Transcript Document

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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Frequency
1
f 
T
2
Angular frequency  
T
  2f
o
sin z  cos(z  90 )
sin t  cos(t  90 )
o
sin(t  90 )  cost
o
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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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Euler’s identity
  t
  2 f
A cos t  A cos2 f t
In Euler expression,
A cos t = Real (A e j t )
A sin t = Im( A e j t )
Any sinusoidal function can
be 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
T imefunction: v1 t   V1 cosωt  θ1 
Phasor: V1  V11
V1  Re(e
j (t 1 )
V1  Re(e
j (1 )
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)
) by droppingt
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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 = 9.8524.0o 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
v1t   20cos(t  45 )


v2 (t )  10cos(t  30 )

V1  20  45

V2  10  30
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Vs  V1  V2
 20  45 10  30
 14.14  j14.14  8.660 j5
 23.06  j19.14
 29.97  39.7
Vs  Ae j
 19.14

A  23.06  ( 19.14)  29.96,   tan
 39.7
23.06
2
2
1
vs t   29.97cost  39.7

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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
VL  jL  I L
Z L  jL  L90

VL  Z L I L
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(a)
(b)
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/jC;
(c)
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(c) an inductor L is represented by
an impedance jL.
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V  Ve jt
dV
d Ve jt
IC
C
 jCVe jt
dt
dt
V
1
I  j CV  
 Zc
I
j C
Zc is defined as the impedance of a capacitor
The impedance of a capacitor is 1/jC. It is simply a mathematical
expression. The physical meaning of the j term is that it will introduce
a phase shift between the voltage across the capacitor and the current
flowing through the capacitor.
As I  j C V, if v  V cos t, then i  CV cos( t  90 )
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As I  j C V, if v  VM cos t, then i   CVM cos( t  90 )
I M   CVM
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I  Ie jt
dI
d Ie jt
VL
L
 jLIe jt
dt
dt
V
V  j LI   j L  Z L
I
ZL is defined as the impedance of an inductor
The impedance of a inductor is jL. It is simply a mathematical
expression. The physical meaning of the j term is that it will introduce
a phase shift between the voltage across the inductor and the current
flowing through the inductor.
As V  j C I, if i  I cos t , then v   LI cos( t  90 )
or i  I cos( t  90 ), and v   LI cos t.
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As V  j C I, if i  I M cos t, then v   LI M cos( t  90 )
or i  I M cos( t  90 ), and v   LI M cos t, VM   LI M
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Complex Impedance in Phasor Notation
VC  ZC I C
1
1
1

ZC 
j

  90
jC
C C
VL  Z L I L
Z L  jL  L90

VR  RI R
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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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Ztotal  100  j (150  50)
 100  j100  141.445
V
10030
I

 0.70730  45

Ztotal 141.445
 0.707  15
VR  100 I  70.7  15 , vR (t )  70.7 cos(500t  15 )
VL  j150 I  15090  0.707  15  106.0590  15
 106.0575 , vL (t )  106.05cos(500t  75 )
VC   j50  I  50  90  0.707  15  35.35  90  15
 35.35  105 , vL (t )  35.35cos(500t  105 )
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1
1
Z RC 

1 / R  1 / Zc 1 / 100  1 /(  j100)
1
1
j j 0.01
As

 
 j 0.01
2
 j100  j100 j  j
Z RC
1
10




70
.
71


45
0.01 j 0.01 0.0141445
 50  j50
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Z RC
Vc  Vs
( voltagedivision)
Z L  Z RC


70
.
71


45
70
.
71


45
 10  90
 10  90
j100  50  j50
50  j50

70
.
71


45

 10  45

10


135
70.7145
vc (t )  10 cos (1000t  135 )   10cos 1000t V
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I
Vs
Z L  Z RC
10  90
10  90


j100  50  j50
50  j50
10  90



0
.
414


135
70.7145
i (t )  0.414 cos (1000t  135 )
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VC
IR 
R
10  135

 0.1  135
100
iR (t )  0.1 cos (1000t  135 )
VC 10  135 10  135

IC 



0
.
1


45
Zc
 j100
100  90
iR (t )  0.1 cos (1000t  45 ) A
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Solve by nodal analysis
V1 V1  V2


 2  90   j 2 eq(1)
10  j5
V2 V2  V1

 1.50  1.5
eq(2)
j10  j5
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V1 V1  V2
V2 V2  V1

 2  90   j 2 eq(1)

 1.50  1.5
10  j5
j10  j5
From eq (1)
1 1 j
j
1
0.1V1  j 0.2V1  j 0.2V2   j 2 As   
  j,
 j
j j j 1
j
(0.1  j 0.2)V1  j 0.2V2   j 2
From eq ( 2)
 j 0.2V1  j 0.1V2  1.5
SolvingV1 by eq(1)  2  eq( 2)
(0.1  j 0.2)V1  3  j 2
3  j2
3.6  33.69


V1 


16
.
1


33
.
69

63
.
43
0.1  j 0.2 0.2236  63.43
 16.129.74
v1  16.1cos(100t  29.74 ) V
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eq( 2)
Vs= - j10, ZL=jL=j(0.5×500)=j250
Use mesh analysis,
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 Vs  VR  VZ  0
 (  j10)  I  250  I  ( j 250)  0
 j10
10  90


I


0
.
028


90

45
250  j 250 353.3345
I  0.028  135
i  0.028cos(500t  135 ) A
VL  I  Z L  (0.028  135 )  25090
 7  45
vL (t )  7 cos(500t  45 ) V
VR  I  R  (0.028  135 )  250  7  135
vR (t )  7 cos(500t  135 )
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5
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-j50
j200
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5
-j50
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j100
-j200
j100
100
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j100
100
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-j200
j100
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AC Power Calculations
P  Vrms I rms cos 
PF  cos 
   v  i
Q  Vrms I rms sin 
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apparentpower  Vrms I rms
P  Q  Vrms I rms 
2
PI R
2
rms
QI
2
rms
2
2
X
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P
Q
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V
2
Rrms
R
V
2
Xrms
X
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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.
Vt  Voc
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.
Voc Vt
Z t

I sc
I sc
I n  Isc
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Maximum Power Transfer
If 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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