ch05 transient analysis 1s09
TRANSCRIPT
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Introduction to DifferentialIntroduction to Differential
EquationsEquations
Transient Analysis ofTransient Analysis of
FirstFirst--Order NetworksOrder Networks
Chapter 5
Department of Electrical and Electronics EngineeringUniversity of the Philippines - Diliman
Department of Electrical and Electronics Engineering EEE 33 - p2
Differential Equations
Definition: Differential equations are equationthat involve dependent variables and theirderivatives with respect to the independentvariables.
02
2
=+ kudx
udSimple harmonicmotion: u(x)
2
22
2
2
2
2
2
2
t
uc
z
u
y
u
x
u
=
+
+
Wave equation in threedimensions: u(x,y,z,t)
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Ordinary Differential Equations
Definition: Ordinary differential equations(ODE) are differential equations that involve onlyONE independent variable.
02
2
=+ kudx
)x(ud
u(x) is the dependent variable
xis the independent variable
Example:
Department of Electrical and Electronics Engineering EEE 33 - p4
Ordinary Differential Equations
We can classify all ODEs according to
order, linearityand homogeneity.
The order of a differential equation is just thehighest differential term involved:
2nd order
3
3
dt
xd
xdt
dx
=
0012
2
2 =++ adt
dya
dt
yda
3rd order
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Linearity
The important issue is how the unknown variable(ie y) appears in the equation. A linear equationmust have constant coefficients, or coefficientswhich depend on the independent variable. Ifyor its derivatives appear in the coefficient theequation is non-linear.
is linear0=+ ydt
dy
02 =+ x
dt
dxis non-linear
is linear02 =+ tdt
dy
02 =+ t
dt
dyy is non-linear
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Linearity - Summary
y2
dt
dy
dt
dyy
dt
dyt
2
dt
dy
yt)sin32( + yy )32( 2
Non-linearLinear
2y )sin(yor
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Department of Electrical and Electronics Engineering EEE 33 - p7
Homogeniety
Put all the terms of the differential equation whichinvolve the dependent variable on the left handside (LHS) of the equation.
Homogeneous: If there is nothing left on theright-hand side (RHS), the equation ishomogeneous. (unforced or free)
Nonhomogeneous: If there are terms left onthe RHS involving constants or the independentvariable, the equation is nonhomogeneous (forced)
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Examples of Classification
0=+ ydx
dy 1st Order
Linear
Homogeneous
)sin()cos(2
2
2
xyxdx
yd=+
2nd OrderNon-linear
Non-homogeneous
)xcos(ydx
yd= 45
3
3 3rd Order
Linear
Non-homogeneous
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Linear Differential Equations
A linear ordinary differential equation describinglinear electric circuits is of the form
)t(vadt
dxa...
dt
xda
dt
xda
n
n
nn
n
n =++++
011
1
1
where
an, an-1,,a0 constants
x(t) dependent variable (current or voltage)
t independent variable
v(t) voltage or current sources
Department of Electrical and Electronics Engineering EEE 33 - p1
Linear Differential Equations
Assume that we are given a network of passiveelements and sources where all currents andvoltages are initially known. At a reference instantof time designated t=0, the system is altered in amanner that is represented by the opening orclosing of a switch.
Our objective is to obtain equations for currentsand voltages in terms of time measured from theinstant equilibrium was altered by the switching.
Department of Electrical and Electronics Engineering EEE 33 - p11
Solution to Differential Equations
In the network shown, theswitch is moved from position
1 to position 2 at time t=0.L
R
E+
-
1
2
0Ridt
diL =+
After switching, the KVL equation is
(1)
dtL
R
i
di=
Re-arranging the equation to separate thevariables, we get
(2)
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Solution to Differential Equations
The constant Kcan be expressed as ln k
where ln means the natural logarithm (base e).
KtL
Ri +=ln
Equation 2 can be integrated to give
(3)
Thus, equation 3 can be written as
kei L/Rt lnlnln += (4)
We know that ln y+ lnz= ln yz
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Solution to Differential Equations
Equation 6 is known as the general solution. Ifthe constant kis evaluated, the solution is aparticular solution.
Equation 4 is equivalent to
)ke(i L/Rt= lnln (5)
Applying the antilogarithm we get(6)L/Rtkei =
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General and Particular Solutions
The general solution refers to a set of solutionssatisfying the differential equation.
A particular
solution fits thespecification of aparticular problem.
Assume in theprevious circuit,R=1k, L=4H.
1.5e-250t
e-250t
0.5e-250t
Transient Analysis ofTransient Analysis of
FirstFirst--Order NetworksOrder Networks
Artemio P. MagaboArtemio P. MagaboProfessor of Electrical EngineeringProfessor of Electrical Engineering
Department of Electrical and Electronics EngineeringUniversity of the Philippines - Diliman
Department of Electrical and Electronics Engineering EEE 33 - p1
First-Order Transients
Consider the homogeneous differential equation
0bxdt
dxa =+
with initial condition x(0)=X0.
where K and s are constants.
stKx =
The solution can be shown to be an exponential ofthe form
0bKasKstst
=+
Substitution gives
Department of Electrical and Electronics Engineering EEE 33 - p17
After canceling the exponential term, we get
0bas =+ ora
bs =
Thus the solution is
ta
b
Kx =
The constant K can be found using the given initialcondition. At t=0, we get
KKX)0(x 00 ===
The final solution is
0tXxt
a
b
0 =
Department of Electrical and Electronics Engineering EEE 33 - p1
Source-Free RL Network
Consider the circuit shown. Leti(0) = I0. From KVL, we get
0Ridt
diL =+
Li
R
The solution can be found to be
tL
R
Ki
=
At t=0, we get
KKI)0(i 00 ===
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Department of Electrical and Electronics Engineering EEE 33 - p19
Note: Every current and voltage in an RLnetwork is a decaying exponential with a
time constant of=L/R.
Substitution givest
L
R
0I)t(i
=
From Ohms Law, we get the resistor voltage.
tL
R
0R RIRiv
==
The voltage across the inductor is given by
R
tL
R
0L vRIdt
diLv ===
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Source-Free RC Network
Consider the circuit shown. LetvC(0) = V0. From KCL, we getfor t 0
0vR
1
dt
dvC C
C =+
i
R
vC+
-C
The solution can be shown to be
tRC
1
C Kv
=
At t=0, we get
KKV)0(v 00C ===
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Substitution givest
RC
1
0C V)t(v
=
R
tRC
1
0CC i
R
V
dt
dvCi ===
The current in the capacitor is is given by
From Ohms Law, we get the resistor current.
tRC
1
0CR
R
V
R
vi
==
Note: Every current and voltage in an RC
network is a decaying exponential with atime constant of=RC.
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The Exponential Function
t1
0X)t(x
=Given the function
When t=0, 00
0 XX)0(x ==
When t=, 01
0 X368.0X)(x ==
When t=2, 02
0 X135.0X)2(x ==
When t=3, 03
0 X050.0X)3(x ==
When t=4, 04
0 X018.0X)4(x ==
When t=5,0
5
0X007.0X)5(x
==
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Plot of the Exponential Function
X0
4321 5 t
.
.
.. . .
.
.
. . .
.
.
... . . . . .
..
.
.
..
.
.
..
...
0teX)t(xt
1-
0 =
Note: As seen from the plot, after t=5, or after 5time constants, the function is practically zero.
Department of Electrical and Electronics Engineering EEE 33 - p2
Comments:
1. When R is expressed in ohms, L in Henrysand C in Farads, the time constant is inseconds.
Note: For practical circuits, the exponentialfunction will decay to zero in less than 1 second.
2. For practical circuits, the typical values ofthe parameters are: R in ohms, L in mH,C in F.
secinRC
msecinRL
=
=3. Typically,
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Department of Electrical and Electronics Engineering EEE 33 - p25
A More General RL Circuit
The circuit shown has several resistors but onlyone inductor. Given
i1(0+)=I0=2 Amps,
find i1, i2, and i3 fort 0.
I0i16
i3i2
3 2
4 0.1H
First, determine theequivalent resistanceseen by the inductor.
6 3 2
4 a b
36
)3(642Rab
+++=
= 8
Department of Electrical and Electronics Engineering EEE 33 - p2
Next, find the time constant of the circuit.
sec80
1
R
L
ab
==
Every current will be described by the exponential
0tK t80
For example, we get
0tKi 80t11 =
At t=0+, i1(0+)=I0=2 Amps. Thus, we get
10
11 KK2)0(i ===+
Department of Electrical and Electronics Engineering EEE 33 - p27
Thus, we find the current i1 to be
0tAmps2i t801 =
The remaining currents, i2 and i3, can be foundusing current division. We get
112 i3
1i
63
3i =
+=
0tAmps3
2i t802 =
or
Similarly, we get
0tAmps3
4
it80
3 =
Department of Electrical and Electronics Engineering EEE 33 - p2
A More General RC Circuit
The circuit shown has several resistors but onlyone capacitor. Given
vC(0+)=V0=20 volts,
find i for t 0. i6k2K
3kvC+
-1F
k3k6
)k3(k6k2Rab
++=
= k4
First, determine theequivalent resistanceseen by the capacitor.
6k2K
3ka
b
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Next, find the time constant of the circuit.
msec4)F1)(k4(CRab ===
Any current or voltage will be described by theexponential
0tK t250
For example, we get
0tKv t250C =
At t=0+, vC(0+)=V0=20 volts. Thus, we get
KK20)0(v 0C ===+
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Thus, we find the Voltage vC to be
0tvolts20v t250C =
Applying current division, we get the current i(t).
0tmAe3.33)-i(k3k6
k6)t(i t250-C =
+=
The current in the capacitor is described by
0tmA5dtdvCit250CC ==
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RL Network with Constant Source
In the circuit shown, theswitch is closed at t = 0.Find current i(t) for t 0.
R
Li
E+
-
t=0
For t 0, we get fromKVL
ERidtdiL =+
The solution of a non-homogeneous differentialequation consists of two components:
1. The transient response
2. The steady-state response
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Transient Response: The solution of the homo-geneous differential equation; that is
0Ridt
diL t
t =+
The transient response for the RL circuit is
tLR
t Ki
=
Steady-State Response: The solution of thedifferential equation itself; that is
ERidt
diL ss
ss =+
Department of Electrical and Electronics Engineering EEE 33 - p33
0dt
diss =
The steady-state response is similar in form to theforcing function plus all its unique derivatives. Forconstant excitation, the steady-state response isalso constant.
Let iss=A, constant
Substitute in the differential equation
ERA0 =+or
R
E
A =
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Complete Response: The sum of the transientresponse and steady-state response.
0tKR
Eii)t(i
tL
R
tss +=+=
Initial Condition: For t
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Steady-State Response
The steady-state response is the solution of theoriginal differential equation.
(3) It is independent of the initial conditions; and
(4) It exists for as long as the source is applied.
(1) It is also called the forced response since itsform is forced on the electrical network by theapplied source;
(2) It is similar in form to the applied source plusall its unique derivatives;
The forced response is the response that will be left afterthe natural response dies out.
Department of Electrical and Electronics Engineering EEE 33 - p3
RC Network with Constant Source
In the circuit shown, theswitch is closed at t = 0.Assume vC(0)=V0. FindvC(t) for t 0.
R
iE
+
-
t=0
v+
-C
For t 0, we get from KVL
EvRi C =+
Since , we get
Evdt
dvRC C
C =+
dt
dvCi C=
Department of Electrical and Electronics Engineering EEE 33 - p39
Transient Response: For an RC network, we get
tRC
1
t,C Kv
=
Steady-State Response: Since the forcingfunction is constant, the steady-state response isalso constant.
Let vC,ss = A, constant
0dt
dv ss,C =
Evdt
dvRC
ss,C
ss,C =+
Substitute in
Department of Electrical and Electronics Engineering EEE 33 - p4
We get
orEA0 =+ EA =
Complete Response: Add the transient responseand steady-state response.
tRC
1
t,Css,CC KEvvv
+=+=
Evaluate K. At t=0+, we get
KEV)0(v 0C +==+
or EVK 0 =
Finally,we get
0t)EV(E)t(vt
RC
1
0C +=
Department of Electrical and Electronics Engineering EEE 33 - p41
L and C at Steady State
With all sources constant, then at steady-state,all currents and voltages are constant.
LI0
+ -vL
iC
V0+ -
C
0dt
dILv 0L ==
If the current isconstant, then
0dt
dVCi 0C ==
If the voltage isconstant, then
Note: With constant sources, L is short-circuitedand C is open-circuited at steady state condition.
Department of Electrical and Electronics Engineering EEE 33 - p4
Example: Find thecurrent and voltagesat steady state.
10
Li24V+
-
vR+ -
vL
+
-
Since the source is constant,the inductor is shorted atsteady state.
10
iss24V+
-
vR,ss+ -
vL,ss
+
-
A4.210
24iss ==
V24v ss,R =
0v ss,L =
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Example: Find thecurrent and voltagesat steady state.
10
Ci24V+
-
vR+ -
vC
+
-
Since the source is constant,the capacitor is open-circuited
at steady state.
10
iss24V+
-
vR,ss+ -
vC,ss
+
-
0iss =
0v ss,R =
V24v ss,C =
Department of Electrical and Electronics Engineering EEE 33 - p4
Example: Find theinductor current andcapacitor voltage atsteady state.
3
CiL
24V+
-vC+
-
L
9
At steady state, short
the inductor and openthe capacitor.
iL
3
24V+
-vC,ss
+
- 9
A212
24i ss,L ==
V18i9v ss,Lss,C ==
Department of Electrical and Electronics Engineering EEE 33 - p45
Example: Find theinductor currentsand capacitorvoltages at
4
iL224V
+
-vC1
+
-C3
iL1
vC3+
-
8
vC2+
-steady state.
Equivalent circuit at steady-state
4
IL224V
+
-VC1+
-
C3
IL1
VC3
+
-
8
VC2
+
-
0I 1L =
A2I 2L =
V16V 3C =
0V 2C =
V16V 1C =
Department of Electrical and Electronics Engineering EEE 33 - p4
Example: The switch isclosed at t=0. Find thecurrent i(t) for t 0.
4
10mH12Vi+
-
t=0
The transient current is
0tKKi t400t
L
R
t ==
The steady-state equivalent circuit for t 0
4
Iss12V
+
-
A34
12Iss ==
Department of Electrical and Electronics Engineering EEE 33 - p47
The complete solution
0tK3ii)t(i t400tss +=+=
Initial condition: At t=0+, i(0+)=0 since theinductor current cannot change instantaneously.
Evaluate K: At t = 0+,
0K30)0(i +==+
Thus, we get
0tA33)t(i t400 =
3K =or
Department of Electrical and Electronics Engineering EEE 33 - p4
RL and RC Networks
With constant sources, L is short-circuited and Cis open-circuited at steady state condition.
The solution of a non-homogeneous differentialequation consists of two components: the transienresponse and the steady-state response
t
L
R
KA)t(i
+=
RL Network withConstant Source
tRC
1
KA)t(v
+=
RC Network withConstant Source
transientresponse
steady-stateresponse
transientresponse
steady-stateresponse
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Example: The switch has been in position 1 for along time. At t=0, the switch is moved to position2. Find the current i(t) for t 0.
5k
1F i12V+
-
vC
+
-
6V+
-
10k
t=0
1 2
The circuit is at steady-statecondition prior to switching.
5k
12V+
-vC,ss
+
-)(0vV12v-
Css,C ==
Department of Electrical and Electronics Engineering EEE 33 - p5
Equivalent circuit for t 0
1F i E=6+
-
10k
From KVL, we get
EidtC
1Ri
t
=+ At t=0+,
E)0(v)0(Ri C =+++
or
R
)0(vE)0(i C
++ =
Since the capacitor voltage cannot changeinstantaneously,
V12)(0v)0(v -CC ==+
Department of Electrical and Electronics Engineering EEE 33 - p51
We get
mA6.0k10
126)0(i =
=+
The transient response is
0tKKi t100t
RC
1
t ==
The steady-state current is zero since the capacitorwill be open-circuited. Thus, the total current isequal to the transient current. Since i(0+)=-0.6 mA,we get
0tmA6.0)t(i
t100
=
VC(0+)
=12Vi 6V
+
-
10k
1uF
Department of Electrical and Electronics Engineering EEE 33 - p5
Comments:
1. The actual current flows in the clockwisedirection. The capacitor supplies the current.The 6-volt source is absorbing power.
2. The voltages across the resistor and capacitorcan be found to be
0tV6)t(Riv t100R ==
0tV66v6v t100RC +==
6V+VC-
i +
-
10k
1uF
- VR +
Department of Electrical and Electronics Engineering EEE 33 - p53
Comments:
3. The energy stored in the capacitor decreasesfrom 72 J to 18 J.
J72)12)(F1()0(Cv)0(W2
212
C21
C ===++
J18)6)(F1()(Cv)(W2
212
C21
C ===
The resistor will dissipate a total energy of 18 J.
J=
==
18dtk10
36dt
R
vW
0
t200
0
2
R
The 6V source will absorb a total energy of 36 J.
J===
36dt)mA6.0(6dtViW0
t100
0R
Department of Electrical and Electronics Engineering EEE 33 - p5
Example: The network has reached steady-statecondition with the switch in position 1. At t=0, theswitch is moved to position 2. Find i, vc1 and vC2for t 0. Assume that capacitor C2 is initiallyuncharged. 10k
5F i100V+
-vC1
+
-
2.5k
t=0
1 2
20FvC2+
-
The circuit is at steady-stateprior to switching.
10k
100V+
-vC1,s
+
-V100v ss,1C =
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Equivalent circuit at t=0+2.5k
i(0+)vC2(0
+)+
-
+
-C1vC1(0
+) C2
From KVL, we get
)0(v)0(Ri)0(v2C1C
+++ +=
V100)0(v 1C =+
0)0(v 2C =+
Substitution gives i(0+) = 40 mA.
Equivalent circuit for t 0
+
--
+
2.5k
i5F 20FF4Ceq =
ms10RCeq ==
Department of Electrical and Electronics Engineering EEE 33 - p5
The current for a source-free RC circuit is given by
0tKK)t(i t100t
RC
1
==
Since i(0+) = 40 mA, we get
0tmA40)t(i t100 =
The voltages are
0tV100)t(Riv t100R ==
++==+
t
02
2C
t
2
2C idtC
1)0(vidt
C
1v
0tV2020 t100 =
Department of Electrical and Electronics Engineering EEE 33 - p57
2CR1C vvv +=
0tV8020 t100 +=
Comments:
1. The current decays to zero but vC1 And vC2 donot decay to zero. At steady-state (t=),
V20VV ss,2Css,1C ==
2. The initial energy stored in C1 and C2
mJ25)0(vC)0(W 21C121
1C ==++
0)0(vC)0(W 22C2212C == ++
Department of Electrical and Electronics Engineering EEE 33 - p5
3. The final energy stored in C1 and C2
mJ1)(vC)(W 21C121
1C ==
mJ4)(vC)(W 22C221
2C ==
4. The total energy lost is 20 mJ.
5. The total energy dissipated by the resistor
mJ20dt4RdtiW0
t200
0
2R ===
Note: At t=0+, vC1=100 volts and vC2=0. CapacitoC1 supplies the current that charges capacitor C2.
The current stops when vC1 = vC2 =20 V.
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The circuit is at steady-state prior to switching.
36V
1k
IL,ss+
-
2k
12V+
-
mA30=
k2
36
k1
12I ss,L +=
Example: The network has reached steady-statecondition with the switch closed. At t=0, the switchis opened. Find i(t) 1k
i12V
+
- 36V
+
-
2kt=0
0.1H
for t 0.
Department of Electrical and Electronics Engineering EEE 33 - p6
Equivalent circuit for t 0 1k
i12V+
-0.1H
The transient current is
t000,10t
L
R
t KKi
==
At steady-state, the inductor is short-circuited.Thus, the steady state current is 12 mA.
The complete response is
0tmA += t000,10K12)t(i
Since i(0+) = 30 mA, we get K = 18 mA. The finalexpression is
0tmA += t000,101812)t(i
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First-Order RL and RC CircuitsGeneral Procedure
4. The solution is:
f(t) = f() + [f(0+) - f()] e-t/
1. Find f(0+), the initial value of the variable tobe solved.
2. Find f(), the final value of the variable to besolved.
3. Simplify the RC or RL circuit to get Req, Ceq or
Leq. The time constant is ReqCeq or Leq/Req.
Note: When solving for the initial and final values, treat thecapacitors as open circuits & the inductors as short circuits.
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3. If a switch changed state (closes or opens) att = t0, then
vC(t0+) = vC(t0
-)
The voltage across acapacitor cannot change
instantaneously.
iL(t0+) = iL(t0
-)
The current through aninductor cannot change
instantaneously.
NOTES:
1. f(t) = f() + [f(0+) - f()] e-t/
forced response natural response
All other voltages and currents can change instantaneously.
2. Req is the thevenin resistance seen by the
capacitor or inductor.
Department of Electrical and Electronics Engineering EEE 33 - p63
Example: In the circuit,vC1(0
-) = 12 Vand vC2(0
-) = 0 V.
t = 0
+
vC1
_
1 k
+
vC2
_
3 uF 6 uF
iR(t)
Find vC1(t), vC2(t) and iR(t).
Step 1: Initial conditions
12V)(0v)(0v C1C1 ==+
0V)(0v)(0v C2C2 ==+
12mA1k 0121k )(0v)(0v)(0iC2C1R ===
++
+
at t = 0+
+12 V_
1 k
+0 V_
3 uF 6 uF
12 mAAt t=0+:
Department of Electrical and Electronics Engineering EEE 33 - p6
After a very long time, iR() = 0.
Therefore, vC1() = vC2() or
2121 Q2Q
6u
Q
3u
Q==
Step 2: Final conditions
24uCQand12uCQ
2QQQQ36uC(12V)(3uF)
21
1121
==
+=+==
Initial charge stored = final charge stored
+
4 V
_
1 k
+
4 V
_
3 uF 6 uF
0 mA
Therefore, vC1() = 4 V
vC2() = 4 V
Department of Electrical and Electronics Engineering EEE 33 - p65
iR(t) = 0 + [12 - 0] e-t / 2ms = 12 e -t / 2ms mA
vC1(t) = 4 + [12 - 4] e-t / 2ms
= 4 + 8 e -t / 2ms V
vC2(t) = 4 + [0 - 4] e-t / 2ms
= 4 - 4 e -t / 2ms V
Step 3: Find the time constant,
Req = 1 k
Ceq = 3 uF in series with 6 uF = 2 uF
Therefore, = ReqCeq = (1 k)(2u) = 2 ms
Step 4: f(t) = f() + [f(0+) - f()] e-t/
Department of Electrical and Electronics Engineering EEE 33 - p6
iR
vC1
vC2
iR(t) = 12 e-t / 2ms mA
vC1(t) = 4 + 8 e-t / 2msV
vC2(t) = 4 - 4 e-t / 2msV
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Department of Electrical and Electronics Engineering EEE 33 - p67
Example: Find the inductorcurrent iL(t) and theinductor voltage vL(t).
Step 2: Final conditions
The inductor will behave like ashort circuit so
vL() = 0 V
iL() = 102400= 4.167 mA
2.4 k
80 uH10 V
t = 0iL(t)
+v
L(t)
_
2.4 k
80 uH10 V+
0 V
_
Step 1: Initial conditions
iL
(0+) = iL
(0-) = 0 vL
(0+) = 10 V
Department of Electrical and Electronics Engineering EEE 33 - p6
iL(t) = 4.167 + [0 - 4.167] e-t/33.33n
= 4.167 - 4.167 e-t/33.33n mA
Step 3: Find the time constant,
Step 4: f(t) = f() + [f(0+) - f()] e-t/
Req = 2.4 k Leq = 80 uH
Therefore = Leq/ Req = 33.33 ns
vL(t) = 0 + [10 - 0] e-t/33.33n V
= 10 e-t/33.33n V
Department of Electrical and Electronics Engineering EEE 33 - p69
1.8 2
x 1 0-4
iLvL
Forced response
iL(t) = 4.167 - 4.167 e-t/33.33n mA
vL(t) = 10 e-t/33.33uV
Transient response
Department of Electrical and Electronics Engineering EEE 33 - p7
Example: If the switch in the network closes att=0, find v0(t) for t>0.
+vo
4
+
- -3A
- + - +
4vA
24V 2vA
Step 1: Initial conditions
vA(0-)=3A(4)= 12V
+vC(0
-)
4
+
- -
3A
- + - +
4vA
24V 2vAAt t=0-
vC(0-)= 2vA+24+
= 60 V
2F
Department of Electrical and Electronics Engineering EEE 33 - p71
Step 2: Final conditions
+vo
4
+
- -3A
- + - +
4vA
24V
2vA
vA,ss = 0
v0,ss = 24V
Step 3: Find the time constant,
Since we have a dependent source, the equivalentresistance seen by the capacitor can be obtainedby finding vOC/iSC
At t=0+,
v0(0+) = vC(0
+) = vC(0-) = 60V
Department of Electrical and Electronics Engineering EEE 33 - p7
Determine vOC
+vOC
4
+
- -
3A
- + - +
4vA
24V 2vA
v0C= 24V
Get iSC
iSC
4
+
-
3A
- + - +
4vA
24V 2vA
From KVL,
2vA + vA = -24vA = -8V
The two resistors are in parallel, thus
2iSC 24 2vA = 0 iSC= 4 A
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Department of Electrical and Electronics Engineering EEE 33 - p73
The time constant is
= ReqC = 6(2F)= 12sec
Step 4: f(t) = f() + [f(0+) - f()] e-t/
v0(t) = 24 + [60 - 24] e-t/12 V
= 24 + 36 e-t/12 V
The equivalent resistance is
Req = vOC iSC = 24V / 4A = 6
Department of Electrical and Electronics Engineering EEE 33 - p7
Unit Step Forcing Function
>
0:
+_5 V
t
5u(t)
5V
Department of Electrical and Electronics Engineering EEE 33 - p7
TranslatedStep Function
>
0 or t < 2 s:
2 u(2 - t)
mA
2 mA
2 - t < 0 or t > 2 s:
2 t
2u(2-t)
2mA
Department of Electrical and Electronics Engineering EEE 33 - p7
Example: The circuit shown is initially at steady-state condition. Formulate the expression forvC(t) and iR(t) for t>0.
Evaluate the forcing function:
+vc
3k
6k+
--
24u(t) 24u(t-4ms) F1
iR
t
24u(t)
124V
t
24u(t-4ms)
24V
4ms
24u(t) - 24u(t-4ms
24V
4ms t
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Department of Electrical and Electronics Engineering EEE 33 - p79
We need to evaluate the circuit using two timeintervals: 0 < t < 4ms , voltage source = 24V
t > 4ms , voltage source = 0
First time interval: 0 < t < 4ms
At t= e305.2)'t(iR
Department of Electrical and Electronics Engineering EEE 33 - p8
1.8 2
x 10-4
iLvL
Transient and Steady-State Response
iL(t) = 4.167 - 4.167 e-t/33.33n mA
iL(0+)=0A iL()=4.167mA
vL(t) = 10 e-t/33.33uV
vL(0+
)=10V vL()=0V
=33.33 ns 5=1.67x10-4 s
Transient
response Steady-
Respo
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Department of Electrical and Electronics Engineering EEE 33 - p85
Example: The circuit shown is initially at steady-state condition. Formulate the expression forvC(t) and iR(t) for t>0.
Evaluate the forcing function:
+vc
3k
6k+
-
-24u(t) 24u(t-4ms) F1
iR
t
24u(t)
124V
t
24u(t-4ms)
24V
4ms
24u(t) - 24u(t-4ms)
24V
4ms t
Department of Electrical and Electronics Engineering EEE 33 - p8
Thus, the expression for vC and iR for t>0
16 16e-500t V, t < 4ms
13.83e-500(t-4ms) V, t > 4msvC(t) =
2.67 2.67e-500t mA, t < 4ms
2.305e-500(t-4ms) mA, t > 4msiR(t) =
Department of Electrical and Electronics Engineering EEE 33 - p87
Graph for vc(t)16 16e-500t V
13.83e-500(t-4ms)
VC(t)
(V)
VC(t)
(V)
t
16 16e-500t V, t < 4ms
13.83e-500(t-4ms) V, t > 4ms
=2 ms5= 10 ms
Department of Electrical and Electronics Engineering EEE 33 - p8
Graph for iR(t)2.67 2.67e-500t mA
2.305e-500(t-4ms) mA
iR(t)
(mA)
t
2.67 2.67e-500t mA, t < 4ms
2.305e-500(t-4ms) mA, t > 4ms
=2 ms5= 10 ms
Department of Electrical and Electronics Engineering EEE 33 - p89
Equivalent of Switching
General
Network
Vu(t-t0)+
_
General
Network+_V
Equivalent circuit
General
Network
Iu(t-t0)
t0 t
i(t)
General
NetworkI
Equivalent circuit
t0 t
v(t)
V
Department of Electrical and Electronics Engineering EEE 33 - p9
+
_
30
50
2 H2 u(t)
100 u(t)
iExample: Find i(t)for t>0.
When t < 0, the sourcesare off, thus i(0-) = 0 A
30
50
2 H
i
At t = 0+, the sourcesturn on
+
_
30
50
2 H2 A
100 V
i
i(0+) = i(0-) = 0 A
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Department of Electrical and Electronics Engineering EEE 33 - p91
Final condition: After a very long time, theinductor will behave like a short circuit
+
_
30
50
2 A
100 V
i
ix
From KCL, i + ix = 2
Thus, i = 2 A and ix = 0i() = 2 A
KVL yields
-100 30ix + 50i = 0
Time constant:
Leq = 2 H
Req = 30 + 50 = 80
= 0.025 s
Finally, i(t) = i() + [i(0+) i()]e-t/
i(t) = 2 + (0 2) e-t/0.025 = 2 - 2 e-40tA
Department of Electrical and Electronics Engineering EEE 33 - p9
Sinusoidal Sources
Consider the network shown.Let v(t)=Vm sin t where Vmand are constant.
R
Lv(t)
+
-
t=0
iFor t 0, we get from KVL
tsinVRidtdiL m =+
The transient response is
0tKit
L
R
t =
Remember: The transient response is independeof the source.
Department of Electrical and Electronics Engineering EEE 33 - p93
The steady-state response is the solution of thedifferential equation itself. Let
tcosKtsinKi 21ss +=
tsinK-tcosKdt
di21
ss =
tsinKL-tcosLK 21
tsinVtcosRKtsinRK m21 =++
Substituting in the original equation
tsinVRidt
diL m =+
gives
Department of Electrical and Electronics Engineering EEE 33 - p9
Substitution gives
tsinKL-tcosLK 21
tsinVtcosRKtsinRK m21 =++
Comparing coefficients, we get
21m KLRKV = 12 KLRK0 +=and
Solving simultaneously, we get
222
m1
LR
RVK
+
=222
m2
LR
LVK
+
=and
Department of Electrical and Electronics Engineering EEE 33 - p95
Substituting K1 and K2
t)cosLtsinR(LR
Vi222
mss
+=
The complete response is
t)cosLtsinR(LR
V)t(i
222
m +
=
0tKt
L
R
+
The steady-state response is
tcosKtsinKi 21ss +=