the v i relationship for a resistor let the current through the resistor be a sinusoidal given as...

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


(( )) 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 voltage lag the current by 90o or the current leading the voltage by 90o


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


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


1

θ = 180 OR = 1A

180= 18 1 1 0 je =

180= 18 1 1 0 je =


Real

Imaginary

1|A|=

θ = 90

j= 1A =

j= 190 90j= 1 1 je A = =


Imaginary Numbers


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 = =


1j

1 jj j

2

j

j

j( 1)

j

OR 01 9= 90

1 1 je

901 je

=1

j


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


(( )) dLv ittt

d

cos() ( )m ii t tI


(( )) dLv ittt

d sin( )m itIL



L

(t)i

( )v t


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


( )= ij te |I| ( )

m= I ij te


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


( )= ij te |I| ( )

m= I ij te


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


(( )) 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


(( )) 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



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


Real

Imaginary

I

i

V

vmV

mI


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 voltage lag the current by 90o or the current leading the voltage by 90o


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

the v i relationship for a resistor let the current through the resistor be a sinusoidal given as...

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