Ch05 Transient Analysis 1s09

8/9/2019 Ch05 Transient Analysis 1s09

1/16

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)


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:


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


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


Linearity - Summary

y2

dt

dy

dt

dyy

dt

dyt

2

dt

dy

yt)sin32( + yy )32( 2

Non-linearLinear

2y )sin(yor


2/16


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)


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


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


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.


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)



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


3/16



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 =


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


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


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 =


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 ===


4/16


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 ===


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 ===


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.


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

==


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.


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,


5/16


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


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 ===+


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 =


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


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 ===+


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 ==


6/16


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


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 =+


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 =


Complete Response: The sum of the transientresponse and steady-state response.

0tKR

Eii)t(i

tL

R

tss +=+=

Initial Condition: For t


7/16


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.


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=


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


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 +=


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.


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 =


8/16


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 =


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 ==


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 =


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 ==


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


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


9/16


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 ==


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 ==+


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


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 +


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


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 =


10/16


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 ==


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 =


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 == ++


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.


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.


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


11/16


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.


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.


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+:


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


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/


iR

vC1

vC2

iR(t) = 12 e-t / 2ms mA

vC1(t) = 4 + 8 e-t / 2msV

vC2(t) = 4 - 4 e-t / 2msV


12/16


Example: Find the inductorcurrent iL(t) and theinductor voltage vL(t).


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

_


iL

(0+) = iL

(0-) = 0 vL

(0+) = 10 V


iL(t) = 4.167 + [0 - 4.167] e-t/33.33n

= 4.167 - 4.167 e-t/33.33n mA


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


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


Example: If the switch in the network closes att=0, find v0(t) for t>0.

+vo

4

+

- -3A

- + - +

4vA

24V 2vA


vA(0-)=3A(4)= 12V

+vC(0

-)

4

+

- -

3A

- + - +

4vA

24V 2vAAt t=0-

vC(0-)= 2vA+24+

= 60 V

2F



+vo

4

+

- -3A

- + - +

4vA

24V

2vA

vA,ss = 0

v0,ss = 24V


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


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


13/16


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


Unit Step Forcing Function

>

0:

+_5 V

t

5u(t)

5V


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


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


14/16


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


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


15/16


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


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) =


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


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


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


+

_

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


16/16


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


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.


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


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


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 +=

Documents

Ch05 Transient Analysis 1s09