sinusoidal steady-state analysis complex number reviews phasors and ordinary differential equations...
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Sinusoidal Steady-state Analysis 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
1, jjyxz
yzxz )Im()Re(
| | jz z e
Complex number
In polar form2 2 1/ 2 1| | ( ) tan
yz x y
x
| |z z or | | cos | | sinx z y z The complex number can be of voltage, current, power, impedance etc..in any circuit with sinusoid excitation.
z
Magnitude
Phase or angle
Operations: Add, subtract, multiply, divide, power, root, conjugate
Phasors and ordinary differential equationsA sinusoid of angular frequency is in the form
)cos( tAm
Theorem
The algebraic sum of sinusoids of the same frequency and of their derivatives is also a sinusoid of the same frequency
Example 1
( ) 2cos(2 60) 4sin 2 2sin 2
2cos 2 cos60 2sin 2 sin 60 4sin 2 4cos 2
cos 2 3 sin 2 4sin 2 4cos 2
5cos 2 (4 3)sin 2 7.6cos(2 48.8)
df t t t t
dtt t t t
t t t t
t t t
Phasors and ordinary differential equations
jmA A e
)3/602cos(2110)( ttv
phasor form
Example 2
/ 3110 2 jA e (120 )( ) Re(110 2 )j tv t e
phasor
( )Re( ) Re( )
Re( cos( ) sin( ))
cos( ) ( )
j t j tm m
m m
m
A e A e
A t jA t
A t x t
Phasors and ordinary differential equationsOrdinary linear differential equation with sinusoid excitation
1
0 1 1.. cos( )........(1)
n n
n mn n
d x d xa a a x A t
dt dt
Lemma: Re[.. ] is additive and homogenous
)](Re[)](Re[)]()(Re[ 2121 tztztztz
1 1 2 2 1 1 2 2Re[ ( ) ( )] Re[ ( )] Re[ ( )]z t z t z t z t
Re( ) Re Re( )j t j t j td dAe Ae j Ae
dt dt
Phasors and ordinary differential equationsApplication of the phasor to differential equation
,j jm mA A e X X e Let
substitute ( ) Re( )j tx t Xe in (1) yields
0 Re( ) .. Re( ) Re( )n
j t j t j tnn
dXe Xe Ae
dt
0Re( ) .. Re( ) Re( )n
j t j t j tnn
dXe Xe Ae
dt
0Re( ( ) ) .. Re( ) Re( )n j t j t j tnj Xe Xe Ae
Phasors and ordinary differential equations 10 1 1Re [ ( ) ( ) .. ( ) ] Re( )n n j t j t
n nj j j Xe Ae
12
10 1 1
10 1 1
2 2 3 22 1 3
[ ( ) ( ) .. ( ) ]
[ ( ) ( ) .. ( ) ]
[( .. ) ( ..) ]
n nn n
n nn n
mm
n n n n
j j j X A
AX
j j j
AX
even power odd power3
1 1 32
2
..tan
..n n
n n
Phasors and ordinary differential equations
)cos(||)Re()( tEEete tjs
Example 3
From the circuit in fig1 let the input be a sinusoidal voltage source and the output is the voltage across the capacitor.
+-
( ) | | cos( )se t E t
L R
C
+
-
( ) | | cos( )c cv t V t
( )se t( )cV t
i(t)
Fig1
Phasors and ordinary differential equations
)()()()( tetvtRitidt
dL sc
2
2
( ) ( )( ) ( )c c
c sd v t dv t
LC RC v t e tdtdt
)cos(||)Re()( tVeVtv ctj
cc
KVL
2[ )( ) ( ) 1] (2)cLC j RC j V E
Particular solution
Phasors and ordinary differential equations
2
2 2 2 1/ 2
12
1
| || |
[(1 ) ( ) ]
tan1
c
c
EV
LC j RC
EV
LC RC
RC
LC
Complete response and sinusoidal steady-state responseComplete reponse
)()()( tytyty ph
)(ty p = sinusoid of the same input frequency (forced component)
( )hy t =solution of homogeneous equation (natural component)
1
( ) i
ns t
h ii
y t k e
(for distinct frequencies)
Complete response and sinusoidal steady-state responseExample 4
)(2cos)( tuttes For the circuit of fig 1, the sinusoid input is applied to the circuit at time . Determine the complete response of theCapacitor voltage. C=1Farad, L=1/2 Henry, R=3/2 ohms.
0t
1)0(,2)0( 0 cL vIi
2
2
( ) ( )1 3( ) cos 2 ( )
2 2c c
cd v t dv t
v t t u tdtdt
From example 3
Initial conditions
2)0()0(
,1)0(
C
i
dt
dvv Lcc
Complete response and sinusoidal steady-state responseCharacteristic equation
2,1,01232
21 sss
tth ekektv 2
21)( 2( ) Re( ) | | cos(2 )j t
pv t Ve V t
2( ) Re( ) cos 2j tse t Ee t
From (2) 2 312 2[ ( ) ( ) 1]j j V E
108.42 31
2 2
10.316
1 31jE
V ejj
Natural component
Forced component
Complete response and sinusoidal steady-state responseThe complete solution is
21 2
( ) ( ) ( )
0.316cos(2 108.4 )
c h p
t t
v t v t v t
k e k e t
1 2
1 2
(0) 1 0.316cos( 108.4 )
1.1cv k k
k k
1 2
1 2
(0) 2 2 0.316 2sin( 108.4 )
2 1.4
cdv k k
dtk k
6.31 k 2 2.5k
Complete response and sinusoidal steady-state response
2( ) 3.6 2.5 0.316cos(2 108.4 )t tcv t e e t
The complete solution is
Complete response and sinusoidal steady-state responseSinusoidal steady-state response
)(ty
In a linear time invariant circuit driven by a sinusoid source, the response
Is of the form
1 21 2( ) .. cos( )ns ts t s t
n my t k e k e k e A t Irrespective of initial conditions ,if the natural frequencies lie in the left-half complex plane, the natural components convergeto zero as and the response becomes close to a sinusoid. The sinusoid steady state response can be calculated by the phasor method.
t
Complete response and sinusoidal steady-state response
40
220
4220
2 2)( sss
1 2 0 3 4 0,s s j s s j
0 01 2 3 4( ) ( ) ( )j t j t
hy t k k t e k k t e
Example 5
Let the characteristic polynomial of a differential a differential equationBe of the form
The characteristic roots are
and the solution is of the form
1 0 1 2 0 2( ) cos( ) cos( )hy t k t k t t In term of cosine
The solution becomes unstable as t
Complete response and sinusoidal steady-state response
0 01 2 0( ) cos( )j t j t
hy t k e k e K t
t0
220s
1 0 2 0,s j s j
( ) ( ) ( )h py t y t y t
Example 6
Let the characteristic polynomial of a differential a differential equationBe of the form
The characteristic roots are
and the solution is of the form
( ) cos( )py t B t and
The solution is oscillatory at different frequencies. If the output is unstable as ( ) cos( )y t At t
Complete response and sinusoidal steady-state responseSuperposition in the steady state
If a linear time-invariant circuit is driven by two or more sinusoidalsources 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.
+-
1 1 1cos( )mA t
L R
C
+
-
( )cV t
i(t)
+-
2 2 2cos( )mA t
Phasors and ordinary differential equations
1 2( ) ( ) ( ) ( ) ( )c s sd
L i t Ri t v t e t e tdt
2
1 1 1 2 2 22
( ) ( )( ) cos( ) cos( )c c
c m md v t dv t
LC RC v t A t A tdtdt
21 1
1 1 1 12
( ) ( )( ) cos( )
p pp m
d v t dv tLC RC v t A t
dtdt
KVL
Differential equation for each source
22 2
2 2 2 22
( ) ( )( ) cos( )
p pp m
d v t dv tLC RC v t A t
dtdt
Phasors and ordinary differential equations
)cos()cos(
)()()(
22221111
21
tVtV
tvtvtv
mm
ppp
The particular solution is
11 1
22 2
( ) 11 2
1 1
( ) 22 2
2 2
1
1
jj m
m
jj m
m
A eV e
LC j RC
A eV e
LC j RC
where
Complete response and sinusoidal steady-state response
Summary
A linear time-invariant circuit whose natural frequencies are all withinthe 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 differentfrom the angular frequency of the input sinusoid, the steady-stateresponse also exists.
The sinusoidal steady state response has the same frequencyas the input and can be obtained most efficiently by the phasor method
Concepts of impedance and admittanceProperties of impedances and admittances play important roles in circuit analyses with sinusoid excitation.
Linear time-invariant circuit in sinusoid
steady stateElement
( ) Re( )j ti t Ie
( ) Re( )j tv t Ve
+
-
Phasor relation for circuit elements
( ) Re( ) | | cos( )j tv t Ve V t V
( ) Re( ) | | cos( )j ti t Ie I t I
Fig 2
Concepts of impedance and admittance
( ) ( ), ( ) ( )
,
v t Ri t i t Gv t
V RI I GV
I V
Resistor
The voltage and current phasors are in phase.
Capacitor
1,
90 , 90 ,
dvi C
dt
I j CV V Ij C
V I I V
The current phasor leads the voltage phasor by 90 degrees.
Concepts of impedance and admittanceInductor
1,
90 , 90 ,
div L
dt
V j L I I Vj L
V I I V
The current phasor lags the voltage phasor by 90 degrees.
Concepts of impedance and admittanceDefinition 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
| || ( ) | , ( )
| |
VZ j Z j V I
I
( ) | ( ) || | cos( )s sv t Z j I t Z I or
The driving point admittance of the one port at the angular frequency is the ratio of the output current phasor I to the input voltage phasor V
| || ( ) | , ( )
| |
IY j Y j I V
V
( ) | ( ) || | cos( )si t Y j V t Y V or
Concepts of impedance and admittance 1
| ( ) | , ( ) ( )| ( |
Z j Z j Y jY j
Angular frequency Z Y
Resistor
Capacitor
Inductor
RG
1
1
j Cj C
j L1
j L
R
Sinusoidal steady-state analysis of simple circuits
0)()()( 321 tvtvtv
0)cos()cos()cos( 332211 tVtVtV mmm
In the sinusoid steady state Kirchhoff’s equations can be written directlyin terms o voltage phasors and current phasors. For example:
If each voltage is sinusoid of the same frequency
1 2 3
1 2 3
1 2 3
1 2 3
Re( ) Re( ) Re( ) 0
Re( ) 0
Re[( ) ] 0
0
j t j t j t
j t j t j t
j t
V e V e V e
V e V e V e
V V V e
V V V
Sinusoidal steady-state analysis of simple circuits
n
n
VVVV
IIII
....
....
21
21
Series parallel connections
In a series sinusoid circuit
Z1 ZnZ2I
1I 2I nI1V+ - 2V - ++ nV -
+
-
V
1
( ) ( )n
ii
Z j Z j
Fig 3
Sinusoidal steady-state analysis of simple circuits
1 2
1 2
....
....n
n
V V V v
I I I I
In a parallel sinusoid circuit
Y1 YnY2
I
1I 2I nI
1V+
-2V
-
++nV
-
+
-
V
1
( ) ( )n
ii
Y j Y j
Fig 4
Sinusoidal steady-state analysis of simple circuitsNode and mesh analyses
)302cos(10)( ttis
Node and mesh analysis can be used in a linear time-invariant circuit todetermine the sinusoid steady state response. KCL, KVL and the conceptsof impedance and admittance are also important for the analyses.
Example 8
Fig 5
In figure 5 the input is a current source Determine the sinusoid steady-state voltage at node 3
1 32
2FCi
-
+++
--1v 2v 3v1W
1W 1W
2H 2W
Lisi
Node and mesh analyses2 30( ) 10cos(2 30 ) Re( ) 10j t j
s s si t t I e or I e
)Re()(),Re()(),Re()( 233
222
211
tjtjtj eVtveVtveVtv
1 1 3 1 24( ) ( ) sV j V V V V I
1 3 1 3 2 1 3( ) ( ) 4( )C CI Y V V j C V V V j V V
222 4
11V
jV
LjVYI LL
KCL at node 1
KCL at node 2 22 1 2 3( ) ( ) 0
4
VV V V V
j
KCL at node 3 33 1 3 24( ) ( ) 0
2
Vj V V V V
Node and mesh analysesRearrange the equations
1 2 3(2 4) 4 sj V V j V I
1 2 31
(2 ) 04
V V Vj
1 2 33
4 ( 4) 02
j V V j V By Crammer’s Rule
3
32
2 4 1
11 2 0
4
4 1 0 2 8
6 11.252 4 1 4
11 2 1
4
4 1 4
s
s
j I
j
j jV I
jj j
j
j j
Node and mesh analyses3010 j
sI e 44
3 6.45 jV e
3 ( ) 6.45cos(2 44 )v t t
Since
and the sinusoid steady-state voltage at node 3 is
Then
Example 9Solve example 8 using mesh analysis
2F
2i
-
++
-Lv 3v
1W
1W 1W
2H 2Wsv 1i 3i
+
-
Fig 6
Node and mesh analyses2 30( ) 10cos(2 30 ) Re( ) 10j t j
s s sv t t V e or V e
2 2 21 1 2 2 3 3( ) Re( ), ( ) Re( ), ( ) Re( )j t j t j ti t I e i t I e i t I e
1 1 2 1 3( ) 4( ) sI I I j I I V
3 31
C CV Z I Ij C
1 24( )L L L LV Z i j LI j I I
KVL at mesh 1
KVL at mesh 2
3 3 2 3 12 ( ) 4( ) 0I I I j I I KVL at mesh 3
2 2 3 3 11
( ) 4( ) 04I I I j I I
j
Node and mesh analyses
1 2 3(2 4) 4 sj I I j I V
1 2 31
(2 ) 04
I I Ij
1 2 34 (3 4) 0j I I j I By Crammer’s Rule
3
2 4 1
11 2 0
4
4 1 0 2 8
12 22.52 4 1 4
11 2 1
4
4 1 3 4
s
s
j V
j
j jI V
jj j
j
j j
Rearrange the equations
Node and mesh analyses
303 310 2j
sV e and V I
443 6.45 jV e
3 ( ) 6.45cos(2 44 )v t t
Since
and the sinusoid steady-state voltage at node 3 is
Then
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.
( ) Re( )j ts si t I e C L
G=1/R
+
-
V
Y
CI LI RI
Impedance, Admittance, Phasors
Fig 7
Figure 7 show a simple parallel resonant circuit driven by a sinusoid source.
( ) Re( ) | | cos( )j ts s s si t I e I t I
Resonance circuit
1( )
1( )
Y j G j Cj L
G j CL
( )Y j
The input admittance at the angular frequency is
The real part of is constant but the imaginary part varies with frequency
1( )B j C
L
At the frequency 0 00
1
2 2 2f
LC
the susceptance is zero. The frequency is called the resonant frequency.0f
Resonance circuit
LCjY
GjY
1
)](Im[
)](Re[
2 2
2 2
Re[ ( )]( )
( )Im[ ( )]
( )
GZ j
G B
BZ j
G B
2 2 2 2
1 1( )
( ) ( )
( )
( ) ( )
Z jY j G jB
G Bj
G B G B
The admittance of the parallel circuit in Fig 7 is frequency dependant
B
C
1
L
0
Susceptance plot
Fig 8
Resonance circuit
0
GRe( )Y
Im( )Y
| |YY
0
Admittance Plane
1( ) ( )Y j G j C
L
1( )
1( )
Z jG j C
L
Locus of Y
Re( )Z
0
Im( )Z
1
2G
1
G
0
Locus of Z
Fig 9
Resonance circuit
CVjIVLj
IGVI CLR
,1
,
SCLR IIII
)Re(cos)( jtss eItti
The currents in each element are
and
If for example1
1 , , 14
R L H C F W 0 , 1 /j
sI e rad s
The admittance of the circuit is
71.6( 1) 1 (1 4) 1 3 10 jY j j j e
The impedance of the circuit is71.61 1
( 1)( 1) 10
jZ j eY j
Resonance circuit
71.61( 1)
10j
sV Z j I e
71.6 18.4 161.61 1 1, ,
10 10 10j j j
R L CI e I e I e
The voltage phasor is
Thus
Re
ImV
LI
CIRI
SI
Fig 10
Resonance circuitsrad
LC/2
10 2( ) cos 2 Re( )j t
s si t t I e
0 , 2 /jsI e rad s
Similarly if and
( 2) 1, ( 2) 1, 1Y j Z j V 90 901, 2 , 2j j
R L CI I e I e
The voltage and current phasors are
Re
Im
V
LI
CI
R sI ISI
Note that it is a resonance and
,L s C sI I I I
Fig 11
Resonance circuit
| || |
| | | |pCL
ps s
RIIQ
I I L
1Q
)1
()(C
LjRjZ
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.
Generally and the voltages or currents in a resonance circuit is very large!
Analysis for a series R-L-C resonance is the very similar
( )( )s
sV
I jZ j
S R L CV V V V | || |
| | | |CL
ss s s
VV LQ
V V R
Power in sinusoidal steady-state
)()()( titvtp
The instantaneous power enter a one port circuit is
The energy delivered to the in the interval is
t
t
o
o
dttpttW ')'(),(
),( tto
Generator
( )i t
( )v t+
-
Fig 12
Power in sinusoidal steady-stateInstantaneous, Average and Complex power
In sinusoidal steady-state the power at the port is
( ) cos( ) Re( )j tmv t V t V Ve
, | |j Vm mV V e V V
where
If the port current is
( ) cos( ) Re( )j tmi t I t I Ie
, | |j Im mI I e I I
where
Power in sinusoidal steady-state
m
1 1m m2 2
( ) ( ) ( )
cos( )cos( )
cos( ) cos(2 )
m
m m
p t v t i t
V I t V t I
V I V I V I t V I
1m2
0
1( ') ' cos( )
T
av mP p t dt V I V It
)(tp
Then
avP)(tv
)(tiFig 13
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
The complex power in a two-port circuit is 12 rms rmsP V I V I
( )12
1 12 2
| || |
| || | cos( ) | || | sin( )
j V IP V I e
V I V I j V I V I
12
Re( ) Re( ) Re( )av rms rmsP P V I V I
2
2 21 12 2
2
| | Re[ ( )] | | Re[ ( )]
Re[ ( )] Re[ ( )]rms rms
avP I Z j V Y j
I Z j V Y j
Average power is additive
Power in sinusoidal steady-state Maximum power transfer
SL ZZ
The condition for maximum transfer for sinusoid steady-state is thatThe load impedance must be conjugately matched to the source imedance
s
sav R
VP
8
|| 2
max 2| |
4s
ss
VP
R
Q of a resonance circuit
00
212
0 212
| |
| |
RQ CR
L
C V
G V
For a parallel resonance circuit
0energy stored
Qaverage power dissipated in the 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.
L
R
Qbandwidthdb
LCRZ 0
00 3,1
,
From
200
1,
RL C and
QR
Let the normalized component values are
0 0 01
1, ,R L Q CQ
Then0 0
0 20 0 0 0 0
1, ,
QZ ZR Z L C
Q Z Z
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
ndesired impedance level
rimpedance level of normalized design
Let
ndesired typical frequency
typical frequency of normalized designW
Then
nnn
nn r
CCL
rLRrR
W
W 0
00 ,,
Impedance and frequency normalizationExample
6
2
1
2
1
1
I
E
Fig. 14 shows a low pass filter whose transfer impedance
1I1.5F 0.5F
1.333H
1W+
-
E2
2
1
E
I
1
2
10Rad/s
1
The gain of the filter is 1 at 0 And at 1 Design the circuit to have an impedance of 600 ohms at 0 and equal to 1/ 2 at 3.5 kHz then
600nr 3 42 3.5 10 2.199 10n W and
2/1
Impedance and frequency normalization
0
0 4
011 4
022 4
600
6001.333 36.37
2.199 10
1.50.1137
600 2.199 10
0.50.0379
600 2.199 10
n
n
n
n n
n n
R r R
rL L mH
C FC F
r
C FC F
r
W
W
W