Sinusoidal Steady-state Analysis
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Complex number reviews
Phasors and ordinary differential equations
Complete response and sinusoidal steady-state
response
Concepts of impedance and admittance
Sinusoidal steady-state analysis of simple circuits
Resonance circuit
Power in sinusoidal steady –state
Impedance and frequency normalization
Complex number reviews
Complex number
z  x  jy , j   1
Re( z)  x Im( z)  y
Magnitude
Phase or angle
In polar form
z | z | e
or
j
z | z | 
| z | ( x  y )
2
2 1/ 2
  tan
1
x | z | cos  y | z | sin 
z
y
x
The complex number
can be of voltage, current, power, impedance etc..
in any circuit with sinusoid excitation.
Operations: Add, subtract, multiply, divide, power, root, conjugate
Phasors and ordinary differential equations
A sinusoid of angular frequency  is in the form
Am cos( t   )
Theorem
The algebraic sum of sinusoids of the same frequency and of their
derivatives is also a sinusoid of the same frequency
Example 1
d
2sin 2t
dt
 2 cos 2t cos 60  2sin 2t sin 60  4sin 2t  4 cos 2t
f (t )  2 cos(2t  60)  4sin 2t 
 cos 2t  3 sin 2t  4sin 2t  4 cos 2t
 5cos 2t  (4  3)sin 2t  7.6cos(2t  48.8)
Phasors and ordinary differential equations
Re( Am e jt )  Re( Am e j (t  ) )
 Re( Am cos(t   )  jAm sin(t   ))
 Am cos(t   )  x(t )
A
phasor
Ame j
Example 2
v(t )  110 2 cos(2 60t   / 3)
A  110 2 e j /3
v(t )  Re(110 2 e j (120 t ) )
phasor form
Phasors and ordinary differential equations
Ordinary linear differential equation with sinusoid excitation
a0
dnx
dt
n
 a1
d n 1 x
dt
n 1
 ..  an x  Am cos(t   )........(1)
Lemma: Re[.. ] is additive and homogenous
Re[ z1 (t )  z 2 (t )]  Re[ z1 (t )]  Re[ z 2 (t )]
Re[1 z1 (t )   2 z2 (t )]  1 Re[ z1 (t )]  2 Re[ z2 (t )]
d
d
jt
jt 
Re( Ae )  Re  Ae   Re( j Ae jt )
dt
 dt

Phasors and ordinary differential equations
Application of the phasor to differential equation
A
Let
substitute
0
dn
dt n
dn
dt n
Ame j , X
X me j
x(t )  Re( Xe jt )
in (1) yields
Re( Xe jt )  ..   n Re( Xe jt )  Re( Ae jt )
Re( 0 Xe jt )  ..  Re( n Xe jt )  Re( Ae jt )
Re(0 ( j)n Xe jt )  ..  Re(n Xe jt )  Re( Ae jt )
Phasors and ordinary differential equations


Re [ 0 ( j )n  1 ( j )n 1  ..   n 1 ( j )   n ] Xe jt  Re( Ae jt )
[ 0 ( j ) n  1 ( j ) n 1  ..   n 1 ( j )   n ] X  A
X 
A
[ 0 ( j ) n  1 ( j ) n 1  ..   n 1 ( j )   n ]
Xm 
Am
[( n   n  2  ..)  ( n 1   n 3  ..) ]
2
2
3
even power
 n 1   n 3  ..
    tan
 n   n  2 2  ..
1
3
2
1
2
odd power
Phasors and ordinary differential equations
Example 3
From the circuit in fig1 let the input be a sinusoidal voltage source and the
output is the voltage across the capacitor.
es (t )  Re(Ee jt ) | E | cos(t   )
L
R
+
Fig1
es (t )
+
-
es (t ) | E | cos(t   )
Vc (t )
C
i(t)
-
vc (t ) | Vc | cos(t  )
Phasors and ordinary differential equations
KVL
d
L i (t )  Ri (t )  vc (t )  es (t )
dt
2
LC
d vc (t )
dt 2
dvc (t )
 RC
 vc (t )  es (t )
dt
Particular solution
vc (t )  Re(Vc e jt ) | Vc | cos(t  )
[ LC)( j)2  RC( j)  1]Vc  E
(2)
Phasors and ordinary differential equations
Vc 
E
1   LC  j RC
|E|
| Vc |
[(1   2 LC ) 2  ( RC ) 2 ]1/ 2
1  RC
    tan
1   2 LC
2
Complete response and sinusoidal steady-state
response
Complete reponse
y(t )  yh (t )  y p (t )
y p (t )
yh (t )
yh (t )
= sinusoid of the same input frequency (forced component)
=solution of homogeneous equation (natural component)
n

ki e si t (for distinct frequencies)
i 1

Complete response and sinusoidal steady-state
response
Example 4
For the circuit of fig 1, the sinusoid input es (t )  cos 2t u (t ) is applied
to the circuit at time t  0 . Determine the complete response of the
Capacitor voltage. C=1Farad, L=1/2 Henry, R=3/2 ohms.
iL (0)  I 0  2, vc (0)  1
From example 3
1 d 2 vc (t ) 3 dvc (t )

 vc (t )  cos 2t u (t )
2
2 dt
2 dt
Initial conditions


dv
(
0
)
i
(
0
)
vc (0  )  1, c
 L
2
dt
C
Complete response and sinusoidal steady-state
response
Characteristic equation
2 3
s
 2 s  1  0, s  1,2
2
vh (t )  k1e t  k2e 2t
1
Natural component
Forced component
v p (t )  Re(Ve j 2t ) | V | cos(2t   )
es (t )  Re( Ee j 2t )  cos 2t
3 ( j)  1]V  E
(
j

)

2
2
E
1
 j108.4
V


0.316
e
1  12  2  j 23  1  j3
From (2)
[1
2
Complete response and sinusoidal steady-state
response
The complete solution is
vc (t )  vh (t )  v p (t )
t
 k1e  k2 e
2t
 0.316 cos(2t  108.4 )
vc (0)  1  k1  k2  0.316cos(108.4 )
k1  k2  1.1
d
vc (0)  2  k1  2k2  0.316  2sin(108.4 )
dt
k1  2k2  1.4
k1  3.6
k2  2.5
Complete response and sinusoidal steady-state
response
The complete solution is
vc (t )  3.6et  2.5e2t  0.316cos(2t 108.4 )
Complete response and sinusoidal steady-state
response
Sinusoidal steady-state response
In a linear time invariant circuit driven by a sinusoid source, the response
y (t )
Is of the form
y(t )  k1e
s1t
 k2 e
s2t
 ..  kn e
snt
 Am cos(t  )
Irrespective of initial conditions ,if the natural frequencies lie in
the left-half complex plane, the natural components converge
to zero as t   and the response becomes close to a
sinusoid. The sinusoid steady state response can be calculated
by the phasor method.
Complete response and sinusoidal steady-state
response
Example 5
Let the characteristic polynomial of a differential a differential equation
Be of the form
( s 2  0 2 ) 2  s 4  20 2 s 2  0 4
The characteristic roots are
s1  s2  j0 , s3  s4   j0
and the solution is of the form
yh (t )  (k1  k2t )e j0t  (k3  k4t )e j0t
In term of cosine
yh (t )  k1 cos(0t  1 )  k2t cos(0t  2 )
The solution becomes unstable as t  
Complete response and sinusoidal steady-state
response
Example 6
Let the characteristic polynomial of a differential a differential equation
Be of the form
s 2  02
The characteristic roots are
s1  j0 , s2   j0
and the solution is of the form
y(t )  yh (t )  y p (t )
yh (t )  k1e j0t  k2 e j0t  K cos(0t   )
and
y p (t )  B cos( t  )
The solution is oscillatory at different frequencies. If 
unstable as t  
y(t )  At cos(t   )
 0 the output is
Complete response and sinusoidal steady-state
response
Superposition in the steady state
If a linear time-invariant circuit is driven by two or more sinusoidal
sources the output response is the sum of the output from each source.
Example 7
The circuit of fig1 is applied with two sinusoidal voltage sources and the
output is the voltage across the capacitor.
L
A2m cos(2t  2 ) +
A1m cos(1t  1 ) +
-
i(t)
R
+
C
-
Vc (t )
Phasors and ordinary differential equations
KVL
d
L i (t )  Ri (t )  vc (t )  e1s (t )  e2 s (t )
dt
LC
d 2 vc (t )
dt 2
dvc (t )
 RC
 vc (t )  A1m cos( 1t  1 )  A2m cos( 2 t  2 )
dt
Differential equation for each source
LC
LC
d 2 v p1 (t )
dt
2
d 2 v p 2 (t )
dt
2
 RC
 RC
dv p1 (t )
dt
dv p 2 (t )
dt
 v p1 (t )  A1m cos( 1t  1 )
 v p 2 (t )  A2 m cos( 2 t  2 )
Phasors and ordinary differential equations
The particular solution is
v p (t )  v p1 (t )  v p 2 (t )
 V1m cos(1t  1  1 )  V2 m cos( 2t  2   2 )
where
V1m e j (1 1 ) 
V2 m e
j (2  2 )
A1m e j1
1  12 LC  j1 RC

A2 m e
j2
1  2 LC  j2 RC
2
Complete response and sinusoidal steady-state
response
Summary
A linear time-invariant circuit whose natural frequencies are all within
the open left-half of the complex frequency plane has a sinusoid steady
state response when driven by a sinusoid input. If the circuit has
Imaginary natural frequencies that are simple and if these are different
from the angular frequency of the input sinusoid, the steady-state
response also exists.
The sinusoidal steady state response has the same frequency
as the input and can be obtained most efficiently by the phasor method
Concepts of impedance and admittance
Properties of impedances and admittances play important roles in
circuit analyses with sinusoid excitation.
Phasor relation for circuit elements
Linear time-invariant
circuit in sinusoid
steady state
+
v(t )  Re(Ve jt )
i (t )  Re( Ie jt )
Element
-
v(t )  Re(Ve jt ) | V | cos(t  V )
i(t )  Re( Ie jt ) | I | cos(t  I )
Fig 2
Concepts of impedance and admittance
Resistor
v(t )  Ri (t ), i (t )  Gv(t )
V  RI , I  GV
I V
The voltage and current phasors are in phase.
Capacitor
dv
iC
dt
I  jCV ,
1
V
I
jC
V   I  90 ,  I  V  90 ,
The current phasor leads the voltage phasor by 90 degrees.
Concepts of impedance and admittance
Inductor
di
vL
dt
V  j L I ,
V
I
I  90 ,
1
j L
V
I  V  90 ,
The current phasor lags the voltage phasor by 90 degrees.
Concepts of impedance and admittance
Definition of impedance and admittance
The driving point impedance of the one port
 at the angular frequency
 is the ratio of the output voltage phasor V to the input current phasor I
| Z ( j ) |
or
|V |
,
|I|
Z ( j )  V 
I
v(t ) | Z ( j) || I s | cos(t  Z 
The driving point admittance of the one port
Is )
 at the angular frequency
 is the ratio of the output current phasor I to the input voltage phasor V
|I|
| Y ( j ) |
,
Y ( j )  I  V
|V |
or
i(t ) | Y ( j) || Vs | cos(t  Y  V )
Concepts of impedance and admittance
| Z ( j ) |
1
,
| Y ( j |
Angular frequency 
Resistor
Capacitor
Inductor
Z ( j )   Y ( j )
Z
Y
R
1
G
R
1
jC
jC
j L
1
j L
Sinusoidal steady-state analysis of simple circuits
In the sinusoid steady state Kirchhoff’s equations can be written directly
in terms o voltage phasors and current phasors. For example:
v1 (t )  v2 (t )  v3 (t )  0
If each voltage is sinusoid of the same frequency

V 1mcos(t  1 ) V 2mcos(t  2 ) V 3mcos(t  3 )  0
Re(V 1 e jt )  Re(V 2 e jt )  Re(V 3 e jt )  0
Re(V 1 e jt  V 2 e jt  V 3 e jt )  0
Re[(V 1 V 2 V 3 )e jt ]  0
V 1 V 2 V 3  0
Sinusoidal steady-state analysis of simple circuits
Series parallel connections
I1  I 2  ....  I n  I
In a series sinusoid circuit
V  V1  V2  ....Vn
I
+ V1 -
Z1
+ V2 -
I1
Z2
+
I2
+ Vn -
Zn
In
Fig 3
V
-
Z ( j ) 
n
 Z ( j )
i
i 1
Sinusoidal steady-state analysis of simple circuits
In a parallel sinusoid circuit
V1  V2  ....  Vn  v
I  I1  I 2  ....I n
I
+
Fig 4
V
I1
+
Y1 V
1
-
-
Y ( j ) 
n
 Y ( j )
i
i 1
I2
+
Y2 V
2
-
In
+
Yn Vn
-
Sinusoidal steady-state analysis of simple circuits
Node and mesh analyses
Node and mesh analysis can be used in a linear time-invariant circuit to
determine the sinusoid steady state response. KCL, KVL and the concepts
of impedance and admittance are also important for the analyses.
Example 8

i
(
t
)

10
cos(
2
t

30
)
In figure 5 the input is a current source s
Determine the sinusoid steady-state voltage at node 3
2F
1
1W
2
+
is
1W
v1
-
iC
1W
3
Fig 5
+
2H
iL
v2
-
+
2W
v3
-
Node and mesh analyses
is (t )  10 cos(2t  30 )  Re( I s e j 2t ) or I s  10e j 30
v1 (t )  Re(V1e j 2t ), v2 (t )  Re(V2e j 2t ), v3 (t )  Re(V3e j 2t )
1
1
I L  YLV2 
V2 
V2
jL
j4
IC  YC (V1  V3 )  jC (V1  V3 )V2  j 4(V1  V3 )
KCL at node 1
V1  j 4(V1  V3 )  (V1  V2 )  I s
KCL at node 2
V2
 (V2  V1 )  (V2  V3 )  0
j 4
KCL at node 3
V3
 j 4(V3  V1 )  (V3  V2 )  0
2
Node and mesh analyses
Rearrange the equations
(2  j 4)V1  V2  j 4V3  I s
1
)V2  V3  0
j4
3
 j 4V1  V2  (  j 4)V3  0
2
V1  (2 
By Crammer’s Rule
V3 
2  j4
1
1
1
2
j4
1
 j4
2  j4
1
 j4
Is
0
0
1
1
j4
1
 j4
2
1
3
2
j4
2  j8

Is
6  j11.25
Node and mesh analyses
Since
I s  10e j 30
V3  6.45e j 44
Then
and the sinusoid steady-state voltage at node 3 is
v3 (t )  6.45cos(2t  44 )
Example 9
Solve example 8 using mesh analysis
2F
i2
Fig 6
1W
1W
+
1W
vs
+
-
i1
2H
vL i3
-
+
2W
v3
-
Node and mesh analyses
vs (t )  10 cos(2t  30 )  Re(Vs e j 2t ) or Vs  10e j 30
i1 (t )  Re( I1e j 2t ), i2 (t )  Re( I 2e j 2t ), i3 (t )  Re( I3e j 2t )
VL  Z L iL  j LI L  j 4( I1  I 2 )
1
VC  ZC I 3 
I3
jC
KVL at mesh 1
I1  ( I1  I 2 )  j 4( I1  I3 )  Vs
KVL at mesh 2
1
I 2  ( I 2  I 3 )  j 4( I 3  I1 )  0
j4
KVL at mesh 3
2I3  ( I3  I 2 )  j 4( I3  I1 )  0
Node and mesh analyses
Rearrange the equations
(2  j 4) I1  I 2  j 4I3  Vs
1
) I 2  I3  0
j4
 j 4I1  I 2  (3  j 4) I3  0
 I1  (2 
By Crammer’s Rule
2  j4
1
I3 
 j4
2  j4
1
 j4
1
Vs
1
2
0
j4
1
0
1
1
2
j4
1
2  j8

Vs
12  j 22.5
 j4
1
3  j4
Node and mesh analyses
Since
Then
Vs  10e j 30 and V3  2 I 3
V3  6.45e j 44
and the sinusoid steady-state voltage at node 3 is
v3 (t )  6.45cos(2t  44 )
The solution is exactly the same as from the node analysis
Resonance circuit
Resonance circuits form the basics in electronics and communications. It is
useful for sinusoidal steady-state analysis in complex circuits.
Impedance, Admittance, Phasors
Figure 7 show a simple parallel resonant circuit driven by a sinusoid source.
IC
is (t )  Re( I s e jt )
IR
IL
G=1/R
C
L
Fig 7
is (t )  Re( I s e
+
V
-
jt
Y
) | I s | cos(t  I s )
Resonance circuit
The input admittance at the angular frequency
Y ( j )  G  jC 
 is
1
j L
1
 G  j (C 
)
L
The real part of Y ( j ) is constant but the imaginary part varies with frequency
1
B ( j )  C 
L


1
At the frequency
f0  0  0 
2 2 2 LC
the susceptance is zero. The frequency
f0
is called the resonant frequency.
Resonance circuit
The admittance of the parallel circuit in Fig 7 is frequency dependant
Re[Y ( j )]  G
Im[ Y ( j )]  C 
1
L
Re[ Z ( j )] 
G
Im[ Z ( j )] 
G 2  B 2 ( )
 B ( )
G 2  B 2 ( )
1
1
Z ( j ) 

Y ( j ) G  jB( )
G
 B( )
 2
j 2
2
G  B ( )
G  B 2 ( )
B
C

0
1
L
Susceptance plot
Fig 8
Resonance circuit
1
Y ( j )  G  j (C 
)
L
1
Z ( j ) 
1
G  j (C 
)
L
Im( Z )
Im (Y )
Admittance Plane


  0
|Y |
Y
G
0
Locus of Y
Re(Y )
0

  0
Re( Z )
1
2G
Fig 9
Locus of Z
1
G
Resonance circuit
The currents in each element are
I R  GV , I L 
and
1
jL
I R  I L  IC  I S
If for example
V , I C  jCV
is (t )  cost  Re(I s e ) R  1W,
jt
I s e j 0 ,   1rad / s
The admittance of the circuit is
Y ( j1)  1  j (1  4)  1  3 j  10e  j 71.6
The impedance of the circuit is
1
1 j 71.6
Z ( j1) 

e
Y ( j1)
10
1
L  H , C  1F
4
Resonance circuit
The voltage phasor is
1
V  Z ( j1) I s 
Thus
IR 
1
10
10
e j 71.6
e j 71.6 , I L 
Im
1
10
e j18.4 , IC 
1
10
e j161.6
V
Fig 10
IC
IR
IS
Re
IL
Resonance circuit
Similarly if
0 
1
 2 rad / s
is (t )  cos 2t  Re( I s e j 2t )
and
LC
I s e j 0 ,   2 rad / s
Y ( j 2)  1, Z ( j 2)  1, V  1
I R  1, I L  2e j 90 , I C  2e j 90
The voltage and current phasors are
Note that it is a resonance and
Im
I L  I s , IC  I s
IS
IC
IR  Is
IL
V
Re
Fig 11
Resonance circuit
The ratio of the current in the inductor or capacitor to the input current is
the quality factor or Q-factor of the resonance circuit.
| I L | | IC |

 Qp
| Is | | Is |
Rp
L
Generally Q  1 and the voltages or currents in a resonance circuit is very large!
Analysis for a series R-L-C resonance is the very similar
1
Z ( j )  R  j (L 
)
C
VS  VR  VL  VC
Vs
I s ( j ) 
Z ( j )
| VL | | VC |

 Qs
| Vs | | Vs |
L
Rs
Power in sinusoidal steady-state
The instantaneous power enter a one port circuit
p(t )  v(t )i(t )
The energy delivered to the
t
W (t o , t ) 
 in the interval (to , t ) is
 p(t ' )dt'
to
i (t )
+
Generator

v(t )
-
Fig 12

is
Power in sinusoidal steady-state
Instantaneous, Average and Complex power
In sinusoidal steady-state the power at the port  is
v(t )  Vm cos(t  V )  Re(Ve jt )
where
V
Vme j
V
, Vm | V |
If the port current is
i(t )  I m cos(t  I )  Re( Ie jt )
where
I
I me j I , I m | I |
Power in sinusoidal steady-state
Then
p(t )  v(t )i (t )
 Vm I m cos(t  V ) cos(t  I )
 12 Vm I m cos( V  I )  12 Vm I m cos(2t  V  I )
T
Pav

1
p (t ')dt '  12 Vm I m cos( V  I )
t
0
p(t )
Pav
Fig 13
v(t )
i (t )
Power in sinusoidal steady-state

Remarks



The phase difference in power equation is the impedance
angle
Pav is the average power over one period and is non
negative. But p(t) may be negative at some t
1
The complex power in a two-port circuit is P
V I  Vrms I rms
P

1
2
1
2
| V || I | e
2
j( V  I )
| V || I | cos( V  I )  j 12 | V || I | sin( V  I )
Pav  Re( P)  Re( 12 V I )  Re(Vrms I rms )
Pav 
1
2
|
I
|
Re[ Z ( j )] 
2
2
1
2
|
V
|
Re[Y ( j )]
2
 I rms Re[ Z ( j )]  Vrms Re[Y ( j )]
2
 Average power is additive
Power in sinusoidal steady-state

Maximum power transfer
The condition for maximum transfer for sinusoid steady-state is that
The load impedance must be conjugately matched to the source
imedance
ZL  ZS

Pav max
| Vs2 |

8Rs
| Vs2 |
Ps 
4 Rs
Q of a resonance circuit
For a parallel resonance circuit
Q
R
 0CR
0 L
 0
1
2
C
|
V
|
2
1
2
G
|
V
|
2
Q  0
energy stored
average power dissipated inthe resistor
(Valid for both series and parallel resonance circuits)
Impedance and frequency normalization
In designing a resonance circuit to meet some specification component
values are usually express in normalized form.
From
0
1
R
Z 0  R, 0 
, 3  db bandwidth 

Q
L
LC
R

 1
L
, C  2 and


0 Q
0 R
Let the normalized component values are
Then
1
R0  1, L0  Q, C0 
Q
R  Z0 , L 
QZ 0
0

Z0
1

,C


Q0 Z 0 Z 002
Impedance and frequency normalization

Popularity of normalized design:


The circuit design can be made at any impedance level and
center frequency
Well-known solutions exist
Let
rn 
desired impedance level
impedance level of normalized design
desired typical frequency
Wn 
typical frequency of normalized design
Then
rn
C0
R  rn R0 , L 
L0 , C 
Wn
rn W n
Impedance and frequency normalization
Example
Fig. 14 shows a low pass filter whose transfer impedance
E2
I1
2

1
1 6
E2
I1
1.333H
1
+
I1
1.5F
0.5F 1W
E2
1
2
0
The gain of the filter is 1 at
  0 And 1 /
2 at
1
Design the circuit to have an impedance of 600 ohms at
and equal to 1/
1
0
2 at 3.5 kHz then
rn  600 and
Wn  2  3.5 103  2.199 104

Rad/s
Impedance and frequency normalization
R  rn R0  600
rn
600
L
L0 
1.333  36.37 mH
4
Wn
2.199  10
C01
1.5 F
C1 

 0.1137  F
4
rn W n 600  2.199  10
C02
0.5 F
C2 

 0.0379  F
4
rn W n 600  2.199  10
Impedance and frequency normalization
E2
I1
36.37mH
+
I1
0.11378F
6 00W
0.0379F
E2
600
600
2
0
Designed circuit
2x3500

Rad/s