chapter 5 steady-state sinusoidal analysis
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Chapter 5 Steady-State Sinusoidal Analysis. Electrical Engineering and Electronics II. Scott. 2008.10. Main Contents. 1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 5Steady-State Sinusoidal
Analysis
2008.10
Electrical Engineering and Electronics II Electrical Engineering and Electronics II
Scott
1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and complex impedances.
•Main Contents
3. Compute power for steady-state sinusoidal ac circuits.
4. Find Thévenin equivalent circuits for steady-state ac circuits.
5. Determine load impedances for maximum power transfer.
6. Solve balanced three-phase circuits.
•Main Contents
•The importance of steady-state sinusoidal analysis
Electric power transmission and distribution by sinusoidal currents and voltages
Sinusoidal signals in radio communication
All periodic signals are composed of sinusoidal components according to Fourier analysis
5.1 Sinusoidal Currents and Voltages
•Parameters of Sinusoidal Currents and Voltages
•Vm is the peak value, unit is volt
•ω is the angular frequency, unit is radians per second
•f is the frequency, unit is Hertz (Hz) or inverse second.
•θ is the phase angle, unit is radian or degree.
T
2
f 2
90cossin zz
Frequency T
f1
Angular frequency
To uniformity, This textbook expresses sinusoidal functions by using cosine function rather than the sine function. 。
•Parameters of Sinusoidal Currents and Voltages
•Root-Mean-Square Values or Effective Values
dttvT
VT
2
0
rms
1
R
VP
2rms
avg
dttiT
IT
2
0
rms
1
RIP 2rmsavg
Average Power
•RMS Value of a Sinusoid
2rms
mVV
The rms value for a sinusoid is the peak value divided by the square root of two.
However, this is not true for other periodic waveforms such as square waves or triangular waves.
)100cos(100)( ttv
)100cos(100)( ttv A voltage given by is applied to a 50Ω resistance. Find the rms value of the voltage and the average power delivered to the resistance.
70.71V2m
rms
VV
2 2
22
70.71100 W
50
( )( ) 200cos (100 ) W
rmsavg
VP
R
v tp t t
R
Complex Numbers
3 forms of complex numbers
Arithmetic operations of complex numbers
•Rectangular formbaA j
1j
]Re[Aa
]Im[Ab
a
b
Im
Re0
|A|
A
Imaginary unit
Real part
Imaginary part
Magnitude
Angle
Conjugate
22222 baAAba
a
barctan
jA a b A A 2AA A
a
b
Im
Re0
|A|
A
22 baA
a
barctan
|| AA
cos|| Aa
sin|| Ab
Conversion between Rectangular and polar
•Polar form
2 2 arctanb
A a ba
|| AA
•Conversion
baA j cos j sin
cos jsin
A A A
A
j
sinjcos
eA
AA
Euler’s Identity
sinjcosj ea
b
Im
Re0
|A|
A
•Exponential form
Rectangular form
Polar form
Exponential form
baA j
j|| eAA
|| AA
•Three forms
•Arithmetic Operations
baA j dcB jGiven
Identity BA dbca ,
Adding and subtracting
21 jeeBAC
cae 1dbe 2
•Arithmetic Operations
baA j dcB jGiven
Product
C
BA
eC
eBeABAC
j
jj
BAC
)j(j BAC ee
BAC
B
A
eB
eABAD
j
j
B
ABAD )j(j BAD ee BAD
)j(j BAD eB
AeDD
•Arithmetic Operations
baA j dcB jGiven
Dividing
•Phasor Definition
111 cos :function Time θtωVtv
1 1 1 peak Phasor: V V
Sinusoidal steady-state analysis is greatly facilitated if currents and voltages are represented as vectors or Phasors.
•rms Phasor or effective phasor.
•In this book, if phasors are not labeled as rms, then they are peak phasors.
5.2 PHASORS
正弦量可以表示为在复平面 complex plane上按逆时针方向旋转的相量的实部 real part。
Angular velocity
•Phasors as Rotating Vector
•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.
θ
•Phase Relationships
•Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ≤180o as the smaller angle between the two phasors.)
•Then when standing at a fixed point, if V1 arrives first followed by V2 after a rotation of θ , we say that V1 leads V2 by θ .
•To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise.
•Phase Relationships
•Phase Relationships
•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.
1 20cos 45 Vv t t
2 10sin 60 Vv t t
1 20 45 V V 2 10 30 V V
•Example:
1 2Find ?sv v v
Solution:
•Adding Sinusoids Using Phasors
1 2
20 45 10 30
14.14 j14.14 8.660 j5
23.06 j19.14
29.97 39.7 V
sV V V
29.97cos 39.7 Vsv t t
•Adding Sinusoids Using Phasors
•Exercise
•Answers
1
1
2
1. ( ) 10cos( ) 10sin( )
2. ( ) 10cos( 30 ) 5sin( 30 )
3. ( ) 20sin( 90 ) 15cos( 60 )
o o
o o
v t t t
i t t t
i t t t
1
1
2
1. ( ) 14.14cos( 45 )
2. ( ) 11.18cos( 3.44 )
3. ( ) 30.4cos( 25.3 )
o
o
o
v t t
i t t
i t t
•Adding Sinusoids Using Phasors
5.3 COMPLEX IMPEDANCES
•By using phasors to represent sinusoidal voltages and currents, we can solve sinusoidal steady-state circuit problems with relative ease.
•Sinusoidal steady-state circuit analysis is virtually the same as the analysis of resistive circuits except for using complex arithmetic.
•Sinusoidal response of resistance• Relationship between voltage and current
Uu
i
I
0 t
Ru
icos( )mi I t
cos( )
cos( )m
m
v Ri RI t
V t
rms rmsV RI
rms m
rms m
V VR
I I
then:
• 2.Phasor relation and diagram
Substituting for current and voltage phasors
Rv
i
IV
O0
RV
I
mV RI RI
rms rms rmsV RI RI
rms
rms
V VR
I I
Phasor diagram
cos( )mi I t
mI I
rmsrmsII
•Sinusoidal response of resistance
Plot the phasor diagram
IV
00m mI I I
00m mV RI RI
VR
I phasors expression in
Ohm’s Law
•Sinusoidal response of resistance
p vi
2 cos 2 cosrms rmsV t I t
22 cos cos 2rms rms rms rms rms rmsV I t V I V I t
No doubt, p≥0,Resistance always absorbs energy.
• Power
Instantaneous power
•Sinusoidal response of resistance
0t
vi
p
VrmsIrms
2VrmsIrms
• Power
22 cos (1 cos 2 )rms rms rms rmsp V I t V I t
•Sinusoidal response of resistance
Average Power
——Average value of instantaneous power in a period
0 0
1 1(1 cos 2 )
T T
rms rmsP pdt V I t dtT T
2
2 rmsrms rms rms
VV I I R
R
Average Power represents the consumed power, is also named as
Real power.
• Power
•Sinusoidal response of resistance
u、 i have the same frequency, u leads i by 90o
i
u π 2π0 ωt
•Sinusoidal response of Inductance• Relationship between voltage and current
sinmi I t
0cos sin( 90 )m m
div L LI t V t
dt
rms rmsV LI m mV LI
2rms mL
rms m
U UX L fL
I I
DefinitionReactance电抗-感抗
sinmi I t 0sin( 90 )mv V t
• Effective Value
•Sinusoidal response of Inductance
cosmi I t
0cos( 90 )mv V t
AII 00
i
v L
jωL
I
V
0 090 0m L m L m LV V jX I jX I jX I
j jL
VZ X L
I
Phasor diagram:
I
LV jX I
• Phasor relations
•Sinusoidal response of Inductance
Complex impedance
• Power
sinmi I t
sin sin( 90 )L m mp p vi V I t t
sin cos
sin 2m m
rms rms
V I t t
V I t
0sin( 90 )mv V t
0t
ui
pVrmsIrms
•Sinusoidal response of Inductance
0
ti
UI
0
t
p
释放能量
吸收能量
u
释放能量
吸收能量
i
u
i
u
i
u
i
u
——magnitude of magnetic field increases, inductance absorbs energy
——magnitude of magnetic field decreases, inductance supplys energy
increases 0
( ) 0, increasesL
dii
dtp t W
, ,
decreases 0
( ) 0, decreasesL
dii
dtp t W
, ,
• Power
•Sinusoidal response of Inductance
0
0
1
1sin 2 0
T
T
rms rms
P pdtT
V I tdtT
•Inductance is an energy-storage element rather than an energy-consuming element
Average PowerAverage Power
• Power
•Sinusoidal response of Inductance
The power flows back and forth to inductances and capacitances is called reactive power Q, it is the peak instantaneous power。
22Lrms
Lrms Lrms L LrmsL
VQ V I X I
X Unit: Var ( 乏 )
Reactive PowerReactive Power
Reactive power is important because it causes power dissipation in the lines and transformers of a power distribution system. Specific Charge is executed by electric-power companies for reactive power.
•Sinusoidal response of Inductance
i
v C
sinmv V t
0
0
sin( 90 )
sin( 90 )
m
m
dvi C C U t
dt
I t
Current leads voltage by 90°
2m m mI CV fCV u
i 20
•Sinusoidal response of capacitance• Relationship between voltage and current
i
v C
2rms rms rmsI CV fCV
2m m mI CV fCV
1 1
2rms m
Crms m
V UX
I I C fC
Definition
Reactance电抗-容抗
• Relationship between voltage and current
•Sinusoidal response of capacitance
sinmv V t0sin( 90 )mi I t A 090mI I A
00mV V V
i
v C
I
U -jωC
• Phasor diagram
U
CX
UjI
0
0
0j j
90 jm m m
Cm m m
V V VUX
I I II
•Sinusoidal response of capacitance
0sin sin( 90 )
sin 2m m
rms rms
p vi V tI t
V I t
0t
ui
pUrmsIrms
•PowerPower•Instantaneous power
•Sinusoidal response of capacitance
0
t
UI
0
t
p
释放能量
吸收能量
ui
释放能
量
吸收能量
i
u
i
u
i
u
i
u
increases increases
0 ( ) 0
u Wc
dup t
dt
, ,
,
——capacitance charges,it absorbs energy
——capacitance discharges,it supplys energy
decreases decreases
0 ( ) 0
u Wc
dup t
dt
, ,
,
•PowerPower
•Sinusoidal response of capacitance
0 0
1 1sin 2 0
T T
rms rmsP pdt V I tdtT T
Average PowerAverage Power
•PowerPower
•Capacitance is an energy-storage element rather than an energy-consuming element.
•Sinusoidal response of capacitance
• Assuming the same current acts on the inductance and capacitance respectively, the initial phase is 0, then
sinmi I t sin( 90 )C mv V t
2C Crms Crms Crms CQ V I I X
0sin sin( 90 )
sin 2m m
rms rms
p vi V tI t
V I t
Reactive PowerReactive Power•PowerPower
•Sinusoidal response of capacitance
It is given that the frequency of sinusoidal source is 50Hz, the rms or effective value is 10V, capacitor is 25μF, determine the rms value of current. If the frequency is 5000Hz, then what is the rms value of current now?
i
u C
when f =50Hz
6
1 1127.4
2 2 3.14 50 (25 10 )CX fC
mAAX
UI
C
rms
rms78078.0
4.127
10
When f =5000Hz
6
1 11.274
2 2 3.14 5000 (25 10 )CX fC
AX
UI
C
rms
rms8.7
274.1
10
The higher frequency is under fixed rms value of voltage, the bigger the rms value of current flowing through capacitor.
Solution
•Summaryele
ment
circuitRelationship
between u and i
Ohm’s Law
Complex impedance
Phasors diagram
R
L
C
R
i
v v Ri
div L
dt
dvi C
dt
IRU
IjXU L
IjXU C
I
UR
I
UjXC
I
UjX L
U
I
U
I
U
I
i
v C
i
v L
•Inductance
•Capacitance
In phase
•Resistance
•True or False1.1. 220 sin( 45 )Vv ω t
e V 45220V ?)A30(sin24 tω ?
RMS e A j304I3.3. ComplexComplex
j45
)A60(sin10 tωi ?Peak ValuePeak Value
RMS 100VV ?RMS e j15100 V
V ? minusminus
2. RMS 10 60 AI4.4.
RMS V 100 15V
•Complex Impedances
Complex Impedance
Two-terminal circuit of
zero sources
+
-V
I
ii
( )vv
V ΨV VZ Ψ Ψ Z
I I Ψ I
MagnitudeV
ZI
Angle ivΨ Ψ
j jLL L
L
VZ L X
I
:L
:C
RR
R
VZ R
I
:R | | 0RZ R
| |2LZ L
1| |
2CZ C
1j
jC
C CC
VZ X
I C
•Complex Impedances of R L C
•The Laws of Phasor Form
(KCL): 0i 0I
(KVL): 0v 0V
v iR V IZ
Voltage-DividerVoltage-Divider1
11 2
ZV V
Z Z
ZZeqeq ==ZZ11++ZZ22Equivalent ImpedanceEquivalent Impedance
22
1 2
ZV V
Z Z
ZZ11
ZZ22
++
––
++
++
–– ––
V1V
2V
I
ZZ
++
––
V
I
•Complex Impedances in Series
IZZ
ZI
21
21
ZZ
++
––
V
I
V
I
ZZ11 ZZ22
++
––
1I 2I
1 2
1 1 1
eqZ Z Z
IZZ
ZI
21
12 Current-DividerCurrent-Divider
Equivalent ImpedanceEquivalent Impedance
•Complex Impedances in Parallel