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Transient Analysis of Transient Analysis of HigherHigher -- Order Netw orksOrder Netw orks
BY: ArtemioBY: Artemio P. MagaboP. MagaboProfessor of Electrical Engineering, UP EEEIProfessor of Electrical Engineering, UP EEEI
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p2
Second-Order Transients
The solution can be shown to be an exponential of the form
stKx =
Consider the homogeneous differential equation
0cxdtdx
bdt
xda 2
2
=++
with initial conditions x(0)=X 0 and =X 0 .)0(xtdx
where K and s are constants. Substitution gives
0cKbsKKas ststst2 =++
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After canceling the exponential term, we get thecharacteristic equation
0cbsas 2 =++Using the quadratic formula, we get the two roots
a24ac-bb-
s,s2
21=
Assuming the roots are real and distinct, thesolution will consist of two exponentials. Thus
ts
2
ts
121
KK)t(x +=K1 and K 2 can be evaluated using x(0) and (0). dtdx
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p4
Series RLC Netw orkConsider the circuitshown. From KVL, weget for t 0
R
iE+
-
t=0v C+
-
L
C
EidtC1Ri
dtdiL =++
Differentiating, we get
0iC1
dtdi
Rdtid
L 22
=++This is a homogeneous second-order differentialequation.
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p5
The characteristic equation is
0C1RsLs 2 =++
0LC1s
LRs 2 =++
or
From the quadratic formula, we get the two roots
LC1
L2R
L2R
s,s2
21
=
Note: There are three types of root depending onthe value of the term inside the square root sign.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p6
1. Overdamped Case: The roots are real anddistinct when
LC1
L2R
2
>
ts2
ts1
21 KK)t(x +=The solution is the sum of two exponential terms
2. Critically Dam ped Case: The roots are real butrepeated when
LC1
L2R
2
=
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st21 )KtK()t(x +=
The solution can be shown to be
LC1
L2R
2
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CIRCUITS 1 p9
Example: The capacitor is initially uncharged. Att=0, the switch is closed. Find
)0(dtdi + R
i12V+
-
t=0vC+
-
1H
C
i(0+) and .
From KVL, we getfor t 0,
EidtC1
Ridtdi
L =++ At t=0 + , i(0 + )=0. Also, we are given that v C(0 + )=0.Substitution gives
A/s12LE)0(
dtdi ==+
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p10
F161
10
i12V+
-
t=0v C+
-
1HExample: The capacitoris initially uncharged. Att=0, the switch isclosed. Find i(t)for t 0.
The characteristic equation is thus
016s10s 2 =++
From KVL, we get for t 0,
12idt16i101 =++ dtdi
Differentiating theequation, we have 0i16dt
di10 =++2
2
dtid
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p11
F161
10
i12V+
-v C+
-
1HThe characteristicequation is
016s10s 2 =++
whose roots are s 1=-2 and s 2=-8. Thus, we get8t-
22t-
1 KK)t(i +=and
8t-2
2t-1 K8K2dt
di =
From the previous example, weve found that att=0 + ,
A/sdtdi
12LE
)0( ==+i(0 + )=0, v C(0 + )=0 and
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p12
The initial conditions are i(0 + )=0 and =12A/sec. Substitution gives
)0(dtdi +
21 KK0)0(i +==+
21 8K--2K12)0(dtdi ==+
Solving simultaneously, we get K 1=2 and K 2=-2.Thus,
0t Amp 2-2)t(i 8t-2t- =Note: For an over-damped case, the solutionconsists of two distinct exponential terms.
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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F161
8
i12V+
-
t=0v C+
-
1HExample: The capacitor
is initially uncharged. Att=0, the switch isclosed. Find i(t)for t 0.
The characteristic equation can be shown to be
016s8s 2 =++whose roots are s 1=-4 and s 2=-4. Thus, we get
4t-3
4t-2
4t-1 KKK)t(i =+=
A single exponential solution will not work since theoriginal differential equation is second-order.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p14
Assume . Differentiating twice, we get4t-)t(y)t(i =4t-4t- )t('y)t(y-4
dtdi +=
4t-4t-4t-2
2
)t(''y)t('y8-)t(y16dt
id +=
The original differential equation is
0i16dtdi
8dt
id2
2
=++
-4t-4t-4t )t(''y(t)8y'-)t(y160 +=-4t-4t4t- 16y(t))t('y832y(t)- ++
Substitution gives
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p15
Integrating twice, we get
1K)t('y =or21 KtK)t(y +=
Simplifying, we get
0)t(''y 4t- =0(t)'y' =
or
4t-
)t(y)t(i =
Finally, the solution is
4t-2
4t-1 KtK)t(i +=
or
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p16
Differentiating the solution, we get
4t-2
4t-1
4t-1 4K-Kt-4Kdt
di +=
We get K 1=12 and K 2=0. Thus
0t Amp t12)t(i 4t- =
2K00)0(i +==+
21 4K-K012)0(dtdi
+==+
The initial conditions are i(0 + )=0 and =12A/sec. Substitution gives
)0(dtdi +
4t-2
4t-1 KtK)t(i +=
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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F161
6
i12V+
-
t=0v C+
-
1HExample: The capacitor
is initially uncharged. Att=0, the switch isclosed. Find i(t)for t 0.
The characteristic equation can be shown to be
016s6s 2 =++whose roots are s 1 , s 2=-3 j2.65. Thus, we get
j2.65)t-(-32
j2.65)t(-31 KK)t(i += +
or)KK()t(i t j2.65-2
t j2.651
3t- +=
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p18
Eulers Identities:
xsin jxcos jx +=(1)sin x j-xcos = jx(2)
To prove the first identity, let y=cos x + j sin x.Differentiating, we get
xcos jx-sindxdy +=
x)sin jx(cos j +=xcos jxsin j
dxdy 2 +=
and since , the equation can be re-writtenas
1 j =
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p19
We get y jdx
dy =
Integrating both sides, we get
Kx jyln +=
dx jdyy1 =
or
Evaluate K. When x=0, y=1.
0KorK j01ln =+=Thus we get ln y = jx, or xsin jxcos jx +=Note: The other Eulers identity can be verifiedfollowing the same analysis.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p20
Back to the expression for the current
)KK()t(i t j2.65-2t j2.65
13t- +=
tsin jKtcosK[)t(i 11t3 +=
]tsin jKtcosK 22 +
From Eulers identities, we get
where = 2.65. Combining the two cosine termsand the two sine terms, we get
]tsinKtcosK[)t(i 43t3 +=
where K 3 = K 1+K 2 and K 4 = j(K 1-K 2 ).
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p21
3K0)0(i ==+
.544KorK12)0(dtdi
44 ===+
We get
0t Amp t2.65sin54.4)t(i t3 =
The initial conditions are i(0 + )=0 and =12A/sec. Substitution gives
)0(dtdi +
Differentiate to get
]tcosKtsinK[dtdi 43
t3 +=
]tsinKtcosK[3 43t3 +
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p22
R=8
R=10
R=6
P lot of the Currents
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p23
P arallel RLC Netw ork
Differentiating, we get
0v
L
1
dt
dv
R
1
dt
vdC 22
=++
This is a homogeneous second-order differentialequation.
Consider the circuitshown. From KCL, weget for t 0
IvdtL1v
R1
dtdvC =++
RI v+
-CL
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p24
0LC1s
RC1s 2 =++
or
Note: We get three types of root depending on thevalue of the term inside the square root sign.
The characteristic equation is
0L1s
R1Cs 2 =++
From the quadratic formula, we get the two roots
LC1
RC21
2RC1
-s,s
2
21
=
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Higher-Order Transients
The solution can be shown to be an exponential of
the form stKx =where K and s are constants.
Consider the homogeneous differential equation
0xadtdxa...
dtxda
dtxda 011n
1n
1nn
n
n =++++
with initial conditions x(0)=X 0 , =X 0)0(xtdx
'''''''01n
1n
X)0(dx
xd =
''02
2
X)0(dt
xd = ,
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p26
Substitution gives
0KasKa...KsaKsa st0st
1st1n
1nstn
n =++++ After canceling the exponential term, we get thecharacteristic equation.
0asa...sasa 011n
1nn
n =++++ This is a polynomial of n th order and there will be nroots. The type of response will depend on the
values of these roots. Assuming all the n roots arereal and distinct, the solution can be shown to be
tsn
ts1n
ts2
ts1
n1n21 KK...KKx ++++=
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The roots of the characteristic equation can beshown to be s 1=-2, s 2=-4 and s 3 =-8.
Example: Consider the differential equation
0v64dtdv56
dtvd14
dtvd
2
2
3
3 =+++
with initial conditions v(0)=7 volts, (0)=-24 v/sdtdv
and (0)=112 v/s 2 . Find v(t).22
dtvd
064s56s14s 23 =+++The characteristic equation is
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p28
-8t3
-4t2
-2t1 KKK)t(v ++=
Since the roots are real and distinct, the solution is
Differentiating twice, we get
8t-3
4t-2
2t-1 K8K4K2dt
dv =
8t-3
4t-2
2t-12
2
K64K16K4
dt
vd ++=
Evaluate the expressions for v, and at t=0and use the initial conditions. dt
dv2
2
dtvd
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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321 K8K4K224)0(dtdv ==
3212
2
K64K16K4112)0(dt
vd ++==
Solving simultaneously, we get K 1=4, K 2 =2 andK3=1. The final solution is
V 124)t(v -8t-4t-2t ++=
321 KKK7)0(v ++==We get
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p30
Example: Consider the differential equation
0i32dtdi
32dt
id10
dtid
2
2
3
3
=+++
032s32s10s 23 =+++The characteristic equation is
The roots of the characteristic equation can beshown to be s 1=-2, s 2=-4 and s 3 =-4. The solution
is -4t3-4t
2-2t
1 KtKK)t(i ++=The constants K 1 , K 2 and K 3 can be evaluated if the values of i, di/dt and d 2 i/dt 2 are known at t=0.
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Getting the Differen tial EquationUsing nodal analysis or loop analysis, write theKCL or KVL equations that describe the circuit.
1.
Differentiate the equations, if necessary, toeliminate any integral expressions.
2.
In every equation, replace the derivatives withoperators.
3.
Eliminate all variables, except one, using anyappropriate method.
4.
Simplify as necessary and replace the operatorswith the corresponding derivative terms.
5.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p32
=++t
1222
0dt)ii(16i10dtdi
Mesh 2:
Mesh 1: =+t
211 )t(vdt)ii(16i8First, write the mesh equations for the circuit.
Then, differentiate the mesh equations to eliminatethe integrals.
Example: Find thedifferential equationsthat describe themesh currents i 1 andi2 in the networkshown.
1H
v(t)+
-
8
i1 i2 10 F161
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Next, multiply equation (1) by 16 and equation (2)by (8D+16), then add the resulting equations. Thiswill eliminate the current variable i 1 .
0i)16D10D(i16 22
1 =+++ (2)
)t(vdtdi16i16
dtdi8 211 =+
We get
(a)
0i16dtdi
10dt
idi16 222
22
1 =+++ (b)
)t(Dvi16i)16D8( 21 =+ (1)
Using operators, let D= . Substitution givesdtd
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p34
We get
)t(Dv16i)D288D96D8( 223 =++
which simplifies to
)t(v2i)36D12D( 22 =++
The differential equation for the current i 2 is
)t(v2i36dtdi12
dtid
22
22
2
=++
0i)16D10D(i16 22
1 =+++ (2))t(Dvi16i)16D8( 21 =+ (1)16
(8D+16)
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)t(v)16D10D(Di)D288D96D8( 2123 ++=++
which simplifies to
)t(v)16D10D(i)36D12D( 281
12 ++=++
The differential equation for current i 1 is
2
2
11
21
2
dt
)t(vd125.0i36
dt
di12
dt
id =++
)t(v2dt
)t(dv25.1 ++
Similarly, if we multiply equation (2) by 16 andequation (1) by (D 2 +10D+16), then add theresulting equations, we will eliminate the currentvariable i 2 . We get
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p36
22
85
22
2
161
1 idtdi
dtid
i ++=
Alternative Procedure: First, solve for i 1 from (b)and differentiate the resulting equation. We get
Next, substitute the equations in (a). We get
)t(v2i36dtdi
12dtid
22
22
2
=++A similar procedure, applied on equation (a), willresult in the differential equation for current i 1 .
dtdi
dtid
dtid
dtdi 2
22
2
85
32
3
1611 ++=
and
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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Example: The switch is moved from b to a at t=0.
Find the differential equations that describe thevoltages v 1 and v 2 for t 0.4
E+
-
t=0
a
1H
bv(t)
+
-F
416
REF
+v 1
+v 2
From KCL, we getfor t 0
0dtv6
v-v1
21 =+ Node 1:0
dtdv
41
6vv
4v(t)-v 2122 =++Node 2:
(1)
(2)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p38
Differentiate (1) and re-write the equations. We get
0dt
dv61v
dtdv
61 2
11 =+
)t(v41
v125
dtdv
41
v61
- 22
1 =++
)t(v3v5dt
dv3v2- 22
1 =++
0dt
dvv6
dtdv 2
11 =+
Simplify into
0Dv6)v(D 21 =+
)t(v3)v5(3D2v- 21 =++
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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For t 0, we get from KVL
)t(vdt)ii(20i4 211 =+ 0dt)ii(20
dtdi
122 =+
or
dt)t(dv
i20i20dtdi
4 211 =+
0i20i20dt
id122
22 =+
3
t=0
1Hv(t)
+
-
F2011
i1 i2
Example: At t=0, the switch is opened. Find the
differential equations that describe the currents i 1and i 2 for t 0.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p42
)t(Dvi20i)20D4( 21 =+Using operators, we get
(1)
0i)20D(i20 22
1 =++ (2)Multiply equation (1) by (D 2+20) and (2) by 20,then add the resulting equations. This willeliminate the variable i 2 . We get
)t(v)5D(i)20D5D( 241
12 +=++
The differential equation for i 1 is
)t(v5dt
)t(vdi20
dtdi
5dt
id2
2
41
11
21
2
+=++
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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Multiply equation (1) by 20 and (2) by (4D+20),then add the resulting equations. This willeliminate the variable i 1 . We get
)t(v5i)20D5D( 22 =++
The differential equation for i 2 is
)t(v5i20dtdi
5dt
id2
222
2
=++
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p44
Complete Response:1. Steady-state Response2. Transient Response
W hy Get Initial Conditions?
Transient Response: General form is exponentialts
nts
1nts
2ts
1n1n21 KK...KKx ++++=
where K 1 , K 2 , Kn are arbitrary constants.
Answer: The initial conditions are necessary in thedetermination of the numerical values of thearbitrary constants K 1 , K 2 , Kn .
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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Evaluating Initial Conditions
5. Use the KVL and KCL equations for t 0 and
their derivatives, plus the inductor currentsand capacitor voltages at t=0 + to evaluatethe required initial conditions.
1. Assume switching operation at t=0.
2. Evaluate the inductor currents and capacitorvoltages at t=0 - .
3. Find inductor currents and capacitor voltagesat t=0 + .
4. Write the KVL and KCL equations describing thenetwork for t 0.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p460)0(v
A1.0)0(i
C
ss,L
=
==
1k
100
Example: The circuit has reached steady-statecondition with the switch in position a. At t=0, theswitch is moved to position b. If the capacitor isinitially uncharged, find i(0 + ),
).0(dt
id )0(
dtdi
2
2++ and
1k
i100V+
-
t=0
a
1H0.1 F
b
The circuit is at steadystate for t
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0idtC1Ri
dtdiL =++ (1)
whose derivative is
0iC1
dtdi
Rdt
idL 2
2
=++(2)
At t=0 + ,
A1.0i)0(iss,L
==+
0)0(v C =+
From KVL, we get for t 0,1k
i 1H0.1 F
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p48
From (2), we get
)]0(iC1)0(
dtdiR[
L1-)0(
dtid2
2 +++ +=
2kA/s-900=
0From (1), we get at t=0 +
0)0(v)0(Ri)0(dtdi
L C =++ +++
which gives
A/s100)0(iLR
)0(dtdi == ++
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A430
120)0(i -L ==
V80)0(i20)0(v L-
C ==
t=0
i1 i2
10 120V
+
- 1H
20
1 F
20
and .)0(dtdi 2 +
Example: The network is initially at steady-state
condition with the switch open. At t=0, the switchis closed. Find i 1(0 + ), i 2(0 + ),
)0(dtdi1 +
At t=0 - , we get
iL(0 -)
10
120V
+
-
20 +-
vC(0 - )
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p50
For t 0, we get from KVL
i1 i2120V+
- 1H
20
1 F
20
At t=0 + , we get
A4)0(i)0(i -L1 ==+
V80)0(v)0(v-
CC ==+
120i20dtdi
11 =+(1)
120dti10i20t
26
2 =+ (2)
From equation (2), we get
A2)]0(v120[)0(i C201
2 == ++
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From equation (1), we get
A/s40)0(i20120)0(dtdi
11 == ++
To get an equation involving , differentiate (2).We get dt
di 2
0i10dtdi
20 262 =+
At t=0 + , we get
kA/sec -100)0(dtdi 2 =+
120i20dt
di1
1 =+(1) 120dti10i20t
26
2 =+ (2)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p52
dtdi1
Example: The network is initially unenergized. Att=0, the switch is closed. Determine i 1(0 + ), i 2 (0 + ),
(0 + ) and (0 + ).dtdi 2
i1E+
-
t=0
i2R1 L
R2C
For t 0, we get fromKVL,
E)ii(RdtiC
12111
=+
(1)
0iRi)RR(dtdi
L 112212 =++(2)
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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Since the circuit is initially unenergized, we knowthat v C(0 + )=0 and i L(0 + )=0. Thus
0)0(i2 =+
From (1), we get
E)0(iR)0(iR)0(v 2111C =+ +++
or
11 R
E)0(i =+
E)ii(RdtiC
12111 =+ (1)
0iRi)RR(dtdiL 11221
2 =++(2)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p54
0)0(i2 =+1
1 RE
)0(i =+
From (2), we get
0)0(iR)0(i)RR()0(dtdi
L 112212 =++ +++
E)ii(RdtiC1
2111 =+ (1)0iRi)RR(
dtdi
L 112212 =++(2)
LE)0(i
LR)0(
dtdi
112 == ++
which gives
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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
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For t
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At t=0 + , we get from (1)
mA9)]0(v)0(v27[R1
)0(i 2C1C2
1 == +++
Since i 1 (0+
) - i 3 (0+
)=i L(0+
), then0)0(i)0(i)0(i L13 == +++
++= dtiC1
dtiC1
iR27 32
21
12(1)
(2) += dtiC1)ii(R0 2
1121
(3) += dtiC1
)ii(dtd
L0 32
13
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p60
For resistor R 1 , we get
)0(v)]0(i)0(i[R)0(v 1C2111R++++ ==
or
mA3R
)0(v)0(i)0(i
1
1C12 ==
+++
++= dtiC1dtiC1iR27 322112(1)(2) += dtiC
1)ii(R0 2
1121
(3) += dtiC1
)ii(dtd
L0 32
13
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Differentiate equations (1) and (2). We get
32
21
12 iC
1i
C1
dtdi
R0 ++=(4)
(5) 21
11
21 iC
1dtdiR
dtdiR0 +=
At t=0 + , we get from (4)
s /A3=
)]0(iC1)0(iC1[R1)0(dtdi 322
12
1 +++ +=
++= dtiC1
dtiC1
iR27 32
21
12(1)
(2) += dtiC1)ii(R0 21121
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p62
At t=0 + , we get from (5)
s /A5.4=
)0(iCR
1)0(dtdi)0(
dtdi
211
12 +++ =
At t=0 + , we get from (3)
s /A9=
)0(v
L
1)0(
dt
di)0(
dt
di2C
13 +++ =
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dtdi
Example: The network has reached steady-state
condition with the switch closed. At t=0, the switchis opened. Find i(0 + ), v(0 + ),
(0 + ) and (0 + ).dtdv
t=0
i60V
+
- 1H 10 -3 Fv
+
-
10 20
Equivalent circuit at t=0 -
iL(0 -)60V
+
-+
-
10 20
v C(0 -)
A32060
)0(iL ==
V60)0(v C =
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p64
Equivalent circuit for t 0
i1H 10 -3 Fv
+
-
10 20 vi20dtdi =+(1)
vidt10i10t3 =+ (2)
V30)0(v =+or
At t=0 + , we get
A3)0(i)0(i L == +
V60)0(v)0(v CC == +
)0(v)0(v)0(i10 C+++ =
From equation (2), we get
+vC-
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dtdv
6vv
4v120 2
41122 +=(2)
Equivalent circuit for t 0 4
1H120V
+
-F
416
REF
+v 1
+v 2
= dtv6vv
112(1)
From KCL, we get
0)0(i6)0(v)0(v L21 == +++From (1), we get
At t=0 + ,V36)0(v)0(v C2 == ++
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p68
From (2), we get
V/s60)0(dt
dv 2 =+
Differentiate equation (1). We get
121 v6
dt
dv
dt
dv =At t=0 + ,
)0(v6)0(dt
dv)0(dt
dv1
21 +++ = V/s60=
dtdv
6vv
4v120 2
41122 +=(2) = dtv6
vv1
12(1)
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Solving the Differential Equation
Find the steady-state response x ss . This issimilar in form to the forcing function g(t) plusall its unique derivatives.
1.
Consider the n th -order differential equation
)t(gxadtdx
a...dt
xda
dtxd
a 011n1n
1nn
n
n =++++
Find the transient response x t . This is generally
an exponential of the form
2.
tsn
ts2
ts1t
n21 K...KKx +++=
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p70
)0(xtdx )0(
dxxd1n
1n
)0(dt
xd2
2Evaluate the initial conditions. We need the3.
values of x(0), , , .
Find the total response. Add the steady-stateresponse and transient response.
4.
tsn
ts2
ts1ss
n21 K...KKxx(t) ++++=Differentiate the total response (n-1) times.5.
Using the expressions for x(t) and its (n-1)derivatives in step 5, and the initial conditions instep 3, find the arbitrary constants K 1 , K 2 , Kn .
6.
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First, get the differential equations that describecurrents i 1(t) and i 2 for t 0. The mesh equationsare
24i8i12dtdi
211 =+ (1)
0i8i12dtdi2 12
2 =+ (2)
Example: The network
is initially unenergized.At t=0, the switch isclosed. Find currentsi1(t) and i 2 (t) for t 0.
4
i124V+
-
t=0 1H
i28 2H
4
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p72
24i8i12dtdi
211 =+
0i8i12dtdi
2 122 =+
Using operators we get(D+12)i 1 8i 2 = 24 (a)
-8i 1 + (2D+12)i 2 = 0 (b)
To eliminate i 2 , multiply (a) by (2D+12) and (b)by 8 and add the resulting equations. We get
(D 2 + 18D + 40)i 1 = (D+12)12
144i40dtdi
18dt
id1
121
2
=++or
(1)
(2)
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(D 2 + 18D + 40)i 2 = 96
Similarly, we eliminate i 1 by multiplying (a) by 8
and (b) by (D+12) and adding the equations. Weget
96i40dtdi
18dt
id2
222
2
=++or
Alternatively, we can solve for i 2 in equation (1)and differentiate the resulting equation. We get
(3)3idtdi
i 1231
81
2 +=
dtdi
dtid
dtdi 1
23
21
2
812 += (4)
and
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p74
Substitute (3) and (4) in equation (2). We get
144i40dtdi
18dt
id1
121
2
=++
This is the required differential equation for i 1 .
2232
41
1 idtdii += (5)
Solve for i 1 in equation (2) and differentiate theresulting equation. We get
dtdi
dtid
dtdi 2
23
22
2
411 += (6)and
Substitute (5) and (6) in equation (1). We get
96i40dtdi
18dt
id2
222
2
=++This is the required differential equation for i 2 .
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Next, we determine the steady-state response i 1,ssand i 2,ss which are both constant. We get from
144i40dtdi
18dt
id1
121
2
=++
A6.340
144i ss,1 ==
and from
96i40dtdi
18dt
id2
222
2
=++
A4.24096i ss,2 ==
Since theforcing functionis a constant(24V) thesteady-stateresponse of anycurrent orvoltage shouldalso be aconstant. Thus,
i1,ss = A
di 1,ss / dt = 0d 2 i1,ss / d t 2 = 0
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p76
We can also draw the equivalent circuit at steadystate. We get 4
24V+
-i2,ss8
4
i1,ss24i8i12 ss,2ss,1 =0i12i8 ss,2ss,1 =+
which gives i 1,ss = 3.6 Ampsand i 2,ss = 2.4 Amps.
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Next, find the transient response. The differential
equations are
Solving for the roots, we get s 1 =-2.6 and s 2 =-15.4.
144i40dtdi
18dt
id1
121
2
=++
040s18s2
=++
Setting the right-hand side of the equation to zeroand changing operators, we get the characteristicequation to be
96i40dtdi
18dt
id2
222
2
=++
Only one characteristicequation defines anycurrent or voltage inthe circuit.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p78
-15.4t2
-2.6t1t1 KKi +=
Thus, we get the transient response.
-15.4t4
-2.6t3t2 KKi +=
Next, we find the initial conditions. We need i 1 (0 + ),
dtdi 2i2(0 + ), (0 + ) and (0 + ).dt
di1
0)0(i)0(i 21 == ++Since the circuit was initially unenergized, we get
From (1), we get A/s24)0(dtdi 1 =+
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From (2), we get 0)0(dt
di 2 =+
Next, determine the total response. We get-15.4t
2-2.6t
11 KK6.3)t(i ++=-15.4t
4-2.6t
32 KK4.2)t(i ++=
whose derivative are
15.4t-2
2.6t-1
1 K4.15K6.2dtdi =
15.4t-4
2.6t-3
2 K4.15K6.2dtdi =
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p80
Evaluate the constants K 1 and K 2 . At t=0 + , we get
211 KK6.30)0(i ++==+
211 K4.15K6.224)0(
dtdi ==+
Solving simultaneously, we get K 1=-2.46 andK2=-1.14. The final expression for current i 1 is
A 14.146.26.3)t(i -15.4t-2.6t1 =
Evaluate the constants K 3 and K 4 . We get K 3 =-2.89and K 4=0.49. Thus
A 49.089.24.2)t(i -15.4t-2.6t2 +=
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I 1
I2
P lot of the Currents
4
i124+
-
t=0 1
i28 2
4
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p82
4
60V+
-
t=0
a
1H
b120V
+
-F
416
REF
+v 1
+v 2
Example: The switch has been in position b for along time. At t=0, the switch is moved to a. Findv 1(t) and v 2(t) for t 0.
In a previous example, we got the differentialequations that describe the voltages v 1 and v 2 . Inanother example, we derived the initial conditions.
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0v10dt
dv7dt
vd1
1212 =++
720v10dt
dv7dt
vd2
222
2
=++
The differential equations are
0)0(v 1 =+
V36)0(v 2 =+
V/s60)0(dt
dv)0(dt
dv 21 == ++
with initial conditions
(1)
(2)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p84
Steady-state respo nse:
V7210720
v ss,2 ==
We can also draw the equivalent circuit at steadystate.
v 2,ssv 1,ss
120V+
-
+
-
6
4
+
-
0v ss,1 =
V72)120(106v ss,2 ==
v 1,ss = A
dv 1,ssdt
= 0
d 2 v 1,ssdt 2
= 0
v 2,ss = B
dv 2,ssdt
= 0
d 2 v 1,ssdt 2
= 0
0v10dtdv
7dtvd
1121
2
=++ 720v10dtdv
7dtvd
2222
2
=++
0v ss,1 =From the differential equations (1) and (2), we get
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Transient response:
whose roots are s 1=-2 and s 2=-5. We get
-5t
2
-2t
1t1 KKv +=-5t
4-2t
3t2 KKv +=
010s7s 2 =++The characteristic equation is
0v10dtdv
7dtvd
11
21
2
=++
720v10dt
dv7
dtvd
22
22
2
=++
The differential equations are
(1)
(2)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p86
Complete response: We get -5t2
-2t11 KK)t(v +=
-5t4
-2t32 KK72)t(v ++=
The derivatives are
5t-2
2t-1
1 K5K2dt
dv =
5t-4
2t-3
2 K5K2dt
dv =
At t=0 + , we get
211 KK0)0(v +==+ 211 K5K260)0(dtdv ==+
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Also at t=0 + , we get
432 KK7236)(0v ++==+
432 5K2K60)(0
dtdv ==+
Solving simultaneously, we get K 3=140 andK4=-68. The final expressions are
0t V 2020)t(v -5t-2t1 =
0t V 4 40 72(t)v -5t-2t2 +=
Solving simultaneously, we get K 1=20 and K 2=-20.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p88
V1
V2
P lot of the Voltages
4
60V+
-
t=0
1120
+
-F
416
REF
+v 1
+v 2
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Example: The network
is initially unenergized.At t=0, the switch isclosed. Find currenti2(t) for t 0. Assumev(t)=20 cos 4t volts.
8
i1v(t)+
-
t=0 2H
i24 1H
For t 0, the mesh equations are
)t(vi4i12dtdi
2 211 =+ (1)
0i4i4dtdi
122 =+ (2)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p90
22
41
1 idtdi
i += (3)
Solve for i 1 in equation (2) and differentiate theresulting equation. We get
dtdi
dtid
dtdi 2
22
2
411 += (4)
and
Substitute (3) and (4) in equation (1). We get
)t(v2i16dtdi10
dtid 2222
2
=++
where v(t)=20 cos 4t volts.
(5)
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Forced Response: Since the source is sinusoidal,
4tsinB4tcosAi ss,2 +=
4tcosB44tsinA4dt
di ss,2 +=
4tsin16B-4tcosA16dtid
2ss,2
2
=
Substitute in the differential equation (5). We get
4tsin40A-4tsin16B-4tcos16A4tcos404tsin16B4tcos16A4tcos40B =+++
4tcos4016idtdi
10dt
id2
222
2
=++ (5)
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p92
A16B40A1640 ++=B16A40B160 +=
Comparing coefficients, we get
Solving simultaneously, we get A=0 and B=1. Thus
t4sini ss,2 =Transient response: The characteristic equation is
016s10s2
=++The roots are s 1=-2 and s 2 =-8. Thus
t82
t21t,2 KKi
+=
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Complete response:t8
2t2
12 KK4tsin)t(i
++=t8
2t2
12 K8K2-4tcos4
dtdi =
dtdi 2At t=0 + , i 1 (0 + )=i 2 (0 + )=0. From (2), get (0+)=0.
21 KK0 +=
21 K8K2-40 =Solving simultaneously, we get K 1=-K 2=-2/3. Thus
0t A -4tsin)t(i t832t2322 +=
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p94
P lot of the Current & Voltage
i2
V(t) = 20 cos (4t)
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Equivalent circuit for t 0 4
1Hv(t)
+
-F
416
iLv C+
-
KVL:dtdi1i6v LLC +=
KCL: LC
41C i
dtdv
4v)t(v +=
In matrix form, we get
C
L
vi
C
L
v
i
v(t)= +10
1416
with initial conditions i L(0 + )=6 A and v C(0 + )=36 V.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p98
Example: The circuit hasreached steady-statecondition with the switchclosed. At t=0, the switchis opened. Find v C(0 + ),iL(0 + ) and the stateequations for t 0.
3
t=0
1H12V
+
-
F2011 v C iL
+
-
Equivalent circuit at t=0 -1
12V+
-iL(0 -)
+
-vC(0 -)
)0(v0)0(v C-C +==
)0(iA12)0(i L-
L+==
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Equivalent circuit for t 0 4
1H12V+
-F
201v C iL
+
-KVL:
dtdi1v LC =
KCL: LC
201C i
dtdv
4v12 +=
In matrix form, we get
C
L
vi
C
L
v
i
12V= +50
52010
with initial conditions i L(0 + )=12 Amps and v C(0 + )=0.
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p100
Solution of the State EquationThe Euler method for integrating a first-orderdifferential equation is of the form
t)t,x(f )t(x)tt(x0t00
++where x=f(x,t). The method can be extended tothe case when x is a vector.
.
Substitution givest)]t(uB)t(xA[)t(x)tt(x 0000 +++)t(utB)t(x)tAI( 00 ++
uBxA)t,x(f x +==
Consider the stateequation
Identity matrix Step-size
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Example: The network is initially unenergized. At
t=0, the switch is closed. t=0 8
iL1v(t)+
-
2H
iL24 1H
Use the Euler methodwith t=0.02 sec tofind i L1 and i L2 for t 0.Let v(t)=20 cos 4t V.
For t 0, we get from KVL,
)t(vi4i12dtdi
2 211 =+
0i4i4dtdi
122 =+
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p102
In matrix form, we get
2L
1L
ii
2L
1L
i
iv(t)= +
05.0
4426
with initial conditions i L1(0 + ) = i L2(0 + ) = 0.
Recall the Euler method
)t(utB)t(x)tAI()tt(x 000 +++
= +44
26
0.02I+ tA
1001
=92.008.004.088.0
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We get
= +iL1(t+ t)iL2(t+ t) 92.008.0
04.088.0 iL1(t)
iL2(t)
0.2cos 4t
0
At t=0,
= +iL1(0.02)iL2(0.02) 92.008.0
04.088.0 iL1(0+
)iL2(0 + )
0.20 0
0.2=
where
)t(utB)t(x)tAI()tt(x 000 +++
=I+ tA 92.008.004.088.0
and t=0.2
Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World
CIRCUITS 1 p104
At t=0.04 second:
= +iL1(0.04)
iL2(0.04) 92.008.004.088.0 0.3754
0.016 0.0448
0.5284=
0.1974
0
= +iL1(t+ t)iL2(t+ t) 92.008.0
04.088.0 iL1(t)iL2(t)
0.2cos 4t0
and
Weve found
=iL1(0 + )iL2(0 + ) 0
0=
iL1(0.02)iL2(0.02) 0
0.2
At t=0.02 second:
= +iL1(0.04)
iL2(0.04) 92.008.004.088.0 0.2
0 0.016
0.3754=
0.1994
0
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Comparison o f ResultsFrom a previous example, we got
0t A -4tsin)t(i t832t2
32
2 +=
Time Actual Euler Error
0.02 0.00748 0.0 0.00748
0.04 0.02801 0.016 0.01201
0.06 0.05894 0.04475 0.014190.08 0.09800 0.08344 0.01456
Comparing the actual value with the estimate,
P lot of the Current