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Fundamentals ofElectrical Engineering
Circuits with Capacitors and Inductors
Solving circuits for sinusoidal sources
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Using Impedances The circuit consists of sources and any number
of
resistors, capacitors and inductors
Pretend the sources are complex exponentials havinga
frequencyf
Consider each element an impedance
Use voltage divider, current divider, series/parallelrules to
relate output variables complex amplitude to
the complex amplitude of the source
element impedance
R R
C 1
j2
f CL j2f L
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Sinusoidal Sources
vin+
R
C vout
+
Vin+
Vout
+
ZC
ZR
Voutej2ft = 1
j2fRC+ 1Vine
j2ft
vin(t) =A cos2f0t =1
2
Ae
j2f0t +Aej2f0t
vin(t) =Aej2f0t = vout(t) =
1
j2f0RC+ 1Aej2f0t
vin(t) =Aej2f0t = vout(t) =
1j2f0RC+ 1
Aej2f0t
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Sinusoidal Sources
vin+
R
C vout
+
Vin+
Vout
+
ZC
ZR
vin = Vinej2ft
Since the circuit elements and KVL/KCL are linear,superposition
applies
vin(t) =A cos2f0t =1
2
Aej2
f0t +Aej2f0t
vout(t) = 12
1
j2f0RC+ 1Aej2f0t + 1
j2f0RC+ 1Aej2f0t
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Sinusoidal Sources Note that
vin(t) =12
Aej
2f0t +Aej2f0t
vin(t) = Re[Aej2f0t]
vout(t) =1
2
1
j2f0RC+ 1Aej2f0t +
1
j2f0RC+ 1Aej2f0t
vout(t) = Re
1j2f0RC+ 1
Aej2f0t
vout(t) = Re
1
42f20R2C2 + 1ejtan
1 2f0RCAej2f0t
= Re
A
42f20R2C2 + 1
ej(2f0ttan1 2f0RC)
=
A42f20R2C2 + 1 cos(2
f0t
tan
1
2
f0RC)
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Output for a Sinusoidal Source
vin+
R
C vout
+
vout(t) = A
42f2
0R2C2 + 1
cos(2f0t tan1 2f0RC)
vout(t) = Re 1j2f0RC+ 1Ae
j2f0t
Vout =1
j2f0RC+ 1Vin
vin(t) =A cos(2
f0t) = Re[Aej2f0t
]
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Sinusoidal SourcesIn general, vin(t) =A cos(2f0t+ ) = Re[Ae
jej2f0t]
Vin
Using impedances, find that Vout=H(f0) Vin
The transfer function captures the amplitude
change and phase shift that the circuit imposes
H(f0) =Vout
Vin
vout(t) = ReH(f0)Vine
j2f0t
=|H(f0)| |Vin| cos
2f0t+ + H(f0)
H(f0) = 1
j2f0RC+ 1 =
1
42f20R2C2 + 1
ej tan1
2f0RC
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Sinusoidal SourcesNote that if vin(t) =A sin(2f0t+ ) = Im[Ae
jej2f0t]
vout(t) = ImH(f0)Vine
j2f0t
=|H(f0)| |Vin|sin
2f0t+ + H(f0)
cos 2ft = sin
2ft+
2
= Im
ej
2 ej2ft
You can use eitherthe real or imaginary part in
yourcalculations
sin 2ft = cos
2ft
2
= Re
e
j
2 ej2
ft
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Using Impedances The circuit consists of sources and any number
of
resistors, capacitors and inductors
Pretend the sources are complex exponentials havinga
frequencyf
Consider each element an impedance
Use voltage divider, current divider, series/parallelrules to
relate output variables complex amplitude tothe complex amplitude
of the source (the transferfunction)
Express the source as the real (or imaginary) part of acomplex
exponential
Output is the real (or imaginary) part of the transferfunction
times the complex exponential representingthe source
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An Example+
+
vout
1
2 4vin vin(t) = 10 sin(t/2)
Whats the transfer function?
Whats the output for the given input voltage?
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An Example
Vin
+
+
Vout4s
1/s
2
s =j2f
+
+
vout
1
2 4vin
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An Example
Vin
+
+
Vout4s
1/s
2
s =j2f
+
+
vout
1
2 4vin
vin
(t) = 10 sin(t/2)
H(s) = 4s2
4s2 + 2s+ 1 orH(f) =
162f2
162f2 +j4f+ 1
2f=
1
2=
f=
1
4 H 1
4
=
1
1 +j + 1 =j
vout(t) = Im[10jejt/2] = Im
10ej(t/2+
2)
= 10 sint/2 +
2
= 10 cos(t/2)
= Im 10ejt/2
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WhatNotto Do
+
+
vout
1
2 4vin
H(f) = 162f2
162f2 +j4f+ 1
vin(t) = 10 sin(t/2)
vout(t) = Im
H
1
4
10 sin(t/2)
vout(t) = ImH 1
4
10ejt/2
vout(t) =H
1
4
10 sin(t/2)
!
= Im 10ejt/2
