csd2_pid

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......Control System Design-II

Dr. Abdul Qayyum Khan

Room No. SE-302Department of Electrical Engineering,

Pakistan Institute of Engineering and Applied Sciences,P.O. Nilore Islamabd Pakistan

Email: [email protected]://faculty.pieas.edu.pk/aqayyum/

Copyrighted@AQKhan. All rights reserved 1 / 46

http://faculty.pieas.edu.pk/aqayyum/

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PID Controllers I

PID controllers were occasionally discussed

PID Controllers

comprises more than 50% industrial controllers

are adjusted on-site −→ many tuning methods have been proposed

are useful, especially when the plant model is unknown

are useful, but not necessarily to be optimal in many cases


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Tuning Rules for PID Controllers I

The transfer function is

Gpid = Kp

(I + Tds +

1

Ti s

)where the PID parameters


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Tuning Rules for PID Controllers II

Kp : Proportional GainTd : Derivative Time ConstantTi : Integral Time Constant

Tuning the PID parameters =⇒ Ziegler and Nichols methodsFirst method:

Obtain a step response of the plant to be controlledIf there is no integrator and no complex conjugate poles (dominant),then the unit step response looks like S-shapedNote down the characteristics of the S-shaped curve.(1) Delay time L, (2) Time Constant TThe transfer function of the plant is approximated as

Y (s)

U(s)=

Ke−Ls

Ts + 1

Then the parameters of the PID controller can be set according to thefollowing table


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Tuning Rules for PID Controllers III

Type of Controller Kp Ti Td

P TL

∞ 0

PI 0.9TL

L0.3

0

PID 1.2TL

2L 0.5L

The PID controller according to the 1st method of Z&N

Gpid(s) = Kp

(I + Td s +

1

Ti s

)= 1.2

T

L

(I + 0.5Ls +

1

2Ls

)= 0.6T

(s + 1

L

)2s

One pole at the origin;double zeros at s = − 1

L

Second method

This method applies when the response is not S-shapedPut Ti = ∞, Td = 0. Vary Kp unless sustained oscillation starts.Note down that value of Kp = Kcr


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Tuning Rules for PID Controllers IV

Also determine, the corresponding period Pcr

Then the parameters of the PID controller can be set according to thefollowing table

Type of Controller Kp Ti Td

P 0.5Kcr ∞ 0

PI 0.45KcrPcr1.2

0

PID 0.6Kcr 0.5Pcr 0.125Pcr

The PID controller according to the 1st method of Z&N

Gpid(s) = Kp

(I + Td s +

1

Ti s

)= 0.6Kcr

(I + 0.125Pcr s +

1

0.5Pcr

)

= 0.075KcrPcr

(s + 4

Pcr

)2

s

One pole at the origin;double zeros at s = − 4

Pcr


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Tuning Rules for PID Controllers V

Comments:

Z&N rules have been widely used for tuning PID controller in theprocess industryIf the mathematical model of the plant to be controlled is known e.g.Transfer function; then root locus is used to find Kcr and Pcr ; that is,

Pcr =2π

wcr

If the root-locus parameter does not cross jw− axis, this method willnot applyFor the plant with known mathematical model, many other analyticaland graphical methods can be used besides Z&N.Z&N and all methods proposed in literature provide initial guess. Theparameters may further be tuned for desirable response.


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Tuning Rules for PID Controllers VI

Table: Effects of increasing a parameter independently

Parameter Rise time Overshoot Settling time Steady-state error StabilityKp Decrease Increase Small change Decrease DegradeKi Decrease Increase Increase Eliminate DegradeKd Minor change Decrease Decrease No effect in theory Improve if Kd small


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Modification OF PID Control Schemes I

Figure: PID controlled system and equivalent block diagram


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Modification OF PID Control Schemes II

R(s) : Reference Signal

E (s) : Error Signal

D(s) : Disturbance Signal

Y (s) : Output Signal

N(s) : Noise Signal

B(s) : Observed Signal

Notice that for R(s) : Step function; U(s) will involve delta functiondue to Tds term. Due to this reason; Tds is replaced by Td s

1+γTd s

Advantage: U(s) will involve sharp pulses instead of impulses. Thisphenomenon is known as set-point kick


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PI-D Control I

To avoid set-point kick phenomenon

Derivative action in the feedback path


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PI-D Control II

The transfer functions Y (s)R(s)

Y (s)

R(s)

∣∣∣∣D=0,N=0

=

(1 +

1

Ti s

)KpGp(s)

1 + KpGc(s)Gp(s)

The transfer functions Y (s)D(s)

Y (s)

D(s)

∣∣∣∣R=0,N=0

=Gp(s)

1 + KpGc(s)Gp(s)

where

Gc(s) =

(I + Tds +

1

Ti s

)


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I-PD Control I

PID & PI-D (R(s) = step, U(s) will involve step)

NOT desirable in many occasion

Advantageous if move KP+ Derivative action into feedback path


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I-PD Control II

Only feedback signal is affected

The transfer functions Y (s)R(s)

Y (s)

R(s)

∣∣∣∣D=0,N=0

=

(1

Ti s

)KpGp(s)

1 + KpGc(s)Gp(s)

The transfer functions Y (s)D(s)

Y (s)

D(s)

∣∣∣∣R=0,N=0

=Gp(s)

1 + KpGc(s)Gp(s)

where

Gc(s) =

(I + Tds +

1

Ti s

)


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Two Degrees of Freedom PID Control I

PI-D: Derivative control action is in feedback path

I-PD: Proportional+derivative control in feedback path

PI-PD: to move some portion of the controllers

PI-PD Control: Characteristics of this control lie between PID controland I-PD control

PID-PD Control: One control is Feed forward path and one is inFeedback path.


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Two Degrees of Freedom PID Control II

In the above figure

D(s) : Process disturbance


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Two Degrees of Freedom PID Control III

N(s) : Measurement noise

R(s) : Reference Input

Gp(s) : Plant Transfer function (fixed and uncertain)

Gc(s) : Controller transfer function

Notice that

Gyr =Y (s)

R(s)=

GcGp

1 + GcGp

Gyd =Y (s)

D(s)=

Gp

1 + GcGp

Gyn =Y (s)

N(s)= − GcGp

1 + GcGp= −Gyr


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Two Degrees of Freedom PID Control IV

Also

Gyr =Gp − Gyd

Gp

Gyn =Gyd − Gp

Gp

Notice that if Gyd is known, then Gyr and Gyn are fixed. This means thatby only computing Gyd , Gyr and Gyn can be determined. Hence the abovesystem is one-degree of freedom. Consider another example


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Two Degrees of Freedom PID Control V


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Two Degrees of Freedom PID Control VI

Gyr =Y (s)

R(s)=

Gc1Gp

1 + GcGp

Gyd =Y (s)

D(s)=

Gp

1 + GcGp

Gyn =Y (s)

N(s)= − GcGp

1 + GcGp

where

Gc = Gc1 + Gc2

Also

Gyr = Gc1Gyd

Gyn =Gyd − Gp

Gp


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Two Degrees of Freedom PID Control VII

For Gyd is known, Gyn is fixed. Even then Gyr is unknown due to Gc1. Itmeans that the two loops are independent. Hence the system is atwo-degrees of freedom control system. Take a look at the following figure


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Two Degrees of Freedom PID Control VIII

Gyr =Y (s)

R(s)=

Gc1Gp

1 + Gc1Gp+

Gc2Gp

1 + Gc1Gp

Gyd =Y (s)

D(s)=

Gp

1 + Gc1Gp

Gyn =Y (s)

N(s)= − Gc1Gp

1 + Gc1Gp

Also

Gyr = Gc2Gyd +Gp − Gyd

Gp

Gyn =Gyd − Gp

Gp

If Gyd is given, Gyn is fixed. Gyr is not fixed. Gc1 can be tunedindependently from Gc2.

Comments


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Two Degrees of Freedom PID Control IX

The degrees of freedom means the control loops which are tuned toget the desired response.

If two control loops are to be tuned independently, then the controlsystem will be two degrees of freedom. The same arguments can beused for one degree of freedom and more degrees of freedom

Sometimes the name degrees of freedom is given based on theindependent control variables to be tuned


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Zero placement approach to improve responsecharacteristics I

It is useful techniques when it is desired to minimize steady stateerror in systems subjected to changing inputs (step input, Rampinput, acceleration input).

In high performance control, zero steady state error is always desired.

Example Problem

Consider the figure 2, design a Control System that will exhibit no steadystate error for ramp and acceleration input and force the response to stepdisturbance input to approach zero as quickly as possible.


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Zero placement approach to improve responsecharacteristics II

Figure: Two-degrees of freedom control system


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Zero placement approach to improve responsecharacteristics III

Solution:

Let Gp(s) = K A(s)B(s) be minimum phase and

A(s) = (s + z1) (s + z2) · · · (s + zm)

B(s) = sN (s + pN+1) (s + pN+2) · · · (s + pn)

where N = 0, 1, 2 and n ≥ m.Let Gc1 be a PID controller followed by a filter 1

A(s) ; that is,

Gc1 =α1s

2 + β1s + γ1s

1

A(s)


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Zero placement approach to improve responsecharacteristics IV

and Gc2 be PID,PI,PD,I,D or P controller followed by a filter 1A(s) ; that is,

Gc2 =α2s

2 + β2s + γ2s

1

A(s)

where α2, β2, γ2 might be zero depending upon the choice of control.Recall

Gyr =Y (s)

R(s)=

Gc1Gp

1 + GcGp

Gyd =Y (s)

D(s)=

Gp

1 + GcGp

where

Gc = Gc1 + Gc2


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Zero placement approach to improve responsecharacteristics V

we need

Gc = Gc1 + Gc2 =αs2 + βs + γ

s

1

A(s)

Then

Gyd =Y (s)

D(s)=

Gp

1 + GcGp

=K A(s)

B(s)

1 + αs2+βs+γs

1

��A(s)K��A(s)

B(s)

KsA(s)

sB(s) + K (αs2 + βs + γ)


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Zero placement approach to improve responsecharacteristics VI

if D(s) = ds : Step input with magnitude d , then

Y (s) = �sKA(s)

sB(s) + K (αs2 + βs + γ)× d

�s

assuming the system is stable, using final value theorem

Y (∞) = lims→0

sKdA(s)

sB(s) + K (αs2 + βs + γ)= 0

Notice that

The denominator of YR and Y

D is the same which means that thecharacteristics equation is the same

The poles can be adjusted to improve the response

What is the importance of adjusting zeros of the closed loop system


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Zero placement I

Consider a system

Y

R=

p(s)

sn+1 + ansn + an−1sn−1 + · · ·+ a2s2 + a1s + a0

If we choose p(s) = a2s2 + a1s + a0, then the steady state error will be

zero for step, ramp, and acceleration input.Define E (s) = R(s)− Y (s), then

E (s) = R(s)− a2s2 + a1s + a0

sn+1 + ansn + an−1sn−1 + · · ·+ a2s2 + a1s + a0R(s)

=

[1− a2s

2 + a1s + a0sn+1 + ansn + an−1sn−1 + · · ·+ a2s2 + a1s + a0

]R(s)


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Zero placement II

For R(s) = 1s : step input

E (∞) = lims→0

[sE (s)]

= lims→0

s

[1− a2s

2 + a1s + a0sn+1 + · · ·+ a2s2 + a1s + a0

]1

s

= lims→0

�s

[1− a2s

2 + a1s + a0sn+1 + · · ·+ a2s2 + a1s + a0

]1

�s

= 0


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Zero placement III

For R(s) = 1s2

: Ramp input

E (∞) = lims→0

[sE (s)]

= lims→0

s

[1− a2s

2 + a1s + a0sn+1 + · · ·+ a2s2 + a1s + a0

]1

s2

= lims→0

[1− a2s

2 + a1s + a0sn+1 + · · ·+ a2s2 + a1s + a0

]1

s

= lims→0

[sn+1 + · · ·+ s3

sn+1 + · · ·+ a2s2 + a1s + a0

]1

s

= lims→0

s

[sn + · · ·+ s2

sn+1 + · · ·+ a2s2 + a1s + a0

]1

s

= lims→0

�s

[sn + · · ·+ s2

sn+1 + · · ·+ a2s2 + a1s + a0

]1

�s

= 0


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Zero placement IV

For R(s) = 1s3

: Acceleration input

E (∞) = lims→0

[sE (s)]

= lims→0

s

[1− a2s

2 + a1s + a0sn+1 + · · ·+ a2s2 + a1s + a0

]1

s3

= lims→0

[1− a2s

2 + a1s + a0sn+1 + · · ·+ a2s2 + a1s + a0

]1

s2

= lims→0

[sn+1 + · · ·+ s3

sn+1 + · · ·+ a2s2 + a1s + a0

]1

s2

= lims→0

��s2[

sn−1 + · · ·+ s

sn+1 + · · ·+ a2s2 + a1s + a0

]1

��s2

= 0


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Zero placement V

Determination of Gc1

Consider again

Y (s)

R(s)=

a2s2 + a1s + a0


As we have

Y (s)

R(s)= Gc1

Y (s)

D(s)

=sGc1KdA(s)

sB(s) + K (αs2 + βs + γ)

=sGc1KdA(s)



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Zero placement VI

Since Gc1 is PID controller and is given by

Gc1 =α1s

2 + β1s + γ1s

1

A(s)

Now

Y (s)

R(s)=

K(α1s

2 + β1s + γ1)


Therefore we choose Kα1 = a2, Kβ1 = a1, Kγ1 = a0, so that

Gc1 =a2s

2 + a1s + a0Ks

1

A(s)


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Zero placement VII

Determination of Gc2:As

Gc1 + Gc2 =αs2 + βs + γ

s

1

A(s)

Gc2 =αs2 + βs + γ

s

1

A(s)− a2s

2 + a1s + a0Ks

1

A(s)

=(Kα− a2) s

2 + (Kβ − a1) s + (Kγ − a0)

Ks

1

A(s)


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Design Problem I

Consider two degrees of freedom control system shown in Figure. Theplant transfer function is

Gp(s) =10

s(s + 1)

Design controllers Gcl(s) and Gc2(s) such that the maximum overshoot inthe response to the unit-step reference input be less than 19%, but morethan 2%, and the settling time be less than 1 sec. It is desired that thesteady-state errors in following the ramp reference input and accelerationreference input be zero.The response to the unit-step disturbance inputshould have a small amplitude and settle to zero quickly.


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Design Problem II


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Design Problem III


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Design Problem IV

0 1 2 3 4 5 6 7 8 9 10−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time [second]

SS

E

Steady State Error

Acceleration Rampstep


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Design Problem V

Step Response

Time (seconds)

Am

plitu

de

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90

0.2

0.4

0.6

0.8

1

1.2

1.4

System: sysPeak amplitude: 1.18Overshoot (%): 17.9At time (seconds): 0.106


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Design Problem I

Consider two degrees of freedom control system shown in Figure. Theplant transfer function is

Gp(s) =5

(s + 1)(s + 5)

Design controllers Gcl(s) and Gc2(s) such that

The maximum overshoot in the response to the unit-step referenceinput be less than 25% and the settling time be less than 2 sec.

It is desired that the steady-state errors in following the rampreference input and acceleration reference input be zero.

The response to the unit-step disturbance input should have a smallamplitude and settle to zero eventually.


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Design Problem II


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Design Problem III

0 0.5 1 1.5 2 2.5 3−0.2

0

0.2

0.4

0.6

0.8

1

1.2

Time [Seconds]

SS

E

SSE

StepRampAcceleration


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Design Problem IV

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

Time[Seconds]

Output

Step Reference input Response


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Design Problem V

0 1 2 3 4 5 6 7 8 9 10−0.005

0

0.005

0.01

0.015

0.02

0.025

0.03Response to Step Disturbance

Time [Second]

Response


csd2_pid

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