Transcript Document
ECE
8443
– PatternContinuous
Recognition
EE
3512
– Signals:
and Discrete
LECTURE 26: FEEDBACK CONTROL
• Objectives:
Typical Feedback System
Feedback Example
Feedback as Compensation
Proportional Feedback
Applications
• Resources:
MIT 6.003: Lecture 20
MIT 6.003: Lecture 21
Wiki: Control Systems
Brit: Feedback Control
JC: Crash Course
Wiki: Root Locus
Wiki: Inverted Pendulum
CJC: Inverted Pendulum
URL:
Audio:
A Typical Feedback System
Feed Forward
Feedback
• Why use feedback?
Reducing Nonlinearities
Reducing Sensitivity to Uncertainties and Variability
Stabilizing Unstable Systems
Reducing Effects of Disturbances
Tracking
Shaping System Response Characteristics (bandwidth/speed)
EE 3512: Lecture 26, Slide 1
Motivating Example
• Open loop system: aim and shoot.
• What happens if you miss?
• Can you automate the correction
process?
EE 3512: Lecture 26, Slide 2
• Closed-loop system: automatically
adjusts until the proper
coordinates are achieved.
• Issues: speed of adjustment,
inertia, momentum, stability, …
System Function For A Closed-Loop System
• The transfer function of this
system can be derived using
principles we learned in
Chapter 6:
E ( s) X ( s) R( s) X ( s) G ( s)Y ( s)
Loop
Y ( s) H ( s) E ( s) H ( s)X ( s) G ( s)Y ( s)
Y ( s)
H ( s)
Q( s )
X ( s) 1 G( s) H ( s)
• Black’s Formula: Closed-loop transfer function is given by:
Forw ardGain
1 LoopGain
Forward Gain: total gain of the forward path from the input
to the output, where the gain of a summer is 1.
Loop Gain: total gain along the closed loop shared by all systems.
Y ( s)
A' B
A
A'
X ( s) 1 A' BC
1 A
AB
1 A ABC
EE 3512: Lecture 26, Slide 3
The Use Of Feedback As Compensation
• Assume the open loop
gain is very large
(e.g., op amp):
KP ( j )
1 KP ( j )G ( j )
1
Independent of P(s)
G ( j )
Q ( j )
Q( s )
1
R R2
1
G( j )
R1
• The closed-loop gain depends only on the passive components (R1 and R2)
and is independent of the open-loop gain of the op amp.
EE 3512: Lecture 26, Slide 4
Stabilization of an Unstable System
• If P(s) is unstable, can we
stabilize the system by
inserting controllers?
• Design C(s) and G(s) so that
the poles of Q(s) are in the LHP:
Q( s )
C ( s) P( s)
1 C ( s)G ( s) P( s)
• Example: Proportional Feedback (C(s) = K)
1
s2
C (s) K
G (s) 1
P(s)
• The overall system gain is:
K
K
Q( s ) s 2
K
s2 K
1
s2
EE 3512: Lecture 26, Slide 5
• The transfer function is stable for K > 2.
• Hence, we can adjust K until the system
is stable.
Second-Order Unstable System
• Try proportional feedback:
1
P( s) 2
C ( s) K G( s) 1
s 4
K
2
K
s
4
Q( s)
2
K
s
4 K
1 2
s 4
K 4
0,
One of the poles is at p1 4 K
j K 4, K 4
Unstable for all values of K.
• Try damping, a term proportional
to d / dt :
K1 K 2 s
2
K K2s
s
4
Q( s )
2 1
K K s s K 2 s K1 4
1 1 2 2
s 4
• This system is stable as long as:
K2 > 0: sufficient damping force
K1 > 4: sufficient gain
EE 3512: Lecture 26, Slide 6
• Using damping and feedback, we
have stabilized a second-order
unstable system.
The Concept of a Root Locus
• Recall our simple control system
with transfer function:
C ( s) P( s)
Q( s )
1 C ( s)G ( s) P( s)
• The controllers C(s) and G(s) can be
designed to stabilize the system, but that
could involve a multidimensional optimization. Instead, we would like a
simpler, more intuitive approach to understand the behavior of this system.
• Recall the stability of the system depends on the poles of 1 + C(s)G(s)P(s).
• A root locus, in its most general form, is simply a plot of how the poles of our
transfer function vary as the parameters of C(s) and G(s) are varied.
• The classic root locus problem involves a simplified system:
Closed-loop poles
are the same.
EE 3512: Lecture 26, Slide 7
Example: First-Order System
• Consider a simple first-order system:
1
C (s) K G(s) 1
s2
K
K
C ( s) H ( s)
s2
Q( s )
s2 K
1 C ( s )G ( s ) H ( s ) 1 K
s2
H ( s)
• The pole is at s0 = -(2+K). Vary K from 0 to :
Becomes more stable
Becomes less stable
• Observation: improper adjustment of the gain can cause the overall system to
become unstable.
EE 3512: Lecture 26, Slide 8
Example: Second-Order System With Proportional Control
• Using Black’s Formula:
K
( s 100)(s 1)
Q( s)
K
1
( s 100)(s 1)
K
( s 100)(s 1) K
• How does the step response
vary as a function of the gain, K?
• Note that as K increases, the
system goes from too little gain
to too much gain.
EE 3512: Lecture 26, Slide 9
How Do The Poles Move?
Q( s)
K
( s 100)(s 1) K
2
101 101
p1, 2
100 K
2
2
K 0000: p1, 2 1,100
K 2000: p1, 2 29.3,71.7
K 2450: p1, 2 50.0,51.0
K 2450: p1, 2 j
Desired Response
• Can we generalize this analysis to systems of arbitrary complexity?
• Fortunately, MATLAB has support for generation of the root locus:
num = [1];
den = [1 101 101]; (assuming K = 1)
P = tf(num, den);
rlocus(P);
EE 3512: Lecture 26, Slide 10
Example
G( s) H ( s)
s2
s( s 1)
s2
rlocus1 K
s(s 1)
EE 3512: Lecture 26, Slide 11
s2
rlocus1 K
s(s 1)
Feedback System – Implementation
EE 3512: Lecture 26, Slide 12
Summary
• Introduced the concept of system control using feedback.
• Demonstrated how we can stabilize first-order systems using simple
proportional feedback, and second-order systems using damping (derivative
proportional feedback).
• Why did we not simply cancel the poles?
In real systems we never know the exact locations of the poles. Slight errors
in predicting these values can be fatal.
Disturbances between the two systems can cause instability.
• There are many ways we can use feedback to control systems including
feedback that adapts over time to changes in the system or environment.
• Discussed an application of feedback control involving stabilization of an
inverted pendulum.
EE 3512: Lecture 26, Slide 13
More General Case
• Assume no pole/zero cancellation in G(s)H(s):
G(s) H (s)
Q( s )
1 KG ( s) H ( s)
• Closed-loop poles are the roots of:
1
1 KG ( s) H ( s) 0 G ( s) H ( s)
K
• It is much easier to plot the root locus for high-order polynomials because we
can usually determine critical points of the plot from limiting cases
(e.g., K = 0, ), and then connect the critical points using some simple rules.
• The root locus is defined as traces of s for unity gain:
KG(s)H (s) 1 and KG(s)H (s) (2n 1)
• Some general rules:
At K = 0, G(s0)H(s0) = s0 are the poles of G(s)H(s).
At K = , G(s0)H(s0) = 0 s0 are the zeroes of G(s)H(s).
Rule #1: start at a pole at K = 0 and end at a zero at K = .
Rule #2: (K 0) number of zeroes and poles to the right of the locus point
must be odd.
EE 3512: Lecture 26, Slide 14
Inverted Pendulum
• Pendulum which has its mass above its pivot point.
• It is often implemented with the pivot point mounted
on a cart that can move horizontally.
• A normal pendulum is stable when hanging
downwards, an inverted pendulum is inherently
unstable.
• Must be actively balanced in order to remain upright,
either by applying a torque at the pivot point or by
moving the pivot point horizontally (Wiki).
EE 3512: Lecture 26, Slide 15
Feedback System – Use Proportional Derivative Control
• Equations describing the physics:
M mass, I momentof inertia
• The poles of the system are inherently
d 2 x(t )
d 2 (t )
m glsin (t ) m l
cos (t ) I
unstable.
dt 2
dt 2
assume is small: sin cos 1
• Feedback control can be used to
stabilize both the angle and position.
d 2 (t )
d 2 x(t )
I
m gl (t ) m l
2
dt
dt 2
• Other approaches involve oscillating
m ls2
the support up and down.
( s) 2
X ( s) (polesare unstable)
ls m gl
EE 3512: Lecture 26, Slide 16