basic properties of feedback

8/6/2019 Basic Properties of Feedback

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mF

bv

bv&

r

K G(s)

w

++ xu

Time (sec.)

Amplitude

Step Response

0 10 20 30 40 50 60 70 80

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

UUV Open-Loop Control

ME 531 - Design of Control Systems

Basic Properties of Feedback

Control of Translational Motion of a UUV

( )mv bv F

x v

ms b V s F s

sX s V s

&

&

( ) ( )

( ) ( )

+ ==

+ =

=

( ) ( )X s

F s s ms b s sG s

m

bm

( )

( )( )=

+=

+=

1 1

Open-Loop Control:

Without feedback, the position of the vehicle is very difficult to control.

If the vehicle is given a step input in r (force), what will happen to velocity? to position?

Note: response is rounded due to ( )s bm+

and integrates the step due to s.


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ME531 - Design of Control Systems


Page 2

r

D(s) G(s)

w

++ xu

+

-

e

Closed-Loop Control

For proportional control: (the control effort, u, is linearly proportional to the D s K p( ) =error, e)

( )

( )

X s

R s

D s G s

D s G s

s s

s s

s s

K

m

bm

K

m

bm

K

m

bm

K

m

p

p

p

p

( )

( )

( ) ( )

( ) ( )=

+=

+

++

=+ +1

1

2

Characteristic Equation: s sb mK

mp2

0+ + =

Poles: ( )s bm

bm

K

mp

1 2 2

2

2, =

for , the roots are on the real axis( )b m K mp2 2

for , the roots are complex.( )b m K mp2 2


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Page 3

-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

Real Axis

ImagA

xis

Locus of Roots for Proportional Control

Performance criteria: we want good transient response

< <


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Page 4

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

position(m)

time (sec)

Kp = 0.2

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1

position(m)

time (sec)

Kp = 0.5

0 20 40 60 80 1000

0.5

1

1.5

position(m)

time (sec)

Kp = 1

0 20 40 60 80 1000

0.5

1

1.5

position(m)

time (sec)

Kp = 2

0 20 40 60 80 1000

0.5

1

1.5

position(m)

time (sec)

Kp = 5

0 20 40 60 80 1000

0.5

1

1.5

2

position(m)

time (sec)

Kp = 20

Closed-Loop UUV System Response

Closed-loop step response:

low gains result in slow response

high gains result in high overshoot

settling time remains the same for each case - this is because the real part of the complex

closed-loop poles is unchanging with increasing Kp


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Page 6

r

D(s) G(s)

w

++ xu

+

-

e

Proportional/Derivative Control

Previously, our closed-loop system had too much overshoot, too slow of a rise time, and

too long of a settling time.

Basically, we need higher proportional gain (to speed up the response and decrease the

steady-state error) and more damping.

Try a combination of proportional and derivative control:

D s k k s u k e k de

dt p d p d ( ) = + = +

The control effort in this case is the sum of a term that is proportional to the error

and a term that is proportional to the derivative of the error.

( )( )

( )( )

( )

( )

( )

G ss s

X

R

DG

DG

s s

s ss s

m

bm

k s k

m

bm

k s k

m

bm

k s k

m

b k

m

k

m

d p

d p

d p

d p

( ) =+

=+

=+

+ +

=

+ +

+

+

+

+

1

21

1

Characteristic Equation (PD Control):

sb k

ms

k

m

d p20+

++ =


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Page 7

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Real Axis

ImagAxis

Locus of Roots for PD Control, Kp = 20, Kd variable

0

0

25

25

50

50

75

75

100

100 125125

Poles:

sb k

m

b k

m

k

m

d d p

1 2

2

2 2, = + +

Zero:

sk

k

p

d

=

Question: For a fixed value of proportional gain (say ), how does the value of affectkp = 20 kdthe location of the closed-loop poles?


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Page 8

Notice that higher values of cause the roots to move toward the negative real axis.Kd

> increases damping, decreases Mp

Notice that both and are increased. n

> faster and fastertr ts

Characteristic equation:

sb

ms

k

ms

k

m

km

s s bm

s km

ks

s s

d p

d p

d

m

bm

km

b

2

2

1

2

0

0

1 0

+ + + =

+ + + =

++ +

=

Note: the form of this last equation will help us in the future when we learn how to

construct the root-locus diagram (it is called Evans form).

Closed-Loop step response:

Note how the behavior changes with increasing :kd< decreased overshoot

< faster rise time

< faster settling time

This corresponds with the pole locations for changing .kd


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Page 9

0 20 40 600

0.5

1

1.5

2

position(m)

time (sec)

Kp = 20, Kd = 0

0 20 40 600

0.2

0.4

0.6

0.8

1

1.2

1.4

position(m)

time (sec)

Kp = 20, Kd = 50

0 20 40 600

0.2

0.4

0.6

0.8

1

1.2

1.4

position(m)

time (sec)

Kp = 20, Kd = 75

0 20 40 600

0.2

0.4

0.6

0.8

1

1.2

1.4

position(m)

time (sec)

Kp = 20, Kd = 100

0 20 40 600

0.2

0.4

0.6

0.8

1

1.2

1.4

position(m)

time (sec)

Kp = 20, Kd = 125

0 20 40 600

0.2

0.4

0.6

0.8

1

1.2

1.4

position(m)

time (sec)

Kp = 20, Kd = 150


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Page 10

Response to a step disturbance:

Note: the zero steady-state error criteria is still not satisfied!

X s

W s s s

m

b k

m

k

md p

( )

( )=

+ ++1

2

The DC gain is , which has been unchanged by the addition of derivative feedback.1

kp

* PD control has greatly improved the transient response, but disturbance rejection is still

unacceptable.


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Page 11

Proportional/Integral/Derivative Control (PID)

Integral feedback has the following form:

u t k e d It

t

( ) = 0

or:U s

E sD s

k

s

I( )

( )( )= =

Integral feedback can provide a finite value for the control even when the error is zero. This is

advantageous because it allows disturbances to be canceled with zero error.

the error, e, no longer needs to be finite to produce a control u to counter a constant

disturbance w.

control is a function of all past values ofe, rather than the current value only.

past errors charge up the integrator to the point that error are driven to zero (control

remains non-zero even though the error goes to zero).

For PID control:

u k e k de

dt

k e d

D s k k sk

s

p d I

t

t

p d

I

= + +

= + +

0

( )


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Page 12

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Axis

ImagAxis

Locus of Roots for PID Control, Kp = 30, Kd = 200, Ki variable

Root-Locus for PID control:


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Page 13

0 10 20 30 400

20

40

60

80

100

position(m)

time (sec)

Kp = 30, Kd = 200, Ki = 0.5

0 10 20 30 400

50

100

150

200

position(m)

time (sec)

Kp = 30, Kd = 200, Ki = 1

0 10 20 30 40

0

50

100

150

200

250

position(m)

time (sec)

Kp = 30, Kd = 200, Ki = 2

0 10 20 30 40

0

50

100

150

200

250

300

position(m)

time (sec)

Kp = 30, Kd = 200, Ki = 5

0 10 20 30 400

100

200

300

400

position(m)

time (sec)

Kp = 30, Kd = 200, Ki = 15

0 10 20 30 40-400

-200

0

200

400

600

800

position(m)

time (sec)

Kp = 30, Kd = 200, Ki = 50

Closed-Loop step response for PID control:


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Page 14

Disturbance response:

Equations of Motion:

( ) ( )

mx bx u w

mx bx k r x k d

dtr x k r x d w p d I

t

t

&& &

&& & ( )

+ = +

+ = + + + 0

Clearing the integral:

mx bx k x k x k x k r k r k r wd p I d p I

&&& && && & && & &+ + + + = + + +

The dynamics are only dependent on (not w).&w For a constant w, and there will be no induced error.&w = 0


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Page 15

( ) ( )

( )

D s

k s k s k

s

X s

R s s s

X s

R s

s s

s s s

d p I

k s k s k

m

bm

k s k s k

m

km

k

mk

m

bm

km

k

mk

m

d p I

d p I

d p I

d p I

( )

( )

( )

( )

( )

=

+ +

=+ +

=+ +

+ + + +

+ +

+ +

2

2

2

3 2

2

2

Characteristic Equation:

( )

( )[ ]

s s s

k

ms

b

m

k

ms

k

ms

ks s s

bm

km

k

mk

m

I d p

I

m

bm

km

k

m

d p I

d p

3 2

3 2

1

2

0

0

1 0

+ + + + =

+ + +

+

=

++ + +

=

DC gain of .X

R= 1

What about disturbance rejection?

( )X s

W s s s sH s

x sX s sH s W s ss

H s

sm

bm

km

k

mk

m

ss s s s

d p I

( )

( )( )

lim ( ) lim ( ) ( ) lim ( )

=+ + + +

=

= = = =

3 2

0 0 0

10

therefore we have zero steady-state error!

Conclusion: we can satisfy all transient response and disturbance rejection criteria by using a PID

controller.


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Page 16

Integrator Advantages:

Zero steady state error in response to constant reference inputs

Zero steady state error in response to constant disturbances

Integrator Disadvantages:

De-stabilizing influence on dynamic behavior

< increases overshoot

< increases settling time

Integrator windup

Integrator Windup and Anti-Windup Techniques

What is windup and how is it caused?

Windup is a condition where the integrator becomes over-charged inadvertently -

or wound-up.

Windup occurs when an integrator integrates when it shouldnt.

< Typically this happens when the actuator saturates.

< The actuator is trying as hard as it can, but is unable to track the reference

command. The integrator keeps adding up the error.

The effect is best illustrated by example. Consider the UUV translation control:

Assume that the thrusters saturate at 50 N. Give the vehicle a step input of 10 m.

Tether snags causing the integrator to windup.

Saturation occurs.


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Page 17


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Page 19

Integrator Wind-up Prevention:

Turn off integrator when saturation of actuator occurs

Note: When exceeds (saturation), the feedback loop around the integrator isuc umaxactivated which quickly drives to zero - effectively turning off the integrator.e1

Limit the output of the integrator

When the integrator output is bounded, its damaging effects on the dynamics are limited,

but its ability to overcome disturbances is also limited.

Turn on the integrator only after transients have died out.

When giving smooth trajectory commands, activate integrator wen desired

velocity = 0.

Turn on when t ts> >4 6.

Turn on when the error is less than some , where the appropriate value of is

chosen by taking into account the DC gain.


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Page 24

r

D(s) G(s)

w

++ xu

+

-

e

H(s)

Steady-State Tracking and System Type

Inputs:

Input Polynomial System Type

Step 1(t) 0th order

Ramp t 1st order

Parabolic t2 2nd

System type describes the order of the input polynomial going into a system for whihc the

tracking error will be constant.

Consider the following system:

[ ]

Y s

R s

DG

DGHT s

E s R s Y s R s T s R s R s T s

E s

R sT s

( )

( )( )

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

( )

( )( )

= + =

= = =

=

1

1

1


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Page 25

Consider the general input:

r tt

kt

k

( )!

( )= 1

If: unit stepk = 0ramp with unit slopek = 1parabola with unit 2nd derivativek = 2

Laplace transform of :r t( )

R ss

k( ) = +1

1

Applying the Final Value Theorem to this general case:

[ ]

[ ]

[ ]

e sE s

e sR s T s

e ss

T s

es

T s

sss

sss

sss

k

sss

k

=

=

=

=

+

lim ( )

lim ( ) ( )

lim ( )

lim ( )

0

0

01

0

1

1 1

11

The result of evaluating this limit can be either zero, a non-zero constant, or infinite.

The system type is the value ofkfor which the above is a non-zero constant.ess

For example: if and is a constant, then the system is type zero - which meansk = 0 essthat it has a constant steady-state error to a step input.

If , then the system is type I, and it has a constant for a ramp and zerok = 1 essfor a step input.ess


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Page 26

r x

+

-

e

K

A

s( ) +11

s

Example: Determine the system type for the following system


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Page 27

r x

+

-

eK T s

T sp DI

11

+ +

12Js

Example: Determine the system type for the following system


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Page 28

Stability

Bounded Input - Bounded Output (BIBO) Stability:

A system is said to be BIBO stable if every bounded input results in a bounded output -

independent of what might be happening to some internal states of the system!

Definition: A system with impulse response h(t) is BIBO stable if and only if the integral

h d( ) <

Asymptotic Internal Stability:

A system is asymptotically internally stable if the output and all internal variables never become

unbounded and go to zero as time goes to infinity for sufficiently small initial conditions.

What does this imply?

Consider the LCC system with the characteristic equation:

s a s a s an n n

n+ + + + =

1

1

2

20K

With characteristic roots such that:pi

( )( ) ( )s a s a s a s p s p s pn n n

n n+ + + + = =

1

1

2

2

1 2 0K K

The solution to the differential equation whose characteristic equation is given above is:

where are the roots, and values depend on ICs y t K eip t

i

n

i( ) ==

1

pi Ki

The system is stable if and only if every term goes to zero as .t for allep ti 0 pi

This will happen only when all the poles are strictly in the left half plane (LHP) where{ }Re pi < 0

In other words, a system is stable if the roots of its characteristic equation lie in the LHP.

This can easily be determined with MATLAB or with any polynomial root solver.


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Page 29

Rouths Stability Criterion

This was especially useful before computer-aided analysis was available.

It is still useful for determining ranges of CE coefficients for stability.

It allows us to make statements about a systems stability without solving for the roots.

Consider the CE for a system of order n:

a s s a s a s a s an n n

n n( ) = + + + + + =

1

1

2

2

1 0K

A necessary condition for stability is that all of the coefficients must be positive (if{ }aiis negative or zero, CE will have roots in the RHP){ }ai

Q: If all of the coefficients are positive, is the system stable? - not necessarily!

A stronger test is needed. The following test was proposed by Routh in 1879:

Calculate a triangular array based on . A necessary and sufficient condition for{ }aistability is that all of the elements in the first column of the Routh array be positive.

How to construct the Routh array: (see pp 215-216 in text)

The first row contains all the even numbered coefficients and the second row contains all

the odd numbered coefficients of the CE.

Routh Array:

n s a a

n s a a a

n s b b b

n s c c c

s

s

s

n

n

n

n

:

:

:

:

1

1

2

3

2

1 0

0

2 4

1

1 3 5

2

1 2 3

3

1 2 3

2

0

K

K

K

K

M M M M M M


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Page 30

where:

b

aa a

ab

aa a

a

c

a a

b b

bc

a a

b b

b

1

2

1 3

1

2

4

1 5

1

1

1 3

1 2

1

2

1 5

1 3

1

1 1

= =

= =

For the Routh array, the number of sign changes in the first column is equal to the number of

RHP poles.

Example:

a s s s s s s( ) = + + + + + =5 4 3 22 4 3 2 2 0


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Page 31

See pp 217-219 in text for a discussion of special cases for Rouths test:

Special Case I: zero in 1st column

Special Case II: row of zeros

Look over example 4.20 and 4.21: how to use Rouths test to determine stability versus

parameter range.

basic properties of feedback

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