Chapter 5 Steady-State Sinusoidal Analysis

A branch of Mathematics known as Fourier Analysis tell usthat All signals are composed of Sinusoidal Components

1. Frequency, Angular Frequency, Peak , rms & phase

2. Steady-State AC Circuits with Phasors & Complex Impedances.

3. Power for Steady-State AC Circuits.

4. Thévenin and Norton equivalent circuits.

5. Load Impedances for Maximum Power Transfer.

Goal

Sinusoidal Currents & Voltage

v(t) = Vm cos(ωt +θ)

Vm : Peak valueω : Angular frequency (Radians per second)θ : Phase angle

Frequency T

f 1=

Tπω 2

= fπω 2=Angular frequency

v(t) = Vm cos(ωt +θ)

θ=-45o

Root-Mean-Square Values

( )dttvT

VT

2

0rms

1∫=

RIR

VP 2rms

2rms

avg ==

( )dttiT

IT

2

0rms

1∫=

2rmsmVV =

The rms value for a sinusoid is the peak value divided by the square root of two. This is not true for other periodic waveforms such as square waves or triangular waves.

for v(t) = Vm cos(ωt +θ)

Phasors Definition

( ) ( )111 cos :function Time θtωVtv +=

111 :Phasor θ∠=VV

2222111111 , θθ θθ ∠==∠== ZeZZeZ jj ZZ

Vector in Complex Plane

e.g.

)( 2121)(

2121212121 θθθθθθ +∠==×=× + ZZeZZeZeZ jjjZZ

)(//// 2121)(

2121212121 θθθθθθ −∠=== − ZZeZZeZeZ jjjZZ

)sinsin(coscos 22112211

212121

θθθθ

θθ

ZZjZZeZeZ jj

+++=+=+ZZ

Using Phasors to Add Sinusoids

( ) ( )o45cos201 −= ttv ω( ) ( )o60cos102 += ttv ω

o45201 −∠=Vo30102 −∠=V

o

oo

7.3997.2914.1906.23

5660.814.1414.1430104520

21s

−∠=

−=−+−=

−∠+−∠=

+=

jjj

VVV

( ) ( )o7.39cos97.29 −= ttvs ω

Sinusoids can be visualized as the real-axis projection of vectors rotating in the complex plane. The phasor for a sinusoid is a snapshot of the corresponding rotating vector at t = 0.

Phasors as Rotating Vector

( ) ( )θω += tVtv m cos

θ∠= mVV

Phase Relationships

V1 leads V2 by 60o . Alternatively, V2 lags V1 by 60o.

( ) ( )o40cos31 += ttv ω ( ) ( )o20cos42 −= ttv ω

Time function Phasors Diagram

o90∠== LLjZL ωωLLL Z IV =

COMPLEX IMPEDANCES

( ) ( )θω += tIti mL sin ( ) ( )θω += tVtv m cos( )

dttdiLtv L

L =)(

θθω ∠=∠= mmL VLIV)90( o−∠= θmL II

ZL : Impedance of Inductance Pure Imaginary Part : Reactance

Inductor

CCC Z IV = o90111−∠==−=

CCjCjZC ωωω

( ) ( )θω += tVtv mc cos ( ) ( )θω += tIti mc sin( )

dttdvCti c

C =)(

)90( o+∠=∠= θθω mmC ICVIθ∠= mC VV

Capacitor

RRR Z IV =

Resistor

o0∠= RZR

( ) ( )θω += tVtv mR cos ( ) ( )θω += tIti mR cos

θθ ∠=∠= mmR IVR1Iθ∠= mR VV

( )tvR

ti RR1)( =

V

IR

IC

IL

I

VR

VL

VC

RRR Z IV = o0∠= RZR

CCC Z IV = o9011−∠==

CCjZC ωω

o90∠== LLjZL ωωLLL Z IV =Summary of Impedance

Kirchhoff’s Laws in Phasor Form

Direct Application of KVL to Phasors Voltage :“The sum of the phasor voltages equals zero for any closed path.”

Direct Application of KCL to Phasors Current :“The sum of the phasor currents entering a node must equal the

sum of the phasor currents leaving.”

Σ Vk=0

Σ Ik=0

Serial Circuit : Constant Current

∑∑ ===k

kk

k ZZIZIZV

Resistor ∑=k

kRR Capacitor ∑=k

kCC /1

∑∑ ===k kk k Z

ZZV

ZVI 1

∑=k

kRR /1 ∑=k

kCC

)1(1C

LjRCj

RLjCRL ωω

ωω −+=++=++= ZZZZ

Inductor ∑=k

kLL

Parallel Circuit : Constant Voltage

Resistor Capacitor

CjRLjCRL ωω ++=++= /1/1/1/1/1 ZZZZ

Inductor ∑=k

kLL /1

)1(/1L

CjRω

ω −+=

Circuit AnalysisUsing Phasors and Impedances

1. Replace the Time Function of the Voltage & Current sources with the corresponding Phasors.

2. Replace Inductances by Complex Impedances ZL = jωL. Replace Capacitances by Complex Impedances ZC = 1/(jωC). Resistances equal to their resistances. ZR = R

3. Analyze the Circuit Equation using the techniques in Chapter 2.Performing the Calculations with Complex Arithmetic.

Note: All of the sources must have the same frequency

Step 1. Time Function to Phasors.

Steady State AC Analysis of Series Circuit

( ) ( )o30500cos100 += ttvso30100∠=sV

Step 2. Complex Impedances

Ω=∠= 1000oRZR

Ω−=××

−== − 501040500

116 jj

CjZC ω

Ω=×== 1503.0500 jjLjZL ω

o45210010010050150100 ∠=+=−+=++= jjjZZZZ CLReq

o

o

o

152/1452100

30100−∠=

∠∠

==eq

s

ZVI

Step 3. Analyze the Circuit Equation

o152/100 −∠=×= IV RR

ooo

o

752/150152/190150

90

∠=−∠×∠=

×∠=×= IIV LLjL ωω

ooo

o

1152/50152/19050

9011

−∠=−∠×−∠=

×−∠=×−= IIVCC

jC ωω

Series/Parallel Combination of Impedances

Ω=∠= 1000oRZR

Ω−=××

−== − 10010101000

116 jj

CjZC ω

Ω=×== 1001.01000 jjLjZL ω

o9010 −∠=sV

R & C is Parallel

505002/10045201.0

01)100/(1100/1

1/1/1

1 jjZR

ZC

RC −=∠=∠

∠=

−+=

+= o

o

o

50505050100 jjjZZZ LRCeq +=−+=+=

Phasors form

Steady State AC Node-Voltage Technique

KCL at Node 1 & 2

o902510

211 −∠=−−

+jVVV

o05.1510

122 ∠=−−

+jjVVV

o7.291.161 ∠=V

( ) ( )o7.29100cos1.161 += ttv

Power in AC Circuits

θθ

−∠=∠∠

== mm IZ

VZ

o0VI

)()( titvPPower =

Current, Voltage & Power in Impedance

)cos()( tVtv m ω=

)cos()( tIti m ω=

)(cos)()()( 2 tIVtitvtp mm ω==

)cos()( tVtv m ω=

)sin()( tIti m ω=

)2sin(2

)sin()cos()()()(

tIVttIV

titvtp

mm

mm

ω

ωω

=

==

)cos()( tVtv m ω=

)sin()( tIti m ω=

)2sin(2

)sin()cos()()()(

tIVttIV

titvtp

mm

mm

ω

ωω

−=

−==

Resistor Inductor Capacitor

Always Positive Power Half Positive & Half Negative PowerReactive Power : Average Power = Zero

General AC Power Calculations( ) ( )tVtv m ωcos= ( ) ( )θω −= tIti m cos

( )[ ] ( )tIVtIVtp mmmm ωθωθ 2sinsin2

2cos1cos2

)( ++=

( ) ( )θθ coscos2 rmsrms

mm IVIVPavg ==

( )θcosPF = iv θθθ −=

General Power

Power Factor Power angle

( ) ( )θθ sinsin2 rmsrms

mm IVIVQ ==

rmsrmsPower Apparent IV= ( )2rmsrms22 IVQP =+

Reactive Power

Apparent Power

Unit : VAR

RIP 2rms=

XIQ 2rms=

RVP R

2rms=

XVQ X

2rms=

Power Triangle

THÉVENIN/NORTON Equivalent Circuits

Thévenin voltage : Open-circuit phasor voltage of the circuit

ocVV =t

Thévenin impedance : Open-circuit Voltage / Short-circuit Current.by zeroing the independent sources and determining the impedancelooking into the circuit terminals

scsc

oc

IV

IV t

tZ == scII =n

Balanced 3 Phase Circuits

Much of the power used by business and industry is supplied by 3-phase distribution systems. Plant engineers

need to be familiar with 3-phase power.

Phase SequenceThree-phase sources can have either a positive or negative phase sequence. The direction of rotation of certain three-phase motors can be reversed by changing the phase sequence.

Wye–Wye Connection

Three-phase sources and loads can be connected either in a Wye configuration or in a Delta configuration.

Why Wye Connection?

Direct Connection Requires 6 wire Connections.

( ) ( )θcos3 rmsrmsavg LY IVtpP ==

( ) ( )θθ sin3sin2

3 rmsrms LYLY IVIVQ ==

=

YZZ 3=∆