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AERO 422: Active Controls for Aerospace Vehicles
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Proportional, Integral & Derivative Control Design
Raktim Bhattacharya
Laboratory For Uncertainty Quantification
Aerospace Engineering, Texas A&M University.
Proportional Control
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional Control
.
r
+
−
y
e
C
u
P
y
C(s) = K, constant gain
u(t) = Ke(t)
Use Routh’s criterion to determine range for K
Verify if the system is stabilizable with C(s) = K
AERO 422, Instructor: Raktim Bhattacharya
3 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Second order system
.
r
+
−
y
e
C
P (s) =
u
P
y
A
(τ1 s+1)(τ2 s+1) , C(s)
=K
Characteristic equation: zeros of 1 + P C
τ1 τ2 s2 + (τ1 + τ2 )s + AK + 1 = 0
Open loop poles at −1/τ1 , −1/τ2
Closed loop poles at
√
τ1 + τ2 + τ1 2 − 2 τ1 τ2 + τ2 2 − 4 AK τ1 τ2
p1 = −
2 τ1 τ2
√
2
τ1 + τ2 − τ1 − 2 τ1 τ2 + τ2 2 − 4 AK τ1 τ2
p2 = −
2 τ1 τ2
AERO 422, Instructor: Raktim Bhattacharya
4 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Routh Stability Criterion
.
+
−
y
r
P (s) =
e
C
u
y
P
Range for stabilization K > 0
A
(τ1 s+1)(τ2 s+1) , C(s)
=K
Root Locus
1.5
Imaginary Axis (seconds−1)
1
s2 τ1 τ2 AK
s 1 τ1 + τ2
0
0
s
AK
0
0.5
Poles move towards each 2
2)
other till K = K ∗ = (τ4τ1 −τ
1 τ2 A
0
K > K ∗ , poles are purely
imaginary
−0.5
−1
−1.5
−2.5
−2
−1.5
−1
Real Axis (seconds−1)
−0.5
0
0.5
K > K ∗ , ωn ↑, ζ ↓
AERO 422, Instructor: Raktim Bhattacharya
5 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional Control
Steady State Error
.
r
+
−
y
e
C
u
P
y
System
P (s) =
A
, C(s) = Kp
(τ1 s + 1)(τ2 s + 1)
Transfer Function
Gyr (s) :=
AKp
PC
=
.
1 + PC
τ1 τ2 s2 + (τ1 + τ2 )s + AKp + 1
Corresponding ODE
τ1 τ2 ÿ + (τ1 + τ2 )ẏ + (AKp + 1)y = AKp r
Steady state
ÿ = ẏ = 0 =⇒ y(t) =
AKp
r(t).
1 + AKp
AERO 422, Instructor: Raktim Bhattacharya
6 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional Control
Summary
1
Root Locus
1.5
0.9
0.8
1
Step Response
1.5
Amplitude
1
0.6
Mp
Imaginary Axis (seconds−1)
0.7
0.5
0.5
0
0.4
−0.5
0.3
0.5
0.2
−1
0.1
0
0
2
4
6
8
10
Time
12
14
16
18
(a) Step Response
K > K ∗ , ωn ↑, ζ ↓
tr =
1.8
ωn , ts
=
4.6
σ
20
−1.5
−2.5
−2
−1.5
−1
−0.5
0
0.5
0
0
0.1
0.2
0.3
0.4
Real Axis (seconds−1)
(b) Root Locus
0.5
0.6
0.7
0.8
0.9
1
ζ
(c) Mp vs ζ
Effect of Proportional Control
Reduces rise time Good!
Increases overshoot Bad!
Large gain ⇒ small steady state error
Amplifies noise and disturbances Bad!
▶ Look at frequency characteristics of all the transfer
functions (later)
AERO 422, Instructor: Raktim Bhattacharya
7 / 31
Integral Control
P
......
I
.....
PI
.....
PD
......
PID
........
Integral Control
.
r
+
−
y
e
C
u
P
y
C(s) = KI /s
∫
u(t) = KI
t
e(t)
0
Use Routh’s criterion to determine range for KI
Verify if the system is stabilizable with C(s) = KI /s
AERO 422, Instructor: Raktim Bhattacharya
9 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Second Order System
.
r
+
−
y
e
u
C
P (s) =
P
y
A
(τ1 s+1)(τ2 s+1) , C(s)
= KI /s
Characteristic equation: τ1 τ2 s3 + (τ1 + τ2 )s2 + s + AKI
Root Locus
2
1.5
s3
s2
s1
s0
Imaginary Axis (seconds−1)
1
0.5
0
−0.5
τ1 τ2
τ1 + τ2
τ1 +τ2 −AKI τ1 τ2
τ1 +τ2
AKI
1
AKI
0
0
−1
−1.5
−2
−2
0 < KI <
−1.5
−1
−0.5
0
Real Axis (seconds−1)
0.5
1
1.5
2
τ1 + τ2
Aτ1 τ2
AERO 422, Instructor: Raktim Bhattacharya
10 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Integral Control
Steady State Error
.
r
+
−
y
e
C
u
P
y
System
P (s) =
A
, C(s) = KI /s
(τ1 s + 1)(τ2 s + 1)
Transfer Function
Gyr (s) :=
PC
AKI
=
.
1 + PC
τ1 τ2 s3 + (τ1 + τ2 )s2 + AKI
Corresponding ODE
...
τ1 τ2 y + (τ1 + τ2 )ÿ + AKI y = AKI r
Steady state
...
y = ÿ = 0 =⇒ y(t) = r(t).
AERO 422, Instructor: Raktim Bhattacharya
11 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Integral Control
Summary
.
r
+
−
y
e
u
C
P
y
Root Locus
KI ↑ =⇒ ζ ↓ =⇒ Mp ↑
2
1.5
KI ↑ =⇒ ωn ↑ =⇒ tr ↓
Imaginary Axis (seconds−1)
1
KI >
0.5
unstable.
Zero steady state error
0
−0.5
Needs anti windup
mechanism for saturated
actuators discussed later
−1
−1.5
−2
−2
τ1 +τ2
Aτ1 τ2
−1.5
−1
−0.5
0
Real Axis (seconds−1)
0.5
1
1.5
2
Good disturbance rejection
property read book
AERO 422, Instructor: Raktim Bhattacharya
12 / 31
PI Control
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional-Integral Control
.
r
+
−
y
e
C
u
P
y
System
P (s) =
(
)
A
1
, C(s) = K 1 +
(τ1 s + 1)(τ2 s + 1)
TI s
TI is called the integral, or reset time
1/TI is reset rate, related to speed of response
u(t) is a mixture of two signals
)
(
∫ t
1
e(t)dτ .
u(t) = K e(t) +
T I t0
u(t) ̸= 0 even when e(t) = 0, because of integral action
AERO 422, Instructor: Raktim Bhattacharya
14 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional-Integral Control
Steady State Error
.
r
+
−
y
System
e
C
u
P
y
(
)
A
1
P (s) =
, C(s) = K 1 +
(τ1 s + 1)(τ2 s + 1)
TI s
Transfer Function
PC
AK(TI s + 1)
Gyr (s) :=
=
.
3
1 + PC
TI τ1 τ2 s + TI (τ1 + τ2 )s2 + TI (1 + AK)s + AK
Corresponding ODE
...
(∗) y + (∗)ÿ + (∗)ẏ + AKy = AKr + (∗)ṙ
Steady state
...
y = ÿ = ẏ = ṙ = 0 =⇒ y(t) = r(t).
AERO 422, Instructor: Raktim Bhattacharya
15 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional-Integral Control
Independent control over two of the three terms
.
r
+
−
y
e
C
u
P
y
System
(
)
A
1
P (s) =
, C(s) = K 1 +
(τ1 s + 1)(τ2 s + 1)
TI s
Transfer Function
Gyr (s) :=
PC
AK(TI s + 1)
=
.
3
1 + PC
TI τ1 τ2 s + TI (τ1 + τ2 )s2 + TI (1 + AK)s + AK
Characteristic Equation
TI τ1 τ2 s3 + TI (τ1 + τ2 )s2 + TI (1 + AK)s + AK = 0.
AERO 422, Instructor: Raktim Bhattacharya
16 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional-Integral Control
Two tuning variables
.
r
+
−
y
e
u
C
P
y
Routh’s Table
s3
s2
s1
s0
T I τ1 τ2
TI (τ1 + τ2 )
TI + AKTI − AKτ1 +
AK
AKτ1 2
τ1 +τ2
TI (AK + 1)
AK
0
0
Constraints
K > 0,
TI >
τ1 τ2
AK
.
1 + AK τ1 + τ2
AERO 422, Instructor: Raktim Bhattacharya
17 / 31
Derivative Control
P
......
I
.....
PI
.....
PD
......
PID
........
Derivative Control
.
r
+
−
y
e
C
u
P
y
C(s) = KD s
u(t) = KD ė(t)
Use Routh’s criterion to determine range for KD
Verify if the system is stabilizable with C(s) = KD s
Almost never used by itself usually augmented by proportional control
Derivative control is not causal depends on future values
e(t + ∆t) − e(t)
ė ≈
∆t
▶
▶
▶
▶
Knows the slope ← known from future values of e(t)
e(t) = t, ė = 1.
e(t) = t2 , ė = 2t.
e(t) = sin(t), ė = cos(t) = sin(t + π/2).
AERO 422, Instructor: Raktim Bhattacharya
19 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Approximate Derivative Control
Approximation over Frequency range
C(s) = KD s
u(t) = KD ė(t)
Not implementable in applications
Approximate as
KD s
, α >> 1 pole at far left
s/α + 1
≈ KD s, for small s
C(s) =
What is the effect of this approximation?
Look at a step response
AERO 422, Instructor: Raktim Bhattacharya
20 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Approximate Derivative Control
Effect of the Approximation
Step Response
1.5
Amplitude
1
0.5
0
0
2
4
6
8
10
Time
12
14
16
18
20
e(t) = r(t) − y(t) = 1 − y(t)
ė(t) = −ẏ(t)
AERO 422, Instructor: Raktim Bhattacharya
21 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Approximate Derivative Control
Effect of the Approximation (contd.)
0.6
True
alp=1
alp=10
E r r or D e r i v at i v e
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
0
2
4
6
8
10
T im e
12
14
16
18
AERO 422, Instructor: Raktim Bhattacharya
20
22 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
Proportional-Derivative Control
.
r
+
−
y
e
C
u
y
P
Controller Structure
(
C(s) = K 1 + TD
s
s/α + 1
s
= KP + KD
s/α + 1
)
Tune KP and KD to get desired response
Use Routh’s table to determine range for stable values
Derivative control increases damping
reduces overshoot
AERO 422, Instructor: Raktim Bhattacharya
23 / 31
PID Control
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Basic Idea
Controller Structure
C(s) = KP +
KI
s
+ KD
s
s/α + 1
Tune KP , KI and KD to get desired response
Use Routh’s table to determine range for stable values
Proportional term increases ωn and decreases ζ.
▶
▶
Improves rise time
Needs large gain to reduce steady-state error
Integral term increases ωn and decreases ζ.
▶
▶
Zero steady-state
May make the system unstable
Derivative control increases damping – reduces overshoot
and oscillations
AERO 422, Instructor: Raktim Bhattacharya
25 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Example
System
1
+ 0.5s + 1
Study behavior of this system with various control strategies.
P (s) =
s2
Proportional (P)
Proportional Integral (PI)
Proportional Derivative (PD)
Proportional Integral and Derivative (PID)
AERO 422, Instructor: Raktim Bhattacharya
26 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Amplitude
Example: P
2
Amplitude
0
Amplitude
0
5
10
15
Time (seconds)
20
25
2
Open Loop
K=2
1
0
0
5
10
15
Time (seconds)
20
25
30
2
Open Loop
K=10
1
0
Amplitude
Open Loop
K=1
1
0
5
10
15
Time (seconds)
20
25
2
Open Loop
K=50
1
0
0
5
10
15
Time (seconds)
20
25
AERO 422, Instructor: Raktim Bhattacharya
27 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Example: PI
Step Response
1.4
P
I
PI
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
Time (seconds)
25
30
35
40
AERO 422, Instructor: Raktim Bhattacharya
28 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Amplitude
Example: PI (contd.)
1.5
Amplitude
0
Amplitude
0
5
10
15
20
25
30
Time (seconds)
35
40
45
50
1.5
Open Loop
Kp=2
Kp=2, Ki = 1
1
0.5
0
0
10
20
30
40
Time (seconds)
50
60
70
2
Open Loop
Kp=10
Kp=10, Ki = 5
1
0
Amplitude
Open Loop
Kp=1
Kp=1, Ki = 0.5
1
0.5
0
50
100
150
Time (seconds)
200
250
2
Open Loop
Kp=50
Kp=50, Ki = 25
1
0
0
200
400
600
Time (seconds)
800
1000
1200
AERO 422, Instructor: Raktim Bhattacharya
29 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Example: PD
Step Response
1.5
Open loop
P
D
PD
Amplitude
1
0.5
0
0
2
4
6
8
10
12
Time (seconds)
14
16
18
20
AERO 422, Instructor: Raktim Bhattacharya
30 / 31
P
......
I
.....
PI
.....
PD
......
PID
........
PID Control
Example: PID
Step Response
1.5
Open loop
PID
P
PD
Amplitude
1
0.5
0
0
2
4
6
8
10
12
Time (seconds)
14
16
18
20
AERO 422, Instructor: Raktim Bhattacharya
31 / 31