first and second-order transient...
TRANSCRIPT
FIRST AND SECOND-ORDER
TRANSIENT CIRCUITS
IN CIRCUITS WITH INDUCTORS AND CAPACITORS VOLTAGES AND CURRENTS
CANNOT CHANGE INSTANTANEOUSLY.
EVEN THE APPLICATION, OR REMOVAL, OF CONSTANT SOURCES CREATES A
TRANSIENT BEHAVIOR
LEARNING GOALS
FIRST ORDER CIRCUITS
Circuits that contain a single energy storing elements.
Either a capacitor or an inductor
SECOND ORDER CIRCUITS
Circuits with two energy storing elements in any combination
THE CONVENTIONAL ANALYSIS USING MATHEMATICAL MODELS REQUIRES THE DETERMINATION OF (A SET OF) EQUATIONS THAT REPRESENT THE CIRCUIT.ONCE THE MODEL IS OBTAINED ANALYSIS REQUIRES THE SOLUTION OF THE EQUATIONS FOR THE CASES REQUIRED.
FOR EXAMPLE IN NODE OR LOOP ANALYSIS OF RESISTIVE CIRCUITS ONE REPRESENTS THECIRCUIT BY A SET OF ALGEBRAIC EQUATIONS
WHEN THERE ARE INDUCTORS OR CAPACITORS THE MODELS BECOME LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODEs). HENCE, IN GENERAL, ONE NEEDS ALL THOSE TOOLS IN ORDER TO BE ABLE TO ANALYZE CIRCUITS WITH ENERGY STORING ELEMENTS.
ANALYSIS OF LINEAR CIRCUITS WITH INDUCTORS AND/OR CAPACITORS
THE GENERAL APPROACH CAN BE SIMPLIFIED IN SOME SPECIAL CASES WHEN THE FORMOF THE SOLUTION CAN BE KNOWN BEFOREHAND. THE ANALYSIS IN THESE CASES BECOMES A SIMPLE MATTER OF DETERMINING SOMEPARAMETERS.TWO SUCH CASES WILL BE DISCUSSED IN DETAIL FOR THE CASE OF CONSTANT SOURCES. ONE THAT ASSUMES THE AVAILABILITY OF THE DIFFERENTIAL EQUATION AND A SECOND THAT IS ENTIRELY BASED ON ELEMENTARY CIRCUIT ANALYSIS… BUT IT IS NORMALLY LONGER
A METHOD BASED ON THEVENIN WILL BE DEVELOPED TO DERIVE MATHEMATICAL MODELSFOR ANY ARBITRARY LINEAR CIRCUIT WITH ONE ENERGY STORING ELEMENT.
WE WILL ALSO DISCUSS THE PERFORMANCE OF LINEAR CIRCUITS TO OTHER SIMPLE INPUTS
THE MODEL
AN INTRODUCTION
INDUCTORS AND CAPACITORS CAN STORE ENERGY. UNDER SUITABLE CONDITIONS THIS ENERGYCAN BE RELEASED. THE RATE AT WHICH IT IS RELEASED WILL DEPEND ON THE PARAMETERSOF THE CIRCUIT CONNECTED TO THE TERMINALS OF THE ENERGY STORING ELEMENT
With the switch on the left the capacitor receives
charge from the battery.
Switch to the right
and the capacitor
discharges through
the lamp
dxxfetxetxet
tTH
xtt
)(1
)()(
0
0
0
GENERAL RESPONSE: FIRST ORDER CIRCUITS
0)0(; xxfxdt
dxTH
Including the initial conditionsthe model for the capacitorvoltage or the inductor current will be shown to be of the form
Solving the differential equationusing integrating factors, one tries to convert the LHS into anexact derivative
t
TH efxdt
dx 1/*
TH
ttt
fexedt
dxe
11
TH
tt
fexedt
d
1
t
t0
dxxfetxetx
t
t
TH
xttt
)(1
)()(
0
0
0
0)0();()()( xxtftaxtdt
dx
THIS EXPRESSION ALLOWS THE COMPUTATIONOF THE RESPONSE FOR ANY FORCING FUNCTION.WE WILL CONCENTRATE IN THE SPECIAL CASEWHEN THE RIGHT HAND SIDE IS CONSTANT
t
e
/*
circuit
the of speed reaction the on ninformatio
tsignifican provide to shown be willit
constant." time" the called is
.switchings sequentialstudy
to used be can expression general
The arbitrary. is , time, initial The ot
FIRST ORDER CIRCUITS WITHCONSTANT SOURCES
dxxfetxetxt
tTH
xttt
)(1
)()(
0
0
0
0)0(; xxfxdt
dxTH
If the RHS is constant
dxef
txetxt
t
xt
TH
tt
0
0
)()(0
xtxt
eee
dxeef
txetxt
t
xt
TH
tt
0
0
)()(0
t
t
xt
TH
tt
eef
txetx
0
0
)()(0
00
)()(0
ttt
TH
tt
eeeftxetx
0
)()(0
tt
THTHeftxftx
0tt
The form of the solution is
021;)(
0
tteKKtxtt
Any variable in the circuit is ofthe form
021;)(
0
tteKKtytt
Only the values of the constantsK_1, K_2 will change
TRANSIENT
TIMECONSTANT
EVOLUTION OF THE TRANSIENT AND INTERPRETATION OF THE TIME CONSTANT
A QUALITATIVE VIEW:
THE SMALLER THE THE TIME
CONSTANT THE FASTER THE
TRANSIENT DISAPPEARS
With less than 2% error
transient is zero
beyond this point
Drops 0.632 of initial
value in one time constant
Tangent reaches x-axis in one time constant
CRTH
THE TIME CONSTANT
The following example illustratesthe physical meaning of timeconstant
vS
RS a
b
C
+
vc
_
Charging a capacitor
THCC
TH vvdt
dvCR
The model
0)0(, CSS vVv
Assume
The solution can be shown to be
t
SSC eVVtv
)(
CRTH
transient
For practical purposes thecapacitor is charged when thetransient is negligible
0067.0
0183.00498.0135.0
5
432
368.0
t
et
With less than 1%error the transientis negligible afterfive time constants
dt
dvC C
S
SC
R
vv
0
S
SCc
R
vv
dt
dvC
: KCL@a
1. THE CIRCUIT HAS ONLY CONSTANT INDEPENDENT SOURCES
THE DIFFERENTIAL EQUATION APPROACH
CIRCUITS WITH ONE ENERGY STORING ELEMENT
CONDITIONS
2. THE DIFFERENTIAL EQUATION FOR THE VARIABLE OF INTERESTIS SIMPLE TO OBTAIN. NORMALLY USING BASIC ANALYSIS TOOLS;e.g., KCL, KVL. . . OR THEVENIN
3. THE INITIAL CONDITION FOR THE DIFFERENTIAL EQUATIONIS KNOWN, OR CAN BE OBTAINED USING STEADY STATE ANALYSIS
SOLUTION STRATEGY: USE THE DIFFERENTIAL EQUATION AND THE INITIAL CONDITIONS TO FIND THE PARAMETERS ,, 21 KK
( )
1 2
FACT: WHEN ALL INDEPENDENT SOURCES ARE CONSTANT
FOR ANY VARIABLE, ( ), IN THE CIRCUIT THE
SOLUTION IS OF THE FORM
( ) ,Ot t
O
y t
y t K K e t t
If the diff eq for y is knownin the form
Use the diff eq to find twomore equations by replacingthe form of solution into thedifferential equation
0
01
)0( yy
fyadt
dya
We can use thisinfo to find the unknowns
feKKaeK
a
tt
210
21
0
110a
fKfKa
0
120
1 0a
aeKa
at
t
eK
dt
dy
2
0,)( 21 teKKty
t
21)0( KKy
Use the initial condition to getone more equation
12 )0( KyK
SHORTCUT: WRITE DIFFERENTIAL EQ.IN NORMALIZED FORM WITH COEFFICIENTOF VARIABLE = 1.
00
101
a
fy
dt
dy
a
afya
dt
dya
1K
ASSUME FIND2
)0(.0),( SVvttv
)(tv @KCL USE 0.t FORMODEL
0)()(
tdt
dvC
R
Vtv S
2/)0( SVv condition initial
(DIFF. EQ. KNOWN, INITIAL CONDITION KNOWN)
STEP 1 TIME CONSTANT
fydt
dy
Get time constant ascoefficient of derivative
STEP 2 STEADY STATE ANALYSIS
value)state(steady ,t and 0for
IS SOLUTION
1
21 0,)(
Kv(t)
teKKtv
t
IN STEADY STATE THE SOLUTION ISA CONSTANT. HENCE ITS DERIVATIVEIS ZERO. FROM DIFF EQ.
SVvdt
dv 0
Steady state valuefrom diff. eq.
SVK
1
values)statesteady (equating
fKfydt
dy 1 THEN ISMODEL THE IF
STEP 3 USE OF INITIAL CONDITION
1221 )0()0(
0
KvKKKv
t
AT
fvK )0(2
2/2/)0( 2 SS VKVv
0,)2/()(
teVVtv RC
t
SS :ANSWER
LEARNING EXAMPLE
sVtvtdt
dvRC )()(
R/*
)0();(
0,)(
211
21
xKKxK
teKKtx
t
0),( tti FIND
0t FORKVL USE MODEL.
Rv
Lv)(ti
KVL
)()( tdt
diLtRivvV LRS
0)0()0()0(
0)0(0
i
ii
it
inductor
CONDITIONINITIAL
STEP 1R
Vtit
dt
di
R
L S )()(R
L
STEP 2 STEADY STATER
VKi S 1)(
STEP 3 INITIAL CONDITION
21)0( KKi
RL
t
S eR
Vti 1)( :ANS
LEARNING EXAMPLE
)0();(
0,)(
211
21
xKKxK
teKKtx
t
1 2( ) , 0t
i t K K e t
0t FORKCL MODEL.
)()(
tiR
tvIS
)(tv
)()( tdt
diLtv )()( tit
dt
di
R
LIS
STEP 1
STEP 2 SS IKIi 1)(
STEP 3 210)0( KKi
RL
t
S eIti 1)( :ANS
0)0( i :CONDITIONINITIAL
R
L
LEARNING BY DOING
1 2( ) , 0t
i t K K e t
0t FORMODEL
2
)()(
R
tvti
IT IS SIMPLER TO DETERMINE MODELFOR CAPACITOR VOLTAGE
INITIAL CONDITIONS
VvVkk
kvC 4)0(4)12(
63
3)0(
0)(
)(
||;0)(
)()(
21
21
P
P
R
tvt
dt
dvC
RRRR
tvt
dt
dvC
R
tv
kkkRP 26||3
sFCRP 2.0)10100)(102( 63
STEP 1
STEP 2 0)( 1 Kv
STEP 3 VKVKKv 44)0( 221
0],[4)( 2.0
tVetv
t
0],[3
4)( 2.0
tmAeti
t
:ANS
0,)( 21
teKKti
t
0t FOR STATESTEADY IN CIRCUIT
Vtvtdt
dvO
O 6)()(5.0
][3)()(5.0
12)(4)(2
Atitdt
di
titdt
di
])[(2)( VtitvO
0),( ttvO FIND
KVL(t>0)
)(ti
KVL USE 0.t FORMODEL
0)()()( 311 tiRtdt
diLtiRVS
5.0 STEP 1
0,)( 21
teKKtv
t
O
LEARNING EXAMPLE
STEP 2: FIND K1 USING STEADY STATEANALYSIS
Vvtvtdt
dvOO
O 6)(6)()(5.0
1)( KvO
VK 61
FOR THE INITIAL CONDITION ONE NEEDSTHE INDUCTOR CURRENT FOR t<0 ANDUSES THE CONTINUITY OF THE INDUCTORCURRENT DURING THE SWITCHING .
THE NEXT STEP REQUIRES THE INITIALVALUE OF THE VARIABLE, )0( Ov
THE STEADY STATE ASSUMPTION FOR t<0SIMPLIFIES THE ANALYSIS
)0();(
0,)(
211
21
xKKxK
teKKtx
t
FOR EXAMPLE USE THEVENINASSUMING INDUCTOR IN STEADYSTATE
12||2THR
04412 1 I1I
KVL
KVL ][442 1 VIVV OCTH
][41 AI
][3
4)0()0( AiiL
0,)( 21
teKKtv
t
O
][3
8)0(
3
4)0( Vvi O
0,3
53)( 5.0
teti
t
a
b
0],[3
106)( 5.0
tVetv
t
O
CIRCUIT IN STEADY STATE (t<0)
)(tiL FIND MUST
3
106
3
82221 KKKK
Li
0t
0),( ttvO FIND
C
1R
2R
KCL USE 0.t FORMODEL
0)()(0)( 21
21
cCCC vt
dt
dvCRR
RR
vt
dt
dvC
sFCRR 6.0)10100)(106()( 6321 STEP 1
)(3
1)(
42
2)( tvtvtv CCO
STEP 20,)( 21
teKKtv
t
C 01 K
INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t<0
)0(Cv V)12(9
6
][88)0( 221 VKKKvC
STEP 3
0],[8)( 6.0
tVetv
t
C
0],[3
8)( 6.0
tVetv
t
O
LEARNING EXTENSION
)(tvc DETERMINE
)0();(
0,)(
1211
21
iKKvK
teKKtv
C
t
C
0),(1 tti FIND
KVL USE 0.t FORMODEL
0)(18 11 ti
dt
diL
L
0)()(9
11
1 titdt
di
)0();(
0,)(
12111
211
iKKiK
teKKti
t
STEP 1 s9
1
STEP 2 01 K
FOR INITIAL CONDITIONS ONE NEEDSINDUCTOR CURRENT FOR t<0
)0(1 i
CIRCUIT IN STEADY STATEPRIOR TO SWITCHING
AV
i 112
12)0(1
STEP 3
][1)0()0( 22111 AKKKii
0],[][)( 991
1
tAeAetit
t
:ANS
)(1 ti
Lv
LEARNING EXTENSION
USING THEVENIN TO OBTAIN MODELS
Obtain the voltage across the capacitor or the current through the inductor
Circuitwith
resistancesand
sources
InductororCapacitor
a
b
Representation of an arbitrarycircuit with one storage element
Thevenin
VTH
RTH
InductororCapacitor
a
b
VTH
RTH a
b
C
+
vc
_
Case 1.1Voltage across capacitor
VTH
RTH a
b
L
iL
Case 1.2Current through inductor
KCL@ node a
ciRi
0 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 viR
dt
diL
TH
THL
L
TH R
vi
dt
di
R
LSCi
EXAMPLE
66
6
6
H3
V24
)(tiO
0t
0t;(t)iFindO
The variable of interest is theinductor current. The model is
TH
TH
O
O
THR
vi
dt
di
R
L
And the solution is of the form
0;)(21
teKKtit
O
66
6
6V24
0t
Thevenin for t>0at inductor terminals
a
b
TH
v 0 TH
R ))66(||6(6
sH
R
L
TH
3.010
3
0;03.0 tidt
diO
O
03.0
3.0 3.0
21
3.02
tt
eKKeK
0;)( 3.0
2
teKtit
O 0
1K
Next: Initial Condition
66
6
6V24
)0()0( OOii
0t
Since K1=0 the solution is
0;)( 3.0
2
teKtit
O
Evaluating at 0+6
322K
0;6
32)( 3.0
teti
t
O
66
6
6
H3
V24
)(tiO
0t
Circuit for t<0
1i
2i
3i
0)(6)(6621311 iiiii
0)(6)(6243212 iiii
0)(6)(62313 iiii
3)0( ii
C
mA6
32)(0i:solution
C
currentinductor
of continuity and assumption
statesteady Use Determine ).0( Oi
1v
806
24
661
111
vvvv
66
24)0( 1viO
Loop analysis
Node analysis
+
- 0t
k6
k6
k6
k6
F100
V12
)(tiO
0t(t),iFindO
EXAMPLE
C
v
6kv
i0tFor CO
Hence, if the capacitor voltageis known the problem is solved
Model for v_c
THC
C
THvv
dt
dvCR
+
- 0t
k6
k6
k6
k6V12
)(tiO
a b
THv
kkkRTH 36||6
VvTH 6
sF 3.010*100*10*3 63
63.0 CC
C
vdt
dv
vforModel
3.021
t
C eKKv
65.1
5.1 3.021
5.12
tt
eKKeK
61K
Now we need to determinethe initial value v_c(0+)using continuity and thesteady state assumption
+
- 0t
k6
k6
k6
k6V12
)(tiO
circuit in steady statebefore the switching
)0(C
v
VvC
6)0(
Continuity of capacitor voltage
VvC
6)0(
)0(21
C
vKK
0621 KK
0;6)( tVtvC
0;16
)( tmAk
vti C
O
Diff EqApproach