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7/28/2019 Chap 09 Sinusoids and Phasors-Rev
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Chap 09 Sinusoids and Phasors
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Chap 09 Sinusoids and Phasors 2
Outline
Introduction Sinusoids
Phasors
Phasor Relationships for Circuit Elements
Impedance and Admittance
Kirchhoffs Laws in the Frequency Domain
Impedance Combinations
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Chap 09 Sinusoids and Phasors 3
Introduction
Why sinusoidal waveforms are useful to engineers?
1. Appears everywhere Vibration of a string, ripples of ocean surface, and natural response
of underdamped second-order systems
2. Easily generated
3. Using Fourier analysis, everypracticalperiodic
signal can be represented by a linear combination
of sinusoidal signals
4. Easily analyzed, such as differentiate and integrate
Sinusoid: a signal that has the form of the sine or cosinefunction.
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Periodic Function
A periodic function is one that satisfies f(t) =f(t+ nT),for all tand for all integers n.
Theperiod Tis the number of seconds per cycle
The cyclic frequencyf= 1/Tis the number of cycles
per second
Chap 09 Sinusoids and Phasors 4
T
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Chap 09 Sinusoids and Phasors 5
Sinusoids
Consider the sinusoidal voltagev(t) = Vm cos(t+)
where
Vm : the amplitude of the sinusoid
: the angular frequency in radians/st+ : the argumentof the sinusoid
:phase ofv(t) in [-180, 180] or [-, ] rads/sec
The period Tis given as
2 2 1; in Hertz (Hz)
2fT
f f
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Chap 09 Sinusoids and Phasors 6
Phase Lead or Lag
Two sinusoids with the same frequency.
1 1 1 1
2 2 2 2
( ) cos( ), 0
( ) cos( ), 0
m m
m m
v t V t V
v t V t V
1 2
1 2
1 2 1 21 2
1 2
1 2
1 2 1 2 2 1
( ) and ( ) are .
( ) and ( ) are .
, ( ) ( ) by
0
0 leads
0 lag, ( ) ( ) b
:
0 :
ys
in phav t v t
v
se
out of phaset v t
v t v t
v t v t
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Phase Lead or Lag (cont .)
Chap 09 Sinusoids and Phasors 7
1
2
2 1
( ) sin( )( ) sin( )
( ) leads ( ) by .
m
m
v t V t v t V t
v t v t
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Trigonometric Identities
Chap 09 Sinusoids and Phasors 8
sin( ) sin cos cos sin
cos( ) cos cos sin sin
A B A B A B
A B A B A B
sin( 180 ) sin
cos( 180 ) cos
sin( 90 ) cos
cos( 90 ) sin
t t
t t
t t
t t
2 2 1
cos sin cos( )
, tan
A t B t C t
BC A B
A
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Chap 09 Sinusoids and Phasors 9
Example 9.1
Q: Find the amplitude, phase, period, and frequency ofthe sinusoid
v(t) = 12cos(50t+ 10)
Sol
The amplitude is Vm = 12V.
The phase is = 10
The period
The frequency is
2 20.1257 s.
50T
17.958 Hz.f
T
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Chap 09 Sinusoids and Phasors 10
Example 9.2
Q: Calculus the phase angle between v1 = -10 cos (t+50) and v2 = 12 sin (t- 10). State which sinusoid is
leading.
Sol
v1 = -10 cos (t+ 50) = 10 cos (t+ 50 - 180)
v1 = 10 cos (t- 130)
and
v2 = 12 sin (
t- 10
) = 12 cos (
t- 10
- 90
)v2 = 12cos (t- 100)
1=-130, 2=-100; 2-1=30
v2 leads v1by 30.
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Chap 09 Sinusoids and Phasors 11
Phasors
Phasor: A phasor is complex number which representsthe amplitude andphase of a sinusoid.
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Chap 09 Sinusoids and Phasors 12
Complex Number Representations
Complex number representations:
j
rerz
jyxz
Polar form:
Rectangular form:
Exponential form: j
z
z x j
r
y
z
r
e
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Chap 09 Sinusoids and Phasors 13
Complex Plane
)sin(cos
jr
ryjxz
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Chap 09 Sinusoids and Phasors 14
Basic Operations of Complex Numbers
Addition:
Subtraction:
Multiplication:
Division:
)()( 212121 yyjxxzz
)()( 212121 yyjxxzz
)(212121
rrzz
)( 212
1
2
1 r
r
z
z
B i O ti f C l N b
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Chap 09 Sinusoids and Phasors 15
Basic Operations of Complex Numbers
(cont .)
Reciprocal:
Square Root:
Complex Conjugate:
)(11
rz
)2/( rz
j
rerjyxz
)(*
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Chap 09 Sinusoids and Phasors 16
Example 9.3
Q: Evaluate these complex numbers:
Sol
1/ 2(a) (40 50 20 30 )
10 30 (3 4)(b)
(2 4)(3 5)*
j
j j
1/ 2
40 50 20 30 47.72
40 50 20 30 6
40 50 25.71 30.64
20 30 17
(a) 40(cos50 sin50 )
20 cos( 30 ) sin( 30 )
43.0
.32 1
3 25.63
1
20
2
.
.
64
.91 81
0
j
j
j
j
j
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Chap 09 Sinusoids and Phasors 17
Example 9.3 (cont .)
8.66 5 (3 4 )(b)
(2 4)(3 5)
11.33 9 14.73 37.66
1
10 30 (3 4)
(2 4)(3 5)
4 22 26.08 122.4
16
*
0 05
7
.56 .13
j j j
j j
j
j j
j
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Chap 09 Sinusoids and Phasors 18
Sinusoids and Complex Number
Euler's identity: cos sinje j
Thus ( ) can be written a
(
s
Re( )) j t
v t
v et V
cos Re( )
sin Im( )
j
j
e
e
( )cos( ) Re( )( )
Re( )jm
j t
m m
j t
V t V
e
v e
V e
t
Here is the of ( ).jm m phasor representation v tV e V V
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Chap 09 Sinusoids and Phasors 19
Representation ofVejt
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Chap 09 Sinusoids and Phasors 20
Phasors
cos( )( ) m mV tv Vt V
(Time-domainrepresentation)
(Phasor-domainrepresentation)
Phasor: A phasor is a complex number which
represents the amplitude andphase of a sinusoid.
Given the frequency , the phasor can equivalentlyrepresent the signal v(t).
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Chap 09 Sinusoids and Phasors 21
Sinusoid-Phasor Transformation
Time domainrepresentation
Phasor domainrepresentation
)cos( tVm
)(sin tVm
)cos( tIm
)sin( tIm
mV
)90( mV
mI
)90( mI
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Chap 09 Sinusoids and Phasors 22
mV
Phasor Diagram
Vm
Im
Vm V
Im
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Chap 09 Sinusoids and Phasors 23
Example 9.4
Q: Transform these sinusoid to phasors:
Sol
(a) 6cos(50 40 ) A
(b) 4sin(30 50 ) V
i t
v t
(a) 6cos(50 40 ) has the phasor
(b) Since sin cos( 90 )
4sin(30 50 ) 4cos(30 50 90 )
4cos(30 140 ) V
The phasor of i
6 40
4 140s
A
V
i t
A A
v t t
t
v
I
V
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Chap 09 Sinusoids and Phasors 24
Example 9.5
Q: Find the sinusoid representation by these phasors:
Sol
20
(a) 3 4 A
( ) 8 Vj
j
b j e
I
V
(a) 3 4 5 126.87
( ) 5cos( 126.87 A)i
j
t t
I
(b) 1 90 ,
8 20 (1 90 ) (8 20 )
8 90 20 8 70 V
( ) 8cos( 70 ) V
j
j
v t t
V
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Chap 09 Sinusoids and Phasors 25
Example 9.6
Q: Given i1(t) = 4cos(
t+30
) A and i2(t) = 5sin (
t-20) A, find their sum.
Sol
1
2
1
1
2
1 2
2
4cos( 30 )
5sin( 20 ) 5cos( 20 90 ) 5cos( 110 )
4 30 5 110
3.464 2 1.71 4.698 1.754 2.698
3.218 56.97 A
4 30
5 110
( ) 3.218cos( t
i t
i t t t
i i i
j j j
i t
I
I
I I I
56.97 A)
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Chap 09 Sinusoids and Phasors 26
Differentiation and Integration
)Re()Re(
)90cos()sin(90 tjjjtj
m
mm
ejeeeV
tVtVdt
dv
V
dvj
dt
V
(Time domain) (Phasor domain)
(Time domain) (Phasor domain)
90
sin( ) cos( 90 )cos( )
Re Re
m mm
j t j j
j tm
V t V t vdt V t dt
V e e e ej
V
vdtj
V
( ) cos( )mv t V t
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Chap 09 Sinusoids and Phasors 27
Differentiation and Integration (cont.)
V
Vjdt
dv
jvdt V
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Chap 09 Sinusoids and Phasors 28
Example 9.7
Q: Using the phasor approach, determine the current i(t)
in a circuit described by the integral and differential
equation
Sol
4 8 3 50cos(2 75 )di
i idt t dt
84 3 50 75
2, so
(4 4 6) 50 75
Converting this to the time domain
( ) 4.642cos(2 143 A
,
.2 )
i t
j
j
t
j
j
II I
I
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ch09_Sinusoids and Phasors 29
Remarks
The phasor technique provides a way torepresent or compute the operations (including
additions, differentiations, and integrations) of
sinusoids. Note that the sinusoidal frequency is assumed
to be fixed to a common value.
Phasor Relationships for Circuit
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Chap 09 Sinusoids and Phasors 30
Phasor Relationships for CircuitElements: Resistor
By Ohm's la
cos( )
,
cos( )
wm
m
m
m
i I t
v iR
I
RI RRI t
I
V I
Time domain Phasor domain Phasor diagram
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Chap 09 Sinusoids and Phasors 31
Inductor in Phasor Domain
Time domain Phasor domain Phasor diagram
cos( )
cos( 90 )
90
m
m
m
mi I t
div L LI
I
j L
tdt
LI
I
V I
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Chap 09 Sinusoids and Phasors 32
Capacitor in Phasor Domain
Phasor diagramTime domain Phasor domain
cos( )
cos( 90 )
90
m
m
m
mv V t
dvi C CV t
dt
C
V
V j C
V
I V
Summary of Voltage-Current
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Chap 09 Sinusoids and Phasors 33
Summary of Voltage-Current
Relationships
Element Time domain Frequencydomain
R V=Ri V =RI
L V =jLI
C
dt
diLv
dt
dvCi
Cj
IV
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Chap 09 Sinusoids and Phasors 34
Example 9.8
Q: The voltage v = 12 cos(60t+ 45
) is applied to a 0.1-Hinductor. Find the steady-state current through the inductor.
Sol
12 45 V , where 60rad/s.Hence,
12 45 12 452 45
( ) 2cos(60 4 A5 )
A60 0.1 6 90
Converting,
j L
i t
j
t
j L
V I
VI
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Chap 09 Sinusoids and Phasors 35
Impedance and Admittance
or VZ V ZII
The impedance Z of a circuit is the ratio of the phasorvoltage V to the phasor current I, measured in ohms ().
Impedance and Admittances of
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Chap 09 Sinusoids and Phasors 36
Impedance and Admittances of
Passive Elements
Element Impedance Admittances
R Z =R
L Z =jL
C Y =jCCj
1Z
R
1Y
Lj
1Y
Equivalent Circuits at DC & High
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Chap 09 Sinusoids and Phasors 37
Equivalent Circuits at DC & High
Frequencies
0
L j LZ
1C
j CZ
General Passive Circuit In Phasor
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Chap 09 Sinusoids and Phasors 38
General Passive Circuit In PhasorDomain
R jX Z Z
2 2 1
where
, tan
and
cos Re( ) : of ,
Resistance
Reactancsin Im( ) : of e
R
X
XR X R
Z
Z Z Z
Z Z Z
Inductive and Capacitive
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Chap 09 Sinusoids and Phasors 39
Inductive and CapacitiveImpedance
X> 0: inductive impedance
X=0: pure resistor
X
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Chap 09 Sinusoids and Phasors 40
Admittance
1 Y
Z
I
V
G Bj Y
G = Re Y is called the conductance.
B = Im Y is called the susceptance.
The admittance Y of a circuit is the reciprocal of theimpedance, measured in siemens (S).
Ad itt d I d
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Chap 09 Sinusoids and Phasors 41
Admittance and Impedance
2
2
2 2 2
1
1
1
,
G jB
R jXR jX R jX
G jBR
G BR X
R
jX R
X R X
jX R jX
YZ
E l 9 9
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Chap 09 Sinusoids and Phasors 42
Example 9.9
Q: Find v(t) and i(t) in the circuit.
E l 9 9 ( )
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Chap 09 Sinusoids and Phasors 43
Example 9.9 (cont .)
From the voltage source 10 cos 4t, = 4,
The impedance is
Hence the current
V010 sV
1 15 54 0.1
5 2.5
j C j
j
Z
2 2
10 0 10(5 2.5)
5 2.5 5 2.5
1.6 0.8 1.789 26.57 A
s j
j
j
VI
Z
E l 9 9 ( )
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Chap 09 Sinusoids and Phasors 44
Example 9.9 (cont .)
The voltage across the capacitor is
Converting I and V, we get
1.789 26.57
4 0.1
1.789 26.574.47 63.43 V
0.4 90
j C j
C
IV IZ
A( ) 1.789cos(4 26.57 )
( ) 4.47cos(4 63.43 ) V
i t t
v t t
Kirchhoffs Law in the Frequency
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Kirchhoff s Law in the FrequencyDomain
The KCL and KVL are still applicable in thefrequency domain.
Chap 09 Sinusoids and Phasors 45
1 2 1 2
1 2 1 2
KVLTime do F
0 0
KC
requency domain
L
m n
0 0
ain n
n n
v v v
i i i
V V V
I I I
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Chap 09 Sinusoids and Phasors 46
KVL and KCL in the Phasor Domain
1 1
1
1
2 2
2
2
For , let , ,..., , be the voltages around a closed loop.
0
In the sinusoidal steady state, each voltage may be written in cosine
cos( ) cos(
form.
o
V
) s
K
c
L
m m n
n
m
n
v
V t V t
v v
v v v
V
1 2
1 2
1
1 2
1 2
1 2
2
This can be written as
Re( ) Re( ) Re( ) 0
Re 0
Re 0; ( )
; (
( ) 0
0
n
n
k
jj jj t j t j t
m m mn
jj j j t
m m mn
jj
n
n
t
n K mk
V e e V e e V e e
V e V e V e
V
t
e
e e
V V
V V
V
V V
KVL hold
0
s f
)
or phasor!
j te t
1 2KCL holds forSimilarily, pha or s 0n I I I
S i C t d I d
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Chap 09 Sinusoids and Phasors 47
Series-Connected Impedances
eq 1
1 2 1 2
2
( )N
N
N
V V V V I Z Z Z
VZ Z Z
IZ
V lt Di i i P i i l
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Chap 09 Sinusoids and Phasors 48
Voltage-Division Principle
then,andSince 2211
21
IZVIZV
ZZ
VI
1 21 2
1 2 1 2
Voltage division princile ,:
Z Z
Z Z ZV V
ZV V
P ll l C t d I d
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Chap 09 Sinusoids and Phasors 49
Parallel-Connected Impedances
eq
1
2
2
2
1
1
1 1 1
1 1
1
1
N
N
N
I I I I VZ Z Z
I
VZ Z Z Z
eq 1 2 N Y Y Y Y
C t Di i i P i i l
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Chap 09 Sinusoids and Phasors 50
Current-Division Principle
2211eq
21
21
2121eq
eq/11/
111
ZIZIIZV
ZZ
ZZ
ZZYYYZ
2 1
1 2
1 2
1 2
Current-division principle: ,
Z Z
Z Z ZI I
ZI I
Delta ()-to-Wye (Y) and
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Chap 09 Sinusoids and Phasors 51
Delta () to Wye (Y) andWye (Y)-to-Delta () Transformations
Delta () to Wye (Y) Transformation
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Chap 09 Sinusoids and Phasors 52
Delta ()-to-Wye (Y) Transformation
-Y Conversion
1
2
3
b c
a b c
c a
a b c
a b
a b c
Z
Z Z Z
Z
Z Z Z
Z
Z Z
Z
ZZ
Z
Wye (Y) to Delta () Transformation
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Chap 09 Sinusoids and Phasors 53
Wye (Y)-to-Delta () Transformation
Y- Conversion
1 2 2 3 3 1
1
1 2 2 3 3 1
2
1 2 2 3 3 1
3
a
b
c
Z Z Z Z Z Z
Z
Z Z Z Z Z Z
Z
Z Z Z Z ZZ
Z
Z
Z
Z
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Chap 09 Sinusoids and Phasors 54
A delta () or wye (Y) circuit is said to be balancedif it has equal impedances in all three branches.
or1
3
3Y Y Z Z Z Z
.321 andwhere cbaY ZZZZZZZZ
Example 9 10
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Chap 09 Sinusoids and Phasors 55
Example 9.10
Q: Find the input impedance of the circuit. Assume that
the circuit operations at = 50 red/s.
Example 9 10 (cont )
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Chap 09 Sinusoids and Phasors 56
Example 9.10 (cont .)
31
3
32
1 1
50 2 10
1 13 3
50 10 10
8 8 50 0.2
10
(3 2)
(8 10)
j
j
j C j
j C j
j L j j
Z
Z
Z
Z1
Z3
Z2
Example 9 10 (cont )
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Chap 09 Sinusoids and Phasors 57
Example 9.10 (cont .)
in 1 2 3
2 2
(3 2)(8 10)10
11 8
(44 14)(11 8)10 10 3.22 1.07
11 83.22 11.07
j jj
j
j jj j j
j
Z Z Z Z
The input impedance is
Z1
Z2
Z3
Example 9 11
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Chap 09 Sinusoids and Phasors 58
Example 9.11
Q: Determine v0(t) in the circuit.
Example 9 11 (cont )
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Chap 09 Sinusoids and Phasors 59
Example 9.11 (cont .)
3
20cos(4 15 ) 20 15 V, 4
1 110mF 25
4 10 10
5 H 4 5 20
s sv t
jj C j
j L j j
V
Z1 Z2
Example 9 11(cont )
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Chap 09 Sinusoids and Phasors 60
Example 9.11(cont .)
Let
Z1 = Impedance of the 60- resistor
Z2 = Impedance of the parallel combination of the 10-
mF capacitor and the 5-H inductor
Then Z1 = 60 and
2
25 2100
025 20
25 20
j jj jj
j j
Z
Z1Z2
Example 9 11 (cont )
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Chap 09 Sinusoids and Phasors 61
Example 9.11 (cont .)
By the voltage-division principle,
Convert this to the time domain and obtain
2
1 2
100(20 15 )
60 100
(0.8575 30.96 ) (20 15 17.15 1) . V5 96
o s
j
j
Z
ZV
ZV
( ) 17.15cos(4 15.96 V)ov t t
Example 9 12
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Example 9.12
Chap 09 Sinusoids and Phasors 62
4(2 4) 4(4 2)(1.6 0.8)
4 2 4 8 10
4(8) 8(2 4)3.2 , (1.6 3.2)
10 10
an
bn cn
j j jj
j j
j jj j
Z
Z Z
62
.Find I-Y transformation
Example 9 12 (cont )
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Example 9.12 (cont .)
Chap 09 Sinusoids and Phasors 63
(1.6 0.8) ; 3.2 ; (1.6 3.2)an bn cnj j j Z Z Z
63
.Find I-Y transformation
12 3 || 6 8
13.6 1 13.64 4.2045
3
0 0
13.64 4.204
.666 4.204
an bn cnj j
j
Z
VI
Z
Z Z Z
Application: Phase Shifters
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Chap 09 Sinusoids and Phasors 64
Application: Phase Shifters
i
iio
RCCR
RC
RCj
RCj
CjR
R
V
VVV
1tan
1
11
1
222
Leading
output
Phase Shifters (cont )
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Chap 09 Sinusoids and Phasors 65
Phase Shifters (cont.)
i
iio
RCCR
RCj
CjR
Cj
V
VVV
1
222 tan1
1
11
1
1
Lagging
output
Example 9 13
7/28/2019 Chap 09 Sinusoids and Phasors-Rev
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Chap 09 Sinusoids and Phasors 66
Example 9.13
Q: Design anRCcircuit to provide a phase of 90
leading.Z
Example 9 13
7/28/2019 Chap 09 Sinusoids and Phasors-Rev
67/67
Example 9.13
Using voltage division,
20(20 20)2 12 40 (20 20) 40 20
jj j j
Z
1
1
1
2
45 V3
190
3
Z 12 4
Z 20 12 24
and
20 245
20 20 2
2 245 45
2 3
i i
o
i
i
j
j j
j
V
V
V
V V
V
V
Z
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