Sinusoids and Phasors

Instructor: Chia-Ming TsaiElectronics Engineering

National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.

Contents• Introduction

• Sinusoids

• Phasors

• Phasor Relationships for Circuit Elements

• Impedance and Admittance

• Kirchhoff’s Laws in the Frequency Domain

• Impedance Combinations

• Applications

Introduction• AC is more efficient and economical to

transmit power over long distance.

• A sinusoid is a signal that has the form of the sine or cosine function.

• Circuits driven by sinusoidal current (ac) or voltage sources are called ac circuits.

• Why sinusoid is important in circuit analysis?– Nature itself is characteristically sinusoidal.– A sinusoidal signal is easy to generate and transmit.– Easy to handle mathematically

Sinusoids

)( 2

2

seconds. every itself repeats sinusoid The

sinusoid theofargument the

)(radians/sfrequency angular the

sinusoid theof amplitude the

where

sin)(

voltagesinusoidal heConsider t

T:periodω

TωT

T

ωt

V

tVtv

m

m

)(

)2sin(

)2

(sin

)(sin)(

:Proof

)()(

tv

ntVω

ntV

nTtVnTtv

tvnTtv

m

m

m

Sinusoids (Cont’d)• A period function is one that satisfies

f(t) = f(t+nT), for all t and for all integers n.– The period T is the number of seconds per cycle– The cyclic frequency f = 1/T is the number of cyc

les per second

(Hz) hertz :

(rad/s) secondper radians:

where

2

1

f

fT

f

Phase :

Argument : )(

where

)sin()(

asgiven is expression general moreA

t

tVtv m

Sinusoids (Cont’d)

by

by

say that We

21

12

vlagsv

vleadsv

0 if, are and

0 if, are and

say that We

21

21

seout of phavv

in phasevv

Sinusoids (Cont’d)• To compare sinusoids

– Use the trigonometric identities

– Use the graphical approach

BABABA

BABABA

sinsincoscos)cos(

sincoscossin)sin(

:identities ricTrigonomet

sin)90cos(

cos)90sin(

cos)180cos(

sin)180sin(

tt

tt

tt

tt

The Graphical Approach

tan where

)cos(sincos

1

22

A

BBAC

tCtBtA

)1.53cos(5

sin4cos3

t

tt

Phasors• Sinusoids are easily expressed by using phasors

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

• Phasors provide a simple means of analyzing linear circuits excited by sinusoidal sources.

of the:

of the: where,

form lExponentia :

formPolar :

formr Rectangula :

it.represent to ways threearetheir

,number complex a gConsiderin

zphase

zmagnituder

re

r

jyx

z

z

j

Phasors (Cont’d)

)sin(cos

sin ,cos

as and obtain can we, and know weIf

tan ,

as and get can we, and Given

form lExponentia :

formPolar :

formr Rectangula :

122

jrrjyxz

ryrx

yxrx

yyxr

ryx

re

r

jyx

zj

Important Mathematical Properties

rjyxz

rz

rz

r

r

z

z

rrzz

yyjxxzz

yyjxxzz

2

11

)(

)(

)()(

)()(

212

1

2

1

212121

212121

212121

:ConjugateComplex

:Root Square

:Reciprocal

:Division

:tionMultiplica

:onSubstracti

:Addition

22222

11111

rjyxz

rjyxz

rjyxz

Phasor Representation

.)( sinusoid theof

tionrepresentaphasor theis

)Re()(

)Re()Re(

)cos()(

)Im(sin

)Re(cos

sincos

)(

tv

VeV

etv

eeVeV

tVtv

e

e

je

mj

m

tj

tjjm

tjm

m

j

j

j

V

V

V

Phasor Representation (Cont’d)

Phasor Diagram

mVV

mII

Sinusoid-Phasor Transformation

jvdt

jdt

dv

ej

eeVeeeeV

tVtVdt

tdv

VtVtv

tj

tjjm

jjjtjm

mm

mm

V

V

V

V

Similarly,

)Re(

)(Re)Re(

)90cos()sin()(

)cos()(

9090

Phasor Relationships for Resistor

IV

I

RRItRIiRv

ItIi

mm

mm

)cos(

law, sOhm'By

)cos(

isresistor rough thecurrent th theIf

Time domain Phasor domain Phasor diagram

Phasor Relationships for Inductor

Time domain Phasor domain Phasor diagram

IV

I

LjtLIdt

diLv

ItIi

m

mm

)90cos(

isinductor theacross voltageThe

)cos(

isinductor rough thecurrent th theIf

Phasor Relationships for Capacitor

Phasor diagramTime domain Phasor domain

VI

V

CjtCVdt

dvCi

VtVv

m

mm

)90cos(

iscapacitor rough thecurrent th The

)cos(

iscapacitor theacross voltage theIf

Impedance and Admittance

1

1

1

AdmittanceImpedanceElement

currentphasor theis

tagephasor vol theis re whe

(S) 1

:Admittance , )( :Impedance

LjCj

C

CjLjL

RRR

YZ

YZ

YZ

I

VZ

YI

VZ

Impedance and Admittance (Cont’d)

Cj1

Z

LjZ

0

0

Impedance and Admittance (Cont’d)

voltageleadscurrent since

leadingor capacitive: voltagelagscurrent since

laggingor inductive:

thenpositive, is If

negative is when capacitive

positive is when inductive

be tosaid is impedance The

reactance:

resistance: where

jXR

jXR

X

X

X

X

R

jXR

Z

Z

Z

sin

cos and

tan where

1

22

Z

Z

Z

ZZ

X

R

R

XXR

jXR

Impedance and Admittance (Cont’d)

22

22

22

11

esusceptanc:

econductanc: where

1

XR

XB

XR

RG

XR

jXR

jXR

jXR

jXRjXRjBG

B

G

jBGZ

Y

KVL and KCL in the Phasor Domain

0)Re(

)Re()Re(

as written becan This

0)cos(

)cos()cos(

form. cosinein

writtenbemay geeach volta

state,steady sinusoidal In the

0

loop. closed a around voltages

thebe , ,..., ,let KVL,For

2121

2211

21

21

tjjmn

tjjm

tjjm

nmn

mm

n

n

eeV

eeVeeV

tV

tVtV

vvv

vvv

n

0

phasor.for holds KCL

manner,similar aIn

!!phasor!for holds KVL

0

,any for 0 Since

0Re

then,Let

0

Re

21

21

21

2121

n

n

tj

tjn

jmkK

tj

jmn

jm

jm

te

e

eV

eeV

eVeV

k

n

III

VVV

VVV

V

Series-Connected Impedance

)(

gives KVL Applying

21

21

n

n

ZZZI

VVVV

VZ

ZV

Z

VI

ZZZI

VZ

eq

kk

eq

neq

,

21

Parallel-Connected Impedance

)111

(

gives KCL Applying

21

21

n

n

ZZZV

IIII

IY

YI

Y

IV

YYYV

IY

eq

kk

eq

neq

,

21

Y- Transformations

3

133221

2

133221

1

133221

Z

ZZZZZZZ

Z

ZZZZZZZ

Z

ZZZZZZZ

c

b

a

ion:Δ ConversY

cba

ba

cba

ac

cba

cb

ion:Y ConversΔ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

3

2

1

Example 1

07.1122.3811

1082310

108||2310

2.08||10

13

2

1

|| H2.08F103F2in

jj

jjj

jjj

jmjmj

mm

ZZZZ

rad/s. 50

for Find in

Z

Example 2(t). Find ov

)96.154cos(15.17)(

96.1515.17

152096.308575.0

152010060

100

25||2060

25||20

4 , 1520

)154cos(20

ttv

j

j

jj

jj

tv

Sol:

o

so

s

s

VV

V

Example 3. Find I

-Y transformation

204.4666.3 204.464.13

050

204.464.1316.13

86||3

12

2.36.110

)42(8

2.310

)8(4

8.06.1

8424

)42(4

Z

VI

ZZ

ZZ

Z

Z

Z

j

jj

jj

jj

j

jj

jj

Sol:

cnbn

an

cn

bn

an

Applications: Phase Shifters

i

iio

RCCR

RC

RCj

RCj

CjR

R

V

VVV

1tan

1

11

1

222

Leadingoutput

Phase Shifters (Cont’d)

i

iio

RCCR

RCjCj

R

Cj

V

VVV

1

222tan

1

1

1

11

1

Laggingoutput

Example

903

145

3

245

2

245

2

2

2020

20

453

2

2412

412

20

4122040

)2020(20)2020(||20

11

1

io

iii

j

j

j

j

jj

jj

Sol:

VVVV

VVVZ

ZV

Z

leading.. 90

of phase a provide to

circuit an Design

RC

Applications: AC Bridges

21

3132

321

2

32

21

21

21 :condition Balanced

ZZ

ZZZZZZ

ZZ

Z

ZZ

Z

VZZ

ZVV

ZZ

ZV

VV

xxx

x

sx

xs

AC Bridges (Cont’d)

sx LR

RL

2

1 sx CR

RC

2

1

Bridge for measuring L Bridge for measuring C

Summary• Transformation between sinusoid and phasor i

s given as

• Impedance Z for R, L, and C are given as

• Basic circuit laws apply to ac circuits in the same manner as they do for dc circuits.

CjLjR CLR

1 , , ZZZ

mm VtVtv V )cos()(