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
s2
C (s)  K
G (s)  1
P(s) 
• The overall system gain is:
K
K
Q( s )  s  2 
K
s2 K
1
s2
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
s2
K
K
C ( s) H ( s)
 s2 
Q( s ) 
s2 K
1  C ( s )G ( s ) H ( s ) 1  K
s2
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) 
s2
s( s  1)

s2 
rlocus1  K
s(s  1) 

EE 3512: Lecture 26, Slide 11

s2 
rlocus1  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