Improve the speed of the response
Disadvantage
• May have effects from high-frequency noise
• Generate large signals which may damage the system
Lag Compensator Transfer function:
Frequency response:
The gain will be determined from the specification
1( ) , 1
1ww
wC K
KssE
20log K
20log 20logK
1
1
Lag Characteristic
Comparison:
Key Idea: The lag compensator will
reduce the gain crossover frequency to where the phase margin is satisfied
The zero corner frequency is set to 1 decade
below the new gain cross over frequency
new oldPhase Phase
5 -10 deg.
1/
Design Procedure (Lag Comp.)
1. Determine the gain to satisfy the requirement on
2. Find the new gain cross-over frequency such that
3. Choose the corner frequency one decade below
4. Magnitude reduction from the lag comp. at is equal to , then is determined from
5. Obtain the lag comp.
6. If all the requirements are satisfied,
KssE
gc
gc desiredPM PM
1
gc
gc
20log
20log | ( 20log) |gcKG j
1(
1)
wK
wC w
2 1
1
( ) ( ) | zw
T z
wC z C w
Discussion (Lag Comp.)
Advantage
• Low-frequency characteristics is improved or maintained
• Stability margins are improved
• BW is reduced reduce effects from high-freq. noise
Disadvantage
• BW is reduced slower rise time
• Numerical problems with controller coefficients may result in bad control performance
Lag-Lead Compensator
Objective : Cascade a phase-lag compensator with a phase-
lead compensator to change the overall system characteristics
lead sectionLag section
1 1
1 1
w wC w K K
w w
Lag compensator :
increase the low-frequency gain
Lead compensator :
increase BW and stability margin
Approximation of a PID controller