D1
Compensation Techniques
• Performance specifications for the closed-loop system
• Stability
• Transient response
Æ T , M (settling time, overshoot)
or phase and gain margins
• Steady-state response Æ e (steady state error)
s
s
ss
• Trial and error approach to design
Performance specifications
Root-locus or
Frequency
response
techniques
Synthesis
Analysis of closed-loop system
No
Are specifications
met?
Yes
D2
Basic Controls
1.
Proportional Control
+
E(s)
K
M(s)
G(s)
-
M (s)
= K
E (s)
m (t ) = K ⋅ e (t )
Integral Control
2.
+
E(s)
K
s
M(s)
G(s)
-
M ( s) K
=
s
E (s)
m ( t ) = K ∫ e ( t ) dt
Integral control adds a pole at the origin for the open-loop:
• Type of system increased, better steady-state performance.
•
Root-locus is “pulled” to the left tending to lower the
system’s relative stability.
D3
3.
Proportional + Integral Control
+
E(s)
M(s)
K
Kp + i
s
G(s)
-
M ( s)
K K p s + Ki
= Kp + i =
E ( s)
s
s
m (t ) = K p e (t ) + ∫ K i e (t )dt
A pole at the origin and a zero at
4.
−
Ki
K p
are added.
Proportional + Derivative Control
+
E(s)
K p + K d s M(s)
G(s)
-
M ( s)
= K p + Kd s
E ( s)
m ( t ) = K p e (t ) + K d
de (t )
dt
• Root-locus is “pulled” to the left, system becomes more
stable and response is sped up.
• Differentiation makes the system sensitive to noise.
D4
5.
Proportional + Derivative + Integral (PID) Control
M(s)
+
E(s)
K p + Kd s +
Ki
s
G(s)
-
M ( s)
K
= K p + Kd d + i
E ( s)
s
m ( t ) = K p e (t ) + K
d
de (t )
+ K
dt
i
∫ e (t )dt
• More than 50% of industrial controls are PID.
• More than 80% in process control industry.
• When G(s) of the system is not known, then initial values
for Kp, Kd, Ki can be obtained experimentally and than
fine-tuned to give the desired response (Ziegler-Nichols).
6.
Feed-forward compensator
D5
+
E(s)
Gc (s )
M(s)
G(s)
-
Design Gc(s) using Root-Locus or Frequency Response
techniques.
D6
Frequency response approach to
compensator design
Information about the performance of the closed-loop system,
obtained from the open-loop frequency response:
• Low frequency region indicates the steady-state behavior.
• Medium frequency (around -1 in polar plot, around gain
and phase crossover frequencies in Bode plots) indicates
relative stability.
• High frequency region indicates complexity.
Requirements on open-loop frequency response
• The gain at low frequency should be large enough to give
a high value for error constants.
• At medium frequencies the phase and gain margins should
be large enough.
• At high frequencies, the gain should be attenuated as
rapidly as possible to minimize noise effects.
Compensators
• lead:improves the transient response.
• lag: improves the steady-state performance at the expense
of slower settling time.
• lead-lag: combines both
D7
Lead compensators
1
Ts + 1
T
= Kc
Gc (s ) = K c a
1
aTs + 1
s+
aT
s+
T > 0 and 0 < α < 1
• Poles and zeros of the lead compensator:
−
1
aT
−
1
T
• Frequency response of Gc(jω):
The maximum phase-lead angle φm occurs at ωm , where:
1−α
sin φm =
and
1+α
1
1
logωm = logT + log 
2
aT
Æ
ωm =
1
a T
D8
Since
1 + jω T
1 + jω aT
=
ω =ω m
1
a
the magnitude of Gc(jω) at ωm is given by:
Gc ( jωm) = Kc a
Polar plot of a lead network
a ( jω T + 1)
( jωaT + 1)
is given by
where 0 < a < 1
D9
Lead compensation based on the
frequency response
Procedure:
1. Determine the compensator gain Kcα satisfying the given
error constant.
2. Determined the additional phase lead φm required (+
10%~15%) for the gain adjusted (KcαG(s)) open-loop
system.
3. Obtain α from
sin φ m =
1− a
1+ a
4. Find the new gain cross over frequency c from
Kc a G( jωc ) = 10 log a
5. Find T from c and transfer function of Gc(s)
T =
1
a ωc
Gc (s ) = K c a
and
Ts + 1
aTs + 1
General effect of lead compensator:
• Addition of phase lead near gain crossover frequency.
• Increase of gain at higher frequencies.
• Increase of system bandwidth.
D10
Example:
Consider
+
Gc(s)
G(s)
-
where
G (s) =
4
s (s + 2 )
Performance requirements for the system:
Steady-state:
Transient response:
Kv = 20
phase margin >50°
gain margin >10 dB
Analysis of the system with Gc(s) = K
For Kv = 20
Æ
K = 10
phase margin § 17°
gain margin § ’ G%
This leads to:
Design of a lead compensator:
G c (s ) = K c a
Ts + 1
aTs + 1
4K a
c
1. K v = lim sGc (s )G(s ) = 2 = 2K c a = 20
s →0
Æ K α=10
c
D11
2. From the Bode plot of KcαG(jω), we obtain that the
additional phase-lead required is:
50° - 17° = 33°.
We choose 38° (~33° + 15%)
1− a
1+ a
Æ α = 0.24
3.
sin φ m = sin 38 , =
4.
Since for m, the frequency with the maximum phase-lead
angle, we have:
1 + jωmT
1
=
1 + jωmaT
a
We choose c , the new gain crossover frequency so that
ωm = ωc
and
G c (s )G (s ) s = jω = 1
c
This gives that:
K c aG( jωc ) =
40
jωc ( jωc + 2)
has to be equal:
−1
 1 
 = −6.2dB

a


From the Bode plot of KcαG(jω) we obtain that
40
= − 6 .2 dB at
jω c ( jω c + 2 )
c = 9 rad/sec
D12
5.
This implies for T
1
1
ωc =
=
= 9rad/ sec
aT
0.24T
and
Kc =
Æ
1
= 4.41
T
20
= 41 .7
2a
G c (s ) = 41 . 7
s + 4 . 41
s + 18 . 4
The compensated system is given by:
+
41.7(s + 4.41)
s + 18.4
4
s(s + 2)
-
The effect of the lead compensator is:
•
•
•
•
Æ
Æ
Phase margin: from 17° to 50°
better transient response
with less overshoot.
c : from 6.3rad/sec to 9 rad/sec the system response is
faster.
Gain margin remains \ .
acceptable steady-state response.
Kv is 20, as required
Æ
D13
Bode diagram for
K c ⋅ a ⋅ G ( jω ) =
40
jω (jω + 2)
Bode diagram for the compensated system
G c ( j ω )G ( j ω ) = 41 . 7
j ω + 4 . 41
4
⋅
j ω + 18 . 4 j ω (j ω + 2)
D14
Lag compensators
1
Ts + 1
T
G c (s ) = K c β
= Kc
1
β Ts + 1
s+
βT
s+
T>0, >1
Poles and zeros:
1
T
1
βT
Frequency response:
Polar plot of a lag compensator K.(jT+1)/(jT+1)
D15
Bode diagram of a lag compensator with Kc 1
βT
1
T
-20log
Magnitude of (jT+1)/(jT+1)
D16
Lag compensation based on the
frequency response
Procedure:
1. Determine the compensator gain Kcβ to satisfy the
requirement for the given error constant.
2. Find the frequency point where the phase of the gain
adjusted open-loop system (KcβG(s)) is equal to -180° +
the required phase margin + 5°~ 12°.
This will be the new gain crossover frequency c.
3. Choose the zero of the .425038,947
octave to 1 decade below c .
% ,9 ,-4:9 4. Determine the attenuation necessary to bring the
magnitude curve down to 0dB at the new gain crossover
frequency
− K c β G ( j ω c ) = − 20 log β
Æ
5. Find the transfer function Gc(s).
General effect of lag compensation:
• Decrease gain at high frequencies.
• Move the gain crossover frequency lower to obtain
the desired phase margin.
D17
Example:
Consider
+
Gc(s)
G(s)
-
where
G (s ) =
1
s (s + 1 )(0 . 5 s + 1 )
Performance requirements for the system:
Steady state:
Transient response:
Kv =5
Phase margin > 40°
Gain margin > 10 dB
Analysis of the system with Gc(s) = K
K v = lim KG (s ) = K = 5
s →0
for K = 5, the closed-loop system is unstable
Design of a lag compensator:
1
T = K β Ts + 1
G c (s) = K c
c
1
β Ts + 1
s+
βT
s+
D18
1. K v = lim Gc (s )G(s ) = K c β = 5
s→0
Æ
2. Phase margin of the system 5G(s) is -13°
the closed-loop system is unstable.
From the Bode diagram of 5G(jω) we obtain that the
additional required phase margin of 40° + 12° = 52° is
REWDLQHG DW rad/sec.
The new gain crossover frequency will be:
c = 0.5 rad/sec
3. % rad/sec( at about 1/5 of c).
4. The magnitude of 5G(jω) at the new gain crossover
frequency c =0.5 rad/sec is 20 dB. In order to have c as
the new gain crossover frequency, the lag compensator
must give an attenuation of -20db at ωc.
Therefore
Kc =
5.
- 20log = - 20 dB
5
= 0 .5 ,
β
pole :
and
G c (s ) = 0 . 5
s + 0 .1
s + 0 . 01
Æ
= 10
1
= 0 . 01
βT
D19
Bode diagrams for:
• G1(jω) = 5G(jω) (gain-adjusted KcβG(jω) open-loop
transfer function),
• Gc(jω)/K = Gc(jω)/5 (compensator divided by gain Kcβ =
5),
• Gc(jω)G(jω) (compensated open-loop transfer function)
The effect of the lag compensator is:
•
•
•
•
•
The original unstable closed-loop system is now stable.
The phase margin § ƒ
acceptable transient response.
The gain margin § G%
acceptable transient response.
acceptable steady-state response.
Kv is 5 as required
The gain at high frequencies has been decreased.
Æ
Æ
Æ
D20
Lead-lag compensators
1
1
s+
T1
T2
β sT1 + 1 sT2 + 1
Gc (s) = Kc
⋅
= Kc
⋅
T1
1
γ
γ
s+
s+
s + 1 sβT2 + 1
aT1
βT2
γ
s+
T1, T2 > 0 , > 1 and γ > 1
Frequency response:
Bode diagram of a lag-lead compensator given by
1

s+
T1
Gc (s ) = K c 

γ
s+
 T1
1

 s +
T2


1
 s +
βT2







with Kc = 1, = = 10 and T2 = 10 T1
D21
Polar plot of a lag-lead compensator given by
1

s+
T1
Gc (s ) = K c 

γ
s+
 T1
1

 s +
T2


1
 s +
βT2







with Kc = 1 and = Comparison between lead and lag compensators
Lead compensator
o High pass
o Approximates
derivative plus
proportional control
o Contributes phase lead
o Increases the gain
crossover frequency
o Increases bandwidth
Lag compensator
o Low pass
o Approximates integral plus
proportional control
o Attenuation at high
frequencies
o Moves the gain-crossover
frequency lower
o Reduces bandwidth