ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENT CONSTANT INDEPENDENT SOURCES A STEP-BY-STEP APPROACH THIS APPROACH RELIES ON THE KNOWN FORM OF THE SOLUTION BUT FINDS THE CONSTANTS USING BASIC CIRCUIT ANALYSIS TOOLS AND FORGOES THE DETERMINATION OF THE DIFFERENTIAL EQUATION MODEL , , 2 1 K K 0 , ) ( 2 1 t e K K t x t stat steady in circuit the analyzing determined be can and variable the of value state steady the is 1 K 2 1 , ) 0 ( K K x constants the compute to equation second the provides and condition initial the is element storing energy the across Thevenin using determined be can and constant time the is
ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENT CONSTANT INDEPENDENT SOURCES A STEP-BY-STEP APPROACH THIS APPROACH RELIES ON THE KNOWN FORM OF THE.
Dec 17, 2015
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ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENTCONSTANT INDEPENDENT SOURCES
A STEP-BY-STEP APPROACH
THIS APPROACH RELIES ON THE KNOWN FORM OF THE SOLUTIONBUT FINDS THE CONSTANTS USING BASIC CIRCUIT ANALYSIS TOOLS AND FORGOES THE DETERMINATION OF THE DIFFERENTIAL EQUATION MODEL
,, 21 KK
0,)( 21 teKKtx
t
statesteady incircuit the analyzing determined be
can and variablethe of valuestatesteady the is 1K
21,
)0(
KK
x
constants the compute to equation
second the provides and condition initial the is
element storingenergy the across Thevenin using
determined be can andconstant time the is
Obtaining the time constant: A General Approach
Circuitw ith
resistancesand
sources
I nductororCapacitor
a
b
R epresentation of an arbitrarycircuit w ith one storage elem ent
TheveninVTH
R TH
I nductororCapacitor
a
b
VTH
R TH a
b
C
+
vc
_
Case 1.1Voltage across capacitor
VTH
R TH a
b
L
iL
Case 1.2Current through inductor
KCL@ node a
ciRi0 Rc ii
dt
dvCi C
c
TH
THCR R
vvi
0
TH
THCC
R
vv
dt
dvC
THCC
TH vvdt
dvCR
Use KVL
Rv
Lv
THLR vvv
LTHR iRv
dt
diLv L
L
THLTHL viRdt
diL
TH
THL
L
TH R
vi
dt
di
R
LSCi
CIRCUITS WITH ONE ENERGY STORING ELEMENT
TH
TH
R
L
CR
Circuit Inductive
Circuit Capacitive
THE STEPS
)0();(
0,)(
211
21
xKKxK
teKKtxt
STEP 1. THE FORM OF THE SOLUTION
STEP 2: DRAW THE CIRCUIT IN STEADYSTATE PRIOR TO THE SWITCHING ANDDETERMINE CAPACITOR VOLTAGE ORINDUCTOR CURRENT
STEP 3: DRAW THE CIRCUIT AT 0+THE CAPACITOR ACTS AS A VOLTAGESOURCE. THE INDUCTOR ACTS AS ACURRENT SOURCE.DETERMINE THE VARIABLE AT t=0+
STEP 4: DRAW THE CIRCUIT IN STEADYSTATE AFTER THE SWITCHING ANDDETERMINE THE VARIABLE IN STEADYSTATE.
)0( x DETERMINE
)(x DETERMINE
STEP 5: DETERMINE THE TIME CONSTANT
inductor one withcircuit
capacitor one withcircuit
TH
TH
R
L
CR
STEP 6: DETERMINE THE CONSTANTS K1, K2
)0(),( 211 xKKxK
0),( tti FIND
0,)( 21 teKKti
t :1 STEP
USE CIRCUIT IN STEADY STATEPRIOR TO THE SWITCHING
mAk
Vi 212
24
][32)2)(2(36)0( VkmAVvc
)0()0( cC vv
)0( i Determine :3 STEP
USE A CIRCUIT VALID FOR t=0+.THE CAPACITOR ACTS AS SOURCE
mAk
Vi
3
16
6
32)0(
capacitor across voltageInitial :2 STEP
NOTES FOR INDUCTIVE CIRCUIT(1)DETERMINE INITIAL INDUCTOR CURRENT IN STEP 2(2)FOR THE t=0+ CIRCUIT REPLACE INDUCTOR BY A CURRENT SOURCE
LEARNING EXAMPLE
)0( i
KVL
constant time Determine :5 STEPCRTH :circuit Capacitive
kkkRTH 5.16||2 FC 100
sF 15.0)10100)(105.1( 63
21,KK Determine :6 STEP
0,)( 21 teKKti
t 1) (STEP
mAi3
16)0(( 3) STEP 21 KK
mAi8
36)( 4) (STEP 1K
)(i Determine :4 STEP
USE CIRCUIT IN STEADY STATEAFTER SWITCHING
mAi8
36)( 0,
6
5
8
36)( 15.0
teti
t
WERFINAL ANS
NOTE: FOR INDUCTIVE CIRCUIT
THR
L
ORIGINAL CIRCUIT
6
5
8
36
3
162 K
USING MATLAB TO DISPLAY FINAL ANSWER
0,6
5
8
36
02)(
15.0 te
tmAti t » help linspace
LINSPACE Linearly spaced vector. LINSPACE(x1, x2) generates a row vector of 100 linearly equally spaced points between x1 and x2. LINSPACE(x1, x2, N) generates N points between x1 and x2. See also LOGSPACE, :.
%example6p3.m%commands used to display funtion i(t)%this is an example of MATLAB script or M-file%must be stored in a text file with extension ".m”%the commands are executed when the name of the M-file is typed at the%MATLAB prompt (without the extension)tau=0.15; %define time constanttini=-4*tau; %select left starting pointtend=10*tau; %define right end pointtminus=linspace(tini,0,100); %use 100 points for t<=0tplus=linspace(0,tend, 250); % and 250 for t>=0iminus=2*ones(size(tminus)); %define i for t<=0iplus=36/8+5/6*exp(-tplus/tau); %define i for t>=0plot(tminus,iminus,'ro',tplus,iplus,'bd'), grid; %basic plot command specifying %color and markertitle('VARIATION OF CURRENT i(t)'), xlabel('time(s)'), ylabel('i(t)(mA)')legend('prior to switching', 'after switching')axis([-0.5,1.5,1.5,6]);%define scales for axis [xmin,xmax,ymin,ymax]
Command used to define linearly spaced arrays
Script (m-file) with commands used. Prepared with the MATLAB Editor
0),( ttv FIND
0,)( 21 teKKtv
t :1 STEP
currentinductor Initial :2 STEP
Use circuit in steady stateprior to switching
kk 3||6
mAI 46
241 136
6)0( IiL
mAiL 3
8)0(
)0( v Determine :3 STEP
Use circuit at t=0+.Inductor is replaced bycurrent source
1V
03
8
1264
24 111 VVV ][
3
201 VV
][3
52][24)0( 1 VVVv
LEARNING EXAMPLE
)(v DETERMINE :4 STEP
USE CIRCUIT IN STEADY STATEAFTER SWITCHING
][24)( Vv
CONSTANT TIME DETERMINE :5 STEP
12||6||4THR
2THR
THR
L :Circuit Inductive
sH
22
4
21,KK DETERMINE :6 STEP4) (step ][24)(1 VvK
3) (step 213
52)0( KKv
][3
2024
3
521 VK
0,243
20)( 2
tetv
t
:ANS
ORIGINAL CIRCUIT
0),( ttvO FIND
0,)( 21 teKKtv
t
o :1 STEP
VOLTAGECAPACITORINITIAL :2 STEP
)0(Cv
0t
1v 2v
2
12
12)0(
)12(3
1
vv
vv
C
V12
022
8
3
12: 1111
vvvv @KCL 6/*
][60488 11 Vvv ][10)0()0(][22 VvvVv CC
)0( Ov DETERMINE :3 STEP
0t
)0(Ov
V10
][5)10(22
2)0( VvO
LEARNING EXTENSION
)(Ov DETERMINE :4 STEP
)(Ov
][5
24)12(
5
2)( VvO
CONSTANT TIME DETERMINE :5 STEPCRTH :Circuit Capacitive
THR5
44||1THR
sFC5
82
21,KK DETERMINE :6 STEP
][5
24)(1 VvK O
][5
1][5)0( 221 VKKKVvO
0];[5
1
5
24)( 5
8
tVetv
t
O :ANS
ORIGINAL CIRCUIT
0,)( 21 teKKtv
t
O :1 STEP
CURRENT INDUCTORINITIAL :2 STEP
)0( LiOv0
22
)4(
2
12 OOO vvv
][3
4)0()0(
][3
8
Aii
Vv
LL
O
)0( Ov DETERMINE :3 STEP
)0( Li
][3
8)0(2)0( Viv LO
)0( Ov
0),( ttvO FIND
)(Ov DETERMINE :4 STEP
ORIGINAL CIRCUIT
)(Ov
][6)12(22
2)( VvO
CONSTANT TIME DETERMINE :5 STEP
THR
L
Circuit Inductive
THR
4THR
s5.04
2
21,KK DETERMINE :6 STEP
4) (step ][6)(1 VvK O
3) (step 213
8)( KKvO
][3
106
3
82 VK
0,3
106)( 5.0
tetv
t
O :ANS
0),( ttvO FIND0,)( 21
teKKtv
t
O :1 STEP
)0( Li DETERMINE :2 STEP
AiA 3
][6V
][18V
][3)0()0( Aii LL
)0( Ov DETERMINE :3 STEP
][18)0( VvO
LEARNING EXAMPLE
)(Ov DETERMINE :4 STEP
ORIGINAL CIRCUIT
Bv
06
)2(
42
36 '
ABBB ivvv
4' BA
vi
12/*
636411 ' AB iv ][5.4],[18 ' AiVv AB
][9V
][27)( VvO
KVL
CONSTANT TIME DETERMINE :5 STEP
THR
L
circuit inductive
SC
OCTH i
vR
sourcesdependent
withCircuit
OPEN CIRCUIT VOLTAGE
AiA 6"
V12
KVL
][361224 VvOC
NOTE: FOR THE INDUCTIVE CASETHE CIRCUIT USED TOCOMPUTE THE SHORT CIRCUIT CURRENT IS THE SAME USE TODETERMINE )(Ov
SHORT CIRCUIT CURRENT
ORIGINAL CIRCUIT
1i
'''221
121
26)(236
4)(236
Aiiii
iii
'''1
2
A
SC
i i
i i
][8
36AiSC
8][8/36
][36TH
SC
OC RAi
VvsHL8
33
21,KK DETERMINE :6 STEP4) (step 127)( KvO
3) (step ][918)0( 221 VKKKvO
0,927)( 83
tetv
t
O :ANS
0),( ttvo FIND0,)( 21
teKKtv
t
O :1 STEP
0t AT VOLTAGECAPACITOR DETERMINE :2 STEP
][12VvA
][24V
KVL][60122424)0( VvC
)0(Cv
)0( Ov DETERMINE :3 STEP
)0()0( CC vv
][60)0( Vvvv OCO )(Ov DETERMINE :4 STEP
)(Ov0i
0Av
][24)( VvO
LEARNING EXTENSION
CONSTANT TIME DETERMINE :5 STEP CRTH circuit capacitive
SC
OCTH i
vR
)(Ov][24)( Vvv OOC
OCv
ORIGINAL CIRCUIT
OPEN CIRCUIT VOLTAGE
SHORT CIRCUIT CURRENT
SCi
2
KVL
02242 ASC vi
SCA iv 2 ][4 AiSC
64
24THR
sF 1226 21,KK DETERMINE :6 STEP
4) (step 24)(1 OvK
3) (step 2160)0( KKvO ][362 VK
0,3624)( 12
tetvt
O :ANS
Inductor example
STEP 1: Form of the solution
t
OeKKtv
21
)(
STEP 2: Initial inductor current
2
2 4
V6
0t
)0( Li
AiL
3)0(
STEP 3: Determine output at 0+(inductor current is constant)
2
2 4
V6
0t
_
)0(
Ov
A3
VvO
6)0(
2
2 4
V6
0t
H2
Ov
0),( ttvO FIND
STEP 4: Find output in steadystate after the switching
2
2 4
V6
0t
_
)(
Ov
0)( Ov
2
2 4
H2
V6
0t
_
)(tvO
STEP 5: Find time constant afterswitch
2
2 4
V6
0t
THR
THR
L
8THR
s25.0
STEP 6: Find the solution
VvKKO
6)0(21
0)(
1
OvK
0;6)( 25.0
tetvt
O
0;6)( 4 tetv tO Step by
StepPulse Response
PULSE RESPONSE
WE STUDY THE RESPONSE OF CIRCUITS TO A SPECIAL CLASS OF SINGULARITYFUNCTIONS
01
00)(
t
ttu VOLTAGE STEP
CURRENT STEP TIME SHIFTED STEP
PULSE SIGNAL
PULSE AS SUM OF STEPS
))](01.0()([10)( mAtututi
LEARN BY DOING
))](2()1([10)( Vtututv
NONZERO INITIAL TIME AND REPEATED SWITCHING
dxxfetxetxt
tTH
xttt
)(1
)()(0
0
0
00 )(; xtxfxdt
dxTH
021 ;)(0
tteKKtxtt
RESPONSE FOR CONSTANT SOURCES
This expression will hold on ANY interval where the sources are constant. The values of the constants may be different and must beevaluated for each interval
The values at the end of one interval will serve as initialconditions for the next interval
LEARNING EXAMPLE 0);( ttvO VOLTAGEOUTPUT THE FIND
0)(0)(0 tvtvt O 0)0( Ov
sFkkCRTH 4.0100)12||6(
Vtvt 9)(0
'1)9(
810
8)( Kvo
t
o eKKtv
'2
'1)(
0)0( '2
`1 KKvo
4.014)(t
o etv
0)(3.0 tvt
'
)3.0("2
"1)(
t
O eKKtv
3.0ot )1(4)3.0( 4.03.0
evo
4.0'
00)( "1 Kvo )(11.2)3.0("
2 VvK o
3.0;11.2)( 4.03.0
tetvt
o
+-
k10
F20
Ov
a
b
V12
THE SWITCH IS INITIALLYAT a. AT TIME t=0 IT MOVES TO bAND AT t=0.5 IT MOVES BACK TO a.FIND 0,)( ttvO
t
O eKKtv
'2
'1)(
'2
'1][12)0( KKVv
'10)( KvO
5.00,12)( 2.0
tetvt
O
sFk 2.0)20)(10(
985.012)5.0()5.0( 2.05.0
evv OO
')5.0(
"2
"1)(
t
O eKKtv
"985.0)5.0( 2"1 KKvO "
112)( KvO
015.1112985.0"2 K
5.0,015.1112)( 2.05.0
tetvt
O
EXAMPLE
)(st5.0
12
)(tvs
)(tvS
Piecewise constant source
Ott
O eKKtv
21)(
FORM THE OF IS OUTPUT THE CONSTANT IS
SOURCE THE WHEREINTERVAL EACH ON
The constants are determinedin the usual manner
)5.00 bt at (switch FOR 0ot
)at (switch FOR at 5.0 5.0ot
%pulse1.m% displays the response to a pulse responsetmin=linspace(-0.5,0,50); %negative time segmentt1=linspace(0,0.5,50); %first segmentt2=linspace(0.5, 1.5,100); %second segmentvomin=12*ones(size(tmin));vo1=12*exp(-t1/0.2); %after first switchingvo2=12-11.015*exp(-(t2-0.5)/0.2); %after second switchingplot(tmin,vomin,'bo',t1,vo1,'rx',t2,vo2,'md'),gridtitle('OUTPUT VOLTAGE'), xlabel('t(s)'),ylabel('Vo(V)')
USING MATLAB TO DISPLAY OUTPUT VOLTAGE
FirstOrder
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