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Systems Analysis and Control
Matthew M. PeetArizona State University
Lecture 17: Compensator Design
Overview
In this Lecture, you will learn:
Lead-Lag Compensation
• Designing Leads
• Designing Lags
• Combining Leads and Lags
Notch Filters
• Providing extra zeros
• Eliminates annoying frequency components.
M. Peet Lecture 17: Control Systems 2 / 28
Recall: Pole-Zero Compensation
Definition 1.
A Pole-Zero Compensator is of the form
K(s) =s+ z
s+ p
Lead Compensation
• p < z
• Replaces Pure Zero
Im(s)
Re(s)
Lag Compensation
• z < p
• Replaces Integrator
Im(s)
Re(s)
M. Peet Lecture 17: Control Systems 3 / 28
Lead CompensationExample
G(s) =1
s(s+ 1)
Asymptotes: ±90◦
Intercept: α = −.5
α =
∑pi −
∑zi
n−m=−1− 0
2− 0= −.5
Break Point: s = −.5
n′d− d′n = 2s+ 1 = 0
Conclusion: At high gain, we get
• High Frequency Oscillation
• Lots of overshoot
• Fixed Settling Time
−1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 4 / 28
Lead CompensationExample
Asymptotes: ±90◦
Intercept: α = −1
α =
∑pi −
∑zi
n−m
=−5 + 4
2− 0+−1− 0
2− 0= −.5− .5 = −1
Break Point: s = −.508
nd′ − n′d= (s+ 5)(s2 + s)− (3s2 + 12s+ 5)(s+ 4)
Conclusion: At high gain, we get
• Improved Settling time
• Slightly less overshoot
Gc(s) =s+ 4
s+ 5
−5 −4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0−8
−6
−4
−2
0
2
4
6
8
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 5 / 28
Lead CompensationPhase Interpretation
The effect of a Lead Compensator
• Add Phase at every point
∠(K(s)G(s)) = ∠K(s) + ∠G(s)
• The change in phase is positive.
∆∠ = θz − θp
Points compensate by moving left.
Im(s)
Re(s)θz
θp
M. Peet Lecture 17: Control Systems 6 / 28
Lead CompensationPole Placement
Lead-Lag can be used to do pole-placement
G(s) =1
s(s+ 1)
Suppose we want:
• 20% Overshoot
• ωn = 2
• Ts < 4
We choose a desired point on the root locus:
• The intersection ofI ωn = 2I σ < −1
s1,2 = −1±√
3ı
−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
1.5
2
2.5
3
Root Locus
Real Axis
Imag
inar
y A
xis
Question: Can we achieve this point exactly using Pole-Zero compensation?
M. Peet Lecture 17: Control Systems 7 / 28
Lead-Lag CompensationPole Placement
Lets start with a basic question:
• Is s already on the root locus?
Lets check:
∠G(s) =∑
∠(s− zi)−∑
∠(s− pi)
Working out the geometry:
∠G(s) = −90◦ − 120◦ = −210◦
Not on the Root Locus!
120o90o
The point s lacks 30◦ of phase.
M. Peet Lecture 17: Control Systems 8 / 28
Lead CompensationPole Placement
To place the point s on the root locus:
• we need to add 30◦ of phase at this point.
Phase is sum of zeros minus poles
• Zeros add phase
• Poles subtract phase.
Im(s)
Re(s)θz
θp
We can add 30◦ if we use a pole-zero combo:
• Add a zero at 60◦
• Add a pole at 30◦
M. Peet Lecture 17: Control Systems 9 / 28
Lead CompensationPole Placement
K(s) =s− zs− p
Use reverse geometry to find p and z.
Zero:
tan 60◦ =
√3
x
x =
√3
tan 60◦= 1
Pole:
tan 30◦ =
√3
x
x =
√3
tan 30◦= 3
60o30o
√3
-1
p = −1− x = −4
z = −1− x = −2
M. Peet Lecture 17: Control Systems 10 / 28
Lead CompensationPole Placement
Now, the root locus passes through s.
To find the gain at this point
• Use rlocfind
• Use k = |d(s)||n(s)| .
For this example,
k = 6.00
−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
1.5
2
2.5
3
Root Locus
Real Axis
Imag
inar
y A
xis
Potential Problem: May adversely affect other poles.
M. Peet Lecture 17: Control Systems 11 / 28
Root Locus Demo 2Pole Placement
Wiley+ Root Locus Demo 2
Make the phase 180◦.
M. Peet Lecture 17: Control Systems 12 / 28
Lead CompensationDeparture Angles
The other big use of lead compensation is to change Departure Angles.Recall the Suspension system problem with integral feedback:
G(s) =s2 + s+ 1
s4 + 2s3 + 3s2 + 1s+ 1
1
s
The poles are
• p1,2 = −.957± 1.23ı
• p3,4 = −.0433± .641ı
At pole p3,4 = −.0433 + .641ı, the phase is −156◦.
Departure Angle:∠dep = ∠G(s) + 180◦ = 24◦
Goal: Increase the departure angle to 90◦ or more.
M. Peet Lecture 17: Control Systems 13 / 28
Lead CompensationDeparture Angles
Suppose we want a departure angle of ∠dep = 100◦.
• Recall∠dep = ∠G(s) + 180◦
• Required Phase ∠G(s)
∠reqG(s) = ∠dep − 180◦ = −80◦
• Required Phase Change:
∆∠G(s) = ∠reqG(s)− ∠G(s) = −80 + 156◦ = 76◦
M. Peet Lecture 17: Control Systems 14 / 28
Lead CompensationDeparture Angles
We need to add 76◦.
• Zero at 90◦.
• Pole at 14◦.
Recall departure point is p3 = −.0433 + .641ıZero:
• θ = 90◦,
• ∆x = 0
• z = −.0433
Pole:
tan 14◦ =.641
∆x
∆x =.641
tan 14◦= 2.57
So p = −.0433−∆x = −2.61.
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imag
inar
y A
xis
Controller:
K(s) =s+ .0433
s+ 2.61
M. Peet Lecture 17: Control Systems 15 / 28
Lag CompensationSteady-State Error
Predict the Steady-State Error.
G(s) =s2 + s+ 1
s4 + 2s3 + 3s2 + 1s+ 1
K(s) =s− zs− p
=nK(s)
dK(s)
ess = lims→0
1
1 +G(s)K(s)
=dK(0)
dK(0) + knK(0)G(0)
=−p
−p+−zkG(0)If
• p is small
• z is large
Then
ess ∼=p
kz
1
G(0)
M. Peet Lecture 17: Control Systems 16 / 28
Lag CompensationSteady-State Error
G(s) =s2 + s+ 1
s4 + 2s3 + 3s2 + 1s+ 1
Say we want steady-state error less than .01.
ess ∼=p
kz
1
G(0)=
p
kz≤ .01
or p ≤ .01kz
• Assume k > 10
• Choose p = .1
• Result: z = 100
Alternatively, p = 1, z = 1000.
• But there are dangers!
−1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4
−3
−2
−1
0
1
2
Root Locus
Real Axis
Imag
inar
y A
xis
−1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 17 / 28
Lag Compensation
Notice some negative effects of Lag
• Asymptotes still at ±90◦
Center of Asymptotes:
α =
∑pi −
∑zi
n−m
=
∑pi,old −
∑zi,old
n−m+
∑pi,new −
∑zi,new
n−m
= αold +p− z
2
Creates a Shift in Asymptotes by
∆α ∼=z
2= 50
for the suspension problem (z = −100).
−120 −100 −80 −60 −40 −20 0 20 40 60−800
−600
−400
−200
0
200
400
600
800
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 18 / 28
Lead-Lag Compensation
To mitigate the effect of lag compensation:
• Add some Lead CompensationI Zero at z = .01I Pole at p = 20
Phase at s1
∠G(s) = −25.8◦
Departure Angle:∠dep = 180 + ∠G(s) = 154.25◦
−1 −0.8 −0.6 −0.4 −0.2 0−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
Root Locus
Real AxisIm
agin
ary
Axi
s
M. Peet Lecture 17: Control Systems 19 / 28
Lead-Lag Compensation
Use rlocfind to pick off
• Maximum stable gain
k = .7768
0 10 20 30 40 50 60 700
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Step Response
Time (sec)
Am
plit
ud
e
−1 −0.8 −0.6 −0.4 −0.2 0−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 20 / 28
Notch-Filters
Conclusion:
• Lead-Lag Improves Performance
• Can’t do everything.
Problem Can’t Stabilize those poles at
s = −.0433± .641ı
Im(s)
Re(s)
One solution is to use a Notch Filter.
Definition 2.
A Notch Filter consists of
• Two Complex ZerosI Used to Capture Troublesome Poles
• Two Real Poles far out in the LHP
M. Peet Lecture 17: Control Systems 21 / 28
Notch-Filter Example
To attack the poles at
s = −.0433± .641ı
Lets use a notch filter at
z1,2 = −.5± .641ı
Poles atp1,2 = −20
Im(s)
Re(s)
K(s) =(s+ .5 + .641ı)(s+ .5− .641ı)
(s+ 20)2
=s2 + s+ .66
s2 + 40s+ 400
M. Peet Lecture 17: Control Systems 22 / 28
Notch-Filter Example
Combining the Notch-Filter with the Lag filter.
• Using rlocfind, we pick off the point
k = 10
−1 −0.8 −0.6 −0.4 −0.2 0−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 23 / 28
Notch-Filter Example
0 1 2 3 4 5 60
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Step Response
Time (sec)
Am
plit
ud
e
Reduced Steady-State ErrorM. Peet Lecture 17: Control Systems 24 / 28
Notch-Filter Example
Don’t Forget about the other poles!
−40 −30 −20 −10 0 10−25
−20
−15
−10
−5
0
5
10
15
20
25
Root Locus
Real Axis
Imag
inar
y A
xis
kmax = 19.4
M. Peet Lecture 17: Control Systems 25 / 28
Notch-Filter Example
What about 30% overhoot?
−2 −1.5 −1 −0.5 0 0.5
−1.5
−1
−0.5
0
0.5
1
1.5
Root Locus
Real Axis
Imag
inar
y A
xis
M. Peet Lecture 17: Control Systems 26 / 28
Notch-Filter Example
What about 30% overhoot?
−40 −30 −20 −10 0 10−25
−20
−15
−10
−5
0
5
10
15
20
25
Root Locus
Real Axis
Imag
inar
y A
xis
Don’t forget those other poles!
M. Peet Lecture 17: Control Systems 27 / 28
Summary
What have we learned today?
Lead-Lag Compensation
• Designing Leads
• Designing Lags
• Combining Leads and Lags
Notch Filters
• Providing extra zeros
• Eliminates annoying frequency components.
Next Lecture: The Frequency Domain
M. Peet Lecture 17: Control Systems 28 / 28