EENG 2610: Circuit AnalysisClass 14: Sinusoidal Forcing Functions, Phasors, Impedance and Admittance

Oluwayomi AdamoDepartment of Electrical EngineeringCollege of Engineering, University of North Texas

AC Steady-State Analysis

Sinusoidal forcing function (f(t) is forcing function)

The natural response xc(t) is a characteristics of the circuit network and it is independent of the forcing function.

The forced response xp(t) depends on the type of forcing function. Why study sinusoidal forcing function?

This is the dominant waveform in electric power industry. Any periodic signal can be represented by a sum of sinusoids

(you will learn it in Fourier analysis) We will only concentrate on the steady-state forced response of

networks with sinusoidal forcing functions. We will ignore the initial conditions and the transient or natural

response

),()()(

tftaxdt

tdx ),()(

)()(212

2

tftxadt

tdxa

dt

txd )()()( txtxtx cp

Sinusoids Definition

XM is the amplitude ω is radian or angular frequency (unit: radian/second) θ is phase angle (unit: radian), T=2π/ω is period (unit: second) f=1/T is frequency (unit: Hertz),

)sin()( tXtx M

180radian

fT

22

In-phase and out-of-phase Any point on the waveform XMsin(ωt+θ) occurs θ radians

earlier in time than the corresponding point on the waveform XMsin(ωt). We say XMsin(ωt+θ) leads XMsin(ωt) by θ radians, or

XMsin(ωt) lags XMsin(ωt+θ) by θ radians.

In the more general situation, if

Then, x1(t) leads x2(t) by (θ1 - θ2) radians, or,

x2(t) lags x1(t) by (θ1 - θ2) radians.

If θ1 = θ2 ,the waveforms are identical and the functions are said in phase; otherwise, it is said out of phase.

)sin()(

)sin()(

22

11

tXtx

tXtx

M

M

Important Trigonometric Identities

)2

cos()sin(

)2

sin()cos(

tt

tt

)sin()sin(

)cos()cos(

tt

tt

sinsincoscos)cos(

sinsincoscos)cos(

sincoscossin)sin(

sincoscossin)sin(

2

2

22

sin21

1cos2

sincos)2cos(

cossin2)2sin(

)sin()sin(

)cos()cos(

180radian

jj

j

ee

ry

rxx

y

yxr

rejyx

1

sin

cos

tan 1

22

Rectangular formPolar form

Sinusoidal and Complex Forcing Functions Forcing function and circuit response

If we apply a constant forcing function (i.e., source function) to a network, the steady state circuit response is also a constant.

If we apply a sinusoidal forcing function to a linear network, the steady state circuit response will also be sinusoidal.

Sinusoidal source function If the input source is v(t)=Asin(ωt+θ), then the output will be

in the same sinusoidal form. For example, i(t)=Bsin(ωt+φ). That means: if the input source is sinusoidal function, we

know the form of the output response, and therefore the solution involves simply determining the values of the two parameters B and φ.

Learning Example

)()()( tvtRitdt

diL :KVL

tAtAtdt

di

tAtAti

tAti

cossin)(

sincos)(

)cos()(

21

21

or , statesteady In

tV

tRAALtRAAL

M

cos

cos)(sin)( 1221

MVRAAL

RAAL

12

21 0

algebraic problem

222221)(

,)( LR

LVA

LR

RVA MM

Determining the steady state solution canbe accomplished with only algebraic tools!

FURTHER ANALYSIS OF THE SOLUTION

)cos()(

cos)(

sincos)( 21

tAti

tVtv

tAtAti

M

write can one purposes comparisonFor

is voltageapplied The

is solution The

sin,cos 21 AAAA

222221)(

,)( LR

LVA

LR

RVA MM

1

222

21 tan,

A

AAAA

R

L

LR

VA M

1

22 tan,)(

)tancos()(

)( 122 R

Lt

LR

Vti M

voltagethe lags WAYScurrent AL the For 0L

90by voltagethe lagscurrent the inductor) (pure If 0R

Phasors

If the forcing function for a linear network is of the form v(t)=VMejωt, Then every steady-state voltage or current in the network will have the

same form and the same frequency ω; for example, a current will be of the form i(t)=IMej(ωt+φ).

In our circuit analysis, we can drop the factor ejωt, since it is common to every term in the describing equations.

}Re{

}Re{

}Re{

)cos()()(

tjM

tjjM

tjM

M

eV

eeV

eV

tVtv

M

M

I

V

I

VPhasors are defined as:

The magnitude of phasors are positive !

Sinusoidal signal:

)(

M

M

V

VV

)sin(cos

)sin(cos

jIeI

jVeV

Mj

M

Mj

M

Phasor Analysis (or Frequency Domain Analysis) The circuit analysis after dropping ejωt term is called phasor analysis

or frequency domain analysis. By phasor analysis, we have transformed a set of differential

equations with sinusoidal forcing functions in the time domain into a set of algebraic equations containing complex numbers in the frequency domain.

The phasors are then simply transformed back to the time domain to yield the solution of the original set of differential equations.

Phasor representation:Time Domain Frequency Domain

)90()sin(

)cos(

AtA

AtA

Phasor Relationships for Circuit Elements We will establish the phasor relationships between voltage and

current for the three passive elements R, L, C.

IV R

eRIeV

eRIeV

tRitv

iv

iv

jM

jM

tjM

tjM

)()(

)()(

iMj

M

vMj

M

IeI

VeVi

v

I

V

Phasor diagram

IV Lj

eLIjeV

eIdt

dLeV

dt

tdiLtv

iv

iv

jM

jM

tjM

tjM

}{

)()(

)()(

iMj

M

vMj

M

IeI

VeVi

v

I

V

9011

190je

j

Voltage leads current by 90°

VI Cj

eCVjeI

eVdt

dCeI

dt

tdvCti

vi

vi

jM

jM

tjM

tjM

}{

)()(

)()(

iMj

M

vMj

M

IeI

VeVi

v

I

V

9011

190je

j

Current leads voltage by 90°

Definition of Impedance (unit: ohms): Impedance is defined as the ratio of the phasor voltage V to the

phasor current I at the two terminals of the element related to one another by the passive sign convention:

It’s important to note that: Resistance R and reactance X are real function of the frequency

of the forcing function ω, thus Z(ω) is frequency dependent. Impedance Z is a complex number; however, it is not a phasor,

since phasors denote sinusoidal functions.

)( ivM

M

iM

vM

z

I

V

I

V

Z

I

V

Z

reactance :

,resistance :

)()()(

X

R

jXR Z

)/(tan 1

22

RX

XRZ

z

z

z

ZX

ZR

sin

cos

Cj

Lj

Cj

Lj

C

L

RRR

11

ImpedanceEquationPhasor Element

Z

Z

IV

IVZIV

Passive element impedance:

Equivalent impedance if impedances are connected in series:

ns ZZZZ ...21

Equivalent impedance if impedances are connected in parallel:

np ZZZZ

1...

111

21

Two terminal input admittance:

esusceptanc : e,conductanc :

)()()(

siemens) :(unit 1

BG

jBG

YV

I

ZY

jXRjBG

11

ZY

2222 ,

XR

XB

XR

RG

In general, R and G are not reciprocals of one another. The purely resistive case is an exception.

KVL and KCL are both valid in frequency domain

Example 8.9: Determine the equivalent impedance of the network. Then compute i(t) for f = 60 Hz and f = 400 Hz.

V )30cos(50)( ttv

SPECIAL APPLICATION:IMPEDANCES CAN BE COMBINED USING THE SAME RULES

DEVELOPEDFOR RESISTORS

I 1V

1Z

2V

2ZI

21 ZZZs 1Z 2Z

V

I I

V

21

21

ZZ

ZZZ p

kks ZZ

kk

p ZZ

11

LEARNING EXAMPLE

current and impedance equivalent Compute

)30cos(50)(,60 ttvHzf

63

1050120

1,1020120

25,3050,120

jZjZ

ZV

CL

R

05.53,54.7 jZjZ CL

51.4525 jZZZZ CLRs

)(51.4525

3050A

jZ

VI

s

)(22.6193.51

3050A

))(22.91120cos(96.0)()(22.9196.0 AttiAI

RZR

LjZL

CjZC

1

LEARNING EXAMPLESERIES-PARALLEL REDUCTIONS

243 jZ

4/1

25.05.025.0

44

4

jYZ

jjjY

2

8

24

)2(44 jjj

jjZ

422622 jjjZ

21

)2(11 j

jZ

5.01

11 j

Z

2434 jZ

342

342234 ZZ

ZZZ

13 j

)(1.03.0

1.02.0

)(2.01.0

234

34

2

SjY

jY

SjY

)(4.08.0

)5.0(1

5.01

1

21

jZ

jZ

973.8847.36.08.32341 jZZZeq

222)4()2(

42

42

1

j

jY

20

24

24

134

j

jY

1.0

1.03.0

1.03.0

11

234234

j

jYZ

AC Steady-State Analysis

For relatively simple circuits (e.g., those with single source), use: Ohm’s law for AC analysis, i.e., V=IZ The rules for combining impedance Z (or admittance Y) KCL and KVL Current division and voltage division

For more complicated circuits with multiple sources, use: Nodal analysis Loop or mesh analysis Superposition Source exchange Thevenin’s and Norton’s theorem Software tools: MATLAB, PSPICE, …

LEARNING EXAMPLE COMPUTE ALL THE VOLTAGES AND CURRENTS

)48||6(4 jjZeq

28

4824832

28

48244

j

jj

j

jZeq

)(964.30604.9036.14246.8

45196.79

28

5656

j

jZeq

)(036.29498.2964.30604.9

60241 A

Z

VI

eq

S

)(036.29498.2036.14246.8

906

28

613 AI

j

jI

)(036.29498.2036.14246.8

565.26944.8

28

4812 AI

j

jI

3221 904906 IVIV

10582.158.1171.206.295.2 321 III

)(1528.7

)(42.7826.16

2

1

VV

VV

21

32

1

Vfor V law sOhm'

IIfor divider current Use

I Compute

,

,

Steady-State Power Analysis

Here we study powers in AC circuits: Instantaneous power Average power Maximum power transfer, Power factor, Complex power.

Device power ratings: Typically, electrical and electronic devices have peak

power or maximum instantaneous power ratings that cannot be exceeded without damaging the devices.

Instantaneous Power

Steady-state voltage and current:

)cos()(

)cos()(

iM

vM

tIti

tVtv

With passive sign convention

Instantaneous power:

)]2cos()[cos(2

)cos()cos(

)()()(

ivivMM

iMvM

tIV

tItV

titvtp

independent of time a function of time

)cos()(

)cos()(

iM

vM

tIti

tVtv

Statesteady In

)()()( titvtp Impedance to

SuppliedPower

ousInstantane

)cos()cos()( ivMM ttIVtp

)cos()cos(2

1coscos 212121

)2cos()cos(2

)( ivivMM tIV

tp

LEARNING EXAMPLE

)(),(

302

),60cos(4)(

tpti

Z

ttv

:Find

:Assume

)(302302

604A

Z

VI

))(30cos(2)( Atti

30,2

60,4

iM

vM

I

V

)902cos(430cos4)( ttp

constant Twice thefrequency

)cos(2

1

)]2cos()[cos(2

1

)cos()cos(1

)(1

0

0

0

0

0

0

ivMM

Tt

t ivivMM

Tt

t iMvM

Tt

t

IV

dttIV

T

dttItVT

dttpT

P

Average PowerSince p(t) is a periodic function of time, the average power:

In the equation, t0 is arbitrary, T=2π/ω is the period of the voltage or current.

For purely resistive circuit (i.e., Z = R+j0 ):R

VRIIVIVP M

MMMivMM

22

2

1

2

1

2

1)cos(

2

1

For purely reactive circuit (i.e., Z = 0 + jX ): 0)90cos(2

1 MM IVP

That’s why reactive elements are called lossless elements

For passive sign convention.

LEARNING EXAMPLE Determine the average power absorbed by each resistor, the total average power absorbed and the average power

supplied by the source

)(4534

45121 AI

WP 183122

14

If voltage and current are in phase

MMiv IVP2

1 2

12

1MRI

R

VM2

2

1

)(57.7136.537.265

4512

12

45122 A

jI

)(36.522

1 22 WP W7.28

Inductors and capacitors do not absorbpower in the average

WPtotal 7.2818

absorbedsupplied PP WP 7.46supplied

Verification

57.7136.545321 III

)(10.6215.8 AI

)cos(2 iv

MM IVP

)10.6245cos(15.8122

1suppliedP

Maximum Average Power Transfer Average power at the load:

LLivLL RIIVPLL

2

2

1)cos(

2

1

LTh

LocL

LTh

ocL

ZZ

ZVV

ZZ

VI

LLL

ThThTh

jXR

jXR

Z

Z

22

2

)()(2

1

LThLTh

LocL XXRR

RVP

For maximum average power transfer:

ThThThL jXR *ZZ

Th

ocThLL

ThocL

R

VRIP

R

82

1

)2/(2

2max,

VI

If the load is purely resistive (i.e., XL = 0): LLLL RjXR Z

0L

L

dR

dP For maximum average power transfer:

22ThThL XRR LLL RIP 2

max, 2

1

Effective or RMS Values The effective value of a periodic current (or voltage) is defined as a constant or

DC value, which would deliver the same average power to a resistor R. The 120 V AC electrical outlets in our

home is the rms value of the voltage: It is common practice to specify the voltage rating of AC electrical devices (such

as light bulb) in terms of the rms voltage.

RIP 2eff

The average power delivered to a resistor by DC effective current:

The average power delivered to a resistor by a periodic current:

Tt

tRdtti

TP

0

0

)(1 2

Effective (or rms) value of a periodic current:

Tt

tdtti

TII

0

0

)(1 2

rmseff

On using the rms values for the sinusoidal voltage and current, the average power:

)cos(rmsrms ivIVP

The power absorbed by a resistor:

R

VRIP

2rms2

rms

)377cos(170)( ttv

Hz60377602

V1201702120 rms

f

V

,2

rmsMV

V 2

rmsMI

I

LEARNING EXAMPLE Compute the rms value of the voltage waveform

32)2(4

210

104

)(

tt

t

tt

tv

3T

3

2

21

0

2

0

2 ))2(4()4()( dttdttdttvT

3

32

3

162)(

1

0

33

0

2

tdttv

)(89.13

32

3

1VVrms

Tt

trms dttx

TX

0

0

)(1 2