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Sinusoids and Phasors
Instructor: Chia-Ming TsaiElectronics Engineering
National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.
Contents• Introduction
• Sinusoids
• Phasors
• Phasor Relationships for Circuit Elements
• Impedance and Admittance
• Kirchhoff’s Laws in the Frequency Domain
• Impedance Combinations
• Applications
Introduction• AC is more efficient and economical to
transmit power over long distance.
• A sinusoid is a signal that has the form of the sine or cosine function.
• Circuits driven by sinusoidal current (ac) or voltage sources are called ac circuits.
• Why sinusoid is important in circuit analysis?– Nature itself is characteristically sinusoidal.– A sinusoidal signal is easy to generate and transmit.– Easy to handle mathematically
Sinusoids
)( 2
2
seconds. every itself repeats sinusoid The
sinusoid theofargument the
)(radians/sfrequency angular the
sinusoid theof amplitude the
where
sin)(
voltagesinusoidal heConsider t
T:periodω
TωT
T
ωt
V
tVtv
m
m
)(
)2sin(
)2
(sin
)(sin)(
:Proof
)()(
tv
ntVω
ntV
nTtVnTtv
tvnTtv
m
m
m
Sinusoids (Cont’d)• A period function is one that satisfies
f(t) = f(t+nT), for all t and for all integers n.– The period T is the number of seconds per cycle– The cyclic frequency f = 1/T is the number of cyc
les per second
(Hz) hertz :
(rad/s) secondper radians:
where
2
1
f
fT
f
Phase :
Argument : )(
where
)sin()(
asgiven is expression general moreA
t
tVtv m
Sinusoids (Cont’d)
by
by
say that We
21
12
vlagsv
vleadsv
0 if, are and
0 if, are and
say that We
21
21
seout of phavv
in phasevv
Sinusoids (Cont’d)• To compare sinusoids
– Use the trigonometric identities
– Use the graphical approach
BABABA
BABABA
sinsincoscos)cos(
sincoscossin)sin(
:identities ricTrigonomet
sin)90cos(
cos)90sin(
cos)180cos(
sin)180sin(
tt
tt
tt
tt
The Graphical Approach
tan where
)cos(sincos
1
22
A
BBAC
tCtBtA
)1.53cos(5
sin4cos3
t
tt
Phasors• Sinusoids are easily expressed by using phasors
• A phasor is a complex number that represents the amplitude and the phase of a sinusoid.
• Phasors provide a simple means of analyzing linear circuits excited by sinusoidal sources.
of the:
of the: where,
form lExponentia :
formPolar :
formr Rectangula :
it.represent to ways threearetheir
,number complex a gConsiderin
zphase
zmagnituder
re
r
jyx
z
z
j
Phasors (Cont’d)
)sin(cos
sin ,cos
as and obtain can we, and know weIf
tan ,
as and get can we, and Given
form lExponentia :
formPolar :
formr Rectangula :
122
jrrjyxz
ryrx
yxrx
yyxr
ryx
re
r
jyx
zj
Important Mathematical Properties
rjyxz
rz
rz
r
r
z
z
rrzz
yyjxxzz
yyjxxzz
2
11
)(
)(
)()(
)()(
212
1
2
1
212121
212121
212121
:ConjugateComplex
:Root Square
:Reciprocal
:Division
:tionMultiplica
:onSubstracti
:Addition
22222
11111
rjyxz
rjyxz
rjyxz
Phasor Representation
.)( sinusoid theof
tionrepresentaphasor theis
)Re()(
)Re()Re(
)cos()(
)Im(sin
)Re(cos
sincos
)(
tv
VeV
etv
eeVeV
tVtv
e
e
je
mj
m
tj
tjjm
tjm
m
j
j
j
V
V
V
Phasor Representation (Cont’d)
Phasor Diagram
mVV
mII
Sinusoid-Phasor Transformation
jvdt
jdt
dv
ej
eeVeeeeV
tVtVdt
tdv
VtVtv
tj
tjjm
jjjtjm
mm
mm
V
V
V
V
Similarly,
)Re(
)(Re)Re(
)90cos()sin()(
)cos()(
9090
Phasor Relationships for Resistor
IV
I
RRItRIiRv
ItIi
mm
mm
)cos(
law, sOhm'By
)cos(
isresistor rough thecurrent th theIf
Time domain Phasor domain Phasor diagram
Phasor Relationships for Inductor
Time domain Phasor domain Phasor diagram
IV
I
LjtLIdt
diLv
ItIi
m
mm
)90cos(
isinductor theacross voltageThe
)cos(
isinductor rough thecurrent th theIf
Phasor Relationships for Capacitor
Phasor diagramTime domain Phasor domain
VI
V
CjtCVdt
dvCi
VtVv
m
mm
)90cos(
iscapacitor rough thecurrent th The
)cos(
iscapacitor theacross voltage theIf
Impedance and Admittance
1
1
1
AdmittanceImpedanceElement
currentphasor theis
tagephasor vol theis re whe
(S) 1
:Admittance , )( :Impedance
LjCj
C
CjLjL
RRR
YZ
YZ
YZ
I
VZ
YI
VZ
Impedance and Admittance (Cont’d)
Cj1
Z
LjZ
0
0
Impedance and Admittance (Cont’d)
voltageleadscurrent since
leadingor capacitive: voltagelagscurrent since
laggingor inductive:
thenpositive, is If
negative is when capacitive
positive is when inductive
be tosaid is impedance The
reactance:
resistance: where
jXR
jXR
X
X
X
X
R
jXR
Z
Z
Z
sin
cos and
tan where
1
22
Z
Z
Z
ZZ
X
R
R
XXR
jXR
Impedance and Admittance (Cont’d)
22
22
22
11
esusceptanc:
econductanc: where
1
XR
XB
XR
RG
XR
jXR
jXR
jXR
jXRjXRjBG
B
G
jBGZ
Y
KVL and KCL in the Phasor Domain
0)Re(
)Re()Re(
as written becan This
0)cos(
)cos()cos(
form. cosinein
writtenbemay geeach volta
state,steady sinusoidal In the
0
loop. closed a around voltages
thebe , ,..., ,let KVL,For
2121
2211
21
21
tjjmn
tjjm
tjjm
nmn
mm
n
n
eeV
eeVeeV
tV
tVtV
vvv
vvv
n
0
phasor.for holds KCL
manner,similar aIn
!!phasor!for holds KVL
0
,any for 0 Since
0Re
then,Let
0
Re
21
21
21
2121
n
n
tj
tjn
jmkK
tj
jmn
jm
jm
te
e
eV
eeV
eVeV
k
n
III
VVV
VVV
V
Series-Connected Impedance
)(
gives KVL Applying
21
21
n
n
ZZZI
VVVV
VZ
ZV
Z
VI
ZZZI
VZ
eq
kk
eq
neq
,
21
Parallel-Connected Impedance
)111
(
gives KCL Applying
21
21
n
n
ZZZV
IIII
IY
YI
Y
IV
YYYV
IY
eq
kk
eq
neq
,
21
Y- Transformations
3
133221
2
133221
1
133221
Z
ZZZZZZZ
Z
ZZZZZZZ
Z
ZZZZZZZ
c
b
a
ion:Δ ConversY
cba
ba
cba
ac
cba
cb
ion:Y ConversΔ
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
3
2
1
Example 1
07.1122.3811
1082310
108||2310
2.08||10
13
2
1
|| H2.08F103F2in
jj
jjj
jjj
jmjmj
mm
ZZZZ
rad/s. 50
for Find in
Z
Example 2(t). Find ov
)96.154cos(15.17)(
96.1515.17
152096.308575.0
152010060
100
25||2060
25||20
4 , 1520
)154cos(20
ttv
j
j
jj
jj
tv
Sol:
o
so
s
s
VV
V
Example 3. Find I
-Y transformation
204.4666.3 204.464.13
050
204.464.1316.13
86||3
12
2.36.110
)42(8
2.310
)8(4
8.06.1
8424
)42(4
Z
VI
ZZ
ZZ
Z
Z
Z
j
jj
jj
jj
j
jj
jj
Sol:
cnbn
an
cn
bn
an
Applications: Phase Shifters
i
iio
RCCR
RC
RCj
RCj
CjR
R
V
VVV
1tan
1
11
1
222
Leadingoutput
Phase Shifters (Cont’d)
i
iio
RCCR
RCjCj
R
Cj
V
VVV
1
222tan
1
1
1
11
1
Laggingoutput
Example
903
145
3
245
2
245
2
2
2020
20
453
2
2412
412
20
4122040
)2020(20)2020(||20
11
1
io
iii
j
j
j
j
jj
jj
Sol:
VVVV
VVVZ
ZV
Z
leading.. 90
of phase a provide to
circuit an Design
RC
Applications: AC Bridges
21
3132
321
2
32
21
21
21 :condition Balanced
ZZ
ZZZZZZ
ZZ
Z
ZZ
Z
VZZ
ZVV
ZZ
ZV
VV
xxx
x
sx
xs
AC Bridges (Cont’d)
sx LR
RL
2
1 sx CR
RC
2
1
Bridge for measuring L Bridge for measuring C
Summary• Transformation between sinusoid and phasor i
s given as
• Impedance Z for R, L, and C are given as
• Basic circuit laws apply to ac circuits in the same manner as they do for dc circuits.
CjLjR CLR
1 , , ZZZ
mm VtVtv V )cos()(