Download Chapter-6

cos t  sin(t  90 )
sin t  cos(t  90 )
 sin t  sin(t  180 )
 cos t  cos(t  180 )
sin(t   )  sin t cos   cos t sin 
cos(t   )  cos t cos  sin t sin 
sin( )   sin 
cos( )  cos 
1 2
2
2
2
x (t )   sin 2 t 
sin 6 t 
sin 10 t 
sin 14 t  ...
2 
3
5
7
  2
  6
  10
  14
The VI Relationship for a Resistor
R

v (t )
v (t )  R i (t )

i (t)
Let the current through the resistor be a sinusoidal given as
i (t )  I m cos(t  i )
v (t )  R i (t )  R I m cos(t  i )  R I m cos(t  i )

v (t )  R I m cos(t



i )
voltage phase 
Is also sinusoidal with amplitude
amplitude V m  RI m
And phase v  i
The sinusoidal voltage and current in a resistor are in phase
The VI Relationship for an Inductor
L

v (t )

v (t )  L di (t )
dt
i (t)
Let the current through the resistor
be a sinusoidal given as
i (t )  I m cos(t  i )
v (t )  L di (t )  LI m sin(t  i )
dt
Now we rewrite the sin function
as a cosine function
v (t )  LI m cos(t  i  90o )
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
The VI Relationship for a Capacitor
C

v (t )

i (t ) C dv (t )
dt
i (t)
Let the voltage across the capacitor
be a sinusoidal given as
v (t ) V m cos(t  v )
i (t ) C dv (t )   CV m sin(t  v )
dt
Now we rewrite the sin function
as a cosine function
i (t )  CI m cos(t  i  90o )
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 Sinusoidal Response
i (t)
R
v S (t )
L


KVL
v s (t )  V m cos(t   )
v s (t )  Ri  L di
dt
L di  Ri V m cos(t   )
dt
This is first order differential equations which has the following solution
i (t ) 
Vm
R 2   2L 2
cos(t   
 L 
were   tan 1 

 R 
)
We notice that the solution is also sinusoidal of the same frequency
However they differ in
amplitude
and
phase

Complex Numbers
A = a + jb
j= 1
j = 
2
Rectangular Representation
j = j j= ()j=  j j = j j = ()() 1
3
2
4
2 2
Imaginary
A
b
a
Re(A) = Re(a + jb) =a
Im(A) = Im(a + jb) = b
Real
Complex Numbers (Polar form)
A = a + jb
j= 1
Rectangular Representation
Imaginary
A
b
2
| A|= a  b
2
| A|
a = |A| cos(θ)
θ
a
b

1
θ  tan 
a




b = | A| sin(θ)
Real
A = a + jb = |A| cos(θ) + j|A| sin(θ) = | A|  cos(θ) + j sin(θ) 
Euler’s Identity
e  j   cos( )  j sin( )
Euler’s identity relates the complex exponential function to the trigonometric function
e  j   cos( )  j sin( )
e  j   cos( )  j sin( )
Adding
e  j  e  j   2cos( )
e  j   cos( )  j sin( )
 e  j    cos( )  j sin( )
Subtracting
e  j   e  j   2 j sin( )
 j
 j
e

e
cos( ) 
2
 j
 j
e

e
sin( ) 
2j
Euler’s Identity
e  j   cos( )  j sin( )
The left side is complex function
cos( )  e
 j
e
2
 j
The left side is real function
The right side is complex function
 j
 j
e

e
sin( ) 
2j
The right side is real function
Complex Numbers (Polar form)
A = a + jb
j= 1
Rectangular Representation
Imaginary
A
b
2
| A|= a  b
2
| A|
a = |A| cos(θ)
θ
a
b

1
θ  tan 
a




b = | A| sin(θ)
Real
A = a + jb = |A| cos(θ) + j|A| sin(θ) = | A|  cos(θ) + j sin(θ) 
= | A| e
 j
=| A| 
Short notation
j
e
Real Numbers
A =1
Rectangular Representation
Imaginary
Polar Representation
A =1
= 1e
j0
= 1 0
θ=0
| A|=1
Real
1
OR
θ =  360
A =1
= 1e
 j 360
= 1  360
A = 1
Rectangular Representation
Imaginary
Polar Representation
A = 1
= 1e
 j 180
= 1 180
θ = 180
| A|=1
1
OR
θ = 180
Real
A = 1
= 1e
 j 180
= 1 180
Imaginary Numbers
A = j= 1
Imaginary
Rectangular Representation
Polar Representation
A = j= 1e
j= 1
 j 90
= 1 90
| A|=1
θ = 90
Real
OR
θ =  270
A = 1A = j = 1e  j 270
= 1   270
A = j
Rectangular Representation
Imaginary
Real
| A|=1
θ =  90
Polar Representation
A =  j= 1e
j= 1
1 1 j  j  j  j
2
(1)
j jj
j
1
1
 j 90

=
1
e
OR
 j 90
j 1e
=1  90
 j 90
= 1   90
Complex Conjugate
A = a + jb = | A| e  j 
A
=| A| A
Complex Conjugate is defined as
A = a  jb
*
= | A| e
 j A
=| A| A
Complex Numbers (Addition)
A = a + jb
B = c + jd
A+B =(a + c)+ j(b + d)
Imaginary
B
b+d
b
Re(A + B) = a +c = Re(A) +Re(B)
Im(A + B) = b+d = Im(A) +Im(B)
A+B
A
B
d
a
c
Real
a +c
Complex Numbers (Subtraction)
A = a + jb
Imaginary
B = c + jd
d
A  B =(a  c)+ j(b  d)
B
A
b
Re(A  B) = a  c = Re(A)  Re(B)
Im(A  B) = b  d = Im(A)  Im(B)
a c
c
a
bd
B
A B
Real
Complex Numbers (Multiplication)
A = a + jb = | A| e  j 
B = c + jd = |B| e
A
 j B
=| A| A
2
| A|= a  b
=|B| B |B|= c  d
2
2
2
b

1
A  tan  a 
 
d

1
B  tan  c 
 
Multiplication in Rectangular Form
AB = (a + jb)(c+ jd) = ac + jad + jbc + j2bd = ac + jad + jbc  bd
= (ac  bd)+ j(ad+bc)
Multiplication in Polar Form
AB = ( | A| e
 j A
)( |B| e
 j B
) = | A| |B|e
j A
e
j B
= | A| |B|e
Multiplication in Polar Form is easier than in Rectangular form
j (A B )
Complex Numbers (Division)
A = a + jb = | A| e  j 
=| A| A
A
2
| A|= a  b
2
B = c + jd = |B| e  j  =|B| B |B|= c  d
2
B
2
b

1
A  tan  a 
 
d

1
B  tan  c 
 
Division in Rectangular Form
A (a + jb)  (a + jb)(c+ jd)  (a + jb)(c jd)  (ac bd) j(bcad)
=
2
2
*
(
c
+
j
d)(c

jd)
B (c+ jd) (c+ jd)(c+ jd)
(c  d )
*

(ac bd)
2
2

j(bc ad)
2
2
(c  d ) (c  d )
Complex Numbers (Division)
A = a + jb = | A| e  j 
B = c + jd = |B| e
A
 j B
=| A| A
2
| A|= a  b
=|B| B |B|= c  d
2
2
2
b

1
A  tan  a 
 
d

1
B  tan  c 
 
Division in Polar Form
 j
A
) = | A| e j (
= = ( | A| e
 j
|B|
B
( |B| e
)
A
A
B )
B
Division in Polar Form is easier than in Rectangular form
Complex Conjugate Identities ( can be proven)
*
2
2
AA* = (a + jb)(a + jb) =(a + jb)(a  jb) = a  jab + jab  b
2
2
2
= a +b =| A |
OR
AA* = ( | A| e
 j A
2
=| A| e
)( | A| e
 j (0)
 j A
2
) =| A|2 e  j  e  j  =| A| e
A
A
2
=| A |
Other Complex Conjugate Identities ( can be proven)
*
* *
(A ) = A
*
* *
(AB) = A B
(A +B) = A* +B*
*
 A  A*
 B  = B*
 
 j (A  A )
+
i R (t )
di (t )
v (t )  L R
R
dt
Let the current through the indictor
be a sinusoidal given as
i (t )  I m cos(t  i )
R
v R (t )
L
di (t )
v (t )  L R
R
dt

+
i I (t )
v (t )  L
I
di (t )
I
dt
v (t )  LI m cos(t  i  90o )
R
Let the current through the indictor
be a sinusoidal given as
i (t )  I m sin(t  i )
I
v I (t )

L
v (t )  L
From Linearity if
I
di (t )
I
dt
v (t )  LI m cos(t  i )
I
i (t )  j I m sin(t  i )
I
then v I (t )  j LI m cos(t  i )
+
V(t )
I(t)  i R (t )  ji I (t )
V(t) 
L
LI m cos(t  i 90o) + j LI m cos(t  i )
I(t )  I m cos(t  i )  j I m sin(t  i )

= |I| e
j (t i )
= Ime
j (t i )
V(t) v R (t )  jv I (t )
V(t) 
LI m cos(t  i 90o) + j LI m cos(t  i )
v (t ) LI m cos(t  i 90o)
The solution which was found earlier
The VI Relationship for an Inductor
L

v (t )

v (t )  L di (t )
dt
i (t)
Let the current through the resistor
be a sinusoidal given as
i (t )  I m cos(t  i )
v (t )  L di (t )  LI m sin(t  i )
dt
Now we rewrite the sin function
as a cosine function
v (t )  LI m cos(t  i  90o )
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
From Linearity if
+
V(t )

I(t)  i R (t )  ji I (t )
L
I(t )  I m cos(t  i )  j I m sin(t  i )
= |I| e
j (t i )
= Ime
j (t i )
V(t) v R (t )  jv I (t )
V(t) 
LI m cos(t  i 90o) + j LI m cos(t  i )
v (t ) LI m cos(t  i 90o)
The solution which was found earlier
v (t )  Re( V(t) )
+
v (t )

i (t )  I m cos(t  i )
L
+
V(t )
I(t)= Im e
j (t i )
L

The solution is the real part of
V(t )
v (t )  Re( V(t) )
This will bring us to the PHASOR method in solving sinusoidal excitation of linear circuit
+
V(t )

I(t)  i R (t )  ji I (t )
L
I(t )  I m cos(t  i )  j I m sin(t  i )
= |I| e
j (t i )
= Ime
j (t i )
= Im e j (t ) e
j (i )
V(t) v R (t )  jv I (t )
LI m cos(t  i 90o) + j LI m cos(t  i )
 LI m cos(t  i 90o) + j LI m sin(t  i 90o)
 LI m [cos(t  i 90o) + j sin(t i 90o)]
V(t) 
= LI me
j (t i 90o )
= LI m e
j (t )
e
j (i 90o )
i (t )
+
v (t )
L
i (t )  I m cos(t  i )
v (t )  L di (t )
dt
v (t )  LI m cos(t  i
 90o )

j (t i )
 Re[ Im e
I(t)
j (t i +90o )
 Re[ LI me
V(t)
]
]
The real part is the solution
Now if you pass a complex current
+
V(t )

I(t)= Im e
L
j (t i )
= Im e j t e
j i
I = Ime
Phasor
j i
You get a complex voltage
V(t) LI me
j (t i 90o )
 j LI me
LI me
j t j i
e
e
j 90o
j
j t j i
e
Phasor
V  j LI me
 j L I
j i
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
e  j   cos( )  j sin( )
We can think of the cosine function as the real part of the complex exponential and the
sine function as the imaginary part
cos( ) {e j  }
sin( )  {e j  }
Because we are going to use the cosine function on analyzing the sinusoidal steady-state
we can apply
j
cos( ) {e }
e  j   cos( )  j sin( )
cos( ) {e j  }
sin( )  {e j  }
v V m cos(t  ) V m {e j (t  )} {V me j e j t }
Moving the coefficient Vm inside
v V m {e j t e j  }
Phasor Transform
V  V me j   P{V m cos(t  )}
Were the notation
P{V m cos(t  )}
Is read “ the phasor transform of
V m cos(t  )
The VI Relationship for a Resistor
R

v (t )
v (t )  R i (t )

i (t)
Let the current through the resistor be a sinusoidal given as
i (t )  I m cos(t  i )
v (t )  R i (t )  R I m cos(t  i )  R I m cos(t  i )

v (t )  R I m cos(t



i )
voltage phase 
Is also sinusoidal with amplitude
amplitude V m  RI m
And phase v  i
The sinusoidal voltage and current in a resistor are in phase
R

i (t )  I m cos(t  i )

v (t )
i (t)
v (t )  R I m cos(t  i )
Now let us see the pharos domain representation or pharos transform of the current and voltage
i
I  I me j   I m
i
V m  RI m
V  RI
V  RI me j 
and
v   i

V
I
 RI m
Vm
i
v
Which is Ohm’s law on the phasor ( or complex ) domain
R

i
V  RI
R
V  RI


V
I
The voltage and the current are in phase
Imaginary
V
Vm
Im
I
 i v   i
Real
The VI Relationship for an Inductor
L

v (t )

v (t )  L di (t )
dt
i (t)
Let the current through the resistor be a sinusoidal given as
i (t )  I m cos(t  i )
v (t )  L di (t )  LI m sin(t  i )
dt
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

v (t )
i (t )  I m cos(t  i )

v (t )  LI m sin(t  i )
i (t)
Now we rewrite the sin function as a cosine function
( remember the phasor is defined in terms of a cosine function)
v (t )  LI m cos(t  i  90o )
The pharos representation or transform of the current and voltage
I  I me j   I m
i (t )  I m cos(t  i )
i
v (t )  LI m cos(t  i
But since
Therefore
 90o )
j 1e j 90  1
o
i
V   L I me
o
i
 j
90o
V  j LI me j   LI me j 90 e j   LI me j (
V m   LI m and
 LI me j  e  j 90  j LI me j 
j ( i 90o )
i
v   i  90o
o
i
i
 90o )
 LI m
Vm
(i  90o )
v
i
j L


V
I
V  j L I
V m   LI m
and
v   i  90o
The voltage lead the current by 90o or the current lagging the voltage by 90o
Imaginary
V
Vm
v
Im
I
i
Real
The VI Relationship for a Capacitor
C

v (t )

i (t ) C dv (t )
dt
i (t)
Let the voltage across the capacitor be a sinusoidal given as
v (t ) V m cos(t  v )
i (t ) C dv (t )   CV m sin(t  v )
dt
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 VI Relationship for a Capacitor
C

i (t )   CV m sin(t  v )
v (t ) V m cos(t  v )

v (t )
i (t)
The pharos representation or transform of the voltage and current
V  V me j   V m
v (t ) V m cos(t  v )
v
v
i (t )   CV m sin(t  v )   CV m cos(t  v  90o )
I   CV me j  e j 90
 j  CV me j 
o
v
j
V 
I
j C
V m  Im
C

I me j 
i
1e j 90  C
j
and
o
v
j ( 90 )
 I me
C
v   i  90o
i
o
I  LV me j (
 j  CV
 Im
C
Vm
(i 90o )
v
v
90o )
1 j C
V 


V
I
j C
V m  Im
C
I
and
v   i  90o
The voltage lag the current by 90o or the current lead the voltage by 90o
Imaginary
V
I
i
Im
Vm
v
Real
Phasor ( Complex or Frequency) Domain
Time-Domain
Imaginary
R

R
i (t)

v (t )  R i (t )

V
Im
V  RI
I
i (t )  I m cos(t  i )
Real
Imaginary


v
Vm

I
Im
i
Real
I
i (t )  I m cos(t  i )
v (t )  LI m sin(t  i )
V  j L I
v (t )  L di (t )
dt
Imaginary
1 j C
C
v (t )
V
V
i (t)

V  j L I
j L
L
v (t )
I
v   i
v (t )  R I m cos(t  i )

V  RI
Vm

v (t )
V
V
I
i

i (t)

i (t ) C dv (t )
dt
v (t ) V m cos(t  v )
i (t )   CV m sin(t  v )

V
I
V 
Im
V 
Vm
I
j C
v
Real
I
j C
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
V  RI
V  j L I
V 
I
j C

1
j C
I
When we compare the relation between the voltage and current , we note that they are all of
form:
V  ZI
Which the state that the phasor voltage is some complex constant ( Z )
times the phasor current
This resemble ) ‫ ( شبه‬Ohm’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
V
Z

Solving for ( Z ) we have
I
The Impedance of a resistor is
ZR  R
The Impedance of an indictor is
ZL  j  L
1
ZC 
j C
The Impedance of a capacitor is
In all cases the impedance is measured
in Ohm’s W
V  j L I
V  RI
Z
Impedance
V 
1
j C
I
V
I
The Impedance of a resistor is
ZR  R
The Impedance of an indictor is
ZL  j  L
1
ZC 
j C
The Impedance of a capacitor is
In all cases the impedance is measured
in Ohm’s W
The imaginary part of the impedance is called “reactance”
The reactance of a resistor is
XR  0
We note the “reactance” is associated
with energy storage elements like the
inductor and capacitor
XL   L
The reactance of a capacitor is
XC  1
C
Note that the impedance in general (exception is the resistor) is a function of frequency
The reactance of an inductor is
At  = 0 (DC), we have the following
ZL  j  L  j (0) L  0
ZC 
1
j C

1
 
j (0) C
short
open
9.5 Kirchhoff’s Laws in the Frequency Domain ( Phasor or Complex Domain)
Consider the following circuit
R2
v 2 (t ) 

R1
Phasor Transformation
+
v 1 (t )
v 3 (t )


L
V 1=V 1e
v 2 (t ) V 2 cos(t  2 )
V 2 =V 2e
v 3(t ) V 3 cos(t  3)
V 3 =V 3e
v 4 (t ) V 4 cos(t  4 )
v 4 (t ) 
j 1
v 1(t ) V 1 cos(t  1)
j 2
j 3
V 4 =V 4e
j 4
v 1(t )  v 2 (t )  v 3(t )  v 4 (t )  0
KVL
C
V 1 cos(t  1)  V 2 cos(t  2 )  V 3 cos(t  3)  V 4 cos(t  4 )  0
j
Using Euler Identity we have {V 1e 1e j t } {V 2e
j
Which can be written as {V 1e 1e j t V 2e
{(V 1e
Factoring e j t
j 1
V 2e
V1
Phasor
So in general
V1 +V 2 + +V n = 0
V2
j 2
j 2 j t
e
V 3e
V3
j 3
j 2 j t
e
j
} {V 3e 3e j t } {V 4e
j
V 3e 3e j t V 4e
V 4e
V4
j 4
j 4 j t
e
)e j t }  0
j 4 j t
e
} 0
} 0
V1 +V 2 +V 3 +V 4 = 0
Can not
be zero
KVL on the phasor
domain
Kirchhoff’s Current Law
A similar derivation applies to a set of sinusoidal current summing at a node
i1(t )  I 1 cos(t  1)
Phasor
Transformation
I 1=I 1e
KCL
j 1
i 2 (t )  I 2 cos(t  2 )
I 2 =I 2e
i n (t )  I n cos(t  n )
j 2
i1(t )  i 2 (t ) 
I 1 +I 2 + +I n = 0
I n =I ne
j n
 i n (t )  0
KCL on the phasor
domain
Example 9.6 for the circuit shown below the source voltage is sinusoidal
v s (t )  750cos(5000t  30o )
(a) Construct the frequency-domain (phasor, complex) equivalent circuit ?
(b) Calculte the steady state current i(t) ?
The source voltage pahsor transformation or equivalent
o
V s =750 e j 30 =75030
o
The Impedance of the indictor is
Z  j L  j (5000)(32 X 10)  j 160 W
The Impedance of the capacitor is
ZC 
L
1
j C

1
  j 40 W
j (5000) (5 X 106 )
v s (t )  750cos(5000t  30o )
To Calculate the phasor current I
I  Vs
Z
ab
o
j 30
750
e

90  j 160  j 40
j 30
75030o
750
e


90  j 120
15053.13o
o
i (t )  5cos(5000t  23.13o ) A
 5  23.13o A
Example 9.7 Combining Impedances in series and in Parallel
i s (t )  8cos(200,000t ) A
(a) Construct the frequency-domain (phasor, complex) equivalent circuit ?
(b) Find the steady state expressions for v,i1, i2, and i3 ? ?
(a)
Ex 6.4:Determine the voltage v(t) in the circuit
Replace: source 2cos  5t  30

with 230
Replace : desired voltage v(t) with
Impedance of capacitor is
Vˆ
1
1
1
1


j
 j2
j C
2
j 5

A single-node pair circuit
1
1

j
 1  


1 
ˆ
2
2
  230
V      j    230  
2 
1
1
 2  
j
Parallel combination
2
2
1
  90
4

 230  0.707  15
1
2  45
2
Hence time-domain voltage becomes
v (t )  0.707 cos(5t  15 ) V
Ex 6.5 Determine the current i(t) and voltage v(t)
Single loop phasor circuit
230
The current Iˆ  230

 0.185  26.31
6  j 12  j 3 6  j 9
By voltage division
Vˆ 
j 12  j 3
990
230 
230  1.6663.69
6  j 12  j 3
6 j9
The time-domain
i (t )  0.185sin(4t  26.31 ) A
v (t )  1.66sin(4t  63.69 ) V
Ex 6.6 Determine the current i(t)
The phasor circuit is
Combine resistor and inductor
(3)( j 3)
990
3
3
3
3 j3


45   j
3  j 3 3 245
2
2
2
3
3
3 j3  j
2
2
Use current division to obtain capacitor current
3
45
3
j
3
2
Iˆ 
10  60 
10  60  13.42  33.43
3
3
3 j 3 j1
 j  j1
2
2
Hence time-domain current is:
i (t )  13.42cos(3t  33.43 ) A
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
Phasor circuit
Voltage of source:
Hence the current
In time-domain
Transformed source
j 6   690  575  30165
30165
ˆ
I 
 6.71101.57
2 j 6 j 2
i (t )  6.71sin(2t  101.57 ) A
Ex 6.9 Find voltage v(t) by reducing the phasor
circuit at terminals a and b to a Thevenin
equivalent
Phasor circuit
Remove the load

Vˆ
OC

Voltage Divsion
VˆOC 
2
210  0.43  67.47
2 j9
4
ZˆTH   j
 ( 2 j 9)
3
parallel
 2.11  25.52  1.91  j 0.91
Vˆ 
j6
j6
VˆOC 
0.43  67.47  0.48  46.94
j 6  1.91  j 0.91
j 6  ZˆTH
1.10  20.53
v (t )  0.48cos(3t  46.94 ) V
v (t )  0.48cos(3t  46.94 ) V
v OC (t )  0.43cos(3t  67.47 ) V
The Thevenin impedance can be modeled as 1.19 resistor in series
with a capacitor with value
1
 0.91
(  3)C
or
C  0.37 F