chapter 10 pid

7/21/2019 Chapter 10 PID

http://slidepdf.com/reader/full/chapter-10-pid 1/36

Chapter 10 – The Design of Feedback Control Systems

PID

Compensation Netorks



Different Types of Feedback Control

On-Off Control

This is the simplest form of control.



Proportional ControlA proportional controller attempts to perform better than the On-off type by

applying power in proportion to the difference in temperature between the

measured and the set-point. As the gain is increased the system responds faster tochanges in set-point but becomes progressively underdamped and eventually

unstable. The final temperature lies below the set-point for this system because

some difference is required to keep the heater supplying power.



Proportional! Deri"ati"e Control

The stability and overshoot problems that arise when a proportionalcontroller is used at high gain can be mitigated by adding a term proportional

to the time-derivative of the error signal. The value of the damping can be

adjusted to achieve a critically damped response.



Proportional#Integral#Deri"ati"e Control

Although ! control deals neatly with the overshoot and ringing problems associated with proportional control it does not cure the

problem with the steady-state error. "ortunately it is possible to eliminate

this while using relatively low gain by adding an integral term to the

control function which becomes



The Characteristics of P! I! and D controllers

A proportional controller #$p% will have the effect of reducing the

rise time and will reduce& but never eliminate& the steady-state

error.

An integral control #$i% will have the effect of eliminating the

steady-state error& but it may make the transient response worse.

A derivative control #$d% will have the effect of increasing thestability of the system& reducing the overshoot& and improving the

transient response.



Proportional Control

'y only employing proportional control& a steady state erroroccurs.

Proportional and Integral Control

The response becomes more oscillatory and needs longer to

settle& the error disappears.

Proportional! Integral and Deri"ati"e Control

All design specifications can be reached.



C$ %&SP'NS& %IS& TI(& ')&%S*''T S&TT$IN+ TI(& S,S &%%'%

-p !ecrease (ncrease )mall Change !ecrease

-i !ecrease (ncrease (ncrease *liminate

-d )mall Change !ecrease !ecrease )mall Change

The Characteristics of P! I! and D controllers



Tips for Designing a PID Controller

+. Obtain an open-loop response and determine what needs to be improved

,. Add a proportional control to improve the rise time

. Add a derivative control to improve the overshoot

. Add an integral control to eliminate the steady-state error

/. Adjust each of $p& $i& and $d until you obtain a desired overall

response.

0astly& please keep in mind that you do not need to implement all three controllers

#proportional& derivative& and integral% into a single system& if not necessary. "ore1ample& if a ( controller gives a good enough response #like the above

e1ample%& then you don2t need to implement derivative controller to the system.

$eep the controller as simple as possible.



num=1;den=[1 10 20];step(num,den)

'pen,$oop Control , &.ample

3 s# %+

,

http://www.engin.umich.edu/group/ctm/extras/step.html

http://www.engin.umich.edu/group/ctm/extras/step.html



Proportional Control , &.ample

The proportional controller #$p% reduces the rise time& increasesthe overshoot& and reduces the steady-state error.

5AT0A' *1ample

$p6447

num68$p97

den68+ +4 ,4:$p97t64;4.4+;,7

step#num&den&t%

Time (sec.)

A m

p l i t u d e

Step Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4From: U(1)

T o : Y ( 1 )

T s# %$p

,⋅

Time (sec.)

A m p l i t u d e

Step Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1From: U(1)

T o : Y ( 1 )

$644 $6+44



Time (sec.)

A m p l i t u d e

Step Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4From: U(1)

T o : Y ( 1 )

$p6447

$d6+47

num68$d $p97

den68+ +4:$d ,4:$p97

t64;4.4+;,7

step#num&den&t%

Proportional , Deri"ati"e , &.ample

The derivative controller #$d% reduces both the overshoot and thesettling time.

5AT0A' *1ample T s# %$d s⋅ $p+

,⋅

Time (sec.)

A m p l i t u d e

Step Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1From: U(1)

T o : Y ( 1 )

$d6+4

$d6,4



Proportional , Integral , &.ample

The integral controller #$i% decreases the rise time& increases boththe overshoot and the settling time& and eliminates the steady-state

error

5AT0A' *1ample

Time (sec.)

A m p l i t u d e

Step Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4From: U(1)

T o : Y ( 1 )

$p647

$i6<47

num68$p $i97

den68+ +4 ,4:$p $i97

t64;4.4+;,7

step#num&den&t%

T s# %$p s⋅ $i+

- ,

⋅ ⋅

Time (sec.)

A m

p l i t u d e

Step Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4From: U(1)

T o : Y ( 1 )$i6<4

$i6+44



)ynta1

rltool

rltool#sys%

rltool#sys&comp%

%$T''$



%$T''$



Consider the following configuration;

&.ample , Practice



The design a system for the following specifications;

= >ero steady state error

= )ettling time within / seconds

= ?ise time within , seconds

= Only some overshoot permitted

&.ample , Practice



$ead or Phase,$ead Compensator /sing %oot $ocs

A first-order lead compensator can be designed using the root locus. A lead compensatorin root locus form is given by

where the magnitude of @ is less than the magnitude of p. A phase-lead compensator

tends to shift the root locus toward the left half plane. This results in an improvement inthe system2s stability and an increase in the response speed.

hen a lead compensator is added to a system& the value of this intersection will be a

larger negative number than it was before. The net number of @eros and poles will be the

same #one @ero and one pole are added%& but the added pole is a larger negative numberthan the added @ero. Thus& the result of a lead compensator is that the asymptotes2

intersection is moved further into the left half plane& and the entire root locus will be

shifted to the left. This can increase the region of stability as well as the response speed.

3 c s# % s @+#



(n 5atlab a phase lead compensator in root locus form is implemented by using the

transfer function in the form

numlead=kc*[1 z];

denlead=[1 p];

and using the conv() function to implement it with the numerator and denominator

of the plant

newnum=conv(num,numlead);

newden6conv#den&denlead%7

$ead or Phase,$ead Compensator /sing %oot $ocs



$ead or Phase,$ead Compensator /sing Freency %esponse

A first-order phase-lead compensator can be designed using the frequency response. A lead

compensator in frequency response form is given by

(n frequency response design& the phase-lead compensator adds positive phase to the systemover the frequency range. A bode plot of a phase-lead compensator looks like the following

3c s# %+ α τ⋅ s⋅+( )

α + τ s⋅+( )⋅ p

+

τ@

+

ατωm @ p⋅ sin φm( )

α +−

α ++




Additional positive phase increases the phase margin and thus increases the stability of

the system. This type of compensator is designed by determining alfa from the amount of

phase needed to satisfy the phase margin requirements& and determining tal to place the

added phase at the new gain-crossover frequency.

Another effect of the lead compensator can be seen in the magnitude plot. The lead

compensator increases the gain of the system at high frequencies #the amount of this gain

is equal to alfa. This can increase the crossover frequency& which will help to decrease the

rise time and settling time of the system.



(n 5atlab& a phase lead compensator in frequency response form is

implemented by using the transfer function in the form

numlead=[aT 1];

denlead=[T 1];

and using the conv() function to multiply it by the numerator anddenominator of the plant


newden6conv#den&denlead%7




$ag or Phase,$ag Compensator /sing %oot $ocs

A first-order lag compensator can be designed using the root locus. A lag compensator in rootlocus form is given by

where the magnitude of @ is greater than the magnitude of p. A phase-lag compensator tends to

shift the root locus to the right& which is undesirable. "or this reason& the pole and @ero of a lag

compensator must be placed close together #usually near the origin% so they do not appreciably

change the transient response or stability characteristics of the system.

hen a lag compensator is added to a system& the value of this intersection will be a smaller

negative number than it was before. The net number of @eros and poles will be the same #one

@ero and one pole are added%& but the added pole is a smaller negative number than the added

@ero. Thus& the result of a lag compensator is that the asymptotes2 intersection is moved closer

to the right half plane& and the entire root locus will be shifted to the right.

3 c s# % s @+#



(t was previously stated that that lag controller should only minimally change thetransient response because of its negative effect. (f the phase-lag compensator is

not supposed to change the transient response noticeably& what is it good forB The

answer is that a phase-lag compensator can improve the system2s steady-state

response. (t works in the following manner. At high frequencies& the lag controller

will have unity gain. At low frequencies& the gain will be @4p4 which is greater

than +. This factor @p will multiply the position& velocity& or acceleration constant#$p& $v& or $a%& and the steady-state error will thus decrease by the factor @4p4.

(n 5atlab& a phase lead compensator in root locus form is implemented by using

the transfer function in the form

numlag=[1 z];

denlag=[1 p];

and using the conv() function to implement it with the numerator anddenominator of the plant

newnum=conv(num,numlag);

newden6conv#den&denlag%7

$ag or Phase,$ag Compensator /sing %oot $ocs



$ag or Phase,$ag Compensator sing Freency %esponse

A first-order phase-lag compensator can be designed using the frequency response. A

lag compensator in frequency response form is given by

The phase-lag compensator looks similar to a phase-lead compensator& e1cept that a is

now less than +. The main difference is that the lag compensator adds negative phase tothe system over the specified frequency range& while a lead compensator adds positive

phase over the specified frequency. A bode plot of a phase-lag compensator looks like

the following

3 c s# % + α τ⋅ s⋅+(

⋅⋅



(n 5atlab& a phase-lag compensator in frequency response form is

implemented by using the transfer function in the form

numlead=[a*T 1];

denlead=a*[T 1];and using the conv() function to implement it with the numerator and

denominator of the plant


newden=conv(den,denlead);

$ag or Phase,$ag Compensator sing Freency %esponse



$ead,lag Compensator sing either %oot $ocs or Freency

%esponse

A lead-lag compensator combines the effects of a lead compensator with those of a lag

compensator. The result is a system with improved transient response& stability and

steady-state error. To implement a lead-lag compensator& first design the lead

compensator to achieve the desired transient response and stability& and then add on a lagcompensator to improve the steady-state response



Exercise - Dominant Pole-Zero Approximations and Compensations

The influence of a particular pole (or pair of complex poles) on the response is mainly determined

by two factors: the real part of the pole and the relative magnitude of the residue at the pole. The

real part determines the rate at which the transient term due to the pole decays; the larger the realpart, the faster the decay. The relative magnitude of the residue determines the percentage of the

total response due to a particular pole.

Investigate (using imulin!) the impact of a closed"loop negative real pole on the overshoot of a

system having complex poles.

T s# % pr ωn

,

⋅s pr +# % s

,, ζ⋅ ωn⋅ s⋅( )+ ωn

,+⋅

#a!e pr to vary ($, %, &) times the real part of the complex pole for different values of ζ ('.%, '.&,

'.).

Investigate (using imulin!) the impact of a closed"loop negative real ero on the overshoot of a

system having complex poles.

T s# %s @r +# %

s,

, ζ⋅ ωn⋅ s⋅( )+ ωn,

+

#a!e r to vary ($, %, &) times the real part of the complex pole for different values of ζ ('.%, '.&, '.).



Exercise - Lead and Lag Compensation

Investigate (using #atlab and imulin!) the effect of lead and lag compensations on the two

systems indicated below. ummarie your observations. *lot the root"locus, bode diagram

and output for a step input before and after the compensations.

+emember

lead compensation: p (place ero below the desired root location or to the left of the first two

real poles)

lag compensation: -p (locate the pole and ero near the origin of the s"plane)

ead /ompensation (use 01.%%, p0$' and 2 01&).



ag /ompensation (use 0'.'3 , and p0'.'1&, 20145 )

ummarie your findings



Problem 10234

!etermine a compensator so that the percent overshoot is less than ,4D and $v

#velocity constant% is greater than E.

chapter 10 pid

Documents