Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: imtiaz.hussain@faculty.muet.edu.pk

URL :http://imtiazhussainkalwar.weebly.com/

Lecture-6Mathematical Modelling of Electrical & Electronic Systems

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Outline of this Lecture

• Part-I: Electrical System

• Basic Elements of Electrical Systems

• Equations for Basic Elements

• Examples

• Part-II: Electronic System

• Operational Amplifiers

• Inverting vs Non-inverting

• Examples

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ELECTRICAL SYSTEMS

Part-I

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Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the resistor is given by Ohm’s law i-e

Rtitv RR )()(

• The Laplace transform of the above equation is

RsIsV RR )()(

Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the Capacitor is given as:

dttiC

tv cc )()(1

• The Laplace transform of the above equation (assuming there is no charge stored in the capacitor) is

)()( sICs

sV cc

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Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the inductor is given as:

dt

tdiLtv

LL

)()(

• The Laplace transform of the above equation (assuming there is no energy stored in inductor) is

)()( sLsIsV LL

V-I and I-V relations

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Component Symbol V-I Relation I-V Relation

Resistor

Capacitor

Inductordt

tdiLtv

LL

)()(

dttiC

tv cc )()(1

Rtitv RR )()(R

tvti

RR

)()(

dt

tdvCti

cc

)()(

dttvL

ti LL )()(1

Example#1

• The two-port network shown in the following figure has vi(t) asthe input voltage and vo(t) as the output voltage. Find thetransfer function Vo(s)/Vi(s) of the network.

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Ci(t)vi( t) vo(t)

dttiC

Rtitv i )()()(1

dttiC

tvo )()(1

Example#1

• Taking Laplace transform of both equations, considering initialconditions to zero.

• Re-arrange both equations as:

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dttiC

Rtitv i )()()(1

dttiC

tvo )()(1

)()()( sICs

RsIsV i

1)()( sI

CssVo

1

)()( sIsCsV o))(()(Cs

RsIsV i

1

Example#1

• Substitute I(s) in equation on left

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)()( sIsCsV o))(()(

CsRsIsV i

1

))(()(Cs

RsCsVsV oi

1

)()(

)(

CsRCs

sV

sV

i

o

1

1

RCssV

sV

i

o

1

1

)(

)(

Example#1

• The system has one pole at

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RCssV

sV

i

o

1

1

)(

)(

RCsRCs

101

Example#2

• Design an Electrical system that would place a pole at -3 ifadded to another system.

• System has one pole at

• Therefore,

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Ci(t)vi( t) v2(t)RCssV

sV

i

o

1

1

)(

)(

RCs

1

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RCpFCandMRif 3331

Example#3

• Find the transfer function G(S) of the following two port network.

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i(t)vi(t) vo(t)

L

C

Example#3• Simplify network by replacing multiple components with

their equivalent transform impedance.

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I(s)Vi(s) Vo(s)

L

C

Z

Transform Impedance (Resistor)

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iR(t)

vR(t)

+

-

IR(S)

VR(S)

+

-

ZR = R

Transformation

Transform Impedance (Inductor)

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iL(t)

vL(t)

+

-

IL(S)

VL(S)

+

-

LiL(0)

ZL=LS

Transform Impedance (Capacitor)

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ic(t)

vc(t)

+

-

Ic(S)

Vc(S)

+

-

ZC(S)=1/CS

Equivalent Transform Impedance (Series)

• Consider following arrangement, find out equivalenttransform impedance.

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L

C

R

CLRT ZZZZ

CsLsRZ T

1

Equivalent Transform Impedance (Parallel)

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CLRT ZZZZ

1111

C

L

RCs

LsRZ T1

1111

Equivalent Transform Impedance

• Find out equivalent transform impedance offollowing arrangement.

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L2

L2

R2R1

Back to Example#3

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I(s)Vi(s) Vo(s)

L

C

Z

LR ZZZ

111

LsRZ

111

RLs

RLsZ

1

Example#3

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I(s)Vi(s) Vo(s)

L

C

Z

RLs

RLsZ

1

)()()( sICs

ZsIsV i

1)()( sI

CssVo

1

Example#4

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VinC

R

LVout

• Find transfer function Vout(s)/Vin(s) of the following electricalnetwork

Example#5

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Vin

C1

R

LVout

• Find transfer function Vout(s)/Vin(s) of the following electricalnetwork

C2

C3

ELECTRONIC SYSTEMS

Part-II

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Operational Amplifiers

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1

2

Z

Z

V

V

in

out

1

21

Z

Z

V

V

in

out

Example#6

• Find out the transfer function of the following circuit.

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1

2

Z

Z

V

V

in

out

Example#7

• Find out the transfer function of the following circuit.

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v1

Example#8

• Find out the transfer function of the followingcircuit.

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v1

Example#9

• Find out the transfer function of the followingcircuit and draw the pole zero map.

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10kΩ

100kΩ

END OF LECTURES-6

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