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Dr Ian R. Manchester
Slide 2 Dr. Ian R. Manchester Amme 3500 : Introduction
Week Content Notes
1 Introduction
2 Frequency Domain Modelling
3 Transient Performance and the s-plane
4 Block Diagrams
5 Feedback System Characteristics Assign 1 Due
6 Root Locus
7 Root Locus 2 Assign 2 Due
8 Bode Plots
9 Bode Plots 2
10 State Space Modeling Assign 3 Due
11 State Space Design Techniques
12 Advanced Control Topics
13 Review Assign 4 Due
Slide 3
• Review mathematical justification for
feedback over open loop control
• Introduce the PID controller
• Performance specification: steady-state error
• Introduction to parameter sensitivity aka
“robustness”
Dr Ian R. Manchester
Slide 4
Desired
Value Output +
-
Feedback
Signal
error
Controller Plant
Control
Signal
r(t)
e(t) = r(t) − y(t)Error:
Dr Ian R. Manchester
(Assuming perfect measurement)
Slide 5
• Perfect feed-forward controller: Gn
• Output and error:
• Only zero when:
(Perfect Knowledge)
(No load or disturbance)
Dr Ian R. Manchester
Slide 6
r
d
unity feedback
controller plant
reference output
A sensor
disturbance
y =GK
1+KGr −
G
1+GKd
Reference to Output
Transfer Function
Disturbance to Output
Transfer Function
Dr Ian R. Manchester
Slide 7
rd
Equivalent Open Loop Block Diagram
Dr Ian R. Manchester
y =GK
1+KGr −
G
1+GKd
Reference to Output
Transfer Function
Disturbance to Output
Transfer Function
Slide 8
• Simplest possible feedback control
• useful for thermostats, similarly simple problems
• Often results in oscillations
• Rapid changes in control input are often infeasible
• Nonlinear: can be very difficult to analyse Dr Ian R. Manchester
Slide 9
• A very large number (at least 95%) of feedback controllers used in practice have the following simple form:
• This is called a “PID”
– Proportional (P) term feeds back on the current error
– Integral (I) feedback can eliminate steady state error
– derivative (D) term can improve dynamic response
Dr Ian R. Manchester
u(t) = KPe +K
Ie(τ)dτ
0
t
∫ +KD
˙ e
Slide 10
• Suppose we just use the “P” term, i.e. proportional control
• We have seen how this form of feedback is able to minimize the effect of disturbances by increasing K
• This may adversely affect other performance, in particular overshoot
K
-
+
R(s) E(s) C(s) G(s)
Dr Ian R. Manchester
Slide 11
• Let’s compute an example of a torsional mass-spring-damper system with proportional control
• J = 1 N m, k = 5 N m/rad and b = 6 N m s/rad
b = 6 N m s/rad
* N.S. Nise (2004) �
Control Systems Engineering
�
Wiley & Sons
k = 5 N m s/rad
Dr Ian R. Manchester
Slide 12
• Recall the differential equation
• The transfer function is
• The closed loop response is
K
- +
R(s) E(s) C(s) G(s)
J ˙ ̇ θ + b ˙ θ + kθ = τ
Dr Ian R. Manchester
Slide 13
K=5
K=10
K=100
K=1000
Dr Ian R. Manchester
Slide 14
• Proportional (P) control acts like a spring
pulling the output towards the reference.
• A very stiff spring (large gain K) can result in
highly oscillatory response.
• Why not add damping? This is PD control
Dr Ian R. Manchester
u(t) = KPe +K
D˙ e Kp+Kds
- +
R(s) E(s) C(s) G(s)
U(s)
Slide 15 Dr Ian R. Manchester
• D term adds damping and differentiates the reference
Slide 16 Dr Ian R. Manchester
P control:
K=100
PD control:
K=100+20s
• Note the anticipatory “kick” at the beginning,
and the damping of oscillations
Same torsional setup as
before
Slide 17
• We have looked at transient response for second order systems
• This can be defined by
– Rise time, Tr
– Settling time, Ts
– Peak time, Tp
– Percentage overshoot.
• These can all be specified by PD control
cfinal=steady state response
* N.S. Nise (2004)
�
Control Systems Engineering
�
Wiley & Sons
Dr Ian R. Manchester
Slide 18
• Why use integral control? (The “I” part
in PID)
• Another important characteristic of
systems in general is their steady state
performance
• Steady-state error is the difference
between the reference input and output
for a particular input as t→∞ time
output
input
Steady state
error
Dr Ian R. Manchester
Slide 43
• Intuitively: if the controller continues to see an
error signal over a long time, then the integral
term “builds up”, and increases control signal
• Slowly it builds up to the level required to
completely kill the error
Dr Ian R. Manchester
u(t) = KPe +K
Ie(τ)dτ
0
t
∫ +KD
˙ e
Slide 44
• For a PD controller, if the error and its
derivative are both zero, then
• So, a zero-steady-state-error response to a step
must result in zero control input.
• This is not possible if there is some force (e.g.
a spring, or gravity, or wind) pulling the
system away from the reference Dr Ian R. Manchester
u = KPe +K
D˙ e = 0