9 Laplace Transform and

Its Application

9.1 INTRODUCTION
Time-domain analysis is the conventional method of analysing a network. For a simple network with first-
order difterential equation of network variable. this method is very useful. But as the order of network variable

becomes very tedious. For such applications, frequency domain

cquation increases, this method of analysis
is very convenient. Time-domain analysis, also known as classical method,
analysis using Laplace transform
IS difficult to apply to a differential equation with excitation functions which contain derivatives. Laplace
transform methods prove to be superior. The Laplace transform method has the following advantages:
() Solution of differential equations is a sysiematie procedure.
(2) Initial conditions are automatically incorporated.
(3) It gives the complete solution, 1.e., both compiementary and particular solution in one step

9.2 LAPLACE TRANSFORMATION

The Laplace transform of a

function (1) is defined as

...(9.1)
Pls) £ lf(]= J S)e"dr
=

where s is the complex Irequeney variable

C

.9.2)
is t 0 instead ofi =
0".
Here. the lower limit of integration =

The function ft) must satisty the lollowing condilion to possess a Laplace ransform.

.(9.3)
9.2 1lec a l Nenoks

where 0 is reál and ositive.

The
merseLaplaceuansorm '|Fo) is
S)= Fs)"ds ..(9.4)

9.3 LAPLACE TRANSFORM OF SOME IMPORTANT FUNCTIONS

9.3.1 Unit Step Function

The unit step function is defined by the equation,
u)=
= 0 0
The Laplace transform of unit step function is,

u 01 =
u0)e"d
Fig 9.1
..(9.5)

9.3.2 Delayed or Shifted Unit Step Function

The delayed or shifted unit step function is defined by the equation
u(-a)=
0
I>a

The Laplace iranstorm ot u (1 -a) Is

L [u (t - a)] = JI.e" dr

Fig. 9.2
..(9.6)

9.3.3 Unit Ramp Function

The unit ramp function is defined by the equation
) = >0
0
r()
The Laplace transtorm of the unit ramp function is

r n] J rt)e" dr
=

Fig. 9.3
=
t e dr = (9.7)
 and s ApplatoIS .
4nice I ransform

9.3.4
Delayed Unit Ramp Function
The
delayed unit ramp function is defined by the cquation
r ( t --u)

r(t-a) = t 1>a

The Laplace transfom of r (t-a) 1s

r(t- a)] = Jyed= ..9.8) 9.4

s
9.3.5 Unit Impulse Function
The unit
impulse function is defined by the
cquation
0 t#0

and
$) dr = 1 I=0

The Laplace transform of the unit impulse funetion is

Fig. 9.5
&[8)] = } &)e "dt = 1 ...(9.9)

9.3.6 Exponential Function (e*")

The Laplace transform of the exponential function is

.10)

9.3.7 Sine Function

We know that sin r = ee].

2
The Laplace transform of sine funetion is

£ Ssin oil = ey)

..(9.11)
(0
9.4 letrh al Achrks

9.3.8 Cosine Function

We Aow that cos (0=le

The Laplace transtomm ot COne function is

t lcos on] = t 0

...(9.12)
+(0

9.3.9 Exponentially Damped Function

Laplace transtorm of an exponentially damped functione"f0) is

le "f0) dl=J f ) e " e dr = | f ( ) e " dr

+a) .9.13)
Thus, the
transtorm of the
function ef) is obtained by putting (s + a) in
place of s in the transform of f().

(e sin or=
t*a)+
9.14)

ile" cos O (s+a +0 ..9.15)

Table 9.1 Loplace Transformation

Sr. NO.
F(s)

Fls)= ] ft) d
af,(0) +bf:) as)+bf,{s)
-a)u (-a) Fs)
( Fs+ a)

d) sFs)-f(0
f)dr F(s)

(Contd)
 F
S. No.

ds

f Fds

-7)

17

19

20 s+0
Sin o
(sta)t 0

(s+ a ) w

THE TRANSFORMED CIRCUIT

94 also be represented in the Irequency domain.
of network elements can
Vollage-current relationships
time domain is
Resistor For the resistor, the v-i relationship in
(9. 16)
" ) = Ri(0
be as
rekation will given
The corresponding frequeney-domain ..9.17)
V (s) = RI (s)
()

VS)

Fig. 9.6 Resistor

9.6Electncal Nenvorks
nductor For the inductor, the v-i
relationships in time domain
are
9.18) L
i ) = v (0) dr +i( (9.19
Transforming both the cquations, we get
V(s) = Ls I (s) - L i(0)
(9.20)

I (s) =vs)+ i0) ..(9.21)

() V()
i V() s (+)
i03
Li (0)

Fig. 9.7 Inductor

Capacitor For capacitor. the v-i relationships in time domain are

v()= itydr +vt0)

...(9.22)

i(=dr
(9.23)
The corresponding frequency-domain relations are given as

V(s) =
s)+ O
...(9.24)
I (s) = Cs V (s) - Cv(0)
..9.25)

+O

(7) C V) (s)

Fig. 9.8 Capacitor

 r a n s f o r m and Trs Appiication .
Laplace

9.5 RESISTOR-INDUCTOR CIRcUIT

Onsider a series R-L circuit as shown in Fig. 9.9. The switch is closed at une E .

-o

ig 9.9
F o r 0 ,the transformed network is shówn in Fig. 9.10
Applying KVL to the Mesh.
RI(s) + Lsl(s) = I ()

I(s) = ..(9.26) Fig.9

By partial-fraction expansion,
D

I(s) +
V
A =s/,=sMR
LAs-0
V

L--RIL

-
LJ

Taking the inverse Laplace transform,

i () =-RL)
R
for 1> ...(9.27)

9.6 RESISTOR-CAPACITOR CIRCUIT

Consider a series R-C circuit as shown in Fig. 9.11. The switch is closed at time f = 0.

- VV

i ()

Fig. 9.11
9.8 Electrical Nenworks

For t> 0, the transformed network is shown in Fig. 9.12

Applying KVL to the Mesh.

RI()+ I(6) =
(s)

T(s) =
Fig. 9.12
RC
Taking the inverse Laplace transtom.

i()=e- I/RC) fort>0 ..(9.28)

9.7 RESISTOR-INDUCTOR-CAPACITO CIRCUIT

Consider a series R-L-C circuit shown in Fig. 9.13. The switch is closed at time i=0.

VT

Fig. .13

For t>0, the transformed network is shown in Fig.

9.l14
Applying KVL to the Mesh,
RI (s) + Ls (s) + C s) =

Ls
I (s)

Fig. 9.14

(s-s,Ms-s)
(9.29)

where s and s, are the roots of the equation s +

= - a+ ya - o =-a+B

A
= -a-ya--a-B
 V.
I r a n s j o i (and l s
Appiicaltons
Aplde

where

VLC
and = y-
By partial-fraction expansion, ol 7(s),
B

A (s-)
=
75),,

B= (s-S2) 75)=s

V
0LG-s)s-s-S2J
Taking the inverse Laplace transtorm,
.9.30)

A +A, e
and are constants to be delermined and s, and s, are the roots of the equation.
where
A A,
the values of s, and s2, We have 3 cases of the response.
Now depending upon
When the roots are real and unequal, it gives an overdamped response.
Case

2 LC
a
In this case, the solution is given by
it)= e" (A, e +Az e-b
for > 0

When the roots are real and equal, it gives a crilically damped response.
Case

2L VLCC
a

In this case, the solution given by

for 1 > 0 I
i()= e (A +A21)-

When the roots are complex conjugate. it gives

an
underdamped response.
Case Il
9.10 lectrical Networks

In this case, the solution is

given by
i) =
A e'W +A: e"
where
S1.2-a ya* -w
et

j o
where

and

Hence
0)=e A e o+ A2 e"10t) -

+An at j5invat) |

=
e [(A, +A) cos
a4 +j(A1 -Az) Sin Al
e ( B , cos f + B, sin o) forri>0

Exaphple 9.1 In the

network of Fig. 9.15, the switch is
moved from a to b at i=0. Determine i (1) and v, (1).
12 12
Su
. /

1O V
6F - 3F Ve () VIS) (

9.15
Fig
Solution Alt
0, the network has attained
=

condition. Hence, the capacitor of 6 F acts as

steady-state
an open circuit.
6F(0) = 10 V
(U 10V
V's F (0) = 0 V6F (0

Since voltage across the

i(0)
capacilor cannot change
instantaneously, Fig.9.16
VsF(0") =
10V

For
V3 F(0)=0
1> 0, the transtormed network is shown in
Fig. 9.17.
Applying KVI, to the Mesh
for t>0,
0
s6s1)-/)-1s 0 Ve )
I ()
10
)+)+)=
OS
Fig. 9.17