cos( ) cos
cos sin( 90 ) t t
sin cos( 90 ) t t
sin sin( 180 ) t t
cos cos( 180 ) t t
sin( ) sin cos cos sin t t t
cos( ) cos cos sin sin t t t
sin( ) sin
1 2 2 2 2( ) sin sin sin sin ...
2 3 511 46
72 0x t t ttt
2 6 10 14
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
amplitude mmV RI And phase iv
The sinusoidal voltage and current in a resistor are in phase
R
(t)i
( )v t
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
L
(t)i
( )v t
Now we rewrite the sin function as a cosine function
ocos(( 90 )) m iL tIv t
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
Now we rewrite the sin function as a cosine function
ocos(( 90 )) m iC ti t I
The Sinusoidal Response
( ) cos( ) s mv t tV
2 22( ) cos( )m t
R LVi t
cos( )mdiL Ri tdt
V
1were tanL
R
L
(t)i
( )S tv
R
KVL ( )sdiv t Ri Ldt
This is first order differential equations which has the following solution
We notice that the solution is also sinusoidal of the same frequency
However they differ in amplitude and phase
jba +A =
Complex Numbers
j= 1
2j = 3 2j = j j= ( )j= j 4 2 2j = j j = ( )( ) 1
Rectangular Representation
A
a
b
Real
Imaginary
Re( ) = AIm( ) =A
= aRe(a + jb)Im(a + jb)= b
A
a
b
Real
Imaginary
jba +A =
Complex Numbers (Polar form)
j= 1 Rectangular Representation
|A|
2 2a b|A|=
θ
1θ tan ba
a = cos(θ)|A| b = sin(θ)|A|
a + jbA = = + j cos(θ) sin(θ) |A| |A| cos(θ)= + s (θ)j n i|A|
Euler’s Identity
cos( ) sin( )j je
cos( ) sin( )j je cos( ) sin( )j je
2cos( )j je e cos( )2
j je e
Euler’s identity relates the complex exponential function to the trigonometric function
cos( ) sin( )j je cos( ) sin( )je j
2 sin( )j je e j sin( )
2j
j je e
Adding
Subtracting
Euler’s Identity
cos( ) sin( )je j
cos( ) 2
j je e
sin( ) 2j
j je e
The left side is complex function The right side is complex function
The left side is real function The right side is real function
A
a
b
Real
Imaginary
jba +A =
Complex Numbers (Polar form)
j= 1 Rectangular Representation
|A|
2 2a b|A|=
θ
1θ tan ba
a = cos(θ)|A| b = sin(θ)|A|
a + jbA = = + j cos(θ) sin(θ) |A| |A| cos(θ)= + s (θ)j n i|A|
= je |A| =|A|
je
Short notation
Real
Imaginary
1|A|=θ = 0
1A =
=1A
Rectangular Representation
1
θ = 360 =1AOR
0= 1 1 0je =
360= 36 1 1 0 je =
Polar Representation
Real Numbers
Real
Imaginary
1|A|=θ = 180
1A =
= 1A
Rectangular Representation
1
θ = 180 OR = 1A
180= 18 1 1 0 je =
180= 18 1 1 0 je =
Polar Representation
Real
Imaginary
1|A|=
θ = 90
j= 1A =
j= 190 90j= 1 1 je A = =
Rectangular Representation
Imaginary Numbers
Polar Representation
θ = 270 OR = 1A 270= j 1 1 270 je A = =
Real
Imaginary
1|A|=
θ = 90
jA =
j= 1
90j= 1 901 je A = =
Rectangular Representation
1j
1 jj j
2
j
j
j( 1)
j
OR 01 9= 90
1 1 je
901 je
=1
j
Polar Representation
jba +A =
Complex Conjugate
= Aje |A| A=|A|
Complex Conjugate is defined as
ja b*A = = Aje |A| A=|A|
A
a
bB
c
d
A+B
a +c
b+d
Real
B
Imaginary
Complex Numbers (Addition)
ja b+A =
jc d+B =
A+B + )+ jc d+ )a b=( ( Re( + ) = A B a +c = Re( +Re( ) A) B
Im( + ) = A B b+d = Im( +Im( ) A) B
A
a
b
B
c
d
A B
a c
b d
Real
B
Imaginary ja b+A =
jc d+B =
A B )+ ja c db ) =( (
Complex Numbers (Subtraction)
Re( ) = A B a c = Re( Re( ) A) BIm( ) = A B b d = Im( Im( ) A) B
Complex Numbers (Multiplication)
ja b+A =
jc d+B =
= Aje |A| A=|A|
= Bje |B| B=|B|
2 2a b|A|=
1A
btan a
2 2c d|B|=
1B
dtan c
+ j )(a b )c+ jd(=AB 2= + j + j +c d bca ja bd = + j +a a bd jc c db
= ( )+ j(a b ac )d d+bc
Multiplication in Rectangular Form
Multiplication in Polar Form
AB = ( )( )A Bj je e |A| |B| = A Bj je e |A| |B| ( )= A Bje |A| |B|
Multiplication in Polar Form is easier than in Rectangular form
Complex Numbers (Division)
ja b+A =
jc d+B =
= Aje |A| A=|A|
= Bje |B| B=|B|
2 2a b|A|=
1A
btan a
2 2c d|B|=
1B
dtan c
+ j )(a b
)c+ jd(=A
B
Division in Rectangular Form
*
*
+ j )( + j )
(
c d
c d+ j )( + )
a
c
b
jd
(2 2
c d c da b( ) j( )
(c )
b a
d
2 2 2 2
c d c d
(c d )
a b( ) j( )
b
a
(c d )
+ j )( j )( + j
c dc d)(ca b
jd)
(
Complex Numbers (Division)
ja b+A =
jc d+B =
= Aje |A| A=|A|
= Bje |B| B=|B|
2 2a b|A|=
1A
btan a
2 2c d|B|=
1B
dtan c
Division in Polar Form
Division in Polar Form is easier than in Rectangular form
=AB
( )= ( )
A
B
j
je
e
|A|
|B|
( ) = A Bje |A| |B|
*AA = *+ j )b )a a + jb( ( + j )a b a j )b=( (
2 2a +b= 2=|A|
( )( )A Aj je e |A| |A| *AA = 2 A Aj je e =|A|( )2 A Aje =|A|
(0)2 je=|A|2=|A|
* *(A ) = A
*(AB) = * *A B
*(A+B) = * *A +B
*A=
B
*
*AB
Complex Conjugate Identities ( can be proven)
OR
Other Complex Conjugate Identities ( can be proven)
2 2a ab a bj + j b =
( )( ) R
R
di tv t L
dt
( ) cos( )m iRi t I t
Let the current through the indictor be a sinusoidal given as
o( ) cos( 90 )m iRv t LI t
L
+
( )Ri t
( )Rv t( )
( ) RR
di tv t L
dt
( )( ) I
I
di tv t L
dt
( ) sin( )m iIi t I t
Let the current through the indictor be a sinusoidal given as
( ) cos( )m iIv t LI t ( )
( ) II
di tv t L
dt
L
+
( )Ii t
( )Iv t
From Linearity if ( ) j sin( )m iIi t I t then ( ) cos( )m iI
v t j LI t
L
+
(t) ( ) j ( )R Ii t i t I
( )tV
(t) ( ) j ( )R Iv t v t V
)( ) cos( sin( )m mi it I t j I t I
( )= ij te |I| ( )
m= I ij te
o cos( )(t) cos( 90 ) + j m iLI tm iLI t V
o)cos( 0) 9( m iL tv t I
The solution which was found earlier
o )(t) cos( 90 ) + j cos( m mi iLI t LI t V
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
L
(t)i
( )v t
Now we rewrite the sin function as a cosine function
ocos(( 90 )) m iL tIv t
L
+
(t) ( ) j ( )R Ii t i t I
( )tV
(t) ( ) j ( )R Iv t v t V
From Linearity if
)( ) cos( sin( )m mi it I t j I t I
( )= ij te |I| ( )
m= I ij te
o )(t) cos( 90 ) + j cos( m mi iLI t LI t V
o)cos( 0) 9( m iL tv t I
The solution which was found earlier
Re( (t) )( )v t V
L
+
( )v t
( ) cos( )m ii t I t
L
+
( )tV
( )m(t)= I ij te I
The solution is the real part of ( )tV
Re( (t) )( )v t V
This will bring us to the PHASOR method in solving sinusoidal excitation of linear circuit
L
+
(t) ( ) j ( )R Ii t i t I
( )tV
(t) ( ) j ( )R Iv t v t V
)( ) cos( sin( )m mi it I t j I t I
( )= ij te |I| ( )
m= I ij te
o )(t) cos( 90 ) + j cos( m mi iLI t LI t V
o o) cos( 90 ) + sin( j 90 m mi itLI t LI
o o) cos( 90 ) + j sin( 9[ ]0m i iLI t t
( 90 )= o
ij tLI em ( )
( 90 )=
oijj t em eLI
( )) m
( = I ij t jee
( )( ) di tv t Ldt
( ) cos( )m ii t I t
o( ) cos( 90 )m iv t LI t L
+
( )i t
( )v t
L
+
( )tV
( )m(t)= I ij te I
( )m[ ] I
(t)
Re ij te
I
o ( +90 ) (t)
[Re ]ij tmLI e
V
o 90 )((t) ij tmLI e V
m= I ijj te e
o 90ijj t jmLI e e e
j
ijj tmLI e ej
m= I ije I
ijmLIj e V
Phasor
Phasor
Lj I
Now if you pass a complex current
You get a complex voltage
The real part is the solution
cos( ) sin( )j je
cos( ) { }je sin( ) { }je
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
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 jmv V e e
{ }j j tmV e e
jmV e V cos( }){ mV tP
Phasor Transform
Were the notation cos( ){ } mV tP
Is read “ the phasor transform of cos( )mV t
Moving the coefficient Vm inside
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
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
i
mjI e I
RV I
cos( ( )) m iR tv t I
Now let us see the pharos domain representation or pharos transform of the current and voltage
i
mjRI e V mRI i mI i
Which is Ohm’s law on the phasor ( or complex ) domain
RV I
m
V
v
m mV RI and iv
R
I
V
R
(t)i
( )v t
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
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
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
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(
Ex 6.4:Determine the voltage v(t) in the circuit
Replace: source 2cos 5 30 with 2 30t
Impedance of capacitor is 1 1 1 1
2 25j
j C jj
ˆReplace : desired voltage v(t) with V
A single-node pair circuit
1 1ˆ 2 302 2
V j
Parallel combination
Hence time-domain voltage becomes
( ) 0.707cos(5 15 ) Vv t t
1 12 2
2 301 12 2
j
j
190
4 2 301
2 452
0.707 15
Ex 6.5 Determine the current i(t) and voltage v(t)
2 30 2 30ˆ 0.185 26.316 12 3 6 9
Ij j j
12 3 9 90ˆ 2 30 2 30 1.66 63.696 12 3 6 9
j jV
j j j
Single loop phasor circuit
( ) 0.185sin(4 26.31 ) A
( ) 1.66sin(4 63.69 ) V
i t t
v t t
The current
By voltage division
The time-domain
Ex 6.6 Determine the current i(t)
The phasor circuit is
(3)( 3) 9 90 3 3 33 3 45
3 3 2 23 2 45 2
jj j
j
Combine resistor and inductor
3 33 3
2 2j j
345
3 3 2ˆ 10 60 10 60 13.42 33.433 33 3 1 12 2
jI
j j j j
Use current division to obtain capacitor current
( ) 13.42cos(3 33.43 ) Ai t t
Hence time-domain current is:
9.7 Source Transformations and Thevenin-Norton Equivalent Circuits
Source Transformations
Thevenin-Norton Equivalent Circuits
Example 9.9
Ex 6.7 Determine i(t) using source transformation
6 6 90 5 75 30 165j
30 165ˆ 6.71 101.572 6 2
Ij j
Phasor circuit Transformed source
Voltage of source:
( ) 6.71sin(2 101.57 ) Ai t t
Hence the current
In time-domain
Phasor circuit
2ˆ 2 10 0.43 67.472 9OCV
j
Ex 6.9 Find voltage v(t) by reducing the phasor circuit at terminals a and b to a Thevenin equivalent
Remove the load
OCV
Voltage Divsion
1.10 20.53
6 6ˆ ˆ 0.43 67.47 0.48 46.94ˆ 6 1.91 0.916
OC
TH
j jV V
j jj Z
4ˆ ( 2 9)3THZ j j
( ) 0.48cos(3 46.94 ) Vv t t
2.11 25.52 1.91 0.91j
parallel
( ) 0.48cos(3 46.94 ) Vv t t
( ) 0.43cos(3 67.47 ) VOCv t t
10.91
( 3)C
0.37 FC
The Thevenin impedance can be modeled as 1.19 resistor in series
with a capacitor with value
or