A sinusoidal current source (independent or dependent) produces a currentThat varies sinusoidally with time
t
t T
2 2
Tt 12 f
OR t
We define the Root Mean Square value of v(t) or rms as
Take the square oot R
calculte the ean M
quare (t)vS
The Root Mean Square value of
Expand using trigonometric identity
1T ms 1 100 0 f HZT
2 2000 /f rad s
20(0) cos( )i 10 co20 s( )
1
2010cos 60o
( ) cos(20002 60 ) A0 oi t t
20
2rmsI 14.44 A
9.2 The Sinusoidal Response
In this chapter, we develop a technique for calculating the response directly without solving the differential equation
2 H
3 ( )i t
( ) Rv t
( ) Lv t
( ) cos(3 410 )0SoV t t
c 40os(3 )10 odiL Ri tdt KVL
( ) cos(31.58 1 56 )3 . oi t t
( ) cos(33 ).1 31.56oRv t t
( ) cos(39 ).5 58.43o
Lv t t
Solution for i(t) should be a sinusoidal of frequency 3
We notice that only the amplitudeand phase change
( )SdiL Ri V tdt
2 H
3 ( )i t
( ) Rv t
( ) Lv t
( )SV t
Time Domain Complex Domain
Deferential Equation Algebraic Equation
cos( ) sin( )j je
cos( ) { }je sin( ) { }je
9.3 The phasor
The phasor is a complex number that carries the amplitude and phase angle information of a sinusoidal function
The phasor concept is rooted in Euler’s identity
Euler’s identity relates the complex exponential function to the trigonometric function
We can think of the cosine function as the real part of the complex exponential and the sine function as the imaginary part
Because we are going to use the cosine function on analyzing the sinusoidal steady-state we can apply
cos( ) { }je
cos( ) sin( )j je cos( ) { }je sin( ) { }je
cos( )mv V t ( ){ }j tmV e { }j t j
mV e e
{ }j j tmv V e e
jmV e cos( ){ } mV tP
Phasor Transform
Were the notation cos( ){ } mV tP
Is read “ the phasor transform of cos( )mV t
We can move the coefficient Vm inside
jmV e The quantity is a complex number define to be the phasor
that carries the amplitude and phase angleof a given sinusoidal function
=V
Summation Property of Phasor
1 2where , , are sinusoida lnv v v
)can be shown(
Since
Next we derive y using phsor method
cos() ( )m ii t tI
cos( )m iR tI
(( )) Rv it t
cos( )m iR tI
The VI Relationship for a Resistor
Let the current through the resistor be a sinusoidal given as
(( )) Rv it t
voltage phase
( ) cos( )m iR tt Iv
Is also sinusoidal with
amplitude mmV RI And phase iv
The sinusoidal voltage and current in a resistor are in phase
R
(t)i
( )v t
cos() ( )m ii t tI
cos( ( )) m iR tv t I
Now let us see the pharos domain representation or pharos transform of the current and voltage
i
mjI e I mI i
RV I
i
mjRI e V mRI i
m
V
v
Which is Ohm’s law on the phasor ( or complex ) domain
R
I
V
R
(t)i
( )v t
cos() ( )m ii t tI Phasor Transform
cos( ( )) m iR tv t I Phasor Transform R= I
RV IR
I
V
The voltage and the current are in phase
Real
Imaginary
I
V
iv
mV
mI
i
The VI Relationship for an Inductor
(( )) dLv ittt
d
cos() ( )m ii t tI Let the current through the resistor be a sinusoidal given as
(( )) dLv ittt
d sin( )m itIL
The sinusoidal voltage and current in an inductor are out of phase by 90o
The voltage lead the current by 90o or the current lagging the voltage by 90o
You can express the voltage leading the current by T/4 or 1/4f seconds were T is the period and f is the frequency
L
(t)i
( )v t
cos() ( )m ii t tI
si( n) ( )m iv t tIL
Now we rewrite the sin function as a cosine function (remember the phasor is defined in terms of a cosine function)
ocos( 90 )( ) m iv t tIL
The pharos representation or transform of the current and voltage
i
mjI e I mI i
o( 90 ) ijmL eI V
o 90 ij jm
jL e eI
i
mjLIj e
cos() ( )m ii t tI
ocos( 90 )( ) m iv t tIL
But since 901
ojj e 1 o90
Therefore i
mjIj L e V 90
o ij jmL e eI ( 90 )
oijmeIL mIL ( 90 )o
i
m
V
v
m mIV L and 90i ov
L
(t)i
( )v t
j L
V
I
j LV I
m mIV L and 90i ov
The voltage lead the current by 90o or the current lagging the voltage by 90o
Real
Imaginary
I
i
V
vmV
mI
The VI Relationship for a Capacitor
C
(t)i
( )v t
(( )) dCi vttt
d
cos(( ) )vmv t tV Let the voltage across the capacitor be a sinusoidal given as
(( )) dCi vttt
d sin( ) vm tVC
The sinusoidal voltage and current in an inductor are out of phase by 90o
The voltage lag the current by 90o or the current leading the voltage by 90o
The VI Relationship for a Capacitor
C
(t)i
( )v t
cos(( ) )vmv t tV sin( )( ) vmi C tt V
The pharos representation or transform of the voltage and current
cos(( ) )vmv t tV v
mjV e V mV v
sin( )( ) vmi C tt V cos ( 90 )vmoC tV o( 90 ) vj
mL eV I
o 90 v
mj
j jC e eV
I v
mjVj C e j C V
j CV I
90 1
o
i
mj
je
e Cj
I
( 90 )
oijme
CI
mC
I ( 90 )o
i
m
mIV
C and 90i ov
m
V
v
m
mIV
C and 90i ov
j CV I
1 j C
V
I
The voltage lag the current by 90o or the current lead the voltage by 90o
Real
Imaginary
I
i
V
v
mV
mI
R
I
V
j L
V
I
RV I
Real
Imaginary
I
V
iv
mV
mI
Real
Imaginary
I
i
V
vmV
mI
j LV I
1 j C
V
I
Real
Imaginary
I
i
V
v
mV
mI
j CV I
cos() ( )m ii t tI
si( n) ( )m iv t tIL
cos() ( )m ii t tI
cos( ( )) m iR tv t I
R
(t)i
( )v t
C
(t)i
( )v t
cos(( ) )vmv t tV
sin( )( ) vmi C tt V
Time-Domain Phasor ( Complex or Frequency) Domain
(( )) Rv it t
(( )) dLv ittt
d
RV I
j LV I
j CV I
(( )) dCi vttt
d
L
(t)i
( )v t
Impedance and Reactance
The relation between the voltage and current on the phasor domain (complex or frequency) for the three elements R, L, and C we have
RV I
j C
V I j LV I
When we compare the relation between the voltage and current , we note that they are all of form:
ZV I Which the state that the phasor voltage is some complex constant ( Z ) times the phasor current
This resemble ( شبه ) Ohm’s law were the complex constant ( Z ) is called “Impedance” (أعاقه )
Recall on Ohm’s law previously defined , the proportionality content R was real and called “Resistant” (مقاومه )
Z VISolving for ( Z ) we have
The Impedance of a resistor is
j C 1 I
Z R R
1Z
C j C
Z L j LThe Impedance of an indictor is
The Impedance of a capacitor is
In all cases the impedance is measured in Ohm’s
The reactance of a resistor is X 0R
1X C C
X L LThe reactance of an inductor is
The reactance of a capacitor is
The imaginary part of the impedance is called “reactance”
We note the “reactance” is associated with energy storage elements like the inductor and capacitor
The Impedance of a resistor is Z R R
1Z
C j C
Z L j LThe Impedance of an indictor is
The Impedance of a capacitor is
In all cases the impedance is measured in Ohm’s
RV I j LV I j C
1V I
Z VI
Impedance
Note that the impedance in general (exception is the resistor) is a function of frequency
At = 0 (DC), we have the following
Z L j L (0)j L 0 short
1Z
C j C(
1 0) j C
open
Time Domain
Phasor (Complex) Domain
9.5 Kirchhoff’s Laws in the Frequency Domain ( Phasor or Complex Domain)
Consider the following circuit
1 1 1( ) cos( )v t V t
2 2 2( ) cos( )v t V t
4 4 4( ) cos( )v t V t
3 3 3( ) cos( )v t V t
KVL 21 3 4( ) 0) ( )( ( )vv t v tt tv
3 3 4 41 1 2 2 cos cos(cos( )co ( s( ) ) 0 ) V V tV ttt V
Using Euler Identity we have 42 313 421{ } { } { } { } 0j j tj j j j jtt j tV e e V eVV e ee ee
Which can be written as 42 313 421{ } 0j j t j jj j t tj jt VVV e Vee ee ee e
Factoring j te 321 4
431 2{( ) } 0j tjj j jV eV ee V eeV
1
V
2
V
3
V
4
V
2( ) v t
4( ) v t
3 ( )
+
v t
L
C
1( ) v t
1R
2R
21 43+ + + = 0V VV V
KVL on the phasor domain
11 1= jV e V
22 2= jV e V
33 3= jV e V
44 4= jV e V
Phasor Transformation
PhasorCan not be zero
So in general n1 2+ + + = 0V V V
Kirchhoff’s Current Law
A similar derivation applies to a set of sinusoidal current summing at a node
n1 2+ + + = 0I I I
1 2( ) ( ) ( ) 0ni t i t i t
1 1 1( ) cos( )i t I t 2 2 2( ) cos( )i t I t ( ) cos( )n n ni t I t
11 1= jI e I 2
2 2= jI e I = nn n
jI e IPhasor Transformation
KCL
KCL on the phasor domain
9.6 Series, Parallel, Simplifications
and Ohm’s law in the phosor domain
Example 9.6 for the circuit shown below the source voltage is sinusoidal
)a (Construct the frequency-domain (phasor, complex) equivalent circuit?
o( ) 750cos(5000 30 )sv t t
LZ j L (5000)(32 X 10 )j 160 j The Impedance of the indictor is
1Z
C j CThe Impedance of the capacitor is 6
1 (5000) (5 X 10 )j
40 j
o30=750sjeVThe source voltage pahsor transformation or equivalent
o=750 30
)b (Calculte the steady state current i(t)?
ab
sIZ
V
To Calculate the phasor current Io30750
90 160 40
jej j
o3075090 120
jej
o
o
750 30
150 53.13
o 5 23.13 A
o( ) 5cos(5000 23.13 ) Ai t t
o( ) 750cos(5000 30 )sv t t
and Ohm’s law in the phosor domain
Example 9.7 Combining Impedances in series and in Parallel
)a (Construct the frequency-domain (phasor, complex) equivalent circuit?
)b (Find the steady state expressions for v,i1, i2, and i3? ?
( ) 8cos(200,000 ) Asi t t
)a(
Delta-to Wye Transformations
to Y
Y to
Example 9.8
9.7 Source Transformations and Thevenin-Norton Equivalent Circuits
Source Transformations
Thevenin-Norton Equivalent Circuits
Example 9.9
Example 9.10
Source Transformation
Since then
Next we find the Thevenin Impedance
KVL
Thevenin Impedance
TT T
T
Find interms of then form the ratio V
I VI
Ta bFind and interms of I I V
12 60
9.8 The Node-Voltage Method
Example 9.11
KCL at node 1
KCL at node 2 Since
)1(
)2(
Two Equations and Two Unknown , solving
To Check the work
9.9 The Mesh-Current Method
Example 9.12
KVL at mesh 1
KVL at mesh 2 Since
Two Equations and Two Unknown , solving
)1(
)2(
9.12 The Phasor Diagram
Top Related