ECE 8443 – Pattern RecognitionEE 3512 – Signals: Continuous and Discrete

• Objectives:Typical Feedback SystemFeedback ExampleFeedback as CompensationProportional FeedbackApplications

• Resources:MIT 6.003: Lecture 20MIT 6.003: Lecture 21Wiki: Control SystemsBrit: Feedback ControlJC: Crash CourseWiki: Root LocusWiki: Inverted Pendulum CJC: Inverted Pendulum

LECTURE 27: FEEDBACK CONTROL

Audio:URL:

EE 3512: Lecture 27, Slide 2

System Function For A Closed-Loop System

• The transfer function of thissystem can be derived usingprinciples we learned inChapter 6:

• Black’s Formula: Closed-loop transfer function is given by:

Forward Gain: total gain of the forward path from the inputto the output, where the gain of a summer is 1.

Loop Gain: total gain along the closed loop shared by all systems.

)()(1

)(

)(

)()(

)()()()()()()(

)()()()()()(

sHsG

sH

sX

sYsQ

sYsGsXsHsEsHsY

sYsGsXsRsXsE

Loop

GainLoop1

GainForward

ABCA

AB

A

AA

BCA

BA

sX

sY

1

1'

'1

'

)(

)(

EE 3512: Lecture 27, Slide 3

The Use Of Feedback As Compensation

• Assume the open loopgain is very large(e.g., op amp):

)(

1

)()(1

)()(

jG

jGjKP

jKPjQ

1

21

)(

1)(

R

RR

jGsQ

• 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.

Independent of P(s)

EE 3512: Lecture 27, Slide 4

Stabilization of an Unstable System• If P(s) is unstable, can we

stabilize the system byinserting controllers?

• Design C(s) and G(s) so thatthe poles of Q(s) are in the LHP:

• Example: Proportional Feedback (C(s) = K)

• The overall system gain is:

)()()(1

)()()(

sPsGsC

sPsCsQ

1)(

)(2

1)(

sG

KsCs

sP

Ks

K

sK

sK

sQ

2

21

2)(• The transfer function is stable for K > 2.

• Hence, we can adjust K until the system is stable.

EE 3512: Lecture 27, Slide 5

Second-Order Unstable System• Try proportional feedback:

One of the poles is at

Unstable for all values of K.

• Try damping, a term proportional to :

• This system is stable as long as:

K2 > 0: sufficient damping force

K1 > 4: sufficient gain

Ks

K

sK

sK

sQ

sGKsCs

sP

44

1

4)(

1)()(4

1)(

2

2

2

2

4,4

4,041 KKj

KKp

dtd /

44

1

4)(12

221

221

221

KsKs

sKK

ssKK

ssKK

sQ

• 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.

)unstablearepoles()()(

)()(

)(

1cossin:smallisassume

)()(cos

)()(sin

inertiaofmoment,mass

2

2

2

2

2

2

2

2

2

2

sXmglls

mlss

dt

txdmltmgl

dt

tdI

dt

tdIt

dt

txdmltmgl

IM