Post on 15-Jan-2016
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Steady-State Sinusoidal Analysis
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Steady-State Sinusoidal Analysis 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.
3. Compute power for steady-state ac circuits.
4. Find Thévenin and Norton equivalent circuits.
5. Determine load impedances for maximum power transfer.
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The sinusoidal function v(t) = VM sin t is plotted (a) versus t and (b) versus t.
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The sine wave VM sin ( t + ) leads VM sin t by radian
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T
2
f 2
Frequency T
f1
Angular frequency
tt
tt
zz
o
o
o
cos)90sin(
)90cos(sin
)90cos(sin
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A graphical representation of the two sinusoids v1 and v2.
The magnitude of each sine function is represented by the length of the corresponding arrow, and the phase angle by the orientation with respect to the positive x axis.
In this diagram, v1 leads v2 by 100o + 30o = 130o, although it could also be argued that v2 leads v1 by 230o.
It is customary, however, to express the phase difference by an angle less than or equal to 180o in magnitude.
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4-6
Euler’s identity
tfAtA
f
t
2coscos
2
In Euler expression,
A cos t = Real (A e j t )A sin t = Im( A e j t )
Any sinusoidal function canbe expressed as in Euler form.
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The complex forcing function Vm e j ( t + ) produces the complex response Im e j (t + ).
It is a matter of concept to make use of the mathematics of complex number for circuit analysis. (Euler Identity)
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The sinusoidal forcing function Vm cos ( t + θ) produces the steady-state response Im cos ( t + φ).
The imaginary sinusoidal forcing function j Vm sin ( t + θ)
produces the imaginary sinusoidal response j Im sin ( t + φ).
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Re(Vm e j ( t + ) ) Re(Im e j (t + ))
Im(Vm e j ( t + ) ) Im(Im e j (t + ))
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Phasor Definition
111 cos :function Time θtωVtv
tdroppingbye
e
V
j
tj
)Re(
)Re(
:Phasor
)(1
)(1
111
1
1
V
V
V
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A phasor diagram showing the sum of
V1 = 6 + j8 V and V2 = 3 – j4 V,
V1 + V2 = 9 + j4 V = Vs
Vs = Ae j θ
A = [9 2 + 4 2]1/2
θ = tan -1 (4/9)
Vs = 9.8524.0o V.
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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.
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Using Phasors to Add Sinusoids
)30cos(10)(
)45cos(20
2
1
ttv
ttv
3010
4520
2
1
V
V
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7.39cos97.29 ttvs
7.3906.23
14.19tan,96.29)14.19(06.23
V
122
A
Aes j
7.3997.29
14.1906.235660.814.1414.14
3010452021s
jjj
VVV
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Phase Relationships
To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise.
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 .
Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ as the smaller angle between the two phasors.)
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To determine phase relationships between sinusoids from their plots versus time, find the shortest time interval tp between positive peaks of the two waveforms.
Then, the phase angle isθ = (tp/T ) × 360°.
If the peak of v1(t) occurs first, we say that v1(t) leads v2(t) or that v2(t) lags v1(t).
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COMPLEX IMPEDANCES
LL Lj IV
90 LLjZ L
LLL Z IV
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(b)(a)
(c)
In the phasor domain,
(a) a resistor R is represented by an impedance of the same value;
(b) a capacitor C is represented by an impedance 1/jC;
(c) an inductor L is represented by an impedance jL.
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ZcCj
Cj
CVejdt
VedC
dt
dC
Ve
tjtj
tj
1
I
VVI
VI
V
Zc is defined as the impedance of a capacitor
)90(cos,cos,VI tCVithentVvifCjAs
The impedance of a capacitor is 1/jC. It is simply a mathematicalexpression. The physical meaning of the j term is that it will introducea phase shift between the voltage across the capacitor and the currentflowing through the capacitor.
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MM
MM
CVI
tCVithentVvifCjAs
)90(cos,cos,VI
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L
tjtj
tj
ZLjLj
LIejdt
IedL
dt
dL
Ie
I
VIV
IV
I
ZL is defined as the impedance of an inductor
.cos),90(cos
)90(cos,cos,IV
tLIvandtIior
tLIvthentIiifCjAs
The impedance of a inductor is jL. It is simply a mathematicalexpression. The physical meaning of the j term is that it will introducea phase shift between the voltage across the inductor and the currentflowing through the inductor.
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MMMM
MM
LIVtLIvandtIior
tLIvthentIiifCjAs
,cos),90(cos
)90(cos,cos,IV
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90
90111
LLjZ
Z
CCj
CjZ
Z
L
LLL
C
CCC
IV
IV
RR RIV
Complex Impedance in Phasor Notation
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Vm Im
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Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.
The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.
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Circuit Analysis Using Phasors and Impedances
1. 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). Resistances remain the same as their
resistances.
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3. Analyze the circuit using any of the techniques studied earlier performing the calculations with complex arithmetic.
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)105500cos(35.35)(,10535.35
159035.3515707.09050I50
)75500cos(05.106)(,7505.106
159005.10615707.090150I150
)15500cos(7.70)(,157.70 I100
15707.0
4530707.0454.141
30100I
454.141100100
)50150(100
ttv
jV
ttv
jV
ttvV
Z
V
j
jZ
L
C
L
L
RR
total
total
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5050
4571.704501414.0
01
01.001.0
1
01.001.0
100
1
100
1
)100/(1100/1
1
/1/1
1
2
j
jZ
jj
j
j
j
jjAs
jZcRZ
RC
RC
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V1000cos10)1351000(cos10)(
135104571.70
4571.704510
5050
4571.709010
5050100
4571.709010
)(
tttv
jjj
divisionvoltageZZ
ZVsVc
c
RCL
RC
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)1351000(cos414.0)(
135414.04571.70
9010
5050
9010
5050100
9010
tti
jjj
ZZ
VsI
RCL
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A)451000(cos1.0)(
451.090100
13510
100
13510
)1351000(cos1.0)(
1351.0100
13510
tti
jZc
VI
tti
R
VI
R
CC
R
CR
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Solve by nodal analysis
)2(5.105.1510
)1(2902510
122
211
eqj
VV
j
V
eqjj
VVV
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V)74.29100cos(1.16
74.291.16
43.6369.331.1643.632236.0
69.336.3
2.01.0
23
23)2.01.0(
)2(2)1(
5.11.02.0
)2(
22.0)2.01.0(
1,
1
1122.02.01.0
)1(
)2(5.105.1510
)1(2902510
1
1
1
1
21
21
211
122211
tv
j
jV
jVj
eqeqbyVSolving
VjVj
eqFrom
jVjVj
jj
jj
j
j
jjAsjVjVjV
eqFrom
eqj
VV
j
Veqj
j
VVV
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Vs= - j10, ZL=jL=j(0.5×500)=j250
Use mesh analysis,
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)135500cos(7)(
1357250)135028.0(
V)45500cos(7)(
457
90250)135028.0(
A)135500cos(028.0
135028.0
4590028.04533.353
9010
250250
10
0)250(250)10(
0
ttv
RIV
ttv
ZIV
ti
I
j
jI
jIIj
VVVs
R
R
L
LL
ZR
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5-j50 j200
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5-j50 j200
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100
j100 -j200
j100
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100
j100 -j200
j100
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AC Power Calculations
cosrmsrmsIVP
cosPF
iv
sinrmsrmsIVQ
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rmsrmspower apparent IV
2rmsrms
22 IVQP
RIP 2rms
XIQ 2rms
R
VP R
2rms
X
VQ X
2rms
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THÉVENIN EQUIVALENT CIRCUITS
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The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.
ocVV t
We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.
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The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.
scsc
oc
I
V
I
V ttZ
scII n
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Maximum Power TransferIf the load can take on any complex
value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.
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