Chapter 7 Laplace Transform & its applications · 2020. 4. 6. · ECE -2103: Network Analysis...

ECE -2103: Network Analysis

Department of Electronics and Communication Engineering, M.I.T. Manipal.

Chapter 7

Laplace Transform & its applications

1.1. The Laplace Transform

Let f (t) be a time function which is zero for t ≤ 0 and which is arbitrarily defined for t > 0. Then

the Laplace transform of f (t), denoted ℒ[f(t)], is defined by

(1)

Thus, the operation ℒ[ ] transforms f(t), which is in the time domain, into F(s), which is in the complex frequency domain, or simply the s-domain. The variable s is the complex number

(σ + jω). While it appears that the integration could prove difficult, it will soon be apparent that

application of the Laplace transform method utilizes tables which cover all functions likely to be

encountered in elementary circuit theory.

There is a uniqueness to the transform pairs; that is, if f1(t) and f2(t) have the same s-domain

image F(s), then f1(t) = f2(t). This permits going back in the other direction, from the s-domain to

the time domain, a process called the inverse Laplace transform, ℒ –1[F(s)] = f(t).

1.2 Laplace Transform of Functions:

The Laplace transform of the unit step function is easily obtained:

1.

2. If υ(t) = Vu(t) then in s-domain, V(s) = V/s.

Learning Outcomes:

At the end of this module, students will be able to:

1. Explain about Laplace Transform.

2. Know how to perform an inverse Laplace transform using partial

fraction expansion.

3. Apply the initial-value and final-value theorems.

4. Know how to use the Laplace transform to obtain the response.



The exponential decay function is another time function which is readily transformed.

3. ℒ[Ae-at

] = A / (s+a)

4.

5.

6.

Integrating by parts,



Table 7.1. Laplace Transform Pairs



EXAMPLE 1.3 Consider a series RL circuit, with R = 5 Ω and L = 2.5 mH. At t = 0, when the

current in the circuit is 2 A, a source of 50 V is applied. The time-domain circuit is shown in 7.2.

Figure 7.2

Kirchhoff's voltage law, applied to the circuit for t > 0, yields the familiar differential equation

(i). This equation is transformed, term by term, into the s-domain equation (ii). The unknown

current i(t) becomes I(s), while the known voltage υ = 50u(t) is transformed to 50/s. Also, di/dt is

transformed into –i(0+) + sI(s), in which i(0+) is 2 A. Equation (iii) is solved for I(s), and the

solution is put in the form (iv) by the techniques of Section 16.6. Then lines 1, 3, and 16 of Table

7.1 are applied to obtain the inverse Laplace transform of I(s), which is i(t). A circuit can be

drawn in the s-domain, as shown in Fig. 7.3. The initial current appears in the circuit as a voltage

source, Li(0+).

http://accessengineeringlibrary.com/browse/schaums-outline-of-electric-circuits-sixth-edition/c9780071830454ch16#ch16fig01



Figure 7.3

1.4 Initial-Value and Final-Value Theorems:

Taking the limit as s → ∞ (through real values) of the direct Laplace transform of the derivative

df(t)/dt,

But e–st in the integrand approaches zero as s → ∞. Thus,

Since f(0+) is a constant, we may write

which is the statement of the initial-value theorem.

EXAMPLE 1.4 In Example 1.3 find i(0+) using initial value theorem.

Hence the initial current, i(0+) = 2 A.

The final-value theorem is also developed from the direct Laplace transform of the derivative,

but now the limit is taken as s → 0 (through real values).



But

and f(0+) is a constant. Therefore,

or

This is the statement of the final-value theorem. The theorem may be applied only when all poles

of sF(s) have negative real parts. This excludes the transforms of such functions as et and cos t,

which become infinite or indeterminate as t → ∞

1.5 Partial-Fractions Expansions

The unknown quantity in a problem in circuit analysis can be either a current i(t) or a voltage

υ(t). In the s-domain, it is I(s) or V(s) which may be of the form

where the polynomial Q(s) is of higher degree than P(s).

Case 1: s = a is a simple pole. When s = a is a non repeated root of Q(s) = 0, the corresponding

principal part of R(s) is

If a is real, so is A; if a is complex, then a* is also a simple pole and the numerator of its

principal part is A*. Notice that if a = 0, A is the final value of r(t).

Case 2: s = b is a double pole. When s = b is a double root of Q(s) = 0, the corresponding

principal part of R(s) is



where the constants B2 and B1 may be found from

B1 may be zero. Similar to Case 1, B1 and B2 are real if b is real. The constants for the double

pole b* (if it exists) are the conjugates of those for b.

The principal part of a higher-order pole can be obtained by analogy to Case 2; we shall assume,

however, that R(s) has no such poles. Once the partial-functions expansion of R(s) is known,

Table 6-1 can be used to invert each term and thus obtain the time function r(t).

EXAMPLE 1.6 Find the time-domain current i(t) if its Laplace transform is

Factoring the denominator,

We see that the poles of I(s) are s = 0 (double pole) and s = ± j (simple poles).

The principal part at s = 0 is

Since

The principal part at s = + j is



Since

It follows at once that the principal part at s = – j is

The partial-fractions expansion of I(s) is, therefore,

and term-by-term inversion using Table 7.1 gives

1.7 Heaviside Expansion Formula

If all poles of R(s) are simple, the partial-fractions expansion and term wise inversion can be

accomplished in a single step:

where a1, a2, …, an are the poles and Q′(ak) is dQ(s)/ds evaluated at s = ak.



1.8 Circuits in the s-Domain

In this section, we extend the use of the complex frequency to transform an RLC circuit,

containing sources and initial conditions, from the time domain to the s-domain.

Table 7.4

Table 7.4 exhibits the elements needed to construct the s-domain image of a given time-domain

circuit.

-

-

-

- -

-



EXAMPLE 7.5 In the circuit shown in Fig. 7.6(a) an initial current i1 is established while the

switch is in position 1. At t = 0, it is moved to position 2, introducing both a capacitor with initial

charge Q0 and a constant-voltage source V2. Draw the s-domain circuit & s-domain equation .

Figure 7.6

The s-domain circuit is shown in Fig. 6.4(b). The s-domain equation is

in which V0 = Q0/C and i(0+) = i1 = V1/R.

EXAMPLE 7.6 Find the current developed in a series RLC circuit in response to the following

two voltage sources applied to it at t = 0: (a) a unit-step, (b) a unit-impulse.

The inductor and capacitor contain zero energy at t = 0–. Therefore, the Laplace transform of the

current is I(s) = V(s)Y(s).

(a) V(s) = 1/s and the unit-step response is



where

(b) V(s) = 1 and the unit-impulse response is

(The unit-impulse response may also be found by taking the time-derivative of the unit-step

response.)

EXAMPLE 7.7 Find the voltage across the terminals of a parallel RLC circuit in response to the

following two current sources applied at t = 0: (a) a unit-step, (b) a unit-impulse.

Again, the inductor and capacitor contain zero energy at t = 0–. Therefore, the Laplace transform

of the current is V(s) = I(s)Z(s).

(a) (s) = 1/s and the unit-step response is

where

(b) I(s) = 1 and the unit-impulse response is



7.8. Find the Laplace transform of e–at cosω t, where a is a constant.

Applying the defining equation to the given function, we obtain

7.9 If ℒ [f(t)] = F(s), show that ℒ = [e–atf(t)] = F(s + a). .

By definition, . Then,

(5)

Example:

as determined in Problem 6.1.

7.10 Find the Laplace transform of f(t) = 1 – e–at, where a is a constant.



Alternative Method:

7.11 Find

Using the method of partial fractions,

and the coefficients are

Hence,



Alternative Method:

7.12 Find

Using the method of partial fractions, we have

Then

Hence,

The corresponding time functions are found from Table 6-1:



7.13 In the series RC circuit of Fig. 7.14, the capacitor has an initial charge of 2.5 mC. At t = 0,

the switch is closed and a constant-voltage source V = 100 V is applied. Use the Laplace

transform method to find the current.

Figure 7.14

The time-domain equation for the given circuit after the switch is closed is

or

(3)

Q0 is opposite in polarity to the charge which the source will deposit on the capacitor. Taking the

Laplace transform of the terms in (3), we obtain the s-domain equation

or

(4)

The time function is now obtained by taking the inverse Laplace transform of (4):

(5)



7.15 In the RL circuit shown in Fig. 7.16, the switch is in position 1 long enough to

establish steady-state conditions, and at t = 0 it is switched to position 2. Find the

resulting current using Laplace Transform method.

Figure 7.16.

Assume the direction of the current is as shown in the diagram. The initial current is then i0 = –

50/25 = –2 A.

The time-domain equation is

(6)

Taking the Laplace transform of (6),

(7)

Substituting for i(0+),

(8)

and

(9)

Applying the method of partial fractions,

(10)

with



Then,

(11)

Taking the inverse Laplace transform of (11), we obtain i(t) = 4 – 6e–2500t.

7.17 In the series RL circuit of Fig.7.18, an exponential voltage υ = 50e–100t is applied by closing

the switch at t = 0. Find the resulting current.

Figure 7.18

The time-domain equation for the given circuit is

(12)

In the s-domain, (15) has the form

(13)

Substituting the circuit constants and the transform of the source, V(s) = 50/(s + 100), in (13), we

get

(14)

By the Heaviside expansion formula,

[

T

y

p

e

a

q

u

o

t

e

50

550



Here, P(s) = 250, Q(s) = s2 + 150s + 5000, Q′(s) = 2s + 150, a1 = –100, and a2 = –50. Then,

7.19 The series RC circuit of Fig. 7.20 has a sinusoidal voltage source υ = 180 sin (2000t + φ)

volts and an initial charge on the capacitor Q0 = 1.25 mC with polarity as shown. Determine the

current if the switch is closed at a time corresponding to φ = 90°.

Figure 7.20

The time-domain equation of the circuit is

(15)

The Laplace transform of (15) gives the s-domain equation

(16)

or

(17)

Applying the Heaviside expansion formula to the first term on the right in (17), we have P(s) =

4.5s2, Q(s) = s3 + 103

s2 + 4 × 106

s + 4 × 109, Q′(s) = 3s2 + 2 × 103

s + 4 × 106, a1 = –j2 × 103, a2 =

j2 × 103, and a3 = –103. Then,



(18)

At t = 0, the current is given by the instantaneous voltage, consisting of the source voltage and

the charged capacitor voltage, divided by the resistance. Thus,

The same result is obtained if we set t = 0 in (18).

7.21. In the series RL circuit of Fig. 7.22, the source is υ = 100 sin (500t + φ) volts.. Determine

the resulting current if the switch is closed at a time corresponding to φ = 0°.

Figure 7.22

The s-domain equation of the series RL circuit is

(19)

The transform of the source with φ = 0° is



Since there is no initial current in the inductance, Li(0+) = 0. Substituting the circuit constants

into (19),

(20)

Expanding (20) by partial fractions,

(21)

The inverse Laplace transform of (21) is

7.23. Rework Problem 7.21 by writing the voltage function as

(22)

Now V(s) = 100/(s – j500), and the s-domain equation is

Using partial fractions,

and inverting,

(23)

The actual voltage is the imaginary part of (22); hence the actual current is the imaginary part of

(23).



7.24. In the series RLC circuit shown in Fig.7.25, there is no initial charge on the capacitor. If

the switch is closed at t = 0, determine the resulting current.

Figure 7.25

The time-domain equation of the given circuit is

(24)

Because i(0+) = 0, the Laplace transform of (27) is

(25)

or

(26)

Hence,

(27)

Expanding (27) by partial fractions,

(28)

and the inverse Laplace transform of (28) is



7.26 In the two-mesh network of Fig.7.27, the two loop currents are selected as shown. Write the

s-domain equations in matrix form and construct the corresponding circuit.

Figure 7.27

Writing the set of equations in the time domain,

(29)

Taking the Laplace transform of (29) to obtain the corresponding s-domain equations,

(30)

When this set of s-domain equations is written in matrix form,



The required s-domain circuit can be determined by examination of the Z(s), I(s), and V(s)

matrices & is as shown.

Figure 7.28

7.29 In the two-mesh network of Fig.7.30, find the currents which result when the switch is

closed.

Figure 7.30

The time-domain equations for the network are

(31)

Taking the Laplace transform of set (31),

(32)

From second set of equation (32) we find



(33)

which when substituted into the first set of equation (32) gives

(34),

Finally, substitute (34) into (33) and obtain

7.31 Apply the initial- and final-value theorems to Problem 7.29.

The initial value of i1 is given by

and the final value is

The initial value of i2 is given by

and the final value is



7.31 In the two-mesh network shown in Fig. 7.32 there is no initial charge on the capacitor. Find

the loop currents i1 and i2 which result when the switch is closed at t = 0.

Figure 7.32

The time-domain equations for the circuit are

The corresponding s-domain equations are

Solving,

Therefore,



7.33. In the network of Fig.7.34 the switch is closed at t = 0 and there is no initial charge on

either of the capacitors. Find the resulting current i.

Figure 7.34

The network has an equivalent impedance in the s-domain

Hence, the current is

Expanding I(s) by partial fractions,

7.36 SUPPLEMENTARY PROBLEMS

Q.75 Find the Laplace transform of each of the following functions.



Q.76 Find the inverse Laplace transform of each of the following functions.

Q.77 A series RL circuit, with R = 10 Ω and L = 0.2 H, has a constant voltage V = 50 V applied

at t = 0. Find the resulting current using the Laplace transform method.

Ans. i = 5 – 5e–50t (A)

Q.78 In the series RL circuit of Fig.Q.78, the switch is in position 1 long enough to establish the

steady state and is switched to position 2 at t = 0. Find the current.

Figure Q.78



Q.79 In the circuit shown in Fig.79, switch 1 is closed at t = 0 and then, at t = t′, switch 2 is

opened. Find the current in the intervals 0 < t < t′ and t > t′.

Ans. i = 2(1 – e–500t) A, i = 1.06e–1500t(t–t′) + 0.667 (A)

Figure Q.79

Q.80 In the series RL circuit shown in Fig.80, the switch is closed at position 1 at t = 0 and then,

at t = t′ = 4, it is moved to position 2. Find the current in the intervals 0 < t < t′ and t > t′.

Ans. i = 0.1(1 – e–2000t)(A), i = 0.06e–2000(t–t′) – 0.05 (A)

Figure Q.80

Q.81 A series RC circuit, with R = 10 Ω and C = 4 μF, has an initial charge Q0 = 800 μC on the

capacitor at the time the switch is closed, which results in applying a constant-voltage source V =

100 V. Find the resulting current transient if the charge is (a) of the same polarity as that

deposited by the source, and (b) of the opposite polarity.

Ans.



Q.82 A series RC circuit, with R = 1 kΩ and C = 20 μF, has an initial charge Q0 on the capacitor

at the time the switch is closed, which results in applying a constant-voltage source V = 50 V. If

the resulting current is i = 0.075e–50t (A), find the charge Q0 and its polarity.

Ans. 500 μC, opposite polarity to that deposited by source

Q.83 In the RC circuit shown in Fig.Q.83, the switch is closed at position 1 at t = 0 and then, at t

= t′ (the time constant), is moved to position 2. Find the transient current in the intervals 0 < t < t′

and t > t′.

Ans. i = 0.5e–200t (A), i = –.0516e–200(t–t′) (A)

Figure Q.83

Q.84 In the circuit of Fig.Q.84, Q0 = 300 μC at the time the switch is closed. Find the resulting

current transient.

Ans.

Figure Q.84



Q.85. In the circuit shown in Fig.Q.85, the capacitor has an initial charge Q0 = 25 μC and

the sinusoidal voltage source is υ = 100 sin (1000t + φ) (V). Find the resulting current if

the switch is closed at a time corresponding to φ = 30°.

Ans. i = 0.1535e–4000t + 0.0484 sin (1000t + 106°) (A)

Figure Q.85

Q.86 A series RLC circuit, with R = 5 Ω, L = 0.1 H, and C = 500 μF, has a constant voltage

V = 10 V applied at t = 0. Find the resulting current.

Ans. i = 0.72e–25t sin 139t (A)

Q.87 In the series RLC circuit of Fig.Q.87, the capacitor has an initial charge Q0 = 1 mC and the switch is in position 1 long enough to establish the steady state. Find the transient current which

results when the switch is moved from position 1 to 2 at t = 0.

Ans. i = e–25t(2 cos 222t – 0.45 sin 222t) (A)

Figure Q.87

Q.88 A series RLC circuit, with R = 5 Ω, L = 0.2 H, and C = 1 F has a voltage source υ

= 10e–100t (V) applied at t = 0. Find the resulting current.

Ans. i = –0.666e–100t + 0.670e–24.8t – 0.004e–0.2t (A)



Q.89 A series RLC circuit, with R = 200 Ω, L = 0.5 H, and C = 100 μF has a sinusoidal

voltage source υ = 300 sin (500t + φ) (V). Find the resulting current if the switch is

closed at a time corresponding to φ = 30°.

Ans. i = 0.517e–341.4t – 0.197e–58.6t + 0.983 sin (500t – 19°) (A)

Q.90 A series RLC circuit, with R = 5 Ω, L = 0.1 H, and C = 500 μF has a sinusoidal voltage

source υ = 100 sin 250t (V). Find the resulting current if the switch is closed at t = 0.

Ans. i = e–25t(5.42 cos 139t + 1.89 sin 139t) + 5.65 sin(250t – 73.6°) (A)

Q.90 In the two-mesh network of Fig.90, the currents are selected as shown in the

diagram. Write the time-domain equations, transform them into the corresponding s-

domain equations, and obtain the currents i1 and i2.

Ans.

Figure Q.90

Q.91 For the two-mesh network shown in Fig.Q.91, find the currents i1 and i2 which

result when the switch is closed at t = 0.

Ans. i1 = 0.101e–100t + 9.899e–9950t (A), i2 = –5.05e–100t + 5 + 0.05e–9950t (A)

Figure Q.91



Q.92. In the network shown in Fig.Q.92, the 100-V source passes a continuous current in the

first loop while the switch is open. Find the currents after the switch is closed at t = 0.

Ans. i1 = 1.67e–6.67t + 5 (A), i2 = 0.555e–6.67t + 5 (A)

Figure Q.92

Q.93 The two-mesh network shown in Fig. 93 contains a sinusoidal voltage source υ =

100 sin (200t + φ) (V). The switch is closed at an instant when the voltage is increasing at

its maximum rate. Find the resulting mesh currents, with directions as shown in the

diagram.

Ans. i1 = 3.01e–100t + 8.96 sin (200t – 63.4°) (A), i2 = 1.505e–100t + 4.48 sin (200t – 63.4°) (A)

Figure Q.93

Q.94 In the circuit of Fig. 6.30, υ(0) = 1.2 V and i(0) = 0.4 A. Find υ and i for t > 0.

Ans:

.



Figure Q.94

Q.95 In the circuit of Fig.Q.95, ig(t) = cos tu(t). Find υ and i.

Figure Q.95

Ans. υ = 0.8305 cos (t – 48.4°), t > 0

i = 0.2626 cos (t – 66.8°), t > 0



Summary

1. In applying the Laplace transform, we convert an integrodifferential equation

in the time domain to an algebraic equation, which includes initial conditions, in the s-domain.

We solve for the unknowns in the s-domain and convert the results back to the time domain.

2. Laplace transform pairs, as listed in the Table, can be used to convert back and forth between the

time and frequency domains.

3. The convolution of two functions in the time domain corresponds to a simple multiplication of the

two functions in the s-domain.

4. The initial and final values of a time-domain function can be obtained from its Laplace transform

in the frequency domain.



Laplace Transform of Periodic and Non periodic signals:

About Lapalce Transform:

Laplace transform is yet another operational tool for solving constant coefficient linear

differential equations. The process of solution consists of three main steps:

The given hard problem is transformed into a simple equation.

This simple equation is solved by purely algebraic manipulations.

The solution of the simple equation is transformed back to obtain the solution of the given problem.

In this way the Laplace transformation reduces the problem of solving a differential equation to

an algebraic problem. The third step is made easier by tables, whose role is similar to that of

integral tables in integration. The above procedure can be summarized by Figure

In this section we introduce the concept of Laplace transform and discuss some of its properties.

The Laplace transform is defined in the following way. Let f(t) be defined for t ≥ 0. Then the

Laplace transform of f(t) which is denoted by L[f(t)] or by F(s), is defined by the following

equation.

dtetfdtetfT

sFtfL St

T

St

00

)()(0

lim)()]([

Learning Outcomes:

At the end of this module, students will be able to:

1. Mathematically define fundamental signals such as step, ramp,

exponential, sinusoidal and other waveforms.

2. Represent complex waveform using fundamental waveforms.

3. Find the Laplace domain representation of complex signals.

4. Find the time response for a given circuit excited by particular input

using Laplace transform technique.



Laplace Tranform of Periodic Signal :

Fundamental signals and their Laplace Transform:

1. Unit step function :

0

0,1)(

tforty

ssY

1)(

2. Exponential function : y(t)=

assY

1)(



3.Impulse Function :

Y(s)=1

4.Sine Function : tty sin)(

Y(s) = Ѡ/(S2+Ѡ2)

5.Ramp Function : )()( ttuty

2

1)(

ssY



List of Lapalce Tranforms for some of the functions



Properties of Laplace Transform:

Laplace transform of complex waveforms:

1.Pulse Function

Pulse wave can be obtained by subtracting the shifted version of unit step function from the unit

step function as shown below.



Hence the pulse function can be mathematically described as y (t) = u(t) − u(t −1)

Applying the lapalce transformation.

s

ee

sssY

SS

111)(



Laplace Transform of Periodic Signal

Let us consider a function x(t) that is periodic as shown in Fig. The function x(t) can be

represented as the sum of time-shifted functions as shown in Fig. below.

Hence ........)()()()()( 4321 txtxtxtxtx

.......)3()3()2()2()()()()( 1111 TtuTtxTtuTtxTtuTtxtxtx

where )(1 tx is the waveform described over the first period of x(t). i.e. )(1 tx is the same as

the function x(t) gated over the interval 0< t < T.

Taking the Laplace transform on both sides of above equation with the time-shift property

applied,

we get ........)()()()( 2

111 SS esXesXsXsX

........)1)(()( 2

1 TSTS eesXsX

STe

sXsX

1

)()( 1

And note that )(1 sX is the Laplace transform of x(t)defined over first period only.



Hence, we have shown that the Laplace transform of a periodic function is the Laplace transform

evaluated over its first period divided by TSe1 .

Example 7.36

Describe the periodic waveform in terms of fundamental functions and then find the

Laplce transform.

For the waveform shown the period is T=2 sec.

For periodic waveform the Lapalce transform is given by

STe

sXsX

1

)()( 1 where )(1 sX is the lapalce transform of x(t) defined over a interval 0< t < 2.

The signal x(t) considered over one period is donoted as )(1 tx and shown in Fig

Now )(*)()(1 tgtxtx A

)1()()(

)1()1()()(

)1()()1()1()1()(

)1()()1()11()(

)1()()1()(

)]1()(][1[

trtrtu

tutttutu

tututututttu

tutututttu

tututtuttu

tutut

Taking the Laplace Transform we get

Sesss

SX 221

111)(

STe

sXsX

1

)()( 1

)1(

122 S

S

es

es



Example 7.37

Here STe

sFsF

1

)()( 1 where F1(s) is the Laplace transform of

function f(t) defined over the interval 0 < t < 2

)2()1(2)()(1 tutututf

Taking the lapalce transform we get

SS es

ess

sF 2

1

121)(

Therefore )1(

)1(

)1(

21)(

2

2

2

2

S

S

S

SS

es

e

es

eesF

Exercise:

Obtain the Laplace Transform of waveforms shown below.

Q.96

Q.97



1.20 Laplace Transform of Non Periodic Signal:

Example 7.38

)4(5)3(3)1(3)(5)( tututututf

Applying the Laplace transform we get

s

eeesF

SSS 43 5335)(

Example 7.39

Express f(t) in terms of singularity functions and then find F(s)

Solution :

To find f(t) for 2)(0 tf

Equation of the straight line 1 is

12

12

1

1

xx

yy

xx

yy

Here, y is f(t) and x is t

Hence,

02

33

0

3)(

t

tf

ttf 33)(



To find f(t) for 3)(2 tf

23

30

2

3)(

t

tf

ttf 93)(

Hence

0

93

33

)( t

t

tf

otherwise

t

t

32

20

[Correct 3 – 9t as 3t – 9]

The above equation may also be written as :

)3()2(93)2()(33)( tututtututtf

= )3(9)3(3)2(6)(3)2(12)(3 tuttuttuttututu

= )3(9)3()33(3)2()22(6)(3)2(12)(3 tututtutttututu

= )3(9)3()3(3)2()2(6)(3)(3 tututtutttutu

Hence )()( tfLsF

2

3

2

2

2

3633)(

s

e

s

e

sssF

SS

+ [9/s]e-3S

Exercise:

Q.98 For the following non periodic waveforms, find the Laplace transform



1.10 Application of Laplace Transform:

Example 7.40: A voltage pulse of 10 V magnitude and 5μs duration is applied to the RC

network shown in Fig. Find the current i(t) if R=10Ω and C=0.05μF.

)(10)(10)( totututv where to=5μs.

Hence )1(10

)( toSes

sV

Assuming all initial conditions to be zero.

Now from the Laplace transformed network shown.

sCR

sVsI

1

)()(

sCRs

esI

toS

1

)1(10)(

1)1(10

)(

sRCs

esCsI

toS

Perfom partial fraction

RCs

e

RCs

RsI

toS

11

110)(

Taking the inverse Laplace transform to get current in time domain.

)(10

)(10

)(

)(

totueR

tueR

ti RC

tot

RC

t

Substitute for to=5μs



Therefore )105(10

)(10

)( 6105.0

)105(

105.0 6

6

6

tueR

tueR

ti

tt

7.41 Find )(

)()( 0

sV

sVsH

i

for the circuit shown in Fig. (b) Determine )(0 tV when the intital

current in the inductor is zero.

The Laplace transformed network with all initial conditions set to zero is shown in Fig

ssIsV 3

0 102150)()(

ss

ssVi

33

3

102150103100

102150)(

Therefore s

s

sV

sVsH

i 5105.2

2105.1

)(

)()(

5

5

0

b) )()()(0 sHsVsV i

ss

ssV

100

5105.2

2105.1)(

5

5

0

On performing partial fraction



50105.0

2060)(

sssV

Taking inverse Laplace transform, we get

)(2060)( 1055.0

0 tuetV t volt

Exercise:Q.99

1. Determine the output v(t) for the circuit shown in figure.(Assumption: all initial

conditions are zero).

Ans: )(2)( tutetv t

Q.100 Using Laplace transform method find the impulse response of the circuit shown in

Fig. below

Ans : )()( tuteth t



Summary

1. Complex wave forms can be represented as a combination of step, ramp, sinusoidal,

exponential and other fundamental signals.

2. The time response of a circuit for complex signal input can be found by

transforming both circuit and input complex signal to Laplace domain first and then

after performing the operations in Laplace domain the equivalent resultant signal in

time domain can be obtained by taking Inverse Laplace Transform.

3. Using Laplace transform technique it is easier to find the circuit response as

compared to finding the solution from differential equations technique.

Chapter 7 Laplace Transform & its applications · 2020. 4. 6. · ECE -2103: Network Analysis...

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Transcript of Chapter 7 Laplace Transform & its applications · 2020. 4. 6. · ECE -2103: Network Analysis...