Download ppt - Chapter 5 Steady-State Sinusoidal Analysis. 5.1 Sinusoidal Currents and Voltages A sinusoidal voltage Peak value Phase angle( 相位 ) Angular frequency (

Chapter 5 Steady-State Sinusoidal Analysis

5.1 Sinusoidal Currents and Voltages

)cos()( tVtv mA sinusoidal voltage

Peak value Phase angle( 相位 )

Angular frequency( 角頻率 , rad/sec)

Tf

1 frequency

( 頻率 , Hz)

T Period( 週期 )

Tf

22

2T

Vm is the peak value

ω is the angular frequency in radians per second

θ is the phase angle (0o~360o)

T is the period

fT

22

Frequency T

f1

Angular frequency

90cossin zz

For uniformity, we express sinusoidal functions by cosine function.

tRadians(0~2)

Degree (0o~360o)

)60200cos(10

)9030200cos(10)30200sin(10)(

t

tttvx

Ex

Root-Mean-Square (均方根 )Values

dttvT

VT

2

0

rms

1

dttiT

IT

2

0

rms

1

Root-Mean-Square (均方根 )Values假設電壓源 (v(t)) 週期為 T, 與電阻連接，計算其平均

功率 avgP

R

tvtp

)()(

2

T

T dttpE0

)(

TT

Tavg dt

R

tv

Tdttp

TT

EP

0

2

0

)(1)(

1

R

dttvT

T2

0

2 )(1

RIR

VP rms

rmsavg

22

RMS Value of a Sinusoid

)cos()( tVtv m

T

mrms dttVT

V0

22 )(cos1

)2cos(2

1

2

1)(cos2 zz

T

mrms dtt

T

VV

0

2

)22cos(12

RMS Value of a Sinusoid T

mrms tt

T

VV

0

2

)22sin(2

1

2

)2sin(

2

1)22sin(

2

1

2

2

TTT

VV m

rms

0

)2sin(2

1)2sin(

2

1

)2sin(2

1)24sin(

2

1)2sin(

2

1)22sin(

2

1

TT

2

2m

rms

VV

RMS Value of a Sinusoid

2rms

mVV

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.

Example 5.1 Power delivered to a resistance by a sinusoidal source

•A voltage is applied to a 50- resistance.Find •the rms value of the voltage.•Find the average power delivered to the resistance.•Find the power as a function of time and sketch to scale.

)100cos(100)( ttv

T

2100 msT 20

100

2

1.


2. The rms value of the voltage

VVm 100 V 71.702

mrms

VV

3. The average power

WR

VP rms

avg 10050

)71.70( 22


4. The power as a function of time is

Wtt

R

tvtp )100(cos200

50

)100(cos100)()( 2

222

))200cos(1(100

))200cos(1(2

200)100(cos200 2

t

tt

5.2 Phasors (相量 )

mVV :Phasor

Phasor DefinitionRefer to Appendix A Complex Numbers

•Phasor 以複數 (complex numbers) 來表示sinusoidal voltages or currents. Ex.

•Magnitude (V1) 代表最大振福 (peak value).

•Angle (1) 代表相位 (phase).

111 cos tVtv

111 :Phasor θV V

A sinusoidal cosine voltage

(Time function, 時間函數 )

其 phasor 為

A sinusoidal sine voltage

)sin()( 222 tVtv

)90cos()sin( zz

)90cos()( 222 tVtv

90222 VV

For sinusoidal currents

)cos( 11 tIi1

)sin( 22 tIi2

111 II

90222 II

phasor

phasor

Adding Sinusoids Using Phasors

Step 1: Determine the phasor for each term.Step 2: Add the phasors using complex arithmetic.Step 3: Convert the sum to polar form.

Step 4: Write the result as a time function.

Using Phasors to Add Sinusoids 45cos201 ttv 60sin102 ttv

45201 V 30102 V

To find vs(t)=v1(t)+v2(t).

1. The phasors

2. Complex arithmetic14.1414.14)45sin(20)45cos(20111 jjjyx oo V

566.8)30sin(10)30cos(10222 jjjyx oo V

14.1980.225660.814.1414.14V jjjs

Using Phasors to Add Sinusoids3. Convert the sum to polar form

14.1980.22V js

77.29)14.19()80.22( 22 mV

01.40)77.29

14.19arctan(

01.4077.29 sV

)01.40cos(77.29)( ttvs

Using Euler’s formula to Add Sinusoids

)Re(2045cos20 )45(1

wtjettv

)Re(10)30cos(1060sin10 )30(2

wtjewtttv

jwtjjs eeetvtvtv )1020(Re)()( 3045

21

01.4077.2901.4077.29 je 3045 1020 jj ee

)11.40cos(77.2977.29Re 01.40 wteetv jwtjs

polar form

Visualization of Sinusoids )cos()( tVtv m

]Re[)( )( tjmeVtv

tVeV mtj

m)(

A sinusoidal voltage

Exponential form

Complex Polar form

•Sinusoids 可視為複數向量 (complex vectors) 隨時間改變在實數軸的投影量。

tVeV mtj

m)(

可視為 Vector with length 以角速度 (angular velocity) 在複數平面以逆時針(counterclockwise) 方向旋轉。

mV rad/s

•v(t) 為在實數軸的投影量。

•v(t) 的 Phasor 為在 t = 0 時的向量。

)( tjmeV

mV)( tj

meV

•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 Vectors

Phase Relationships from Phasors

1. To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise. 2. Then when standing at afixed point, if V1 arrives first followed by V2 after a rotation of θ , we say that V1 leads V2 by θ .3. Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ as the smaller angle between the two phasors.)

Phase Relationships from Phasors

V1 leads V2 by 60o .

To determine phase relationships between sinusoids from their plots versus time, 1.find the shortest time interval tp between positive peaks of the two waveforms. 2.Then, the phase angle isθ = (tp/T ) × 360°. 3.If the peak of v1(t) occurs first, we say that v1(t) leads v2(t) or that v2(t) lags v1(t).

Phase Relationships from Time Functions

v1(t) leads v2(t) by θ (60o).

θ = (tp/T ) × 360°.

t

tp

Exercise 5.5 30cos101 ttv 30cos102 ttv 45sin103 ttv

State the phase relationship between each pair of voltages.

-30o

+30o

-45o

5.3 COMPLEX IMPEDANCES (複數阻抗 )

Inductance假設通過電感 L 的 sinusoidal current 為

)sin()( tIti mL

則電感 L 兩端的 voltage 為

dt

tdiLtv L

L

)()( )cos( tLIm ( 亦為 sinusoid)

InductanceVoltage & current 的 phasors 為

90 mL II

mmL VLIV

Current lags the voltage by 90o

)2

( tT

Im

LIm

Ohm’s Law

IRV

What is the relationship between phaor voltage and phasor current? Does it be similar to Ohm’s law?

90)90( mIL

90 mL II

mL LIV

LIL )90( LILj )90sin90(cos90 jLL ( )

90 LLjZ L 稱為 L 的阻抗 (impedance)

Phasor voltage = impedance phasor current (similar to Ohm’s law)

LLL IZV

Note the impedance of an inductance is an imaginary number 虛數 (called reactance). Resistance is a real number.

Capacitance假設通過電容 C 的 sinusoidal voltage 為

)cos()( tVtv mc

則電容 C 兩端的 current 為

dt

tdvCti c

c

)()( )sin( tCVm

1809090 mmc CVCVI

mC VV

9090 mm ICV)180cos(1

Capacitance

CjCj

CCV

V

I

VZ

m

m

c

cc

11

901

90

( )

mC VV90 mc CVI

Current leads the voltage by 90o

Capacitance

Resistance

RR RIV

The current and voltage are in phase. ( 同相位 )

電容阻抗

電感阻抗

電阻

CjZc

1

LjZL

RZR

Current lags voltage by 90o

Current leads voltage by 90o

Current and voltage are in phase

Kirchhoff’s Laws in Phasor FormWe can apply KVL directly to

phasors. The sum of the phasor voltages equals zero for any closed path.KCL-The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

5.4 Circuit Analysis with Phasors and Complex

Impedances

Circuit Analysis Using Phasors and Impedances1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)2. Replace inductances by their complex impedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC) or –j(1/ωC) . Resistances have impedances equal to their resistances.

3. Analyze the circuit using any of the techniques studied earlier in Chapter 2, performing the calculations with complex arithmetic.

Example 5.4 Steady-state AC Analysis of a Series Circuit

I?VR?VL?VC?

1. Phasors

30100sV

2. Complex impedances

1503.0500 jjLjZ L

501040500

116

jjC

jZC CLeq ZZRZ 10010050150100 jjj

454.141 100

100arctan100100 22 ( )

3. Circuit Analysis

15707.0)4530(4.141

100

454.141

30100VI

eq

S

Z

)15500cos(707.0)( tti

3. Circuit Analysis

157.7015707.0100V IRR

751.106

15707.09015090V

ILILjL

1054.35

15707.090509011

V

IC

IC

jC Note

90,90 jj

Figure 5.13 Phasor diagram for Example 5.4.

Example 5.5 Series and Parallel Combinations of Complex Impedances.

VC?IL?IR?IC?

1. Phasors

9010 sV

2. Complex impedances 1001.01000 jjLjZ L

10010101000

116

jjC

jZC

5050)45sin45(cos71.704571.704501414.0

01

01.001.0

1

)100/(1100/1

1

/1/1

1

jj

jjZRZ

CRC

o

jj

j

jj

4571.7050

50arctan5050

505001.001.0

01.001.0

01.001.0

1

01.001.0

1

22

Or

3. Circuit Analysis

o

j

4571.70

5050

18010

4571.70

4571.709010

5050

4571.709010

5050100

4571.709010

j

jjZZ

ZVV

RCL

RCsC

)1000cos(10)1801000cos(10)( tttvC

o

j

4571.70

5050

1351414.04571.70

9010

5050

9010

5050100

9010

jjjZZ

VI

RCL

s

1801.0100

18010

R

VI C

R

901.090100

18010

100

18010

jZ

VI

C

CC

Example 5.6 Steady-State AC Node-Voltage Analysis

v1(t)?

1. Phasors 902)100sin(2 t05.1)100cos(5.1 t

2. Complex impedances 101.0100 jjjwL

5102000100

116

jjwc

j

3. Circuit Analysis

KCL 9025

VV

10

V 211

j

Node 1

KCL Node 2 05.1510

122 j

VV

j

V

22.0)2.01.0( 21 jVjVj

5.11.02.0 21 VjVj

2.05

1

5

1j

j

j

jj

∵

23)2.01.0( 1 jVj

7.291.1681405.0

4.07.0

2.01.0

2.01.0

2.01.0

23

2.01.0

23V1

jj

j

j

j

j

j

j

)7.29100cos(1.16)(1 ttv

5.5 Power in AC Circuits

阻抗有實數 ( 電阻 ) 或虛數 ( 電感 , 電容 ),功率有無實虛 ?

ZjXRZ

mm I

Z

V

Z

VI Z

VI m

m

1. If X=0 實阻抗 ( 純電阻 )

)cos()( tVtv m )cos()( tIti m

)(cos)()()( 2 tIVtitvtp mm

實功率 (real power)

( 參考Ch5.1)

0

,

RZ

0avgp

1. If X=0 實阻抗 ( 純電阻 )

RIR

VP rms

rmsavg

22

222

2222

2

mmmmrmsrmsavg

rmsrmsrmsrms

avg

IVIVIVP

IVRIR

VP

2m

rms

VV

2. If R=0, X>0 虛阻抗 ( 電感性 )

)cos()( tVtv m )sin()90cos()( tItIti mm

)2sin(2

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

ttIVtitvtp mmmm

虛功率 or 無效功率 (Reactive power)

90

,

jXZ

0avgp

3. If R=0, X<0 虛阻抗 ( 電容性 )

)cos()( tVtv m )sin()90cos()( tItIti mm

)2sin(2

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

ttIVtitvtp mmmm

虛功率 or 無效功率 (Reactive power)

90

,

jXZ

0avgp

)cos()( tVtv m

)cos()( tIti m

)2sin()sin(2

)]2cos(1)[cos(2

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

)cos()cos()(2

tIV

tIV

ttIVtIV

ttIVtp

mmmm

mmmm

mm

Power for a General Load一般狀況 R≠0 且 X≠0, RLC 電路 ( 有實與虛阻抗 ) 9090 , 0 jXRZ

)2sin(2

1)sin()cos( ),2cos(1(

2

1)(cos2 ttttt ( )

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

)cos(2

)2sin()sin(2

)]2cos(1)[cos(2

10

mm

Tmmmm

IV

dtwtIV

wtIV

T

T

avg dttpT

P0

)(1

積分為 0 積分為 0

2,

2rmsrms

mm II

VV

)cos(rmsrmsavg IVP

)cos(rmsrmsavg IVPP

)cos( 為 power factor, PF ( 功率因子 )

iv 為 power angle ( 功率角 ) 代表電流 lags 電壓的角度

rmsrms IV 為 apparent power ( 視出功律 )

)sin(rmsrmIVQ 為 reactive power ( 無效功律 or 虛功率 ), 單位為VAR (volt amperes reactive)

為 real power ( 有效功律 or 實功率 ) 單位為 W

單位為 VA(volt-amperes)

AC Power Calculations

cosrmsrmsIVP

cosPF

iv

sinrmsrmsIVQ

(W)

(VAR)

22 QPIV rmsrms (VA)

rmsrms22 IVQP

RIP 2rms

XIQ 2rms

(W)

(VAR)

(VA)

RIRI

IZ

RVIVIP rms

mrms

mrmsrmsrms

2

22cos

Z

Rcos

2m

rms

VV

RIP 2rms

Proof:

Z

VI

Z

VI m

m ,

XIQ rms2

Proof:

XIXI

IZ

XVIVIQ rms

mrms

mrmsrmsrms

2

22sin

Appendix AComplex Numbers

Basic Complex-Number Concepts

• Complex numbers involve the imaginary number( 虛數 ) j =

• Z=x+jy has a real part (實部 ) x and an imaginary part (虛部 ) y , we can represent complex numbers by points in the complex plane (複數平面 )

• The complex numbers of the form x+jy are in rectangular form (直角座標 )

1

y

x

Imaginary

Real

Z

• The complex conjugate (共軛複數 ) of a number in rectangular form is obtained by changing the sign of the imaginary part.

• For example if then the complex conjugate of is

•

432 jZ

43*2 jZ

12 j

2Z

Example A.1 Complex Arithmetic in Rectangular Form

• Solution:

/Z, ,Z

reduce ,43 and 5 5 Given that

2121 21

21

ZZZZ

jZjZ

924355

is difference The

184355

have wesum, For the

21

21

j)-j)-(j(ZZ

j)-j()j(ZZ

to rectangular form

Example A.1 Complex Arithmetic in Rectangular Form

535

20152015

20152015

4355

get we,product For the

2

21

j

jj

-jj-j

)-j)(j(ZZ

4.12.0 25

355

1612129

20152015

1612129

20152015

43

43

43

55

43

55

r denominato theof conjugatecomplex by ther denominato and

numberator thegmultiplyinby formr rectangula toexpression thisreduce can We

43

55

obtain we, numbers thedivide To

2

2

*2

*2

2

1

2

1

j

j

jj

jj

jjj

jjj

j

j

j

j

Z

Z

j

j

Z

Z

j

j

Z

Z

Complex Numbers in Polar Form(極座標 )

• Complex numbers can be expressed in polar form (極座標 ). Examples of complex numbers in polar form are :

The length of the

arrow that represents a complex

number Z is denoted as |Z|

and is called the magnitude

( 幅值 or 大小 ) of the complex number.

• Using the magnitude |Z| , the real part x, and the imaginary part y form a right triangle ( 直角三角形 ).

• Using trigonometry, we can write the following relationships:

These equations can be used to convert numbers from polar to rectangular form.

(A.3)

(A.4)

(A.1)

(A.2)

Example A.2 Polar-to Rectangular Conversion

formr rectangula to305 Convert Z3

j2.5 4.33 jy x 305 Z

2.5 )sin(30 5 )sin( Z y

4.33 )cos(30 5 )cos( Z x

page) (pre. A.4 and A.3Equation Using

:Solution

3

Example A.3 Rectangular-to-Polar Conversion

form.polar toj510- Zand j5 10 Convert Z 65

11.185(-10)yx Z

11.18510yx Z

numbers theof

each of magnitudes thefind

toA.1Equation using First,

:Solution

2226

266

2225

255

10-10

55

Z6 Z5

11.18153.43o

26.57o

Figure A.4

153.4311.18

j510 Z

153.43 57.26180

)x

yarctan(180

0.5- 10-

5

x

y)tan(

For Z

6

6

66

6

66

6

A.2.Equation use weangles, thefind To

26.5711.18

j510Z

26.57)arctan(0.5

0.5 10

5

x

y)tan(

For Z

5

5

5

55

5

• The procedures that we have illustrated in Examples A.2 and A.3 can be carried out with a relatively simple calculator. However, if we find the angle by taking the arctangent of y/x, we must consider the fact that the principal value of the arctangent is the true angle only if the real part x is positive. If x is negative,

we have:

Euler’s Identities

• The connection between sinusoidal signals and complex number is through Euler’s identities, which state that

and

• Another form of these identities is and)sin()cos( je j

2)cos(

jj ee

j

ee jj

2)sin(

)sin()cos( je j

• is a complex number having a real part of and an imaginary part of

• The magnitude is

je)cos( )sin(

1)(sin)(cos 22 je

• The angle of

• A complex number such as can be written as

• We call the exponential form ( 指數形式 )of a complex number.

)sin()cos(1

)sin()cos(1

je

jej

j

ise j

A)sin()cos()1( jAAAeAA j

jAe

• Given complex number can be written in three forms:– The rectangular form– The polar form– Exponential form

Example A.4 Exponential Form of a Complex Number

• Solution:

66.85)]60sin()60[cos(10)(10

10601060

60

jjeZ

eZj

j

planecomplex in thenumber Sketch the

forms.r rectangula and lexponentiain 6010number Zcomplex theExpress o

Arithmetic Operations in Polar and Exponential Form

• To add (or subtract) complex numbers, we must first convert them to rectangular form. Then, we add (or subtract) real part to real part and imaginary to imaginary.

• Two complex numbers in exponential form:

• The polar forms of these numbers are

22211 and 1 jj eZZeZZ

222111 and ZZZZ

• For multiplication of numbers in exponential form, we have

• In polar form, this is

2121212121

jjj eZZeZeZZZ

2121221121 ZZZZZZ

))sin()(cos(

))cossincos(sinsinsincos(cos

)sin(cos)sin(cos

212121

1221212121

22211121

jZZ

jZZ

jZjZZZProof:

( )

• Now consider division:

• In polar form, this is

21

2

1

2

1

2

1

2

1

j

j

j

eZ

Z

eZ

eZ

Z

Z

212

1

22

11

2

1

Z

Z

Z

Z

Z

Z

Example A.5Complex Arithmetic in Polar Form

• Solution:

Before we can add (or subtract) the numbers, we must convert them to rectangular form.

form.polar in and ,ZZ , find ,455 and 6010Given 21

2

12121 ZZZZZZ

152455

6010

105504556010

2

1

21

Z

Z

ZZ

54.354.3)45sin(5)45cos(5455

66.85)60sin(10)60cos(106010

2

1

jjZ

jjZ

The sum as Zs:

Convert the sum to polar form:

Because the real part of Zs is positive, the correct angle is the principal value of the arctangent.

2.1254.854.354.366.8521 jjjZZZ s

55)43.1arctan( 43.154.8

2.12tan

9.14)2.12()54.8( 22

ss

sZ

559.1421 ZZZ s