Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
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
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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
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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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