lesson 17: combined mode control - engineering | siu

11/30/2015

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ET 438a Automatic Control Systems Technology

Lesson 17: Combined Mode Control

1 lesson17et438a.pptx

Learning Objectives

lesson17et438a.pptx 2

After this presentation you will be able to:

Describe the common control mode combinations used in analog

control systems.

List the characteristics of combined control modes.

Write the time, Laplace and transfer functions of combined

control modes.

Identify the Bode plots of combined control modes.

Design OP AMP circuits that realize theoretical combined

control mode performance.

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Proportional-Integral Control


Control Mode Characteristics:

1. Proportional action produces fast response to large load changes.

2. Integral action drives output to zero steady-state error.

3. Adds one pole and one zero to system transfer function.

4. Used on systems that have large load changes and where proportional only

action fails to reduce steady-state error to acceptable limits.

Proportional-Integral Control


Mathematical Representations

Time Function: 0

0

Ipp vdt )t(eKK)t(eK)t(v

Laplace Function: )s(Es

KK)s(EK)s(V

Ip

p

s

sKK

)s(E

)s(V IpTransfer Function:

Proportional-Integral (PI) Controllers add a pole at s=0 and zero at s=0 to system

transfer function

zero

pole

All initial conditions set

to zero in transfer function

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Proportional-Integral Control Response


Time plots of PI controller

output for piece-wise linear

error input

Proportional mode

gives instantaneous response

to error while integral mode

decrease error over time.

Constant error produces

linearly increasing controller

output

OP AMP Realization of PI Controller


PI Controller Circuit Derive the transfer function from the

generalized gain formula for inverting OP

AMP circuits

sCR

1sCR

sCR

sC

sCsCR

)s(A

RsC

sC

1RsC

R

sC

1R

)s(A

sC

1R)s(Z R)s(Z

)s(Z

)s(Z)s(A

i

f

i

f

V

i

f

i

f

V

ffii

i

fV

Simplify sCR

1

R

R

sCR

1

sCR

sCR)s(A

ii

f

ii

fV

CR

1K

R

RK

i

I

i

fp

Controller introduces a pole at s=0

and zero at s=-1/RfC

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Bode Plot of PI Controller


Let Ri = 10 kW, Rf = 100 kW and C=0.01 mF then produce Bode plot

ri=input('Enter value of input resistance: ');

c=input('Enter value of capacitance: ');

rf=input('Enter value of feedback resistance: ');

% compute transfer function model parameters for

% PI controller

% compute numerator parameter

tau=rf*c;

% compute parameter for denominator

tau1 = ri*c;

% build transfer function

% denominator form a1*s^2+a2s+a3

Av=tf([tau 1],[tau1 0])

%plot graph

bode(Av);

MatLAB Code

Bode Plot of PI Controller


Proportional

Action

Integral

Action

Integral

action is below

1000 rad/sec.

1/RfC set break

point

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PI Controller Design


Example 17-1: Design a PI OP AMP controller with Kp = 100 and an integral break

frequency of 100 rad/sec. Ri = 10kW

W

WW

W

M 0.1R

000,000,1100 000,10K RR R

RK

000,10R 100K CR

1

R

RK

f

pif

i

fp

ip

fi

fp

Answer

F 01.0C

F 101rad/sec 100M 0.1

1C

R

1C

CR

1

rad/sec 100

8

ff

m

W

Answer

Proportional-Derivative Control Mode


Proportional-Derivative (PD) Control combines proportional and

derivative actions. Used in processes that have sudden load changes

that proportional only cannot handle.

Mathematical Representations

Time Function:

Laplace Function:

Transfer Function:

sfrequenciehigh limit to Rate dt

)t(v d

vdt

)t(v dK

dt

)t(e dKK)t(eK)t(v 0ddpp

)s(EsK)s(EsKK)s(EK)s(V ddpp

sK1

sK1K

)s(E

)s(V

d

dp

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OP AMP Realization of PD Controller


OP AMP PD circuit

sK1

sK1K

)s(E

)s(V

d

dp

dI

fp

dI

Idd

RR

RK

RR

R CRK

Note: 1/Kd = the derivative action break point frequency

1/Kd = limiter action break point frequency

0≤≤1

f0 RR

PD Controller Design


Example 17-2: Design a PD control that has a proportional gain of 10

a derivative action break point of 100 rad/sec and a limiter frequency

break point of 1000 rad/sec. C = 0.1 mF. Draw the OP AMP circuit

and place all values on the schematic.

Example 17-2 Solution

Answer

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Example 17-2 Solution (2)


Now compute the alpha value from the rate limit break frequency

Answer



Now find the value of RI

From previous calculations Rd = 100 kW and 0.1

Answer

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Find Rf and Ro use proportional gain and previously computed values

Answer



Draw OP AMP Schematic and label components

Transfer function

Negative sign causes 180 degree phase shift due to inverting configuration. Add

Inverting OP AMP circuit with gain of -1 to remove this

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PD Bode Plot


Use MatLAB to generate the Bode plot of the PD controller designed in Example 17-2

ri=11.11e3;

c=0.1e-6;

rf=1.111e6;

rd=1e5;

% compute transfer function model parameters for

% PD controller

% compute parameters

kd=rd*c;

alpha=ri./(ri+rd);

kp=-rf./(ri+rd);

% build transfer function

% denominator form a1*s^2+a2s+a3

% numerator for b1*s^2+b2s+b3

Av1=kp*tf([kd 1],[alpha*kd 1])

%plot graph

bode(Av1);

MatLAB Code

Define

variables

Compute

parameters

Generate transfer

function and plot

response

PD Bode Plot


Computed transfer function

-0.1 s - 10

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

0.001 s + 1

Derivative

Break f Rate limit

Break f

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Proportional + Integral + Derivative

Controllers


Proportional-Integral-Derivative (PID) Controller Characteristics:

1. Proportional mode provides fast response to large process

load changes

2. Integral mode removes the steady-state (offset) error from

the output

3. Derivative mode improves system stability and improves

response to rapid load changes.

4. Widely used in industry

Mathematical Relationships for PID

Controllers


Time Function:

Laplace Function:

Transfer Function:

0ddp

0

Ipp vdt

)t(v dK

dt

)t(e dKKdt )t(eKK)t(eK)t(v

)s(EsK)s(EsK)s(Es

K)s(EK)s(V

)s(EsK)s(EsKK)s(Es

KK)s(EK)s(V

ddI

p

ddp

Ip

p

2

d

2

dIp

sKs

sKsKK

)s(E

)s(V

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OP AMP Realization of PID Controller


sK1

sK1

s

sKK

)sE

)s(V

d

dIp

ii

I

ddd

d1

1

1i

ip

CR

1K

CRK

RR

R

RR

RK

Single OP AMP realization of PID control

Notice that the parameters are dependent on

each other. Changing alpha will also change

Kp.

2

d

2

ddIIp

sKs

sKs1KKKK

)sE

)s(V

Modify system response by changing the values of

Kp, Kd, KI and .



Multiple OP AMPs allow independent adjustment of PID parameters.

sCR

1

)s(E

)s(V

II

I

RI

Ci

+

-Ei(s) VI(s)

RidCd +

- Vd(s)

Rd

R1 +

- V(s)

Rf

R2

sCR1

sCR

)s(E

)s(V

did

ddd

21

did

dd

II1

f

RR Where

sCR1

sCR

CR

1

R

R

)s(E

)s(V

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Multiple OP AMP Realization

Compute common denominator

2

IdIidII

2

IIdd

2

IdIidII

did

1

f

IIdid

IIdd

didII

did

1

f

did

dd

II1

f

sCCRRsCR

sCRCR

sCCRRsCR

sCR1

R

R

)s(E

)s(V

sCRsCR1

sCRsCR

sCR1sCR

sCR1

R

R

)s(E

)s(V

sCR1

sCR

sCR

1

R

R

)s(E

)s(V

II

I

CRK

1

1

fp

R

RK ddd CRK

didd CRK

Define the following relationships between components and parameters

Multiple OP AMP PID Realization


Divide numerator and denominator by RICI and make defined substitutions

2

IdIid

II

II

II

2

IIdd

II

did

IIII

1

f

sCCRRCR

1sCR

CR

1

sCRCRCR

1sCR

CR

1

CR

1

R

R

)s(E

)s(V

2

d

2

ddIIp

sKs

sKsKKKK

)s(E

)s(V

II

I

CRK

1

1

fp

R

RK ddd CRK

didd CRK

Where

Controller adds two zeros and two poles to

System. Change system response by changing

Parameters Kp KI, Kd .

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PID Controller Design


Example 17-3: Design a PID controller using the circuit above.

Proportional gain is 5. The derivative time constant is 0.5

seconds. The integral gain is 0.143 and =0.1. The capacitor for

the integrator Ci=10 mF and the differentiator, Cd= 1 mF

Rf = 1 MW. Find values for all other components.

Example 17-3 Solution

Define

parameters

Answer



Find the feedback

resistor value for the

differentiator

Answer

Now find the input resistor values for the summing amplifier

Answer

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Find the input resistor

value for the

differentiator

Solve for Rid and

substitute in all

known values

Answer

End Lesson 17: Combined Mode

Control

ET 438a Automatic Control Systems Technology


lesson 17: combined mode control - engineering | siu

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