chapter 5 steady-state sinusoidal analysis. 5.1 sinusoidal currents and voltages a sinusoidal...
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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.
Example 5.1 Power delivered to a resistance by a sinusoidal source
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
Example 5.1 Power delivered to a resistance by a sinusoidal source
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
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