ECE 8443 – Pattern Recognition EE 3512 – Signals: Continuous and Discrete •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 LECTURE 27: FEEDBACK CONTROL Audio: URL:
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ECE 8443 – Pattern Recognition EE 3512 – Signals: Continuous and Discrete Objectives: Typical Feedback System Feedback Example Feedback as Compensation.
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• Using damping and feedback, we have stabilized a second-order unstable system.
EE 3512: Lecture 27, Slide 6
The Concept of a Root Locus
)()()(1
)()()(
sPsGsC
sPsCsQ
• Recall our simple control systemwith transfer function:
• The controllers C(s) and G(s) can bedesigned 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 27, Slide 7
Example: First-Order System
• Consider a simple first-order system:
• The pole is at s0 = -(2+K). Vary K from 0 to :
• Observation: improper adjustment of the gain can cause the overall system to become unstable.
Becomes more stable
Ks
K
sK
sK
sHsGsC
sHsCsQ
sGKsCs
sH
22
1
2)()()(1
)()()(
1)()(2
1)(
Becomes less stable
EE 3512: Lecture 27, Slide 8
Example: Second-Order System With Proportional Control
• Using Black’s Formula:
• How does the step responsevary as a function of the gain, K?
• Note that as K increases, thesystem goes from too little gainto too much gain.
Kss
K
ssKss
K
sQ
)1)(100(
)1)(100(1
)1)(100()(
EE 3512: Lecture 27, Slide 9
How Do The Poles Move?
Desired Response
jpK
pK
pK
pK
Kp
Kss
KsQ
2,1
2,1
2,1
2,1
2
2,1
:2450
0.51,0.50:2450
7.71,3.29:2000
100,1:0000
1002
101
2
101
)1)(100()(
• 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 27, Slide 10
Example
)1(
21
ss
sKrlocus
)1(
21
ss
sKrlocus
)1(
2)()(
ss
ssHsG
EE 3512: Lecture 27, Slide 11
Feedback System – Implementation
EE 3512: Lecture 27, 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 27, Slide 13
More General Case
)12()()(and1)()( nsHsKGsHsKG
• Assume no pole/zero cancellation in G(s)H(s):
• Closed-loop poles are the roots of:
• 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:
• 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.
)()(1
)()()(
sHsKG
sHsGsQ
KsHsGsHsKG
1)()(0)()(1
EE 3512: Lecture 27, 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 27, Slide 15
Feedback System – Use Proportional Derivative Control
• Equations describing the physics:
• The poles of the system are inherentlyunstable.
• Feedback control can be used to stabilize both the angle and position.
• Other approaches involve oscillatingthe support up and down.