Chapter 10

Sinusoidal Steady-State Analysis

Charles P. Steinmetz (1865-1923), the developer of the mathematical analytical tools for studying ac circuits. Courtesy of General Electric Co.

a



jb



r

(cos

 

j

Heinrich R. Hertz (1857-1894). Courtesy of the Institution of Electrical Engineers.

cycles/second Hertz, Hz

Sinusoidal Sources

Amplitude Period = 1/f Phase angle Angular or radian frequency = 2 p f = 2 p /T

Sinusoidal voltage source

v

s 

V

m sin( 

t

  ).

Sinusoidal current source

i

s 

I

m sin( 

t

  ).

Example

v + _ v i

circuit element

i

Voltage and current of a circuit element.

The current

leads

the voltage by  radians The voltage

lags

OR

the current by  radians

Example 10.3-1

v

 3cos 3

t i

  2 sin(3

t

Find their phase relationship   sin



t

 sin(

 p

)

i

 2 sin(3

t

 180   and sin



 cos(



i

 2 cos(3

t

 180 10 90 )  2 cos(3

t

 Therefore the current

leads

the voltage by 100 

Recall

v s



V

0

cos



t



v f



A

cos



t



B

si n



t v f



A

2 

B

2    

A A

2 

B

2 cos 

t



B A

2 

B

2 sin 

t

 

A

2

A

2 

B

2 

B

2



co s 



cos cos 

t





 si n 

t



Triangle for

A

where

C



A

2 and

B



B

2 .

of Eq. 10.3-4,  sin  

B A

;

A

 0  180   tan  1

B A A

2 

B

2 cos  

A

2 

B

2  tan  1

B A

;

A

 0

Example 10.3-2

A B

i

  6cos2 A

t

   8sin 2

t

B

A

  6  0  180   tan  1

B A

 180    180   tan  1 8   6

i



10 cos(2

t



Steady-State Response of an RL circuit

v s



V m

cos



t



i f

 ?

An

RL

circuit.

L di dt



Ri



V m

c o s 

t

From #8

i f

 10.4 1

A

cos 

t



B

sin 

t L

( Substitute the assumed solution into 10.4-1  

A

sin 

B

cos 

t

)  

t



B

sin 

t

) 

V m

cos 

t

Coeff. of cos Coeff. of sin  

LB

  

m

 0 Solve for A & B

A



R

2

RV m

 

2

L

2

and

B



R

2

 

LV m



2

L

2

Steady-State Response of an RL circuit (cont.)

i



A

cos 

t



B

sin 

t Z



R

2   2 2

L



V m Z

cos( 

t

  )   tan  1 

L R

Thus the forced (steady-state) response is of the form

i



I m

cos(   )

I m



V m Z

   

Complex Exponential Forcing Function

v s



V m

cos 

t i



V m Z

cos( 

t

  )

Input Response magnitude phase frequency Exponential Signal Note

v e v s



V e m



V m

cos



t



Re



V e m

Re 

a



jb

 

a

 

Re

 

e

Complex Exponential Forcing Function (cont.)

v s



v e

try We get

L di e dt



Ri e



v e i e



Ae

where ( 

R



V m

   tan  1 

L R

and 

V e m



V m Z e



j



Z



R

2   2

L

2

Complex Exponential Forcing Function (cont.)

Substituting for A

i e



V m Z e



j



e

We expect

i

 Re  

e

 Re

V m e



j



e Z



V m Z

Re 

e



j



e



V m Z

Re 

e j

( ) 

V m Z

cos(



t





)  

Example

2

d i dt

2 

di dt

 12

i

 12 cos 3

t i e

 We replace 

dt

2

d i

2

e



v e di e dt

 12  12

i e e Ae

,

di dt e

  12

e

, 2

d i e dt

2   9

Ae

Substituting

i e j Ae

 12

e A

 12 3 

j

3  2 2 45

Example(cont.)



i e



Ae

 2 2

e

  2 2

e j

(3

t

 p / 4)

j e

The desired answer for the steady-state current Or

i

 Re

 

e





e j

(3

t

 p / 4) 

 2 2 cos(3

t

 2 2 cos(3

t

 p 4 ) interchangeable

Using Complex Exponential Excitation to Determine a Circuit’s SS Response to a Sinusoidal Source

Write the excitation as a cosine waveform with a phase angle

y s



Y m

cos(   ) Introduce complex excitation

v s

 Re 

V e m j

( Use the assumed response

x e



Ae j

( ) Determine the constant A 

A



Be j

 ) 

Obtain the solution

x e



Ae j

( ) 

Be j

( ) The desired response is  Re  

e



B

cos(

  

)

Example 10.5-1

R L



1H

v s



10 sin 3

t

Example 10.5-1(cont.)

v s



10sin 3

t



10cos(3

t v e



10

e j

(3

t L di e dt



Ri e



v e i e



Ae j

(3

t j

(3

t



2

Ae j

(3

t



2

A



10



10

e j

(3

t A



2 10



j

3



10 4



9

e



j

   tan  1 3 2  56.3



Example 10.5-1(cont.)

The solution is

i e

 10 13

e



j

56.3



e j

(3

t

The actual response is  10 13

e j

 

i

 Re  

e

 10 13 cos(3

t

 A

The Phasor Concept

A sinusoidal current or voltage at a given frequency is characterized by its

amplitude

 Re 

I e m j

( and

phase angle

. )  

I m

cos(



t

 )

Magnitude

Thus we may write



Re 

I e m j

(

Phase angle

)

e



unchanged

The Phasor Concept(cont.) A phasor

is a complex number that represents the magnitude and phase of a sinusoid.

phasor I



I e m j

(   )

I m

 

The Phasor Concept

may be used when the circuit is

linear , in steady state,

and

all independent sources

are

sinusoidal

and have the

same frequency.

A real sinusoidal current 

I m

cos(   )  Re 

I e m j

(   ) 

I



I e m j



I m



phasor notation

The Transformation



Y m

cos(   )  Re 

Y e m j

(

Time domain Transformation

) 

Frequency domain Y



Y e m j

  

m



Time domain The Transformation (cont.)

 5sin(100

t

  5cos(100

t

Transformation Frequency domain I

Example

v s L



V m di



Ri



dt

cos(

 

v s

)  10.6-2 Re 

V e m j

( ) 

i

 Substitute into 10.6-2

I m

cos(



t





)  Re 

I e m j

( ( 

RI m

)

e j

( ) 

V e m j

( )

m

Suppress

e

( 

R

)

I e m j

 

V e m j

 )  

I

 (

I V



R

)

V

Example (cont.)

R  L  2 H, 

I



V

 200) 

V m



V m

283 

V m

283 cos(100

t

  100 A

Phasor Relationship for R, L, and C Elements

Time domain

v



Ri

Frequency domain

V



R

I

or

I



V

R

Resistor Voltage and current are

in phase

Inductor

Time domain Frequency domain

v



L di dt

V

 

I Voltage

leads

current by

90 

or

I



V

Capacitor

Time domain

i



C dv dt

Frequency domain

I

 

V

or

Voltage

lags

current by

90 

V



I

Impedance and Admittance Impedance

is defined as the ratio of the

phasor voltage

to the

phasor current

.

Z

 

V I

V m I m









Ohm’s law in phasor notation 

V m I m

phase or

Z



Z



polar  magnitude



Ze j

 exponential

Z

jX

rectangular

Graphical representation of impedance Z



Z





Resistor Inductor Capacitor Z



Z



R

Z

 1

Z

 

R

2 

X

2  tan  1

X R

R  L 1/ C

Admittance

is defined as the reciprocal of

impedance

.

Y

 1

Z



Z

1   

Y

   In rectangular form

Y



1

Z Resistor Inductor



R



1

jX

Y



G



Y

1 

R



R R

2  

jX X

2 G 1 conductance 1/ L

jB

susceptance

Capacitor Y

  C

KVL Kirchhoff’s Law using Phasors V

1 

V

2 

V

3  

V

n

 0

KCL I

1

I

2

I

3

I

n

 0 Both

Kirchhoff’s Laws

hold

in the frequency domain.

and so

all

the techniques developed for resistive circuits

hold

Superposition Thevenin &Norton Equivalent Circuits Source Transformation Node & Mesh Analysis etc.

Z Impedances in series

eq



Z

1 

Z

2 

Z

3  

Z

n

Admittances in parallel Y

eq



Y

1 

Y

2 

Y

3  

Y

n

Example 10.9-1 R = 9



, L = 10 mH, C = 1 mF i = ?

KVL Z

2  

j

1

R

I



Z I

2



Z I

3



V Z

3  1  

j

10

(9 

j

1 

j

10)

I



V

s

I

 or (9

V

s



j

9)

i

 9 2 45  7.86 cos(100

t

 A

s

Example 10.9-2 v = ?

KCL V

10  10 

V

j

10  0.1

V

 (0.05



j

0.05)

V



V



j

10

j

0.1

V

 10 or

V

v

 10 0.158 18.4

   63.3cos(1000

t

 63.3

  18.4

 V

Node Voltage & Mesh Current using Phasors

i s



I m

cos 

t

C  100 μF, L  5 mH   1000 rad/s

v a

= ? v

b

= ?

Z

1  1  

j

10

Y

2 5 1  1 (1 5 

j

)

Z

3  10

KCL at node

a

V

a

Z

1 

V

a



V

b

Z

3 

I

s

KCL at node

b

V

b



V

a

Z

3 

V Z

2

b

 0 Rearranging ( (

Y

 1  3

Y V

3 )

a

Y V

a

 (

Y

2

Y V

3 )

b



Y V

3 )

b



I

s

 0   (

Y

1  

Y

3

Y

3 )

Y

2 

Y

3 

Y

3

V V I

s

Y

matrix

Admittance matrix

If

I m

= 10 A and

I s I m

Using Cramer’s rule to solve for

V

a

V

a

 4 

j

 17   100 17 87.5

  47.7

 

j

) Therefore the steady state voltage

v a v a

 87.5cos(1000

t



is

V

Example 10.10-1 v = ?

v s

 10cos 

t

C   10

m

F, L  0.5 H  10 rad/s use supernode concept as in #4

Z

L



Z

C

 1 

Z

3 

R

3 

Z

L j

  5

j

10

j

5

Example 10.10-1 (cont.) Y

1  1

R

1  1 10

Y

2  1

R

2  1

Z

C

 1 10 (1 

j

)

Y

3  1

Z

3  1 50 (5 

j

5) KCL at supernode

Y V

1 

V

s

) 

Y V

2  3   10

Y V

1 (

s



V

)   0 Rearranging (

Y

1 

Y

( 2

Y

 1

Y

3  10

Y Y V

1 3 )  (

Y

1 ( 

Y

1

Y

2  10 

Y Y Y V

3 1  3 10

s

Y Y

1 3 )  10

Y Y V

1 3 )

s

Example 10.10-1 (cont.) V

 ( (

Y

1

Y

1  

Y

10 2 

Y Y

1 3

Y

3  10

Y Y

1 3 )  2 10 

j j

 10 5  63.4

 Therefore the steady state voltage

v

is

v

 10 5 cos(10

t

 V

Example 10.10-2

i 1

= ?

v s

 10 2 cos( 

t

C   5 mF, L  30 mH  100 rad/s

Z

L



Z

C

 1 

j

3  

j

2

V

s

 10 

j

10

Example 10.10-2 (cont.)

KVL at mesh 1 & 2 (3  (3 

j j

3)

I

1  3)

I

1 

j

3

I

2  

V

s j

2)

I

2  0 Using Cramer’s rule to solve for

I

1

I

1  (10  

j

10)

j

I

1

j



j

(10 

j

10)

j

 1.05 71.6



j j

Superposition, Thevenin & Norton Equivalents and Source Transformations Example 10.11-1 i = ?

v s



i s

 3 A C  10 mF, L  1.5 H

V

s

I

s

Consider the response to the voltage source acting alone =

i 1

Example 10.11-2 (cont.) I

1  5  Substitute

Z

p



R

Z

C



R

Z

C



V

 5(1 

s



Z

p j

) and  L  15

I

1  5 

j

 

j

5)  10 10 

j

10  10 200

Example 10.11-2 (cont.)

Consider the response to the current source acting alone =

i 2

I

2   10 15 (3)   2 A Using the principle of superposition

i

 0.71cos(10

t

2 A   0

Source Transformations V



I I



V

Example 10.11-2

I S

= ?

v s

  10cos( 

t

 100 rad/s

I

s

  10 200

Z

s

 10 

j

10

V

s

Example 10.11-3

Thevenin’s equivalent circuit

?

Z

1

Z

2  

j j

V

OC

Z

t

 

Z

1

I Z

s

1

 

Z

2

Example 10.11-4

Thevenin’s equivalent circuit

V

 

V

10

I

s OC

  3

V V V

O



Z

t

 

j

10

I

40   4

V

j

10    40)

I

Example 10.11-4

Norton’s equivalent circuit

?

Z

t



Z

3 

Z Z Z

1 1  2

Z

2 ( (

Z

 1  2

Z I

2 )

Z I

 (

Z

2

Z I

2 )

SC



Z I

3 )

SC



V

s

 0

Phasor Diagrams

A

Phasor Diagram

is a graphical representation of phasors and their relationship on the

complex plane

. Take

I

as a reference phasor

I

I

The voltage phasors are

V

R

V

L

 

V

C R

I



I

RI

 

LI

 

j



C

I



I



C

Phasor Diagrams (cont.) KVL V

s



V

R



V

L



V

C

For a given

L

and

C

there will be a frequency  that

V

L



V

C



L

 1 

C

or  2  1

LC

Resonance V

s



V

R

Resonant frequency

1

LC

Summary

Sinusoidal Sources Steady-State Response of an RL Circuit for Sinusoidal Forcing Function Complex Exponential Forcing Function The Phasor Concept Impedance and Admittance Electrical Circuit Laws using Phasors