Chapter 10 Sinusoidal Steady-State Analysis

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.

(cos sin )a jb r j

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

cycles/second Hertz, Hz

Sinusoidal voltage source vs Vm sin(t ).

Sinusoidal Sources

Sinusoidal current source is Im sin(t ).

Amplitude

Phase angle

Angular or radianfrequency = 2f = 2/T

Period = 1/f

Example

Voltage and current of a circuit element.

v

i

The current leads the voltage by radians

circuitelement

+

v_

i

The voltage lags the current by radians OR

Example 10.3-1 3cos32sin(3 10 )

v ti t

Find their phase relationship

sin sin( )2sin(3 180 10 )

t ti t

andsin cos( 90 )

2cos(3 180 10 90 )2cos(3 100 )

i tt

Therefore the current leads the voltage by 100

Triangle for A and B of Eq. 10.3-4, where C .22 BA

Recall

0 cos cos si 3 4n 10. -s fv V t v A t B t

2 2

2 2 2 2cos sinf

A Bv A B t tA B A B

2 2 2 2sin cosB A

A B A B

2 2 cos sis i nco s nA B t t

2 2 cosA B t

1

1

tan ; 0

180 tan ; 0

B AA

B AA

Example 10.3-2 6cos2 8sin 2i t t A B

A

B

1

1

180 tan

8180 tan6

180 53 1

6 0

.

BA

A

10cos(2 126.9 )i t

An RL circuit.

Steady-State Response of an RL circuit

cos ?s m fv V t i

c 10.4-o 1smdiL Ri V tdt

cos sinfi A t B t From #8&#9

Substitute the assumed solution into 10.4-1 ( sin cos ) ( cos sin ) cosmL A t B t R A t B t V t

mLB RA V

0LA RB Solve for A & B

Coeff. of cosCoeff. of sin

2 2 2 2 2 2andm mRV LVA BR L R L

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

cos sin

cos( )m

i A t B tV tZ

2 2 2Z R L 1tan L

R

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

cos( )mi I t

mm

VIZ

Complex Exponential Forcing Function

cos

cos( )

s m

m

v V tVi tZ

Input

Response

magnitude phasefrequency

Exponential Signal

cos Re Re

j te m

j ts m m e

v V e

v V t V e v

Note Re a jb a

Complex Exponential Forcing Function (cont.)

s ev v

ee e

diL Ri vdt

try

j tei Ae

( ) j t j tm

jm m

j L R Ae V eV VA e

R j L Z

We get

where1 2 2 2tan andL Z R L

R

Complex Exponential Forcing Function (cont.)

Substituting for Aj j tm

eVi e eZ

We expect

( )

Re Re

Re

Re

cos( )

j j tme

j j tm

j tm

m

Vi i e eZ

V e eZ

V eZ

V tZ

Example2

2 12 12cos3d i di i tdt dt

We replace 312 j tev e

23

2 12 12 j te ee

d i di i edt dt

23 3 3

2, 3 , 9j t j t j te ee

di d ii Ae j Ae Aedt dt

Substituting ie3 3( 9 3 12) 12

12 2 2 453 3

j t j tj Ae e

Aj

Example(cont.)3 ( / 4) 3

(3 / 4)

2 2

2 2

j t j j te

j t

i Ae e e

e

The desired answer for the steady-state current

(3 / 4)Re Re 2 2

2 2 cos(3 45 )

j tei i e

t

Or 2 2 cos(3 )4

t

interchangeable

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

Write the excitation as a cosine waveform with a phase angle

Introduce complex excitation

Use the assumed response

Determine the constant A

cos( )s my Y t

( )Re j ts mv V e

( )j tex Ae

jA Be

( ) ( )j t j tex Ae Be

Obtain the solution

The desired response is

( ) Re cos( )ex t x B t

Example 10.5-121H10sin 3s

RLv t

Example 10.5-1(cont.)

10sin 3 10cos(3 90 )sv t t (3 90 )10 j t

ev e

ee e

diL Ri vdt

(3 90 )j t

ei Ae

(3 90 ) (3 90 ) (3 90 )3 2 103 2 10

10 102 3 4 9

j t j t j t

j

j Ae Ae ej A A

A ej

1 3tan 56.32

Example 10.5-1(cont.)

56.3 (3 90 ) (3 146.3 )10 1013 13

j j t j tei e e e

The solution is

The actual response is

10Re cos(3 146.3 ) A13ei i t

The Phasor Concept

A sinusoidal current or voltage at a given frequency is characterized by its amplitude and phase angle.

( )( ) Re

cos( )

j t

m

m

I

i t I e

t

Magnitude Phase angleThus we may write

( )( ) Re j j tmi t I e e

unchanged

The Phasor Concept(cont.)

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

( )jm mI e I Iphasor

The Phasor Concept may be used when the circuit is linear , in steady state, and all independent sources aresinusoidal and have the same frequency.

A real sinusoidal current

( )( ) cos( ) Re j tm m

jm m

i t I t I e

I e I

I phasor notation

The Transformation

( )( ) cos( ) Re j tm my t Y t Y e

jm mY e Y Y

Time domain

Frequency domain

Transformation

The Transformation (cont.)

( ) 5sin(100 120 )5cos(100 30 )

i t tt

5 30 I

Time domain

Frequency domain

Transformation

Example10.6-2s

diL Ri vdt

( )cos( ) Re j ts m mv V t V e

( )cos( ) Re j tm mi I t I e

Substitute into 10.6-2( ) ( )( ) j t j t

m m mj LI RI e V e Suppress j te

( ) j jm mR I e V ej L I V

( )j L R

VI

( 200 200)0

283 45

45283

m

m

jV

V

VI

Example (cont.)

R 200 , L 2 H, 100 /rad s

( ) cos(100 45 ) A283

mVi t t

Phasor Relationship for R, L, and C Elements

v Ri

Time domain

Frequency domain

R orR

VV I I

Resistor

Voltage and current are in phase

Time domain Frequency domain

div Ldt

j L orj L

VV I I

Inductor

Voltage leads current by 90

Time domain Frequency domain

j C orj C

II V V

dvi Cdt

Capacitor

Voltage lags current by 90

Impedance and Admittance

Impedance is defined as the ratio of the phasor voltage to the phasor current.

m m

m m

V VI I

VZI

Ohm’s law in phasor notation

magnitude

phase

Z

or jZe R jX Z Zpolar exponential rectangular

Graphical representation of impedance

Z Z2 2R X Z

1tan XR

Resistor

Inductor

Capacitor

RZj LZ

1j C

Z

R

L

C

Admittance is defined as the reciprocal of impedance.

1 1

Y YZ Z

In rectangular form

2 2

1 1 R jX G jBR jX R X

Y

Z

conductance

susceptanceResistor

Inductor

Capacitor

1GR

Y

1j L

Y

j CY

L

C

G

Kirchhoff’s Law using Phasors

KVL 1 2 3 0n V V V V

KCL 1 2 3 0n I I I I

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.

Impedances in series

1 2 3eq n Z Z Z Z Z

Admittances in parallel

1 2 3eq n Y Y Y Y Y

Example 10.9-1 R = 9 , L = 10 mH, C = 1 mF i = ?

KVL2 3 sR I Z I Z I V

2

3

11 10

j L j

jj C

Z

Z (9 1 10) sj j I V100 0 7.86 45

(9 9) 9 2 45or 7.86cos(100 45 ) A

s

ji t

VI

Example 10.9-2 v = ?

KCL

10 010 10 10 10j j

V V V

0.1 (0.05 0.05) 0.1 1010 63.3 18.4

0.158 18.4or 63.3cos(1000 18.4 ) V

j j

v t

V V V

V

Node Voltage & Mesh Current using Phasors

va = ? vb = ?

cosC 100 μF, L 5 mH

1000 rad/s

s mi I t

11 10j

j C Z

21 1 1 (1 )5 5

jj L

Y

3 10Z

KCL at node a

1 3

a a bs

V V V IZ Z

KCL at node b

3 2

0b a b

V V VZ Z

Rearranging1 3 3

3 2 3

( ) ( )( ) ( ) 0

a b s

a b

Y Y V Y V I

Y V Y Y V

1 3 3

3 2 3

( )0

matrix

a s

b

Y Y Y V IY Y Y V

Y

Admittance matrix

If Im = 10 A and 0s mI I

Using Cramer’s rule to solve for Va

Therefore the steady state voltage va is

100(3 2 )4

100(3 2 )(4 )17

100 (10 11 )17

87.5 47.7

aj

jj j

j

V

87.5cos(1000 47.7 ) Vav t

Example 10.10-1 v = ?

10cosC 10 F, L 0.5 H

10 rad/s

sv tm

use supernode conceptas in #4

3 3

51 10

5 5

L

C

L

j L j

jj C

R j

Z

Z

Z Z

Example 10.10-1 (cont.)1

1

22

33

1 110

1 1 1 (1 )10

1 1 (5 5)50

C

R

jR

j

Y

YZ

YZ

KCL at supernode 1 2 3 1( ) 10 ( ) 0s s Y V V Y V Y V Y V V

Rearranging

1 2 3 1 3 1 1 3

1 1 3

1 2 3 1 3

( 10 ) ( 10 )( 10 )

( 10 )

s

s

Y Y Y Y Y V Y Y Y VY Y Y VV

Y Y Y Y Y

1 1 3

1 2 3 1 3

( 10 )10 0( 10 )10 10 63.42 5

jj

Y Y YVY Y Y Y Y

Example 10.10-1 (cont.)

Therefore the steady state voltage v is

10 cos(10 63.4 ) V5

v t

Example 10.10-2 i1 = ?

10 2 cos( 45 )C 5 mF, L 30 mH

100 rad/s

sv t

31 2

10 2 45 10 10

L

C

s

j L j

jj C

j

Z

Z

V

Example 10.10-2 (cont.)

KVL at mesh 1 & 2

1 2

1 2

(3 3) 3(3 3) ( 3 2) 0

sj jj j j

I I VI I

Using Cramer’s rule to solve for I1

1(10 10)j j

I

where is the determinant(3 3)( ) 3(3- 3) 6 12j j j j j

1(10 10) 1.05 71.6

6 12j j

j

I

Superposition, Thevenin & Norton Equivalentsand Source Transformations

Example 10.11-1 i = ?10cos10 V3 A

C 10 mF, L 1.5 H

s

s

v ti

10 03 0

s

s

VI

Consider the response to the voltage source acting alone = i1

Example 10.11-2 (cont.)

1 5s

pj L

VI

Z

5(1 ) and L 15Cp

C

R jR

ZZZ

Substitute

110 0

5 15 (5 5)10 10 45

10 10 200

j j

j

I

Consider the response to the current source acting alone = i2

Example 10.11-2 (cont.)

0

210 (3) 2 A15

I

Using the principle of superposition

0.71cos(10 45 ) 2 Ai t

Source Transformations

V I

I V

Example 10.11-2 IS = ?

10cos( 45 ) V100 rad/s

sv t

10 1010 45

s

s

j

ZV

10 45200 4510 0200

s

I

Example 10.11-3 Thevenin’s equivalent circuit

?

1

2

1 jj

ZZ

1

1 2

2 2 451

OC s

t

V I ZZ Z Z

Example 10.11-4 Thevenin’s equivalent circuit

10 20 03 80 0

s

OC

V I

V V V

10 4 ( 10 40)40 10

O

t

j jj

V I V I

Z

Example 10.11-4 Norton’s equivalent circuit

?

1 23

1 2t

Z ZZ Z

Z Z1 2 2

2 2 3

( ) ( )( ) ( ) 0

SC s

SC

Z Z I Z I V

Z I Z Z I

Phasor Diagrams

A Phasor Diagram is a graphical representation of phasorsand their relationship on the complex plane.

Take I as a reference phasor

0I I

The voltage phasors are

090

90

R

L

C

R RIj L LI

j IC C

V IV I

V I

Phasor Diagrams (cont.)

KVLs R L C V V V V

For a given L and C there will be a frequency that

2

1

1 1or

L C

LC

LC LC

V V

Resonant frequency

s RV VResonance

Summary

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