AC Analysis
! DC Analysis: voltage and current are constant with respect to time
! AC Analysis: voltage and current vary with time
EE 411: Circuit Theory
" AC can be sinusoidal, square waves, or arbitrary periodic waveforms
! Sinusoidal is particularly important
" Commonly used, e.g., power systems, communications, etc.
Sinusoidal Steady-State Analysis - I
" Simple periodic function (e.g., derivative and anti-derivative of a
sinusoidal is also a sinusoidal)
" Any periodic function can be represented as the sum of sinusoidal
function " Fourier Series
© Copyright by Margarida Jacome
EE411: SL11
1
Margarida Jacome - UT Austin
EE411
Steady-State Sinusoidal Analysis
We assume that the input signal has been applied for a long
time such that the transient component of the total response
has died out.
The Sinusoidal Source
The Sinusoidal Response
The Phasor
Kirchoff’s Laws in the Frequency Domain
The Passive Circuit Elements in the Frequency Domain
Series, Parallel, and !-to-Y Simplifications
Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
output
steady-state
t
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EE411
Outline - Part 1
!
!
!
!
!
!
!
" In other words, we can ignore the initial condition.
transient
2
3
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EE411
4
Sinusoidal Function
" T = period = time interval of a cycle
v(t) = Vm cos (#t + $)
A cosine or sine function repeats itself every 2%
#T = 2% " T = 2% [sec]
#
amplitude
angular frequency
(radian/sec)
Sinusoidal Function (cont)
phase angle
" Frequency (number of cycles per second)
1
#
=
T
2%
f=
o
% radian = 180
" Root mean square of a periodic function
v(t) = Vm cos (#t + $)
(square root of the mean value of the squared
function) - used in power calculations
amplitude
! Sinusoidal function:
angular frequency
(radian/sec)
phase angle
T
!
t0
t0+T
Vm2cos2(#t + $)dt =
Vm
2
v
v(t)
" Phase angle $ " time shift
1
Vrms =
[#cycles/sec] or Hz
Vm
Vm
cos (#t1 + $) = 1
t1
#t1 + $ = 0
t1 = -$
#
if $ > 0, function
shifts to the left
t
t
$/#
Margarida Jacome - UT Austin
EE411
$/#
T
6
AC Analysis for Sinusoidal Function
- Example
Outline - Part 1
!
!
!
!
!
!
!
The Sinusoidal Source
The Sinusoidal Response
The Phasor
Kirchoff’s Laws in the Frequency Domain
The Passive Circuit Elements in the Frequency Domain
Series, Parallel, and !-to-Y Simplifications
Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
Margarida Jacome - UT Austin
EE411
T
Margarida Jacome - UT Austin
EE411
5
t=0
R
vs(t) = Vm cos (#t + $)
i(t)
vs(t)
L
KVL:
L
di
dt
+ Ri = Vm cos (#t + $)
for t " 0
" Solving the DE we obtain:
i(t)
=
-Vm
R2
+
cos ($ – &) e-t/(L/R) +
#2L2
7
&=
)
cos (#t + $ – &)
R2 + #2L2
transient component (dies off for t ' #)
tan-1(#L/R
Vm
steady-state component (still exist
as long as the switch is closed)
Margarida Jacome - UT Austin
EE411
8
Steady-State Sinusoidal Analysis:
Response Characteristics
Example (cont)
i(t)
=
-Vm
cos ($ – &) e-t/(L/R) +
R2 + #2L2
Vm
steady-state
i(t)
vs(t)
R
( forced response has the same form as the forcing function
R2 + #2L2
transient
t=0
! The steady state response signal is a sinusoidal function
cos (#t + $ – &)
! The frequency (#) of the response signal is identical to the
frequency of the source signal (forcing function)
Vm
R2 + #2L2
i(t)
t
L
" Only true for linear circuits, i.e., when the circuit parameters R, L,
and C are constant.
! Amplitude and phase angle of the response signal are
usually different from amplitude and phase angle of the
source
" So, the steady state analysis task is reduced to finding the amplitude
Margarida Jacome - UT Austin
EE411
9
Outline - Part 1
Steady-State Circuit Analysis
!
!
!
!
!
!
!
The Sinusoidal Source
The Sinusoidal Response
The Phasor
Kirchoff’s Laws in the Frequency Domain
The Passive Circuit Elements in the Frequency Domain
Series, Parallel, and !-to-Y Simplifications
Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
Margarida Jacome - UT Austin
EE411
and phase angle of the response (the waveform and frequency of
Margarida Jacome - UT Austin
10
the response are already known).
EE411
" Phasor
Phasor Analysis:
Analysis:solves
solvesthe
theproblem
problemin
inaasimpler
simplerway
way
Conceptually:
(time domain)
(time domain)
Actual input vs(t)
D.E.
(hard)
Actual response vout(t)
Express in
complex (phasor) form
Transfer
back to
Algebra
“Complex” input Vs
(complex number domain
or frequency domain)
11
(simple)
“Complex” output Vout
(complex number domain
or frequency domain)
Margarida Jacome - UT Austin
EE411
12
Representation of Complex Numbers
(cont)
Representation of Complex Numbers
! Complex number = real number + imaginary number
, j * -1
z=x+jy
(complex plane)
z = 3 + j4
4
(2) Polar Form
y r
z = r )&
z = x + j y = r )&
&
3
r=
&=
tan-1
(3) Exponential form
+
vs(t) = Vpcos (#t + &)
Re
(y/x)
= Re { (Vp ej&) (ej#t )}
y r
&
x
= Re { (Vp )&) (ej#t )} = Re { (Vp )(& + #t)}
Re
phasor
standard polar form
captures the two
unknowns of the
response signal (amplitude and phase angle)
z = re j&
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EE411
Euler identity: e±j& = cos& ± j sin&
= Re {Vp ej(#t + &) }
Im(j)
y2
14
" Suppose a sinusoidal voltage source: vs(t) = Vp cos (#t + &)
x
x2
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EE411
Phasor Notation
y
z = r )&
$
Euler identity: e±j& = cos& ± j sin&
13
Im(j)
(2) Polar Form
Re
x
Re
Representation of Complex Numbers:
Summary
z=x+jy
&
y = Im{z} = r sin(&)
z = 3 + j4
= 5)53.13o
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EE411
(1) Rectangular form
x = Re{z} = r cos(&)
z = x + j y = r cos(&) + j r sin(&) = r (cos(&) + j sin(&)) = re j&
Im(j)
4 r
& = tan-1(y/x)
z* = complex conjugate of z
(x – jy)
(projections)
or | z | )&
x2 + y 2
where r = | z | =
Im(j)
(3) Exponential form z = re j&
Re
3
, j * -1
z=x+jy
Im(j)
(1) Rectangular form
| z | = z z*
! Complex number = real number + imaginary number
15
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EE411
16
Im
Phasor: Summary
c
b
#t1
r
a
&
is
Sinusoidal function
in the time domain
(from previous analysis)
vs(t) = Re { (Vp )&) (ej#t )}
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EE411
17
! Phasor for sinusoidal function v(t) = Vm cos (#t + &)
! Inverse phasor transform
e
T
#T = 2%
T = (2%)/#
r = |Vp |
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EE411
t
= Vm Re{e j& e j#t }
( for hand writing use V instead of V )
R
vs(t)
= Vm cos (#t + &)
vs(t) = Vm cos (#t + $)
i(t) = ?
L
KVL:
L
di
dt
i(t) = Im cos (#t + +) = Re {I e j#t } (2)
where I = Im e j+
transfers the sinusoidal function from the
frequency/complex domain to the time
domain)
(multiply by e j#t and extract the real part)
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EE411
18
Vs= Vm e j$
+ Ri = Vm cos (#t + $)
(1)
" We know that the steady state solution for i is in the form [ same forcing form]
of the forcing function (in this case, cos), and same frequency (#)):
V = Vm cos(&) + Vm j sin(&) rectangular form
m
t2
Usefulness of Phasor Notation to Solve DEs
function from the time domain to the
complex number or frequency domain)
form
j& }
sinusoidal function
is a cosine function
phasor transform (transfers the sinusoidal
V = Vm )& polar
or
b
Re {} is the projection
onto the real axis
ex.: Re {Vp )&)]
Time to go one cycle (i.e.,
complete one circle):
exponential form
or
P-1{V
c
vs(t)
vs(t) = Vp cos (#t + &)
= Re { (Vp )&) (ej#t )}
Definition of Phasor
V= P {Vm cos (#t + &)} = Vm e j&
a
t1
vs(t) = Re {Vp ej#t }
(
ej#t describes the
motion of a point
in the complex plane
Vs = Vp )&
" The sinusoidal function of a phasor Vp )&
t=0
Re
Sinusoidal function
in the complex-number
domain (also called
frequency domain)
" The phasor of a sinusoidal function vs(t) = Vp cos (#t + &) is
graphically: at t = 0,
Vp )& is a point in the
complex plane (a)
19
substituting (2) in (1)
L
d
dt
Re {I e j#t }
(a complex constant capturing
both unknowns)
+ R , Re {I e j#t } = Re {Vs e j#t }
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EE411
Vm cos (#t + $)
(a)
20
Usefulness of Phasor Notation to Solve DEs
(cont)
Usefulness of Phasor
Notation to Solve DEs (cont)
ej#t describes the motion
Im
b
r
(b)
R
vs(t) = Vm cos (#t + $)
i(t) = ?
vs(t)
i(t) = Im cos (#t + +) = Re {I e j#t } (2)
where I = Im e j+
L
L
Re { ( j# L I + R I – Vs ) e j#t } = 0
Vs= Vm e j$
Re {I e j#t }
but
d
dt
+ R , Re {I e j#t } = Re {Vs e j#t }
Re {I e j#t } = Re
d
dt
I e j#t
= Re { j# I e j#t }
substituting
in (a)…
Re { (j# L I + R I – Vs) e j#t Margarida
} = 0 Jacome - UT Austin
Im
Re
(b)
b
same, only the phase
angle changes
ej#t rotates this point about the origin (a)
Re
I =
I = Im e j+
Vs
R + j# L
Vm e j$
where & = tan-1(#L/R )
R2 + #2 L2 e j&
=
Vm
(converted to polar form)
e j($ – &)
same as the steady state solution
derived before solving DE (see SL 6)
R2 + #2 L2
j# L I + R I – Vs = 0
( i(t) = P-1 {I}
t
23
22
Vs= Vm e j$
j# L I + R I – Vs = 0
=
i.e., both the real part and the imaginary part of
this complex constant must be zero.
t
Usefulness of Phasor Notation to Solve DEs
(cont)
(
" In order for that to happen, the point must be
at the origin in the complex plane.
Margarida Jacome - UT Austin
EE411
Re
" at t = t1: {} = r e j(& + # t1) magnitude r remains the
EE411
ej#t describes the motion
Equation (b) says that the real part of {} has to the
zero at all times
(
t1
So, equation (b) says that the real part of {} has to the
Margarida Jacome - UT Austin
zero at all times
21
EE411
Re { ( j# L I + R I – Vs ) e j#t } = 0
Re
at frequency #.
Re { j# L I e j#t } + Re {R I e j#t } = Re {Vs e j#t }
Usefulness of Phasor
Notation to Solve DEs (cont)
t = t0
a
" at t = 0: {} = r e j&
(previous)
a
&
what does it mean?
# j# L I + R I – Vs is just a complex number,
a point on the complex plane
(a)
d
dt
t = t1
"
i(t) =
Vm
cos (#t + $ – & )
R2 + #2 L2
Margarida Jacome - UT Austin
EE411
24
Sum of Sinusoidal Functions
Using Phasor Notation
Outline - Part 1
" Let v = v1 + v2 +… + vn , where all voltages in the right hand side are
sinusoidal voltages of the same frequency #.
Then
V = V1 + V2 +… + Vn
sin (#t +- ) = cos (#t +- – 90o)
Example: v = 250 cos (377t + 30o) – 150 sin (377t + 140o )
o
o
o
= 250 cos (377t + 30 ) – 150 cos (377t + 140 – 90 )
= 250 cos (377t + 30o) – 150 cos (377t + 50o )
Moving to frequency domain:
V = 250 ) 30o – 150 ) 50o = 250 [ cos (30o) + j (sin (30o)] – 150 [ cos (50o) + j (sin (50o)]
= 216.5 + j 125 – 96.4 – j 114.9 = 120.1 + j 10.1
V = 120.1 + j 10.1 = 120.52 ) 4.8o
25
KVL in Phasor Notation
+
vs(t)
R
L
–
R
VL
–
V1 + V2 +… + Vn = 0
(-V
(-Vs s+V
+VRR+V
+VLL))==00
KCL applies to phasor currents:
+
L
KVL applies to phasor voltages:
" KCL: a similar derivation can be done for the sum of currents at a node
%0
+
Vs
26
Re{(-Vs +VR +VL ) e j#t} = 0
(
+ VR –
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EE411
KVL and KCL in the Frequency Domain
" KVL: -vs(t) + vR(t) + vL(t) = 0
+
Re{-Vs e j#t} + Re{VR e j#t} + Re{VL e j#t} = 0
vL(t)
–
–
The Sinusoidal Source
The Sinusoidal Response
The Phasor
Kirchoff’s Laws in the Frequency Domain
The Passive Circuit Elements in the Frequency Domain
Series, Parallel, and !-to-Y Simplifications
Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
" back to time domain: v = 120.52 cos (377t + 4.8o)
Margarida Jacome - UT Austin
EE411
+ vR (t) –
!
!
!
!
!
!
!
I1 + I2 +… + In = 0
KVL in the Frequency Domain
Margarida Jacome - UT Austin
EE411
27
Margarida Jacome - UT Austin
EE411
28
Quest 7
i1(t)
!
!
!
!
!
!
!
Given
R
i2(t)
vs(t)
Outline - Part 1
i3(t)
i2 = 50 sin (500t + 120o)
L1
L2
The Sinusoidal Source
The Sinusoidal Response
The Phasor
Kirchoff’s Laws in the Frequency Domain
The Passive Circuit Elements in the Frequency Domain
Series, Parallel, and !-to-Y Simplifications
Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
i1 = 100 cos (500t + 20o)
Determine i3
Reduce i2 to cos form
sin (#t +- ) = cos (#t +- – 90o)
Move to frequency domain, compute I3:
I3 = I1 – I2
Move to back to domain (from I3, determine i3):
Margarida Jacome - UT Austin
EE411
Margarida Jacome - UT Austin
EE411
29
The V-I Characteristic for a Resistor
Time Domain
R
i (t)
+
v (t)
–
i(t) = Im cos (#t + &i)
Definition of Impedance and Admittance
&i: phase angle of the current
I
" The voltage at the terminals of R is
Impedance:
v(t) = R[ Im cos (#t + &i)] = R Im cos (#t + &i)]
Z
Frequency Domain
R
I
+
V
–
I = Im e
j&i
V=RI
V = R Im e j&i
resistor in the frequency
domain
" Magnitude relationship between voltage and current:
|V|= R|I|
(i.e., Vm = R Im )
+
V
–
Z =
Impedance is measured
in ohms
V
I
(complex values)
Admittance: Y =
I
V
Admittance is measured
in Siemens
If the circuit element is a resistor with resistance R:
" Phase relationship between voltage and current:
(i.e., &v = &i)
)V =)I
(both signals are said to be in phase)
Margarida Jacome - UT Austin
EE411
rate of the circuit’s
voltage phasor to its
current phasor
Frequency Domain
" Assume that the current in R varies sinusoidally with time
30
31
" Z = R (Resistance)
" Y = G (Conductance)
(real values)
Margarida Jacome - UT Austin
EE411
32
The V-I Characteristic for an Inductor
Time Domain
j# L
" The voltage at the terminals of L is
v(t) = L di = L[-#Im sin (#t + &i)]
dt
= -#L Im cos (#t + &i - 90o)
Frequency Domain
I
+
V
–
o
[=#L Im cos (#t + &i + 90o)]
= j# L I m e
o
(physically: current cannot vary
“instantaneously”)
# ' 0 (DC) : | V | = 0 (short circuit)
#'.: |I| =
o
e -j90 = cos (- 90 ) + j sin (- 90 ) = -j
|V|
#L
= 0 (open circuit)
" Phase relationship between V and I
V = j#L I
j = e j90
inductor in the frequency domain
Margarida Jacome - UT Austin
EE411
Frequency Domain
I
+
V
–
| V | = #L | I |
&v = &i + 90
Im
o
L
v, i
ej#t
90
&i
C
v(t) = -#L Im cos (#t + &i - 90o)
o
= #L Im cos (#t + &i + 90 )
+
v (t)
–
v(t)
90o/#
| V | = #L | I |
o
Time Domain
i(t) = Im cos (#t + &i)
+
v (t)
–
t
j#C
ej#t
& + 90o
current “lags behind”
34
voltage by 90o
" Assume that the voltage at the terminals of C varies sinusoidally with time
I
+
V
–
&v: phase angle of the voltage
v(t) = Vm cos (#t + &v)
" The current in C is
i(t) = C dv = C[-#Vm sin (#t + &v)]
dt
= -#CVm cos (#t + &v - 90o)
Frequency Domain
i(t)
Re
&
Margarida Jacome
( - UTv Austin
= i
EE411
The V-I Characteristic for a Capacitor
i (t)
i (t)
( V = j#L Im e j&i = #L Im e j&i e j90o = #L Im e j(&i + 90o)
e j90 = cos (90o) + j sin (90o) = j
Time Domain
V = j#L I
o
o
33
The V-I Characteristic for an Inductor (cont)
j#L
(imaginary values)
" Admittance: Y = 1/j#L
| V | = #L | I |
V = -#L Im e j(&i -90 ) = -#L Im e j(&i)e -j90
o
" Impedance: Z = j#L
V = j#L I
" Magnitude relationship between V and I
o
j&i
V = j#L Im e j&i
I
+
V
–
&i: phase angle of the current
i(t) = Im cos (#t + &i)
+
v (t)
–
j#L
Frequency Domain
" Assume that the current in L varies sinusoidally with time
i (t)
L
The V-I Characteristic for an Inductor (cont)
o
[=#C Vm cos (#t + &i + 90o)]
o
I = -#C Vm e j(&v -90 ) = -#C Vm e j(&v)e -j90
= j#C Vm e j&v
I = j#C V
o
e -j90 = cos (- 90o) + j sin (- 90o) = -j
(current phasor )
|I|
Margarida
Jacome
- UTo Austin
current “lags behind”
voltage
by 90
EE411
35
capacitor in the frequency domain
Margarida Jacome - UT Austin
EE411
36
The V-I Characteristic for a Capacitor (cont)
Frequency Domain
I
+
V
1/j#C
–
The V-I Characteristic for a Capacitor (cont)
Frequency Domain
I = j#C Vm e j&v
" Impedance: Z = 1/j#C
I
+
V
–
(imaginary values)
I = j#CV
" Admittance: Y = j#C
1/j#C
(physically: voltage cannot vary
“instantaneously”)
" Magnitude relationship between V and I
# ' 0 (DC) : | I | = 0 (open circuit)
| I | = #C | V |
# ' . : | V |=
|I|
#C
I = j#CV
| I | = #C | V |
&i = &v + 90o
Im
&v
( I = j#C Vm e j&v = #C Vm e j&v e j90o = #C Vm e j(&v + 90o)
o
e j90 = cos (90o) + j sin (90o) = j
&
& + 90
Margarida Jacome - UT =
Austin
i
v
EE411
o
Z =
= R+jX
I
I
= G+jB
reactance
(S)
V
conductance
susceptance
Margarida Jacome - UT Austin
EE411
v(t)
t
Re
ej#t
38
The Sinusoidal Source
The Sinusoidal Response
The Phasor
Kirchoff’s Laws in the Frequency Domain
The Passive Circuit Elements in the Frequency Domain
Series, Parallel, and !-to-Y Simplifications
Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
–
Admittance: Y =
90o/#
!
!
!
!
!
!
!
V
resistance
i(t)
Outline - Part 1
+
(/)
i(t) = -#CVm cos (#t + &v - 90o)
= #CVm cos (#t + &v + 90o)
Jacome
voltage lags behind Margarida
current by
90-oUT Austin
EE411
37
I
Impedance:
+
v (t)
–
|V|
Terminology
V
v(t) = Vm cos (#t + &v)
| I | = #C | V |
90o
o
C
i (t)
v, i
ej#t
= 0 (short circuit)
" Phase relationship between V and I
j = e j90
Time Domain
39
Margarida Jacome - UT Austin
EE411
40
Series and Parallel Combinations
a
Vab = Z1 I + Z2 I + … + Zn I
(
Zab =
Vab
+
= Z 1 + Z2 + … + Z n
I
Z1
Voltage & Current Divider
Z2
V1 = Z1 I
I
Zn
Vab
V1 =
–
(total impedance)
b
I=
Z1
Z 1 + Z2
+ V1 –
Vab
a
Z 1 + Z2
I
Z1
+
Vab
+
V2 Z2
–
Vab
–
b
1
Zab
=
1
Z1
+
1
Z2
+…+
1
a
+
Zn
or
Vab
if n = 2:
Zab =
Z 1 Z2
Z 1 + Z2
Zn
Zn
Zn
I1 =
–
Yab = Y1 + Y2 + … + Yn
I
Z1 I1 = Z2 I2 = Z2 (I - I1)
Z2
I1
I
I2
Z1
Z 1 + Z2
Z2
b
Margarida Jacome - UT Austin
EE411
Margarida Jacome - UT Austin
EE411
41
42
Thevenin & Norton Equivalents /
Source Transformations
Phasor Domain
a
All other methods and transformations also directly apply in
the phasor domain
! Source Transformations and Thevenin-Norton Equivalent
Circuits
! !-to-Y Simplifications
! Node-Voltage Method and Mesh-Current Methods
Zs
a
0
Vs +–
Zs
b
b
with Is = Vs / Zs
(A)
(B)
Thevenin equivalent " Circuit (A)
(rely on KCL & KVL)
Is
For linear networks only (as before).
Norton equivalent " Circuit (B)
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44
Problem 1
1/
j3/
0.2/
j0.6/
9/
10/
j3/
–
0.2/
9/
Is
j3/
j0.6/
Is =
10/
Vo
-j3/
10/
-j3/
40
1+j3
=
40 (1 - j 3)
1+9
Vs
= 4 - j 12
-j19/
1.8/
j2,4/
0.2/
ig
10/
Yo =
Yo
Imaginary component of
admittance (susceptance)
must be zero
=
(
1
2,104
1
2,104
+
+
1
+ j# (0.25,10-6)
3200 + j#4
3200 – j#4
+ j 0.25,10-6 #
10.24 ,106 + 16#2
10.24 ,106 + 16#2
#2 =
36
,104
+ 0.25,10-6 # = 0
ig
250nF
20k/
4H
Yo
from (a)
=
(Im = 0)
Zo = 1/ Yo = 4k/
" # = 600 rad/s
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12 – j 16
36 – j 12 =
= 36.12 – j 18.84 V
–
46
47
(b) If ig = 0.75 cos #t mA (where # is the
frequency found in part (a)), what is the
steady-state expression for vo?
3.2k/
+
vo
–
Yo (600rad/s)
– 4#
= 1.8 + j 2.4
Problem 2 (cont)
(a) What is the value of # in radians per second?
4H
Vo =
Vo
10 – j 19
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The frequency of the sinusoidal current source
is adjusted until vo is in phase with ig.
250nF
20k/
1+j3+9–j3
(voltage divider)
+
+
–
Problem 2
3.2k/
–
(1 + j 3) (9 – j 3)
j0.6/
-j19/
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Zs =
Vs = Is Zs = (4 - j 12 )(1.8 + j 2.4) = 36 – j 12
–
+
vo
–
Vo
-j19/
Vo= ?
Is = 4 - j 12
Zs
+
1/
9/
Is
-j19/
j0.6/
+
1/
Vo
-j3/
0.2/
Use the concept of source transformations
to find the phasor voltage Vo.
+
40)0o
+
–
Problem 1 (cont)
1
2,104
"
+
Yo =
1
2,104
+
1
+ j# (0.25,10-6)
3200 + j#4
=
1
2,104
+
3200 – j#4
+ j 0.25,10-6 #
10.24 ,106 + 16#2
3200
10.24 ,106 + (16)(36,104)
= 250µS
Vo = Zo Ig = 4000 (0.75,10-3) )0o
oo
VVo ==3)0
V
o 3)0 V
vvo(t)
= 3 cos 600t V
o(t) = 3 cos 600t V
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Problem 3
Problem 3 (cont)
Find the Thevenin equivalent of the circuit with respect to the terminals a and b
Find the Thevenin equivalent of the circuit with respect to the terminals a and b
-j40
12
120)0o
+
Vx
–
+
–
120
100)0
+
–
10Vx
Th
12
100)0o +
–
120)0o +
–
(120)(60)
60
– 72
)0o
I
Vx
+
–
+
–
–
I=
10Vx
100)0o
+
–
+
Vx
–
I
VTh = 120 I + 10Vx
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Find the Thevenin equivalent of the circuit with respect to the terminals a and b
Vx = 100 – 10 I
–
Vx
IT
-j40
10
I1
I2
120
+
–
b
VT – 10Vx
I2 =
VTh
Check: -100 + 10 (-270/25 - j(360/25)) – (j40)(-270/25 - j(360/25)) + 784 - j 288 = 0
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+ VT
–
ZTh = VT / IT
I1 =
b
VTh = 20 ((-270/25) - j(360/25)) + 1000 = 784 - j 288 = 835.22 )-20.17o
51
$
Thevenin Impedance
ZTh = ?
10Vx
I = (-270/25) - j(360/25) = 18 )-126.87o A
Thevenin
voltage phasor
a
+
VTh = 120 I + 10 (100 – 10 I) = 20 I + 1000
10Vx
50
Problem 3 (cont)
120
+
–
-900 (30 + j40)
-900
=
30 – j40
302 + 402
= (-270/25) - j(360/25) = 18 )-126.87o A
b
Find the Thevenin equivalent of the circuit with respect to the terminals a and b
a
100 = (130 – j40)I + 10 (100 – 10 I )
120
Vx
(1)
substituting in (1)
Vx = 100 – 10 I
10Vx
b
Problem 3 (cont)
-j40
= (130 – j40)I + 10Vx
a
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10
100 = 10 I – j40 I + 120 I + 10Vx
120
–
+
+
+
–
-j40
(12//60)=10
b
a
+
o
60
KVL:
-j40
10
a
120
Vx = 10 I1
" I =
2
VT –
VT
1 - j4
120
=
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IT = I1 + I2
VT
10 - j40
-(9 + j4)
120 (1 - j4)
VT
52
Problem 3 (cont)
Find the Thevenin equivalent of the circuit with respect to the terminals a and b
–
Vx
IT
-j40
10
a
ZTh = ?
Thevenin Impedance
+
I1
120
I2
+
–
+ VT
–
10Vx
I1 =
b
IT = I1 + I2 =
VT
10 - j40
1–
(9 + j4)
12
Vx = 10 I1
ZTh = VT / IT
VT
10 - j40
I2 =
-(9 + j4)
120 (1 - j4)
" ZTh = VT / IT = 91.2 - j38.4 /
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$
53
VT