Chap 09 Sinusoids and Phasors-Rev

7/28/2019 Chap 09 Sinusoids and Phasors-Rev

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Chap 09 Sinusoids and Phasors


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Chap 09 Sinusoids and Phasors 2

Outline

Introduction Sinusoids

Phasors

Phasor Relationships for Circuit Elements

Impedance and Admittance

Kirchhoffs Laws in the Frequency Domain

Impedance Combinations


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Introduction

Why sinusoidal waveforms are useful to engineers?

1. Appears everywhere Vibration of a string, ripples of ocean surface, and natural response

of underdamped second-order systems

2. Easily generated

3. Using Fourier analysis, everypracticalperiodic

signal can be represented by a linear combination

of sinusoidal signals

4. Easily analyzed, such as differentiate and integrate

Sinusoid: a signal that has the form of the sine or cosinefunction.


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Periodic Function

A periodic function is one that satisfies f(t) =f(t+ nT),for all tand for all integers n.

Theperiod Tis the number of seconds per cycle

The cyclic frequencyf= 1/Tis the number of cycles

per second


T


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Sinusoids

Consider the sinusoidal voltagev(t) = Vm cos(t+)

where

Vm : the amplitude of the sinusoid

: the angular frequency in radians/st+ : the argumentof the sinusoid

:phase ofv(t) in [-180, 180] or [-, ] rads/sec

The period Tis given as

2 2 1; in Hertz (Hz)

2fT

f f


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Phase Lead or Lag

Two sinusoids with the same frequency.

1 1 1 1

2 2 2 2

( ) cos( ), 0

( ) cos( ), 0

m m

m m

v t V t V

v t V t V

1 2

1 2

1 2 1 21 2

1 2

1 2

1 2 1 2 2 1

( ) and ( ) are .

( ) and ( ) are .

, ( ) ( ) by

0

0 leads

0 lag, ( ) ( ) b

:

0 :

ys

in phav t v t

v

se

out of phaset v t

v t v t

v t v t


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Phase Lead or Lag (cont .)


1

2

2 1

( ) sin( )( ) sin( )

( ) leads ( ) by .

m

m

v t V t v t V t

v t v t


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Trigonometric Identities


sin( ) sin cos cos sin

cos( ) cos cos sin sin

A B A B A B

A B A B A B

sin( 180 ) sin

cos( 180 ) cos

sin( 90 ) cos

cos( 90 ) sin

t t

t t

t t

t t

2 2 1

cos sin cos( )

, tan

A t B t C t

BC A B

A


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Example 9.1

Q: Find the amplitude, phase, period, and frequency ofthe sinusoid

v(t) = 12cos(50t+ 10)

Sol

The amplitude is Vm = 12V.

The phase is = 10

The period

The frequency is

2 20.1257 s.

50T

17.958 Hz.f

T


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Example 9.2

Q: Calculus the phase angle between v1 = -10 cos (t+50) and v2 = 12 sin (t- 10). State which sinusoid is

leading.

Sol

v1 = -10 cos (t+ 50) = 10 cos (t+ 50 - 180)

v1 = 10 cos (t- 130)

and

v2 = 12 sin (

t- 10

) = 12 cos (

t- 10

- 90

)v2 = 12cos (t- 100)

1=-130, 2=-100; 2-1=30

v2 leads v1by 30.


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Phasors

Phasor: A phasor is complex number which representsthe amplitude andphase of a sinusoid.


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Complex Number Representations

Complex number representations:

j

rerz

jyxz

Polar form:

Rectangular form:

Exponential form: j

z

z x j

r

y

z

r

e


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Complex Plane

)sin(cos

jr

ryjxz


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Basic Operations of Complex Numbers

Addition:

Subtraction:

Multiplication:

Division:

)()( 212121 yyjxxzz

)()( 212121 yyjxxzz

)(212121

rrzz

)( 212

1

2

1 r

r

z

z

B i O ti f C l N b


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Basic Operations of Complex Numbers

(cont .)

Reciprocal:

Square Root:

Complex Conjugate:

)(11

rz

)2/( rz

j

rerjyxz

)(*


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Example 9.3

Q: Evaluate these complex numbers:

Sol

1/ 2(a) (40 50 20 30 )

10 30 (3 4)(b)

(2 4)(3 5)*

j

j j

1/ 2

40 50 20 30 47.72

40 50 20 30 6

40 50 25.71 30.64

20 30 17

(a) 40(cos50 sin50 )

20 cos( 30 ) sin( 30 )

43.0

.32 1

3 25.63

1

20

2

.

.

64

.91 81

0

j

j

j

j

j


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Example 9.3 (cont .)

8.66 5 (3 4 )(b)

(2 4)(3 5)

11.33 9 14.73 37.66

1

10 30 (3 4)

(2 4)(3 5)

4 22 26.08 122.4

16

*

0 05

7

.56 .13

j j j

j j

j

j j

j


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Sinusoids and Complex Number

Euler's identity: cos sinje j

Thus ( ) can be written a

(

s

Re( )) j t

v t

v et V

cos Re( )

sin Im( )

j

j

e

e

( )cos( ) Re( )( )

Re( )jm

j t

m m

j t

V t V

e

v e

V e

t

Here is the of ( ).jm m phasor representation v tV e V V


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Representation ofVejt


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Phasors

cos( )( ) m mV tv Vt V

(Time-domainrepresentation)

(Phasor-domainrepresentation)

Phasor: A phasor is a complex number which

represents the amplitude andphase of a sinusoid.

Given the frequency , the phasor can equivalentlyrepresent the signal v(t).


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Sinusoid-Phasor Transformation

Time domainrepresentation

Phasor domainrepresentation

)cos( tVm

)(sin tVm

)cos( tIm

)sin( tIm

mV

)90( mV

mI

)90( mI


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mV

Phasor Diagram

Vm

Im

Vm V

Im


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Example 9.4

Q: Transform these sinusoid to phasors:

Sol

(a) 6cos(50 40 ) A

(b) 4sin(30 50 ) V

i t

v t

(a) 6cos(50 40 ) has the phasor

(b) Since sin cos( 90 )

4sin(30 50 ) 4cos(30 50 90 )

4cos(30 140 ) V

The phasor of i

6 40

4 140s

A

V

i t

A A

v t t

t

v

I

V


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Example 9.5

Q: Find the sinusoid representation by these phasors:

Sol

20

(a) 3 4 A

( ) 8 Vj

j

b j e

I

V

(a) 3 4 5 126.87

( ) 5cos( 126.87 A)i

j

t t

I

(b) 1 90 ,

8 20 (1 90 ) (8 20 )

8 90 20 8 70 V

( ) 8cos( 70 ) V

j

j

v t t

V


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Example 9.6

Q: Given i1(t) = 4cos(

t+30

) A and i2(t) = 5sin (

t-20) A, find their sum.

Sol

1

2

1

1

2

1 2

2

4cos( 30 )

5sin( 20 ) 5cos( 20 90 ) 5cos( 110 )

4 30 5 110

3.464 2 1.71 4.698 1.754 2.698

3.218 56.97 A

4 30

5 110

( ) 3.218cos( t

i t

i t t t

i i i

j j j

i t

I

I

I I I

56.97 A)


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Differentiation and Integration

)Re()Re(

)90cos()sin(90 tjjjtj

m

mm

ejeeeV

tVtVdt

dv

V

dvj

dt

V

(Time domain) (Phasor domain)

(Time domain) (Phasor domain)

90

sin( ) cos( 90 )cos( )

Re Re

m mm

j t j j

j tm

V t V t vdt V t dt

V e e e ej

V

vdtj

V

( ) cos( )mv t V t


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Differentiation and Integration (cont.)

V

Vjdt

dv

jvdt V


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Example 9.7

Q: Using the phasor approach, determine the current i(t)

in a circuit described by the integral and differential

equation

Sol

4 8 3 50cos(2 75 )di

i idt t dt

84 3 50 75

2, so

(4 4 6) 50 75

Converting this to the time domain

( ) 4.642cos(2 143 A

,

.2 )

i t

j

j

t

j

j

II I

I


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ch09_Sinusoids and Phasors 29

Remarks

The phasor technique provides a way torepresent or compute the operations (including

additions, differentiations, and integrations) of

sinusoids. Note that the sinusoidal frequency is assumed

to be fixed to a common value.

Phasor Relationships for Circuit


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Phasor Relationships for CircuitElements: Resistor

By Ohm's la

cos( )

,

cos( )

wm

m

m

m

i I t

v iR

I

RI RRI t

I

V I

Time domain Phasor domain Phasor diagram


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Inductor in Phasor Domain

Time domain Phasor domain Phasor diagram

cos( )

cos( 90 )

90

m

m

m

mi I t

div L LI

I

j L

tdt

LI

I

V I


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Capacitor in Phasor Domain

Phasor diagramTime domain Phasor domain

cos( )

cos( 90 )

90

m

m

m

mv V t

dvi C CV t

dt

C

V

V j C

V

I V

Summary of Voltage-Current


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Summary of Voltage-Current

Relationships

Element Time domain Frequencydomain

R V=Ri V =RI

L V =jLI

C

dt

diLv

dt

dvCi

Cj

IV


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Example 9.8

Q: The voltage v = 12 cos(60t+ 45

) is applied to a 0.1-Hinductor. Find the steady-state current through the inductor.

Sol

12 45 V , where 60rad/s.Hence,

12 45 12 452 45

( ) 2cos(60 4 A5 )

A60 0.1 6 90

Converting,

j L

i t

j

t

j L

V I

VI


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Impedance and Admittance

or VZ V ZII

The impedance Z of a circuit is the ratio of the phasorvoltage V to the phasor current I, measured in ohms ().

Impedance and Admittances of


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Impedance and Admittances of

Passive Elements

Element Impedance Admittances

R Z =R

L Z =jL

C Y =jCCj

1Z

R

1Y

Lj

1Y

Equivalent Circuits at DC & High


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Equivalent Circuits at DC & High

Frequencies

0

L j LZ

1C

j CZ

General Passive Circuit In Phasor


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General Passive Circuit In PhasorDomain

R jX Z Z

2 2 1

where

, tan

and

cos Re( ) : of ,

Resistance

Reactancsin Im( ) : of e

R

X

XR X R

Z

Z Z Z

Z Z Z

Inductive and Capacitive


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Inductive and CapacitiveImpedance

X> 0: inductive impedance

X=0: pure resistor

X


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Admittance

1 Y

Z

I

V

G Bj Y

G = Re Y is called the conductance.

B = Im Y is called the susceptance.

The admittance Y of a circuit is the reciprocal of theimpedance, measured in siemens (S).

Ad itt d I d


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Admittance and Impedance

2

2

2 2 2

1

1

1

,

G jB

R jXR jX R jX

G jBR

G BR X

R

jX R

X R X

jX R jX

YZ

E l 9 9


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Example 9.9

Q: Find v(t) and i(t) in the circuit.

E l 9 9 ( )


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From the voltage source 10 cos 4t, = 4,

The impedance is

Hence the current

V010 sV

1 15 54 0.1

5 2.5

j C j

j

Z

2 2

10 0 10(5 2.5)

5 2.5 5 2.5

1.6 0.8 1.789 26.57 A

s j

j

j

VI

Z

E l 9 9 ( )


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The voltage across the capacitor is

Converting I and V, we get

1.789 26.57

4 0.1

1.789 26.574.47 63.43 V

0.4 90

j C j

C

IV IZ

A( ) 1.789cos(4 26.57 )

( ) 4.47cos(4 63.43 ) V

i t t

v t t

Kirchhoffs Law in the Frequency


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Kirchhoff s Law in the FrequencyDomain

The KCL and KVL are still applicable in thefrequency domain.


1 2 1 2

1 2 1 2

KVLTime do F

0 0

KC

requency domain

L

m n

0 0

ain n

n n

v v v

i i i

V V V

I I I


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KVL and KCL in the Phasor Domain

1 1

1

1

2 2

2

2

For , let , ,..., , be the voltages around a closed loop.

0

In the sinusoidal steady state, each voltage may be written in cosine

cos( ) cos(

form.

o

V

) s

K

c

L

m m n

n

m

n

v

V t V t

v v

v v v

V

1 2

1 2

1

1 2

1 2

1 2

2

This can be written as

Re( ) Re( ) Re( ) 0

Re 0

Re 0; ( )

; (

( ) 0

0

n

n

k

jj jj t j t j t

m m mn

jj j j t

m m mn

jj

n

n

t

n K mk

V e e V e e V e e

V e V e V e

V

t

e

e e

V V

V V

V

V V

KVL hold

0

s f

)

or phasor!

j te t

1 2KCL holds forSimilarily, pha or s 0n I I I

S i C t d I d


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Series-Connected Impedances

eq 1

1 2 1 2

2

( )N

N

N

V V V V I Z Z Z

VZ Z Z

IZ

V lt Di i i P i i l


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Voltage-Division Principle

then,andSince 2211

21

IZVIZV

ZZ

VI

1 21 2

1 2 1 2

Voltage division princile ,:

Z Z

Z Z ZV V

ZV V

P ll l C t d I d


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Parallel-Connected Impedances

eq

1

2

2

2

1

1

1 1 1

1 1

1

1

N

N

N

I I I I VZ Z Z

I

VZ Z Z Z

eq 1 2 N Y Y Y Y

C t Di i i P i i l


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Current-Division Principle

2211eq

21

21

2121eq

eq/11/

111

ZIZIIZV

ZZ

ZZ

ZZYYYZ

2 1

1 2

1 2

1 2

Current-division principle: ,

Z Z

Z Z ZI I

ZI I

Delta ()-to-Wye (Y) and


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Delta () to Wye (Y) andWye (Y)-to-Delta () Transformations

Delta () to Wye (Y) Transformation


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Delta ()-to-Wye (Y) Transformation

-Y Conversion

1

2

3

b c

a b c

c a

a b c

a b

a b c

Z

Z Z Z

Z

Z Z Z

Z

Z Z

Z

ZZ

Z

Wye (Y) to Delta () Transformation


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Wye (Y)-to-Delta () Transformation

Y- Conversion

1 2 2 3 3 1

1

1 2 2 3 3 1

2

1 2 2 3 3 1

3

a

b

c

Z Z Z Z Z Z

Z

Z Z Z Z Z Z

Z

Z Z Z Z ZZ

Z

Z

Z

Z


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A delta () or wye (Y) circuit is said to be balancedif it has equal impedances in all three branches.

or1

3

3Y Y Z Z Z Z

.321 andwhere cbaY ZZZZZZZZ

Example 9 10


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Example 9.10

Q: Find the input impedance of the circuit. Assume that

the circuit operations at = 50 red/s.

Example 9 10 (cont )


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31

3

32

1 1

50 2 10

1 13 3

50 10 10

8 8 50 0.2

10

(3 2)

(8 10)

j

j

j C j

j C j

j L j j

Z

Z

Z

Z1

Z3

Z2



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in 1 2 3

2 2

(3 2)(8 10)10

11 8

(44 14)(11 8)10 10 3.22 1.07

11 83.22 11.07

j jj

j

j jj j j

j

Z Z Z Z

The input impedance is

Z1

Z2

Z3

Example 9 11


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Example 9.11

Q: Determine v0(t) in the circuit.



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3

20cos(4 15 ) 20 15 V, 4

1 110mF 25

4 10 10

5 H 4 5 20

s sv t

jj C j

j L j j

V

Z1 Z2

Example 9 11(cont )


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Example 9.11(cont .)

Let

Z1 = Impedance of the 60- resistor

Z2 = Impedance of the parallel combination of the 10-

mF capacitor and the 5-H inductor

Then Z1 = 60 and

2

25 2100

025 20

25 20

j jj jj

j j

Z

Z1Z2



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By the voltage-division principle,

Convert this to the time domain and obtain

2

1 2

100(20 15 )

60 100

(0.8575 30.96 ) (20 15 17.15 1) . V5 96

o s

j

j

Z

ZV

ZV

( ) 17.15cos(4 15.96 V)ov t t

Example 9 12


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Example 9.12


4(2 4) 4(4 2)(1.6 0.8)

4 2 4 8 10

4(8) 8(2 4)3.2 , (1.6 3.2)

10 10

an

bn cn

j j jj

j j

j jj j

Z

Z Z

62

.Find I-Y transformation



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(1.6 0.8) ; 3.2 ; (1.6 3.2)an bn cnj j j Z Z Z

63

.Find I-Y transformation

12 3 || 6 8

13.6 1 13.64 4.2045

3

0 0

13.64 4.204

.666 4.204

an bn cnj j

j

Z

VI

Z

Z Z Z

Application: Phase Shifters


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Application: Phase Shifters

i

iio

RCCR

RC

RCj

RCj

CjR

R

V

VVV

1tan

1

11

1

222

Leading

output

Phase Shifters (cont )


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Phase Shifters (cont.)

i

iio

RCCR

RCj

CjR

Cj

V

VVV

1

222 tan1

1

11

1

1

Lagging

output

Example 9 13


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Example 9.13

Q: Design anRCcircuit to provide a phase of 90

leading.Z

Example 9 13


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Example 9.13

Using voltage division,

20(20 20)2 12 40 (20 20) 40 20

jj j j

Z

1

1

1

2

45 V3

190

3

Z 12 4

Z 20 12 24

and

20 245

20 20 2

2 245 45

2 3

i i

o

i

i

j

j j

j

V

V

V

V V

V

V

Z

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Chap 09 Sinusoids and Phasors-Rev