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Chapter 10
Sinusoidal Steady-State Analysis
Charles P. Steinmetz (1865-1923), the developer of the
mathematical analytical tools for studying ac circuits.
Courtesy of General Electric Co.
a  jb  r (cos   j sin  )
Heinrich R. Hertz (1857-1894).
Courtesy of the Institution of Electrical Engineers.
cycles/second
Hertz, Hz
Sinusoidal Sources
Amplitude
Period = 1/f
Phase angle
Angular or radian
frequency = 2pf = 2p/T
Sinusoidal voltage source
vs  Vm sin(t  ).
Sinusoidal current source
is  Im sin(t  ).
Example
v
i
+
i
circuit
v
element
_
Voltage and current of a circuit element.
The current leads the voltage by  radians
OR
The voltage lags the current by  radians
Example 10.3-1
v  3cos 3t
i  2sin(3t  10)
Find their phase relationship
 sin  t  sin( t  p )
 i  2sin(3t  180  10)
and
sin   cos(  90)
i  2 cos(3t  180  10  90)
 2 cos(3t  100)
Therefore the current leads the voltage by 100
Recall
vs  V0 cos  t  v f  A cos  t  B sin  t 10.3-4

A
vf  A  B 
cos  t 
2
2
 A B
2
2

sin t 
2
2
A B

B
 A2  B2  cos cos t  sin  sin t 
 A  B  cos t   
2
2
B
Triangle for A and B of Eq. 10.3-4,   tan A ; A  0
where C  A2B2 .
1 B
 180  tan
;A 0
B
A
A
sin  
cos 
A2  B 2
A2  B 2
1
Example 10.3-2 i  6cos2t  8sin2t
A
B
B
A
A  6  0
1 B
   180  tan
A
1 8
 180  tan
6
 180  53.1
i  10cos(2t  126.9)
Steady-State Response of an RL circuit
vs  Vm cos  t  i f  ?
di
L  Ri  Vm cos  t
dt
An RL circuit.
From #8&#9
10.4-1
i f  A cos  t  B sin  t
Substitute the assumed solution into 10.4-1
L( A sin t   B cos t )  R( A cos t  B sin t )  Vm cos t
Coeff. of cos
 LB  RA  Vm
Coeff. of sin
 LA  RB  0
Solve for A & B
RVm
A 2
2 2
R  L
 LVm
and B  2
2 2
R  L
Steady-State Response of an RL circuit (cont.)
i  A cos  t  B sin  t
Vm
 cos( t   )
Z
Z  R2   2 L2
  tan
1
L
R
Thus the forced (steady-state) response is of the form
i  I m cos(t   )
Vm
Im 
Z
  
Complex Exponential Forcing Function
vs  Vm cos  t
i
Input
Vm
cos( t   )
Z
magnitude
Response
phase
frequency
Exponential Signal
ve  Vme
j t
vs  Vm cos  t  Re Vme
Note
Rea  jb  a
j t
  Rev 
e
Complex Exponential Forcing Function (cont.)
die
L
 Rie  ve
dt
try
vs  ve
ie  Ae j t
We get
( j L  R ) Ae
j t
 Vme
j t
Vm
Vm  j
A

e
R  j L Z
where
  tan
1
L
R
and
Z  R  L
2
2 2
Complex Exponential Forcing Function (cont.)
Substituting for A
Vm  j jt
ie 
e e
Z
We expect
Vm  j  jt 
i  Re ie   Re  e e 
Z

Vm

Re e  j e j t 
Z
Vm
j ( t   )

Re e

Z
Vm

cos( t   )
Z
Example
2
d i di
  12i  12 cos3t
2
dt
dt
We replace
2
d ie
 2
dt
j 3t die
ie  Ae ,
dt
Substituting ie
ve  12e
die
j 3t

 12ie  12e
dt
2
d
ie
j 3t
j 3t
 j 3 Ae , 2  9 Ae
dt
j 3t
( 9  j 3  12) Ae  12e
12
A
 2 2  45
3  j3
j 3t
j 3t
Example(cont.)
 ie  Ae j 3t  2 2e  j (p / 4)e j 3t
 2 2e
j (3t p / 4)
The desired answer for the steady-state current

i  Re ie   Re 2 2e
j (3t p / 4)

 2 2 cos(3t  45)
Or
p
interchangeable
 2 2 cos(3t  )
4
Using Complex Exponential Excitation to Determine a
Circuit’s SS Response to a Sinusoidal Source
Write the excitation as a cosine waveform with
a phase angle
ys  Ym cos( t   )
Introduce complex excitation
vs  Re Vm e
j ( t  )
Use the assumed response
xe  Ae
j ( t  )
Determine the constant A
A  Be
 j

Obtain the solution
xe  Ae
j ( t  )
 Be
j ( t    )
The desired response is
x(t )  Rexe   B cos(t     )
Example 10.5-1
R  2
L  1H
vs  10sin 3t
Example 10.5-1(cont.)
vs  10sin 3t  10cos(3t  90)
ve  10e
j (3t 90 )
die
L
 Rie  ve
dt
j 3 Ae
j (3t 90 )
 2 Ae
ie  Ae j (3t 90)
j (3t 90 )
 10e
j (3t 90 )
j 3 A  2 A  10
10
10
 j
A

e
2  j3
49
3
  tan
 56.3
2
1
Example 10.5-1(cont.)
The solution is
10  j 56.3 j (3t 90) 10 j (3t 146.3 )
ie 
e
e

e
13
13
The actual response is
10
i  Re ie  
cos(3t  146.3) A
13
The Phasor Concept
A sinusoidal current or voltage at a given frequency
is characterized by its amplitude and phase angle.
i (t )  Re I m e
j ( t    )

 I m cos( t     )
Magnitude
Phase angle
Thus we may write
i (t )  Re I me
j (   )
e
j t

unchanged
The Phasor Concept(cont.)
A phasor is a complex number that represents the
magnitude and phase of a sinusoid.
phasor
I  I me
j (   )
 I m  
The Phasor Concept may be used when the circuit is
linear , in steady state, and all independent sources are
sinusoidal and have the same frequency.
A real sinusoidal current
i (t )  I m cos( t   )  Re I me j (t  ) 
I  I me j  I m
phasor notation
The Transformation
y (t )  Ym cos( t   )  Re Ym e
j ( t  )
Time domain
Transformation
Frequency domain
Y  Ym e  Ym
j

The Transformation (cont.)
Time domain
i (t )  5sin(100t  120)
 5cos(100t  30)
Transformation
Frequency domain
I  530
Example
di
L  Ri  vs
10.6-2
dt
j ( t  )
vs  Vm cos( t   )  Re Vm e

i  I m cos( t   )  Re I m e
j ( t   )
Substitute into 10.6-2
( j LI m  RI m )e
Suppress e
j ( t   )
j t
( j L  R ) I m e
I
j
 Vme
j ( t  )
 Vm e
j
V
V
 I
( j L  R )

Example (cont.)
R  200 , L  2 H,   100 rad / s
V
I
( j 200  200)
Vm0

28345
Vm

  45
283
Vm
 i (t ) 
cos(100t  45) A
283
Phasor Relationship for R, L, and C Elements
Time domain
v  Ri
Frequency domain
V
V  RI or I 
R
Voltage and current are in phase
Resistor
Inductor
Time domain
di
vL
dt
Frequency domain
V
V  j LI or I 
j L
Voltage leads current by 90
Capacitor
Time domain
dv
iC
dt
Frequency domain
I  jCV or V 
Voltage lags current by 90
I
jC
Impedance and Admittance
Impedance is defined as the ratio of the phasor voltage
to the phasor current.
V
Ohm’s law in phasor notation
Z
I
phase     
Vm Vm

   
I m I m
magnitude Z
or
Z  Z 
polar
 Ze
j
exponential
 R  jX
rectangular
Graphical representation of impedance
Z  Z 
Z  R X
1 X
  tan
R
2
Resistor
Inductor
Capacitor
ZR
Z  j L
1
Z
jC
R
L
1/C
2
Admittance is defined as the reciprocal of impedance.
1
1
Y 
 Y  
Z Z 
conductance
In rectangular form
1
1
R  jX
Y 
 2
 G  jB
2
Z R  jX R  X
Resistor
Inductor
Capacitor
1
YG
R
Y
susceptance
G
1
j L
Y  j C
1/L
C
Kirchhoff’s Law using Phasors
KVL
V1  V2  V3 
KCL
I1  I2  I3 
 Vn  0
 In  0
Both Kirchhoff’s Laws hold in the frequency domain.
and so all the techniques developed for resistive circuits hold
Superposition
Thevenin &Norton Equivalent Circuits
Source Transformation
Node & Mesh Analysis
etc.
Impedances in series
Zeq  Z1  Z2  Z3 
 Zn
Admittances in parallel
Yeq  Y1  Y2  Y3 
 Yn
Example 10.9-1
KVL
Z 2  j L  j1
R = 9 , L = 10 mH, C = 1 mF i = ?
RI  Z2I  Z3I  Vs
(9  j1  j10)I  Vs
Vs
1000
I

 7.8645
(9  j 9) 9 2  45
or i  7.86cos(100t  45) A
1
Z3 
  j10
j C
Example 10.9-2
v=?
KCL
V
V
V


 100
10 10  j10  j10
0.1V  (0.05  j 0.05) V  j 0.1V  10
or
10
V
 63.3  18.4
0.15818.4
v  63.3cos(1000t  18.4) V
Node Voltage & Mesh Current using Phasors
is  I m cos  t
C  100 μF, L  5 mH
  1000 rad/s
va = ? vb = ?
1
Z1 
  j10
j C
Y2 
1
1
1

 (1  j )
5 j L 5
Z3  10
KCL at node a
Va Va  Vb

 Is
Z1
Z3
KCL at node b
Vb  Va Vb

0
Z3
Z2
Rearranging
( Y1  Y3 ) Va  (  Y3 ) Vb  I s
(  Y3 ) Va  ( Y2  Y3 ) Vb  0
 Y3   Va   I s 
( Y1  Y3 )
 
 Y



Y2  Y3   Vb   0 
3

Y matrix
Admittance matrix
If Im = 10 A and I s  I m0
Using Cramer’s rule to solve for Va
100(3  2 j )
Va 
4 j
100(3  2 j )(4  j )

17
100

(10  11 j )
17
 87.5  47.7
Therefore the steady state voltage va is
va  87.5cos(1000t  47.7) V
Example 10.10-1
v=?
vs  10cos  t
C  10 mF, L  0.5 H
  10 rad/s
use supernode concept
as in #4
Z L  j L  j5
1
ZC 
  j10
j C
Z3  R3  Z L  5  j5
Example 10.10-1 (cont.)
1
1
Y1 

R1 10
Y2 
1
1
1

 (1  j )
R2 ZC 10
1
1
Y3 
 (5  j5)
Z3 50
KCL at supernode
Y1 ( V  Vs )  Y2 V  Y3  V  10Y1 ( Vs  V)  0
Rearranging
( Y1  Y2  Y3  10Y1Y3 ) V  ( Y1  10Y1Y3 ) Vs
( Y1  10Y1Y3 ) Vs
V 
( Y1  Y2  Y3  10Y1Y3 )
Example 10.10-1 (cont.)
( Y1  10Y1Y3 )100
V
( Y1  Y2  Y3  10Y1Y3 )
10 j
10


63.4
2 j
5
Therefore the steady state voltage v is
10
v
cos(10t  63.4) V
5
Example 10.10-2
i1 = ?
vs  10 2 cos( t  45)
C  5 mF, L  30 mH
  100 rad/s
Z L  j L  j 3
ZC 
1
  j2
jC
Vs  10 245  10  j10
Example 10.10-2 (cont.)
KVL at mesh 1 & 2
(3  j 3)I1  j 3I 2  Vs
(3  j 3)I1  ( j 3  j 2)I2  0
Using Cramer’s rule to solve for I1
(10  j10) j
I1 

where  is the determinant
  (3  j 3)( j )  j 3(3 - j 3)  6  12 j
(10  j10) j
 I1 
 1.0571.6
6  12 j
Superposition, Thevenin & Norton Equivalents
and Source Transformations
Example 10.11-1
i=?
vs  10cos10t V
is  3 A
C  10 mF, L  1.5 H
Vs  100
I s  30
Consider the response to the voltage source acting alone = i1
Example 10.11-2 (cont.)
Substitute
Vs
I1 
5  j L  Z p
ZC R
Zp 
 5(1  j ) and  L  15
R  ZC
100
I1 
5  j15  (5  j5)
10
10


  45
10  j10
200
Example 10.11-2 (cont.)
Consider the response to the current source acting alone = i2
 0
10
I 2   (3)  2 A
15
Using the principle of superposition
i  0.71cos(10t  45)  2 A
Source Transformations
VI
IV
Example 10.11-2
IS = ?
vs  10cos( t  45) V
  100 rad/s
Z s  10  j10
Vs  1045
1045
Is 
20045
10

0
200
Example 10.11-3 Thevenin’s equivalent circuit
?
Z1  1  j
Z2   j
VOC  I s Z1  2 245
Zt  Z1  Z2  1 
Example 10.11-4 Thevenin’s equivalent circuit
V  10I s  200
 VOC  3V  V  800
VO  j10I  4V  ( j10  40)I
 Zt  40  j10 
Example 10.11-4 Norton’s equivalent circuit
?
Z1Z 2
Zt  Z3 
Z1  Z 2
( Z1  Z2 )I  (  Z2 )I SC  Vs
(  Z2 )I  ( Z2  Z3 )I SC  0
Phasor Diagrams
A Phasor Diagram is a graphical representation of phasors
and their relationship on the complex plane.
Take I as a reference phasor
I  I 0
The voltage phasors are
VR  RI  RI 0
VL  j LI   LI 90
j
I
VC 
I
  90
C
C
Phasor Diagrams (cont.)
KVL
Vs  VR  VL  VC
For a given L and C there will be a frequency  that
VL  VC
1
L 
C
Resonant frequency
1
or  
 
LC
2
Resonance
Vs  VR
1
LC
Summary
Sinusoidal Sources
Steady-State Response of an RL Circuit for
Sinusoidal Forcing Function
Complex Exponential Forcing Function
The Phasor Concept
Impedance and Admittance
Electrical Circuit Laws using Phasors