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Lecture 27: Tue Dec 5, 2017

Reminder:

HW11: assigned but not to be graded, not to be turned in.

final exam is Thursday Dec 14, 8am – 10:50am

Control

1

Quiz 1 + Quiz 2

3

9

6

11

80 100 120 140 160 180 200min = 62

max = 192mean = 152

median = 158

SCORE

2

60

192189189187185184183182181180180178177168163161160158154152149146144143142142139138136116113 90 89 68 62

22

2

Reading

3

Closed-Loop Controllers

Gc( s )r( t ) e( t ) y( t )

+

–

Gp( s )

PPIPDPID

Kp

Kp + Ki/s

Kp + Kds

Kp +Ki/s + Kds

Gc( s )

H( s ) =Gc( s )Gp( s )

1 + Gc( s )Gp( s )

4

P Controller for 1st order Gp(s) =

• P control ⇒ Gc( s ) = Kp

⇒ H( s ) =

• moves pole location from –a to – (a + Kpb)

• Can stabilize an unstable system!

• Nonzero steady-state tracking error y∞= H0 = , can make arbitrarily small

bs a+------------

Kbs a Kb+ +-----------------------------

1

1aKb--------+

-----------------

1

H0

0 t

STEADY-STATE TRACKING ERROR

5

A Plant that IntegratesVery common, e.g. a wheeled vehicle:

⇒ plant is Gp( s ) = = (observe built-in integrator)

Pop Quiz:(a) Is the plant stable?

(b) Can it be stabilized via P controller?

force x( t ) position y( t )

applied force friction

x( t ) – d y( t ) = m y( t ) d2

dt2--------d

dt-----

1/ms s d/m+ ------------------------------ b

s s a+ ---------------------

6

P Control for Gp(s) =

P control:

⇒ H(s) = = =

DC gain is H( 0 ) = 1 ⇒ no steady-state tracking error!

In fact, any time the “open loop transfer function” Gc( s )Gp( s ) has a pole at the origin, the closed-loop system will have zero steady-state tracking error.

This makes you think ... if the plant doesn’t have a pole at s = 0, why not put one in the controller?

bs s 1+ ---------------------

Kbs s 1+ --------------------

1 Kbs s 1+ --------------------+

---------------------------- Kbs s 1+ Kb+------------------------------------- Kb

s2 s Kb+ +-------------------------------

7

PI Controller

Gc( s ) = Kp + :

⇒ input to plant is x( t ) = Kpe( t ) + Ki e( )d

Interesting property: Error need not be nonzero to drive plant!

How to choose the parameters {Kp, Ki}?

to put poles where we want them.

e( t ) x( t )

Ki

s------

Kp

Ki

s------

0

t

8

Closed-Loop with PI and 1st-ord Plant

Suppose Gp( s ) = and Gc( s ) = Kp +

⇒ H( s ) = =

⇒ a 2nd-order system with steady-state step response H0 = 1:

How to choose the parameters {Kp, Ki}?

to get desired overshoot

to get desired settling time, etc.

bs a+-------------

Ki

s------

GG1 GG+-------------------

Kps Ki+ bs s a+ Kps Ki+ b+--------------------------------------------------------------

H0 = 1 PERFECT TRACKING!

9

PI Analysis

H( s ) = = .

Three equivalent questions:

How to choose the parameters {Kp, Ki}?

How to choose the parameters {, n}?

Where to place the poles?

A common strategy: Start with a desired settling time, and desired overshoot:

Choose Kp to get desired settling time ts ≈

Choose Ki = to get desired overshoot

Kp s Ki+ bs2 a P0aKp+ s Ki b+ +------------------------------------------------------------------ Kps Ki+ b

s2 2n s n2+ +

----------------------------------------------

4

n

----------

16

b2ts2---------------

10

Relationships2n = a + bKp

n2 = Kib

⇒ can solve for Ki and Kp as a function of desired n and

11

Example

Gp( s ) = and Gc( s ) = Kp +

Find Kp and Ki so that settling time is ts = 0.5 sec, and overshoot is 4.2%.

4s 3+-------------

Ki

s------

12

Solution

H( s ) = =

1. equate coeffs of s1:

2n = 3 + 4Kp ⇒ Kp = (2n – 3) = ( – 3) = .

2. Equate coeffs of s0, using fact that 4.2% overshoot means that = 0.707:

n2 = 4Ki ⇒ Ki = n

2 = = = .

4 Kp s Ki+ s2 3 4Kp+ s 4Ki+ +-------------------------------------------------------- 4 Kps Ki+

s2 2n s n2+ +

--------------------------------------------

14--- 1

4--- 8ts---- Kp = 3.25

14--- 1

4--- 4

ts--------

2 14--- 4

0.707 0.5 ------------------------------- 2 Ki = 32

13

A Linear Amplifier

-4 -2 2 4

-10

10y( t )

x( t )

y( t ) = 10x( t )

14

A Nonlinear Amplifier

How to get both linear behavior and large outputs?

10tanh( )x( t ) y( t )

-4 -2 2 4

-10

10y( t )

x( t )

A memoryless system that is ≈linear for small inputs (less than ±0.3),amplifying by a factor of 10, yielding small outputs (less than ±3):

15

Open Loop Options

2 4 6 8 10

-10

10 r( t )

t

OPEN 1

LARGE, BUT NOT LINEAR

16

Open Loop Options

OPEN 1

2 4 6 8 10

-10

10 r( t )

OPEN 2

t

LINEAR, BUT NOT LARGE

17

Linearize it using Integral Controller:Ki

r( t ) e( t ) y( t )

+

–

x( t )

s

18

Linearize it using Integral Controller:

OPEN 2

OPEN 1

Kir( t ) e( t ) y( t )

+

–

x( t )

2 4 6 8 10

-10

10 r( t )

t

y( t )(CLOSED)

(Ki = 100)

s

19

Scatter Plots

-1 1

-10

10 CLOSED

OPEN 1

OPEN 2

LOOP

20

Examples of Controllers

Gc( s )R( s ) E( s ) Y( s )

+

–

Gp( s )

PPIPDPID

Kp

Kp + Ki/s

Kp + Kds

Kp +Ki/s + Kds

Gc( s )

Also: On-Off control (not considered)

21

So FarFor a 2nd-order plant:

• P control: imperfect tracking, oscillations & overshoot

• PI control: perfect tracking, oscillations & overshoot

22

Error Prediction?

⇒ PD controller: Gc( s ) = Kp + Kds ... it predicts errors

⇒ input to plant is x( t ) = Kpe( t ) +Kd e( t )

Impact on performance:

reduces overshoot

dampens oscillations

t

t + T

e( t ) TIME

e(t + T ) ≈ e( t ) + T e( t )ddt-----

Linear extrapolation:

ddt-----

23

PD Control: 1st-Order Plant

⇒ H( s ) = =

⇒ a 1st-order system with steady-state step response H0 = :

r( t ) y( t )

+

–

bs + a

Kp + Kds

GcGp

1 GcGp+-----------------------

Kp Kds+ bs a b Kp Kds+ + +---------------------------------------------------

1

1 1KP0

-----------+------------------

1

H0

0 t

STEADY-STATE TRACKING ERROR

24

Proportional-Integral-Derivative (PID) Control

Gc( s ) = Kp + + Kds

⇒ input to plant is x( t ) = Kpe( t ) +Ki e( )d + Kd e( t )

Ki

s------

0

t

ddt-----

25

PID Control: 1st-Order Plant

What is order of closed-loop system?

Is there any steady-state tracking error?

r( t ) y( t )

+

–

bs + a

Kp + Ki/s + Kds

26

PID Tuning Not Easy: Trial & Error!Heuristic strategies common, e.g.:

• Set Ki = Kd = 0

• Increase Kp until oscillates; back off 50%

• Increase Ki to get good steady-state tracking, overshoot OK

• Increase Kd to reduce overshoot, damp oscillations

27

Impact of Increasing Gains

28

PID Demohttps://goo.gl/tF881V

29

Example: Move to Desired Position

x( t )

y( t )

k

mINPUT

OUTPUT

yd DESIRED POSITION

30

Example

x( t )

y( t )

k

mINPUT

OUTPUT

yd

e( t ) ERROR

DESIRED POSITION

CONTROLLER

FEEDBACKCONTROL

31

Example

x( t )

y( t )

my..

= k(x – y)

k

mINPUT

OUTPUT

⇒ Gp( s ) = 02

s2 +02

where 0 = k/m

yd

e( t ) ERROR

DESIRED POSITION

CONTROLLER

32

Equivalent Block Diagram

What controller will both stabilize and provide perfect tracking?

• P control? No, not stable. Merely shifts resonance freq.

• PI? No, not stable. (Missing s2 term in denom ⇒ pole on j axis.)

• PD? It stabilizes, but with tracking error: H0 = .

• PID?

r( t ) = yd y( t )+

–

Gc( s )02

s2+02

11 1/Kp+-----------------------

33

Does P Control Stabilize and Track?

34

Does PI Control Stabilize and Track?

35

Does PD Control Stabilize and Track?

36

Does PID Control Stabilize and Track?

37

PID Does it All

H( s ) = =

=

= .

We have enough degrees of freedom (three “knobs”) to stabilize system.

And with perfect tracking(!):

H0 = = 1

GG1 GG+-------------------

Kp Ki/s Kds+ + 0

2

s2 02+

-------------------

1 ...+---------------------------------------------------------------------

Kps Ki Kds2+ + 0

2

s s2 02+ Kps Ki Kds

2+ + 02+

------------------------------------------------------------------------------------------

Kps Ki Kds2+ +

s3/02 Kds

2 1 Kp+ s Ki+ + +---------------------------------------------------------------------------------

Ki

Ki

------

38

Example: w0 = 2, Kp = Ki = 10, Kd = 1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

TIME (s)

y( t )

39

Example: Electric Water HeaterA 10-V signal applied at time zero to an electric heating element in water leads to the following experimentally measured water temperature response:

Design a control system that “settles” water as quickly as possible to 200°.

0 t

70°

100°89°

10 s

WATER TEMPERATURE