Feedback Control

Lecture 5

Feedback Control Systems

President University Erwin Sitompul FCS 5/1

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


Feedback Control A Feedback Control seeks to bring the measured quantity to

its desired value or set-point (also called reference trajectory) automatically.

There are two possible causes of difference between measured value and desired value: Disturbance and noise Change of set point, where the control system must act to

bring the measured quantity to the new set point

Chapter 4 A First Analysis of Feedback


The Three-Term Controller: PID Control The PID Controller is the most popular feedback control

algorithm used in process control and industries.

It is robust yet simple, easy to understand, and can provide excellent control performance despite the variation of dynamic characteristics of the process.

As the name suggests, the PID Controller consists of three basic terms: the Proportional term, the Integral term, and the Derivative term.

Each term can be activated separately (P, I, or D) or combined to obtain the desired controller.

Depending on the characteristics of the process, the following controller are generally used: P, PI, PD, or PID.



To be controlled

Provides excitation to the process

The control variable or manipulated variable

The input to the controller, the tracking error, will vary due to the change in measured value or desired value.

In response, the controller produces control signal in order to minimize the error.

The Three-Term Controller: PID Control

Measured variable



( ) ( )pU s k E s

( ) ( )pu t k e t

P – Term

Proportional Gain, first step to close a feedback loop

Proportional TermChapter 4 A First Analysis of Feedback

t

( )e t

1

t

( )u t

pk


( ) ( )ikU s E ss

0

( ) ( )t

iu t k e d

I – Term

Integral Gain, to assure zero error to constant

reference and disturbance inputs

Integral TermChapter 4 A First Analysis of Feedback

t

( )e t

1

t

( )u t

ik

1• How if P and I are

combined together?


( ) ( )dU s k s E s

( )( ) d

de tu t k

dt

D – Term

Derivative Gain,to improve (or even realize) stability

and good dynamic response

Derivative TermChapter 4 A First Analysis of Feedback

t

( )u t

dk

t

( )e t

1

1

• How if P and D are combined together?

• How if P, I, and D?


Altogether, the PID Controller may be represented as:

( ) ( )ip d

kU s k k s E s

s

0

( )( ) ( ) ( )

t

p i d

de tu t k e t k e d k

dt

Another way to formulate PID Controller in time domain is:

0

1 ( )( ) ( ) ( )

t

p di

de tu t k e t e d

dt

d p dk k ,pii

kk

Integral time constant Derivative time constant

PID ControlChapter 4 A First Analysis of Feedback


The proportional term makes the response of the system faster and reduces error due to disturbance by increasing the overall gain of the system, but it alone may still allow a steady-state error.

The integral term may eliminate the steady-state error to a constant input, although at the cost of deteriorating the dynamic response of the system by causing slower response of the system to varying set point.

The derivative term typically makes the system faster and better damped, although at the cost of less stability and of high sensitivity to high frequency disturbance and sensor noise.

Note that the effects of each term may not be accurately identifiable because P, I, and D gains are

dependent of each other

Effects of Each TermChapter 4 A First Analysis of Feedback


Problem ExampleConsider the mass-spring-damper system below.

The dynamic model of the process can be derived as:

Taking the Laplace Transform yields:

The transfer function is then given by:

( ) ( ) ( ) ( )mx t bx t kx t F t

2 ( ) ( ) ( ) ( )ms X s bsX s kX s F s

2

( ) 1

( )

X s

F s ms bs k



Let m = 1 kg, b = 10 N.s/m, and k = 20 N/m, while the applied force F = 1 N.

The transfer function can now be written as:

The process will now be controlled by a PID-Controller. The contribution of each controller parameter kp, ki, and kd

will be observed on an effort to obtain an overall system with small rise time, minimum overshoot, small settling time, and zero steady-state error.

Problem Example

2

( ) 1

( ) 10 20

X s

F s s s



The open-loop transfer function of the process is:

Problem Example: Open-Loop

ss0

lim ( ) lim ( )t sx t x s X s

20

1lim ( )

10 20ss F ss s

20

1 1lim

10 20sss s s

1

20

The steady-state output value of the system is:

The gain of the process is 1/20 = 0.05



Problem Example: Open-LoopChapter 4 A First Analysis of Feedback

Time [s]

Dis

plac

emen

t [m

]

0 0.5 1 1.5 20

0.01

0.02

0.03

0.04

0.05

0.06

Settling time is around 1.5 s

No overshoot

The open-loop unit-step response of the process can be shown as:


Problem Example: Closed-LoopChapter 4 A First Analysis of Feedback

The process is now placed into a closed-loop system as follows:

≡

Now, the gain of the process is 1/21 = 0.0476 (decrease)


Problem Example: Closed-LoopChapter 4 A First Analysis of Feedback

Time [s]D

ispl

acem

ent [

m]

0 0.5 1 1.5 20

0.10.20.30.40.50.60.70.80.9

11.1

The closed-loop unit-step response of the system can be shown as:

Time [s]

Dis

plac

emen

t [m

]

0 0.5 1 1.5 200

0.01

0.02

0.03

0.04

0.05

0.06

0.0476

ess = 0.9524

Set point

• Large steady-state error ess• Note: no controller

installed yet


Problem Example: Using P-ControllerChapter 4 A First Analysis of Feedback

2

( )

( ) 10 (20 )p

p

kX s

R s s s k

With the P-Controller in the loop, the transfer function of the system is given as:

The parameter kp can now be tuned to obtain the desired closed-loop response of the overall system.


Problem Example: Using P-ControllerChapter 4 A First Analysis of Feedback

Time [s]

Dis

plac

emen

t [m

]

0 0.5 10

0.2

0.4

0.6

0.8

1

1.2

1.4

Let kp = 300, the closed-loop transfer function of the system is:

2

( ) 300

( ) 10 320

X s

R s s s

The unit-step response is now:

Final value 300/320 = 0.9375 Steady-state error

reduced

Rise time reduced

Settling time slightly reduced

Overshoot increased

Effects ofP-Controller


Problem Example: Using PD-ControllerChapter 4 A First Analysis of Feedback

2

( )

( ) (10 ) (20 )p d

d p

k k sX s

R s s k s k

With the PD-Controller in the loop, the transfer function of the system is given as:

The parameter kp and kd can now be tuned to obtain the desired closed-loop response of the overall system.


Problem Example: Using PD-ControllerChapter 4 A First Analysis of Feedback

Time [s]

Dis

plac

emen

t [m

]

0 0.2 0.4 0.60

0.2

0.4

0.6

0.8

1

1.2

1.4

Let kp = 300, kd = 10, the closed-loop transfer function of the system is:

2

( ) 10 300

( ) 20 320

X s s

R s s s


Final value 300/320 = 0.9375 Unchanged Steady-state error still

0.0625

Rise timealmost unchanged

Settling time reduced even more

Overshoot reduced

Effects ofPD-Controller


Problem Example: Using PI-ControllerChapter 4 A First Analysis of Feedback

3 2

( )

( ) 10 (20 )p i

p i

k s kX s

R s s s k s k

With the PI-Controller in the loop, the transfer function of the system is given as:

The parameter kp and ki can now be tuned to obtain the desired closed-loop response of the overall system.


Problem Example: Using PI-ControllerChapter 4 A First Analysis of Feedback

Time [s]

Dis

plac

emen

t [m

]

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

Let kp = 300, ki = 160, the closed-loop transfer function of the system is:

3 2

( ) 300 160

( ) 10 320 160

X s s

R s s s s


Effects ofPI-Controller

Overshoot increased

Final value 160/160 = 1 Zero steady-state error

Rise timealmost unchanged

Settling time increased Actually, both P and I

terms tend to increase overshoot and reduce rise

time at the same time (double effect)


Problem Example: Using PID-ControllerChapter 4 A First Analysis of Feedback

2

3 2

( )

( ) (10 ) (20 )d p i

d p i

k s k s kX s

R s s k s k s k

With the PID-Controller in the loop, the transfer function of the system is given as:

The parameter kp, ki, and kd can now be tuned to obtain the desired closed-loop response of the overall system.


Problem Example: Using PID-ControllerChapter 4 A First Analysis of Feedback

Let kp = 300, ki = 160, kd = 10, the closed-loop transfer function of the system is:

2

3 2

( ) 10 300 160

( ) 20 320 160

X s s s

R s s s s

Effects ofPID-Controller

Time [s]

Dis

plac

emen

t [m

]

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2Overshoot significantly reduced

Small rise time,almost unchanged

Settling time unchanged


Final value 160/160 = 1 Zero steady-state error


Problem ExampleChapter 4 A First Analysis of Feedback

Time[s]

Dis

plac

emen

t [m

]

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4

P-ControllerPD-ControllerPI-ControllerPID-Controller

Original Process

To ease the comparison, all the unit-step responses are plotted on the same figure.


ConclusionsChapter 4 A First Analysis of Feedback

Proportional control can reduce steady-state errors, but the high gains almost always destabilize the system.

Integral control provides total reduction in steady-state errors, but often makes the system less stable.

Derivative control usually increases damping and improves stability, but has almost no effect on the steady-state error.

These 3 kinds of control combine to form the classical PID-Controllers which provide reasonable control for most industrial processes, provided that the performance demand is not too high.


Control of 1st Order System Using P-ControllerChapter 4 A First Analysis of Feedback

( ) 1( ) 1

11

p

p

p p

kkY s s

kR s s ks

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

( ) ( )1p

p

kR s R s

s k

1

( )1 p

sR s

s k

ss0

lim ( ) lim ( )t se t e s E s

0

1 1lim

1sp

sss k s

For a unit-step reference trajectory, the steady-state error:

ss

10

1 p

ek

In order to reduce ess, kp must be increased


Control of 1st Order System Using P-ControllerChapter 4 A First Analysis of Feedback

1

1TG GD

S Close-loop sensitivity

1

11

TG

pk

s

S1

1 p

s

s k

• At high frequency, S≈1• At low frequency, if kp increased,

S decreases, T increases

• If kp is increased to reduce ess, then the system is less vulnerable to disturbance with low frequency, but more vulnerable to sensor noise with low frequency


Control of 1st Order System Using PI-ControllerChapter 4 A First Analysis of Feedback

( ) 1( )

11

ip

ip

kk

sY s s

kR sk

ss

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

2( ) ( )

(1 )p i

p i

k s kR s R s

s k s k

2

2( )

(1 )p i

s sR s

s k s k

2

( )

( ) (1 )p i

p i

k s kY s

R s s k s k


Control of 1st Order System Using PI-ControllerChapter 4 A First Analysis of Feedback

ss0

lim ( ) lim ( )t se t e s E s

2

20

1lim

(1 )sp i

s sss k s k s

For a unit-step reference trajectory, the steady-state error:

ss 0e No steady-state error

1

1TG GD

S

2

2 (1 )TG

p i

s s

s k s k

S

• At high frequency, S≈1, T≈0• At low frequency, S≈0, T≈1

• The system is very sensitive to disturbance with high frequency and sensor noise with low frequency


Control of 2nd Order System Using P-ControllerChapter 4 A First Analysis of Feedback

2

( )

( )p

p

kY s

R s Ms Bs k

2

2 22n

n ns s

2

p

p

k

MkB

s sM M

2 n

B

M

2 pn

k

M

pn

k

M

2 p

B

k M

( ) ( ) ( )E s R s Y s 2

2

( )p

Bs s

M R skB

s sM M



ss0

lim ( )s

e s E s

For R(s) = 1/s,

2

ss 02

1lims p

Bs s

Me skB s

s sM M

ss 0e No steady-state error

ss0

lim ( )s

e s E s

For R(s) = 1/s2,

2

ss 202

1lims p

Bs s

Me skB s

s sM M

ss 0p

Be

k

The error ess can be reduced, but cannot equal zero



ssp

Be

k

pn

k

M

2 p

B

k M

In order to reduce the steady-state error , Decrease B

ζ decreased Increase kp

ωn increased while ζ decreased In any way, ζ will decrease

maximum overshoot increases21

pM e


Control of 2nd Order System Using PD-ControllerChapter 4 A First Analysis of Feedback

2

( )

( ) ( )d p

d p

k s kY s

R s Ms B k s k

2 ( )

d p

pd

k s k

MkB k

s sM M

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

2

2

( )( ) pd

Bs s

M R skB k

s sM M

( )2 d

n

B k

M

2 pn

k

M

pn

k

M

( )

2d

p

B k

k M



ss0

lim ( )s

e s E s

For R(s) = 1/s2,

2

ss 202

1lim

( )s pd

Bs s

Me skB k s

s sM M

ss 0p

Be

k



ssp

Be

kIn order to reduce the steady-state error ,

Decrease B ζ decreased, but can be compensated by increasing kd

Increase kp ωn increased and ζ decreased, but can be compensated by increasing kd

Steady-state response (ess) can be tuned without affecting transient response (Mp)

pn

k

M

( )

2d

p

B k

k M


Homework 5 No.1.

The performance of P- and PD-Controller to control the 2nd order system are already analyzed on the previous pages.Now, conduct further analysis for PI-Controller. Explain about its ability to reduce steady-state error in the presence of step input and ramp input.

No.2.A motor used for laser surgery requires high accuracy for position and velocity response. Consider the system shown below.



Homework 5 No.2. (continued)

The amplifier gain K must be adjusted so that the steady-state error for a ramp input r(t) = At (where A = 1 mm/s) is less than or equals to 0.1 mm, while a stable response is maintained.(a) Determine the value of K to meet the requirement.(b) In MATLAB Simulink, build the motor system as given

previously. Now, prove the answer of (a), that ess = 0.1 mm can be achieved for that certain value of

K.(c) A technician claims that he can obtain ess = 0.02 mm.

What is your opinion? Is his claim possible to do or not? How? Why? Explain.

Due: 16.10.2013.


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