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 Ω( )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 58.43oLv t t= −

Solution for i(t) should be a sinusoidal of frequency 3

We notice that only the amplitude and phase change

( )SdiL Ri V tdt + =

+− 2 H

3 Ω( )i t

( ) Rv t+ −

( ) Lv t

+

−

( )SV t+−

Time Domain Complex Domain

Deferential Equation Algebraic Equation

cos( ) sin( )j je θ θ θ± = ±

cos( ) je θθ =ℜ sin( ) je θθ = ℑ

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 θθ =ℜ

cos( ) sin( )j je θ θ θ± = ± cos( ) je θθ =ℜ sin( ) je θθ = ℑ

cos( )mv V tω φ= + ( ) j tmV e ω φ+= ℜ j t j

mV e eω φ= ℜ

j j tmv V e eφ ω=ℜ

jmV e φ=cos( ) mV tP ω φ+

Phasor Transform

Were the notation cos( ) mV tP ω φ+

Is read “ the phasor transform of cos( )mV tω φ+

We can move the coefficient Vm inside

jmV e φThe quantity is a complex number define to be the phasor

that carries the amplitude and phase angle of a given sinusoidal function

=V

cos() ( )m ii t tI ω θ= +

cos( )m iR tI ω θ = +

(( )) Rv it t=

cos( )m iR tI ω θ = +

The V−I 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 θω

= + Is also sinusoidal with

amplitude mmV RI= And phase ivθ θ=

The sinusoidal voltage and current in a resistor are in phase

R

(t)i+ −( )v t

cos() ( )m ii t tI ω θ= +

cos( ( )) m iR tv t I ω θ = +

Now let us see the pharos domain representation or pharos transform of the current and voltage

im

jI e θ=I mI= iθ

R=V I

im

jRI e θ=V mRI= iθ

m V

v θ

Which is Ohm’s law on the phasor ( or complex ) domain

R

I+ −V

R

(t)i+ −( )v t

cos() ( )m ii t tI ω θ= + Phasor Transform

cos( ( )) m iR tv t I ω θ = + Phasor Transform R= I

R=V IR

I+ −V

The voltage and the current are in phase

Real

Imaginary

I

V

ivθ θ=

mV

mI

iθ

The V−I Relationship for an Inductor

(( )) dLv ittt d=

cos() ( )m ii t tI ω θ= +Let the current through the resistor be a sinusoidal given as

(( )) dLv ittt d= sin( )m itILω ω θ= − +

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 ω θ= +

si( n) ( )m iv t tIL θω ω= − +

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 tILω ω θ= − + −

The pharos representation or transform of the current and voltage

im

jI e θ=I mI= iθ

o( 90 ) ijmL eI θω −= −V

o 90 ij jm

jL e eI θω −

=−= −

im

jLIj e θω=

cos() ( )m ii t tI ω θ= +

ocos( 90 )( ) m iv t tILω ω θ= − + −

But since 901 ojj e= 1 = o90

Therefore im

jIj L e θω=V 90 o ij jmL e eI θω= ( 90 ) oij

meIL θω += mILω= ( 90 )oiθ +

m V

v

θ

m mIV Lω= and 90i ov θθ = +

L

(t)i−+ ( )v t

j Lω

−+ V

I

j Lω=V I

m mIV Lω= and 90i ov θθ = +

The voltage lead the current by 90o or the current lagging the voltage by 90o

Real

Imaginary

I

iθ

V

vθmV

mI

The V−I Relationship for a Capacitor

C

(t)i

−+ ( )v t

(( )) dCi vttt d=

cos(( ) )vmv t tV ω θ= +Let the voltage across the capacitor be a sinusoidal given as

(( )) dCi vttt d= sin( ) vm tVCω ω θ= − +

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

(t)i

−+ ( )v t

cos(( ) )vmv t tV ω θ= + sin( )( ) vmi C tt Vω ω θ= − +

The pharos representation or transform of the voltage and current

cos(( ) )vmv t tV ω θ= + vm

jV e θ=V mV= vθ

sin( )( ) vmi C tt Vω ω θ= − + cos ( 90 )vmoC tVω ω θ= − + − o( 90 ) vj

mL eV θω −= −I

o 90 vm

j

j jC e eV θω−

−= − I vm

jVj C e θω= j Cω= V

j Cω=V I

90 1 o

im

j

je

e Cj

I θ

ω=

( 90 )

oijme

CI θ

ω−

= mC

Iω= ( 90 )o

iθ −

mmIV Cω= and 90i ov θθ = −

m V

v

θ

mmIV Cω= and 90i ov θθ = −

j Cω=V I

−+

1 j Cω

VI

The voltage lag the current by 90o or the current lead the voltage by 90o

Real

Imaginary

I

iθ

V

vθ

mV

mI

R

I+ −V

j Lω

−+ V

I

Real

Imaginary

Real

Imaginary

−+

1 j Cω

VI

Real

Imaginarycos() ( )m ii t tI ω θ= +

si( n) ( )m iv t tIL θω ω= − +

cos() ( )m ii t tI ω θ= +

cos( ( )) m iR tv t I ω θ = +

R

(t)i+ −( )v t

C

(t)i

−+ ( )v t

cos(( ) )vmv t tV ω θ= +

sin( )( ) vmi C tt Vω ω θ= − +

Time-Domain Phasor ( Complex or Frequency) Domain

(( )) Rv it t=

(( )) dLv ittt d=

R=V I

j Lω=V I

j Cω=V 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

R=V I

j Cω

=V I j Lω=V I

When we compare the relation between the voltage and current , we note that they are all of form:

Z=V 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 Cω= 1 I

Z R R=

1Z C j Cω=

Z L j Lω=The Impedance of an indictor is

The Impedance of a capacitor is

In all cases the impedance is measured in Ohm’s Ω

The reactance of a resistor is X 0R =

1X C Cω

−=

X L Lω=The 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 Cω=

Z L j Lω=The Impedance of an indictor is

The Impedance of a capacitor is

In all cases the impedance is measured in Ohm’s Ω

R=V I j Lω=V I j Cω

= 1V I

Z = VI

Impedance

Note that the impedance in general (exception is the resistor) is a function of frequency

At ω = 0 (DC), we have the following Z L j Lω= (0)j L= =0 short

1Z C j Cω= (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 tω θ= +

2 2 2( ) cos( )v t V tω θ= +

4 4 4( ) cos( )v t V tω θ= +

3 3 3( ) cos( )v t V tω θ= +

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 Vωω θ θθ θω ω +++ + ++ + =

Using Euler Identity we have 42 313 421 0j j tj j j j jtt j tV e e V eVV e ee ee θ ωθ ω ω ωθ θ+ + + =ℜ ℜ ℜ ℜ

Which can be written as 42 313 421 0j j t j jj j t tj jt VVV e Vee ee ee eθωθ θ ωω ωθ+ + + =ℜ

Factoring j te ω321 4

431 2( ) 0j tjj j jV eV ee V eeV θ θθθ ω+ + + =ℜ

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 θV

22 2= jV e θV

33 3= jV e θV

44 4= jV e θV

Phasor Transformation

Phasor

Can 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 tω θ= + 2 2 2( ) cos( )i t I tω θ= + ( ) cos( )n n ni t I tω θ= +

11 1= jI e θI 2

2 2= jI e θI = nn n

jI e θIPhasor Transformation

KCL

KCL on the phasor domain

9.6 Series, Parallel, and Delta-to Wye Simplifications

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 Lω= (5000)(32 X 10 )j −3= 160 j= ΩThe Impedance of the indictor is

1Z C j Cω=The Impedance of the capacitor is 6 1

(5000) (5 X 10 )j −= 40 j= − Ω

o30=750sjeVThe source voltage pahsor transformation or equivalent

o=750 30∠

(b) Calculte the steady state current i(t) ?

ab

sI Z= VTo Calculate the phasor current I

o3075090 160 40

jej j=

+ −

o3075090 120

jej=

+

o

o

750 30150 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)

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

Thevenin Impedance

TT T

TFind interms of then form the ratio

VI V I

Ta bFind and interms of I I V

12 60Ω Ω

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