lecture_5 electrical engineering iit g

Electronics & Communication Engineering

Indian Institute of Technology Guwahati

Department of




Lecture 5

The Forced Solution

By

Dr. Praveen Kumar

Assistant Professor

Department of Electronics & Communication Engineering



The response to sinusoidal

source In Fig.1, the excitation is

(1)

The equilibrium equation is found from the relationship

(2)

Rearranging the terms of eq.2 gives

(3)

The right hand side of eq.3 is simplified as

(4)

The eq.4 clearly indicates that the assumed form of the forced solution must contain both a sine

and a cosine component in order to have left hand side equal to the right hand side.

1( ) 10sin(3 )e t t

2

1 2

1 2 4( ) ( ) 10sin(3 )

4 11 8f

ad

p pi t e t t

Z p p

2 2(4 11 8) ( ) (2 4 )10sin(3 )fp p i t p p t

2(2 4 )10sin(3 ) 180sin(3 ) 120cos(3 )p P t t t




source 1 1/ 2F 2

1

2H

110sin(3 )t

Figure 1: Network with sinusoidal excitation

a b c

d




source The assumed solution is

(5)

(6)

(7)

Using eq.5 to 7 to substitutes the values of if(t), pif(t) and p2if(t) into eq.4 gives

(8)

Equating the coefficients of sin(3t) and cos(3t) in eq.22 gives

(9)

(10)

The complete forced solution is

(11)

2

( ) sin(3 ) cos(3 )

( )( ) 3 cos(3 ) 3 sin(3 )

( ) 9 sin(3 ) 9 cos(3 )

f

f

f

f

i t A t B t

di tpi t A t B t

dt

p i t A t B t

( 28 33 )sin(3 ) (33 28 )cos(3 ) 180sin(3 ) 120cos(3 )A B t A B t t t

28 33 180

33 28 120

4.8 and 1.38

A B

A B

A B

( ) 4.8sin(3 ) 1.38cos(3 )fi t t t



The response to Polynomial

Sources The excitation function in the Fig.2 is

(12)

The function in eq.12 varies linearly with time and has a slope of 2. Because of its sloping

graphical representation, eq.12 is also referred to as ramp function.

The current if(t) is given by

(13)

Rearranging the eq.13 gives

(14)

Simplifying eq.14 gives

(15)

( ) 2f t t

2

2

1 2 4 2( ) ( ) (2 )

4 11 8f

p pi t e t t

Z p p

2 2(4 11 8) ( ) (2 4 2)(2 )fp p i t p p t

2(4 11 8) ( ) 8 4fp p i t t




Sources

Fig.2: Network with sinusoidal excitation

1 1/ 2F 2

11

2H2t

a b c

d




Sources The forced response can be written as the sum of a constant term plus a ramp. Hence

(16)

Substituting if(t) from eq.16 into eq.15 gives

(17)

Equating the like coefficients and solving for the unknown quantities gives

(18)

The complete expression for the forced solution becomes

(19)

( )fi t A Bt

11 8 8 8 4B A Bt t

5 1,

16 2A B

5 1( )

16 2fi t t



The Natural Response (Transient

Solution) The forced solution is the solution that is found to exist when the circuit has settled down in its

response to the disturbing effect of an applied source function. This forced solution is also known

as steady state solution.

The forced or the steady state solution always satisfies the defining differential equation but in

general it is not a valid a solution over the entire time domain.

To illustrate the point consider the circuit shown in Fig.3. In this network it is desired to find the

complete solution for the voltage v appearing across the resistor for all time t after the switch is

closed.

The equation that relates v to the source voltage E is

(20)

(21)

1

1 ( )

v R

LE R pLp

R

L dvv E

R dt




Solution)

Fig.3: An R-L circuit




Solution) The forced solution of eq.21 is

(22)

The solution in eq.22 satisfies eq.21, however this solution does not qualify as a solution of the

governing differential equation in the period immediately following the closing of the switch at

time

At time the current is zero because of the presence of the inductor. Hence v, which is iR, is

also zero. Hence, the solution in eq.22 cannot be taken to be a complete description.

In essence, then, there arises, in this time period, , immediately following the application of

the source function, a need to add a complementary function to the forced solution.

This complementary function will disappear as steady state is reached. It is the purpose of the

complementary function to provide a smooth transition from the initial state of the response in

the presence of energy storing elements to the final state.

fv E

0t

0t

0t




Solution) In Fig.3, the complementary function (natural response) is obtained by solving the equation

(23)

The solution to the eq.23 should satisfy that function, v and its derivative, dv/dt, must be of the

same form to make the left hand side equal to the right hand side.

The plausible solution for eq.23 is

(24)

Substituting v from eq.24 into eq.23 gives

(25)

(26)

0L dv

vR dt

stv Ke

1 0stLs Ke

R

Rs

L




Solution) Hence, the complementary solution becomes

(27)

The complete solution of the original governing differential equation (eq.23) is

(28)

The quantity K is found from the initial condition which requires that v to be zero at t=0+

(29)

Hence, the complete solution is

(30)

A graph of eq.30 is shown in Fig.4. The rapidity with which this transition takes place is entirely

dependent upon the value s. In eq.26 when R is large (or L is small) the transition occurs quickly

because the transient term dies out in little time.

( / )R L tv Ke

0 E K K E

( / )(1 )R L tv E e

( / )R L tv E Ke




Solution) vf=E

V=vf+vt

vt=-Ee-(R/L)t

Fig.4: Complete solution of voltage across the resistor in network 3



Example 1

Consider the network shown in Fig.5. The differential equation governing the voltage V to the

source E, is

(31)

The associated differential equation is

(323)

The forced solution is

(33)

The characteristics equation is

(34)

2 2

1

1 4

1 1 5 4

pCv E E

p LC pRC p pR pL

pC

2

25 4 4

d v dvv E

dt dt

2( ) ( ) 5 4 ( 4)( 1) 0D p D s s s s s

2

0

4

5 4f f

p

v E E v Ep p



Example 1

Fig.5: Network for the example



Example 1

Hence

(35)

Since there are two unknown coefficients, two initial conditions are required. The first is that v

has a zero value at time t=0+, then

(36)

To determine the second initial condition, differentiate the eq. 35

(37)

1 2

4

1 2

4

1 2

4, 1

Hence, the complementary solution is

Hence, the complete solution is

t t

t

t t

f t

s s

v K e K e

v v v E K e K e

1 2(0 ) 0v E K K

4

1 24 t tdvK e K e

dt



Example 1

The term dv/dt is the voltage drop across the capacitor. At time t=0+, the voltage across the

capacitor is zero because the inductor does not allow the current to flow the network

immediately. Hence,

(38)

Solving eq.49 and eq.51 gives

(39)

The complete solution is

(40)

4

1 2

1 2

4 0

4 0

t tdvK e K e

dt

K K

1 2

4,

3 3

EK K E

41 4(1 )

3 3

t tv E e e



-150

-100

-50

0

50

100

150

0 2 4 6 8 10 12

Example 1

41 41

3 3

t tE e e

Time

Volt

age

Fig.6: Response of network shown in fig.5



Example 2

Consider the circuit shown in Fig. 7. The configuration of this circuit is same as in Fig.5 except

for the fact that the resistance is reduced to 2 Ohms.

The voltage v is given by

(41)

The forced component of the solution is obtained by

(42)

The characteristic equation and its roots are

(43)

2

4

2 4v E

p p

2

0

4

2 4f f

p

v E E v Ep p

2

1

2

( ) ( ) 2 4

1 3

1 3

D p D s s s

s j

s j



Example 2

It is important to note that the roots are no longer real and distinct but are complex conjugates.

This should lead us to expect some essential difference in the behaviour of the circuit as it reacts

to the application of the source function.

This difference is seen by writing the expression for the transient response:

(44)

Applying Euler’s formula , eq.44 can be written as

(45)

It should be kept in mind that the left hand side of eq.45 is a real quantity. Moreover, vf is

associated with a differential equation that has real coefficients.

1 2 3 3

1 2 1 21/s t s t i t j t

fv K e K e e K e K e

cos sinie j

1 2 1 21/ cos 3 sin 3fv e K K t j K K t



Example 2

Hence, the right hand side of eq.45 must also necessarily have real coefficients.

This implies that the term j(K1-K2) merely means that K1-K2 must assume such a value as to

lead to a real coefficient.

For ease of handling, the eq.44 can be rewritten as

(46)

The complete expression of the total solution is

(47)

Upon applying initial conditions to eq.47, the values of k1 and k2 can be obtained.

1 21/ cos 3 sin 3fv E v E e k t k t

1 21/ cos 3 sin 3fv e k t k t



Example 2

The first initial condition is

(48)

To find k2, it is necessary to impose a second initial condition.

(49)

Using eq.48, k2 is found as

(50)

Substituting k1 and k2 into eq.47 gives

(51)

The response of the system is shown in Fig.7

1 11 cos 3 sin 3

3v E t t

e

1 1(0 ) 0v E k k E

2 10 3dv

k kdt

12

3 3

k Ek



Example 2

-20

0

20

40

60

80

100

120

140

0 2 4 6 8 10 12

Time

Volt

age

Fig.7: Response of network

lecture_5 electrical engineering iit g

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