4

CHAPTER

LAPLACE TRANSFORM
4.1. INTRODUCTION
The Laplace transform is one of the mathematical tools used for the solutionof
linear ordinary integro-differential equations. (Mostly continuous-time systems
are described by integro-differential equations). In comparison (as shown in figure
4.1.) with the classicial method of solving linear integro-differential equations, the
Laplace transform method has the followingtwo attractive features :
(i) The homogeneous equation and the particular integral of the solution are
obtained in one operation.
(ii) The Laplace transform converts the integro-differential equation into an
algebraic equation in s (Laplaceoperator). It is then possible to manipu-
late the algebraic equation by simple algebraic rules to obtain the expres-
sion in suitable forms. The finel solution is obtained by taking the inverse
Laplace transform.
Intego- Laplace
differential Algebraic Initial
transformation equation in s Conditions
equation
Manipulate the
transformed algebraic
Classical
equation to make the
Method
expression in suitable forms

Inverse Laplace Suitable forms

Solution for inverse Laplace
Transformation
transformation
Time Domain Frequency domain

Fig. 4.I. Comparisonof Classical method

and Laplace transform method
From the 2ndfeature of Laplace
transformationis somewhat similar transform over classical method, Laplace
or quotient of two to logarithmic operation. To find the product
number, we find
(i) logarithm of two
numbers
(ii) add or subtract
(iii) take antilogarithm
to get product or quotient
4.2.DEFINITION
OF
The LaplacetransformTHE LAPLACE TRANSFORM
Jems. We define a method is a powerful
Laplace transform technique for solving circuit
as follows :

64
 Laplace Transform
65
function /tt) which is zero for t < 0 and that
For the titne satisfy the condition

Laplace transform of/(t) is defined

for real and positive o, the as

Coas the Laplace operator, which is complexvari-

Thevariables is referred
the functions /(t) and F(s) are known as
And Laplace trans-
able,i.e.,s = o + jo.
formpair.
LAPLACE TRANSFORMATION
4.3.INVERSE transform F(s), the operation of obtaining flt) is termed
Given the Laplace the
Laplacetransformation and is denoted by
inverse
1
fit) = = cls
20
Thoughone couldevaluate the inverse transform of a function F(s) by using
equation,normally the transform table is used to obtain the inverse trans-
above
formation.

4.4. PROPERTIES OF LAPLACE TRANSFORM

1.Multiplication
by a constant
Let h be a constant and F(s) be the Laplace transform offtt). Then
= kF(s)
2.Sumand Difference
Let PI(s)and F2(s)be the Laplace transform
Then of fl(t) and f2(t), respectively.

3.Differentiation
with respect to "t"
Let F(s)be the
Laplace transform
approaches0. offtt) and let be the value offlt) as t
Then,

C df(t) = BF(s)- limf(t) =

dt sF(s) -fl0+)
Proof;

Let
ftt) u then
and dt .dt du
e-stdt dv ,
then V

is written
instead oral).
66 1713Circuits and Systems

On integrating

udv =

df(t)
= f(t). --e-St
( :. putting the values)

1 df(t) 1 1 df(t)
e-st s
s s
Therefore,
df(t)
= sF(s) - f(0+)
Thus the Laplace transform of the second derivative of f(t) as
d2f(t) d df(t) df(t) df(t)
dt2 dt dt dt
= - f(O+)]-f 1(0+)= s2F(s) - sf(0+)- f l (0+)
(where f l (0+)is the value of the first derivative of f(t) as t approaches 0)
In general,
dif(t) = snF(s)- -s
4. Integration by "t"
If
then, the Laplacetransform of the first integral of flt) is given by

s
Proof :
f (t) dt e-st dt

Let
It/ (t) dt then du = At) dt

and = e-st dt 1
then v =
s
On integrating

= uvl

1
—e-st
8 0 s j: f (t)e -st dt

(fit))
 Laplace Transform 67

In general,
.....dtn

Laplace transform of the indefinite integral is given as

Note: The
F (s) f-1(0+ )
4 Jf(t ) dt s
r is the value of the integral /tt) as t approacheszero)
1(0+)
(where
with respect to "s"
Differentiation
5. differentiation in the s-domain corresponds to the multiplication by "t" in
The
domain, I.e.
the time dF(s)

proof :
d dt— . —e-st dt
ds ¯ ds r0 f
-st — dt = — •flt)J

6.Integration by s
Theintegrationof F(s) in s-domain corresponds to division by "t" in the time
domain, i.e.

Proof :

ds — dt

dt = dt

•——dt=æ f(t)
7. Shifting
Theorem
[a]Shifting
in time: The Laplace
transform of a shifted or delayed function is
Proof: - a) = e-asF(s)
Let t
a = y, then,
dt = dy

= f +a)dy (since U(y) = O ; y < 0)

in frequency
:
The Lapl ace
transformof transform of e-ctttimes a function is equal
that function,
with s is replaced by (s + a).
 68 IOC)Circuits and Systems

Proof :
Cle-at fit)) = dt

-(s+a)tdt = F(s + a)
Note :
S. Initial Value Theorem
If the first derivative of fit) is Laplace transformable, then the initial valueof
a function fit) is given as
1(0+) _ Lim f (t) = S 00
9. Final Value Theorem
The final value of a function f(t) is given as
= Lim f (t) =
t 00
10.Theorem for Periodic Functions
The Laplace transform of a periodic function (wave) with period T is equalto
1
times the Laplace transform of the first cycle of that function (wave).
1 —e ¯Ts
Proof :
Letf (t)f (t)f (t) be the functions describing the first, second, third
cycles of a periodic function fit) whose time period is T. Then
= + f2(t) + f3(t) +
= fl(t) + fl(t- T) - T) + fl(t - 2T) - 2T) +
Now, let = PI(s)
Therefore, by shifting theorem (property 7[a] of L.T.), we get
= + e-Ts.F1(s) + e-2Ts Fl(s) +
= Fl(s) [1 + e-Ts +
1
RI(s)

11. Convolution Integral

Given two functions fl(t) and f2(t), which are zero for t < 0. If efl(t)] = F I(s)
= F2(s), then
l'
= fl(t) * f2(t) is called the convolution of fl(t) and f2(t)and
equal to

Jotå(t— or
12. Scaling
If Laplace transform of f(t) is F(s), then
 Laplace Transform
69
Table 4.1. Laplace TransformPairs

S. No. F(s) = dt

1.

2. $2 ' sn+l
3. 1
1
*itt
4.

1
5. (s

6. sin ot s2 + 02

7. cos ot s2 + 02

8. e-at sin ot (s + +02

9. e-at cos ot (s + + 02
a
10. Sinhat 2

11. coshat 2

12. e±at F(s 7- a)

13.

*Allfit) should be thought of as being multiplied by U(t), i.e., flt) = 0 for t < 0.
Example
4.1.Find the Laplace transform of the following standard signals
(functions)
(a) The unit step function U(t).
(b) The delayed step functionKU(t —a).
(c) Theramp function Kr(t)
(d) Thedelayed unit orKt U(t).
(e) The unit impulse ramp functionr(t —a).
(D The unit doublet function ö(t).
Solution : function ö'(t).
(a)
flt) = U(t)

By the definition
of U(t) given in chapter 2, we have

= e -st dt 1
—s
70 Circuits and Systems

(b) fit) = ICU(t- a)

st dt
F(s) =
U(t —a) given in chapter 2, we have
By the definition of
—st —as

Ice -st dt = IC——

using property 7
(Alternatively we can find directly
(c) fit) = = ICt U(t)
—st

Kte -st dt = K I •— dt
cit =

e-st dt — —st

f(t) = rqt —a) = (t —a) U(t —a)

(t — —a)e -St dt

—st —st

dt = (t | •g— dt

—st —as
e-st dt =
(Alternatively we can find directly using property 7 [al)
(e) fit) = Ö(t)
dt
By the definition of ö(t) given in chapter 2, we have
F(s) = e-st (Since ö(t) = 1 only at t = 0
fit) = Ö'(t)
Example 4.2.Find the Laplace transform of the following functions :
(a) The exponential decay function Ke-at
(b) The sinusoidal function sin (Dt.
(c) The cosine function cos (Dt.
(d) e-at sin ot
(e) e-at cos ot
e—att U(t)
(g) sin h at
(h) cos h at
Solution
(a) = Ke—at

Ke -at e -8t dt = K +a)t dt

—(s+a)t
—(s+a)
 Laplace Trangform 71
nt) gin (Dt
(b)
gin oye- st dt 8', gin(Dt
2j
e-j0)t ) •e-8t dt 1 1
( e jot 1
2j
1 s +j0)—s+j(D
¯ 2j (s+j(D)(s—j(D) — +
82
case
(c) As similar to above
s
flt) = cos (Dt,then, F(s) = + 02
Alternatively for (b) and (c) we know, that
1
C[e-atJ =

puta =jo
1
C[e-joY
J= = cosot —jsinot

1 [Multiplying Numerator and

C[cos ot —j sin otJ =
Denominator by (s —jo)J
s —jo
s2 + 02

Equatingreal and imaginaries on both sides, we have

C[cos otJ = $2 + 02 and C[sin Otl = s2+ 02

(d) flt) = e-atsin ot

(s + +02 Using property 7[b]

(e) fit) = e-atcos ot

(s + +02 Using property 7[bl

e-at t U(t)
1
Using property 7[bJ
(g) We know that

Sinhat = --(eat
-e-at)
1
C[sinh atJ = — eat e¯st
2

1 1 1 a
 72 1717Circuits and Systems

(h) As similar to above case

cos h at —
1
(eat + e—at)
2
s
C[cosh at] = 2

Note : This is to point out that the followingsix functions are all different
(i) fit)
(ii) fit) U(t)
(iii) fit to)
(iv) fit —to) U(t)
(v) fit) - to)
(vi) fit —to) U(t —to)
If fit) = sin Ot, then
(i)fit) = sin ot
(ii) = sin 0t U(t)
(iii)fit — to) = sin — to)
(iv) fit —to) U(t) = sin (D(t—to) • U(t)
(v) flt) U(t — to) = sin ot • U(t — to)
(vi) fit — to) U(t — to) = sin — to) • U(t — to)
The six functions given above are shown graphically in figure 4.2. fromwhich
it is obvious that they are different.
sin ot sin ot U(t)

(it)
sin o(t —to) sin —to) U(t)

t 0
t

(iiO
(iv)
sin U(t- to)
sin (D(t — to)

(v)
Fig. 4.2, Six different sinusoidal
functions.
Example 4.3.Find the Laplace transforms of the
above six functions•
Solution ; (i) Clsin (DtJ=
$2 + (02
 Laplace Transform
CJD73

C[sin ot = C[sin 0)tJ = 82 + 02

(ii)
is for O < t < 00,which is the same
Sincethe Laplace transform for func-
tions (i) and (ii).
Clsin —to] = sin ot cos oto -- cos ojt sin 0)toJ
(iii)
COSOto —
s 2 + 02 •sin Ot

0) cosOt0 —ssin Oto

s2 + 0 2
—to) U(t)J = Elsin —to]
(iv) Elsin
O cosOt0 —s sin Oto
+ 02
Sincethe L.T.is for 0 < t < 00,which is the same for functions (iii) and (iv).
(v) C[sin to)] = Jo sin Ot •U(t —to ) •e-st dt sin Ot •e-st dt

+jo)t —e —(s+jO)t dt

(—s + jO)tO —(s + jO)tO

1 e

2j —s+jo
(-s + jo)to e —(s+ jO)tO 1 e -joto
2j s —jo

e-tos -joto(s —jO)

2j
e -tos jo(e JOt0 + e -JOt0 ) + s(e JüY0 —e -jot() )
s + up

= e-tos OcosOt0 +ssinOt

s2 + 02
(vi) Elsin
—
Usingproperty — to)J
7[aJ,shifting in time

rample4.4. .92+ 02
Ire2.4, Obtain the
\.20, (ii) figure Lapl ace transforms
(vi) 2.7, (iii) f of the waveforms shown in (i) fig-
figure2.22(b), igure
2.17, (iv) figure 2.19, (J.P. Univ., 2001) (v) figure
(vii)rxgure 2.22(c),
(viii) figure 2.22(d), (ix) figure 2.22(e).

et1S

se-s)
s s
74 Circuits and Systems

tn S 2 02
(iii)

—2as 2e —gas+ .
(iv) [1 - 2e -as + '2e .1
(1 —e -as)
—1-
1
1 + e ¯as
= —[1 - 2e -aS (1 - e -as +

as/2 e —as/2

s(e as/2+ e
ways : (Using theorem for periodicfunctions)
Alternatively
1 1 2e -as e -2as
1
—2as -2as S s
1

(Sincefl(t) = U(t)- 2U(t -a) + - 2a)

1 —e -as 1 as
1 1
•—•(l—e-
as )2 _ _ —tanh
¯ s(l + e -CS ) 2
I —e -2as s

1
(v) s s
1
3e-2s + 4e -3s - 2e -5s

(vi) 2 2

K 1
(vii) 2

(viii) ms) = a s
—s e-2s
(ix)
1
= — [1 —2e + e¯2S
s2 s

Example 4.5. Obtain the Laplace Transforms of the periodic waveforms as

shown

in (i) figure 2.23 (ii) figure 2.27.

Solution : (Using theorem for periodic functions)
(i) i'(t) = -4)
= 2t U(t) -e 2U(t) - (2t -2 + 2t -6) - 2) + (2t - 6) -4)
= 2tU(t)- 2U(t)- - 2) - 2) + - 4) - 4) + 2U(t
e —28 e-48 2e-4s
82

2e¯28 •+e e—48)

I 12——(1
2 (1 e-2s 2
82 8
Since Time period, T 4, Therefore,
1 2 2 1 tanh s -1
1 e-48 82 1 + e—2s —
s s
 Laplace Transform 75

1.60 1(t) + Gl 2(t)

1) + 2U(t 2)
U(t)
1
1(1 + 2e¯2s) _ (1 - - '2e-s )
VI(s)
period, T = 2, therefore
SinceTime
1
V(s) = I-e-2s 1 —e -28 s s
Alternative ways - : - 1) + 3U(t - 2) - 3U(t - 3) + .
E(t) = U(t)
I —3e¯S + 3e¯2S —3e—3s+

1 1 1 —2e—s
1-

4.6.Obtain the L.T. of the waveform as shown in figure 2.28.

Example
Solution: flt) = - 1) + - 2) - - 3) + - 4)

1 e 2 e
1 1 1 1
2 2 2 2

l
1 l_e++ 2
1
2

1 1 1 1
S + O.5 | + + 0.5)

Example
4.7.Obtainthe Laplace transform
of the waveform shown in figure
Solution :
i(t) = 1.5(1-e-4t) U(t) -
1.5 [I-e -4(t - - 0.1)
I(s) = 1.5 1 1 e -o.ls e -o.ls
-—- -1.5
s
= 1.5 (I-e-O.IS) 1 1 6(1 —e —O.ls)
4.8.Obtain
the Lapl ace s(s+4)
•
Solution transform of the waveform shown in figure
v(t) R[t2
U(t) 4 (t—l)
V(8) & _ 4e—s 2e-2s

4.9.Obtain
- [1-2
the Laplace t ransform
figure4.3. of the periodic, rectified half-sine
 76 Circuits and Systetns

v(t)

t
772 3772
Fig. 4.3.
Solution : In example 4.4(iii), the Laplace transform of the single half-sine
is

VI(s) | +e 2
s2+02
Using theorem for periodic functions.
1
Then V(s)= = 1-e-Ts .V1(s)

1 1
I—e- Ts m s2+02 m (s 2 +02)
I—e 2

Example 4.10.Determine the Laplace transform of the periodic, rectified full.

wave sine wave shown in figure 4.4.
v(t)

o 3T/2
Fig 4.4.
Solution : From Example 4.9, the Laplace transform of the first cycleofthe
wave is given by

VI(s)= $2 +02 2

Therefore, the Laplace transform of the periodic wave form v(t) of period
given by

1
82

$2 + 02

eT$/4 + e-TS/4
$ 2 + (02 Coth—
 4.11. Determine the Laplace trans-
Example waveform as shown in figure 4.5.
forniof the Using the gate function, we can
solution : cycle of the function as,
first
writethe
2K
fl(t)
-K
2 Fig. 4.5.

t ——U(t)+ U(t-T)
2 2
T 2K
2K t-T+— U(t-T)
2 2
2K tU(t)- U(t-T)
2K + U (t -
—- [tU(t)- (t- T) U (t

2K 1 e-Ts T 1 e-Ts
2

2K T 1
orF1(s) = — — (1 -e-Ts )
Therefore,the Laplace transform of the periodic waveform fit) of period T is
given by
1 2K T 1 (1 e-TS)
1
.F1(s) = 1 e-Ts Ts — (1 + e¯Ts ) —
2
2K T 1+e-Ts 1 2K T e Ts/ 2 +e -Ts/2 1
Ts 2 1 e-Ts Ts 2 eTs/2_ e-Ts/2 s

2K T Ts
—cob —
1

Example
4.12.Determine the initial valuef(O+),if
20+1)
82 •
Solution : ß0+) Lim

2
—Lim 2S(S +1)
= Lim
8-+00 $ 2 28 + S 8-+ 00 1 + —+

Example
minethe 4.13.For the current i(t) = 5 U(t) - 3e-2t,find I(s) and hence to deter-
value of i(O+) and i(oo)
Solution •
i(t) = 5U(t) - 3e-2t
3 28+10
¯ S(S+2)
 78 Circuits and Systems

28+10
Lim Lim
Therefore, 00
= S 00 S 2

10
s
= SLim
00 2
s
% +10
And _ Lim
-5 t
Example 4.14. Find the initial value of the function, f(t) = 9 —2e .
Solution :
/(0+) = Limf(t) = Lim (9-2
5s+3
Example 4.15.Given the functionF(s) = s(s + 1) . Find the initial valuef(0+),final

value f(oo),and the corresponding time functionf(t). (I.P.Univ.,2001)

5s+3
Solution : Initial value , ft0+) = = Lim

5+ 3
= Lim

Final value, Lim[s.F(s)J = Lim 5s+3

And, s(s+l)
3 2
(Using partial fraction expansion)
Therefore, fit) = =
Alternative ways :
fl0+) = Lim/ (t) = Lim (3 + 2 e-t) = 3 5
Lim/ (t) = Lim (3 + 2 e-t) = 3 + 0 = 3
t 00
Example 4.16.Without finding the inverse Laplace transform ofF(s), determine
f(O+)and f(••)for each of the following functions ;
4e -28 (s + 50) 2001)
(ii) (U.P.T.U.,
8 82 + 7
Solution : Sincewe knowthat
flO+) Lim

and neo) Lim s.F (s)

 Laplace Transform 79

• +50) —Lim 4e-2S(s+50)

ft0+) = Lim s =O

Lim 4e-2S(s+50) = 4.(50) = 200

Lim s • = Lim
(ii)
3

s(s2 +6)
n 00)= Lim (S2+7)
s +0

Example4.17.Find the Laplace transform of 5

in figure 4.6.
thetimefunction shown (U.P.T.U., 2001)
Solution :
5
fit) =
2
2 4
5
Fig. 4.6.
2
5 5
2 2
5 5
—r(t—2)—
2 2
5 5
—r(t—2)— —r(t—
2 2
so,
5 e -2s 5 —5e-4s

1—e-2s + 2s e-
Example
4.18.Withoutfinding inverse Laplace Transform of F(s), determine
10+))
andf(oo)for the following function :
5s 3 -1600
(U.P.T.U., 2002)
8(8 3 + 188 2 + 90s + 800)
Solution ; ft0+) = Lim s.F(s)

= Lim • 5s 3 -1600
+18s 2 + +800

5- 1600

s s2
neo) Lim• 1600
80 Circuits and Systems
Example 4.19.Determine Laplace Transform of
the following wave shown in figure 4.7. (3,0
(U.P.T.U., 2002)
Solution: From the wave form shown in figure
4.7,
fit) = 4G03 (t) + '2t + 10) 5(t) 3 5
Fig. 4. Z

= 4 U(t) 3) - '2(t-5+2-2) - 3) + -5) -5)

Therefore, Laplace transform of the waveform is

4' 2e -3g 2 e -5s 4 2e -3g '2 e -5g

2 2 2
s2
Example 4.20.Find the Laplace transforms of the following functions :
(a) e-at U(t)
(b) e-at U(t- b)
(C) e-aa -b) U(t-b) = . e-at U(t-b)
(d) e-a(t - b) -c) = .e-at U(t- c).

Solution : (a) F(s) = — e-at.U(t) - 1

(Alternatively we can find directly using properly 7

e -at - b) —(s
+a)t
dt
(b)
b

+a)t +a)b
¯ —(s+a)b
(Alternatively we can find directly using property 7 [a] and property 7(bl)

(c) -a (t -b) —b) e-stdt

b b

—(8+a)b
ab e e—b8

(Alternatively we can find directly using property 7[a])

(d) F(8) = Je -a (t -b ) e—at. e—stdt

c
ab ab +a)c
-(g+a)
 Laplace Transform 81

Laplace of the waveform shown in figure 4.8(a).

01. Find the

v(t)
21' time.
Fig. 4.8 (a).

Let VI(t)be the first cycle of the waveform of the figure 4.8(a), as
solution: 4.8(b).
figure VI(t)
shownin
VI(t) = -LtGOT(t)
so,

wtJ(t)- -ctU(t-T) Fig.4.80.

-L (t-T) -T) - % U(t-T)

or VI(s)
=% -v —
s
the Laplace transform of the periodic waveform U(t) of period T is
Therefore,
givenby

1 % voe -Ts
=
1 e-Ts •VI(s)
T8 2 s (1- e -T8 )

Example
4.22.Avoltage pulse of magnitude 8 volts and duration 2 seconds ex-
tendingfromt = 2 seconds to t = 4 seconds is applied to
aRies RLcircuit as shown in figure 4.9(a). Obtain the 20
expressionfor the current i(t). (U.P.T.U., 2003 C.O.)
Solution
: Thevoltagepulse is given as [as shown in
figure 4.9(b)J
i(t)

mg.4.90.
or
8 -e 48 )
Applying
KVLin the circuit of figure 4.9(a), we have
di(t) v(t)
TakingLaplace dt
transform,
V(s) = 216)
+ sl(s)

or t
o 2 4
Usingpartial Fig. 4.9(b).
fraction expansion,
I(s) e—28 4 4 4
Or 8 8+2
i(t) = —e-2(t -2)) —4)
—2) e 2(t - 4))
 CHAPTER5

CIRCUIT ANALYSIS BY LAPLACE TRANSF0F

5.1.INTRODUCTION
The circuit analysis in time domain were presented in the third chapter using
classical method. Having introduced the definition of Laplace transformation
learned to obtain the Laplace transform of many functions in previous chapter
shall now be exposed to the remarkable power of the Laplace transform as a ma
ematical tool to find the circuit responses in terms of voltages and currents,
ject to any arbitrary input functions.
5.2.SOLUTION OF DIFFERENTIAL EQUATIONS
The Laplace transformation is used to determine the solution of integro-differen
tial equation. A differential equation of the general form
dnz• dn-lz di
an -1 — + ani = v(t)
dt
becomes,as a result of the Laplace transformation, an algebraic equationwhich
may be solved for the unknown as

initial conditionterms
aos n + als n an-IS + a
where I(s) = Ei(t)
I(s) is a ratio of two polynomialsin s. Let the numerator and denomin
at0t
polynomialsbe designated P(s) and Q(s), respectively, as
ID(s)

Note that Q(s) = 0 is the characteristic

P(s)/Q(s)can equation. If the transform
now be found in a table of canbe
written directly. In general, however, transform pairs, the solution i(t)
broken into simpler terms before the transform expression for I(s) mustbe
any practical transform table can be used'
Next, we factor the denominator polynomial,
Q(s),
Q(s) = aosn + alsn-l +
n
ao(s + 81)
or very compactly,

n
Q(s) = ao Il (s +sj)
Where indicates a ofth?
product of factors, and Sl, s2, are the n roots
characteristic equation
Q(s) = 0. Now the possible ..... s these roots are
forms of

84
 Circuit Analysis by Laplace Trangforrn 85

(i) partial Fraction Expansion When All the Roots of Q(8)are Simple :
simple, then
If all roots of Q(s) = 0, are
PCs)
(S +82 s +81 8 +8 2
K1,1<2,
where the Rs are real constants called residues. Any of the residues
denominator
K can be found by multiplying I(s) by the corresponding
+ sj) equal to zero, i.e.
factor and setting (s
s = —s..As
PCs)
K = (S+Sj)—
Q(s)
of Q(s) are of Multi-
(ii) Partial Fraction Expansion When Some Roots
ple Order :
i If a root of Q(s) = 0, is if multiplicity r, then
Kli 1<12 +
I(s) = QI(s)
(s + SIT
(S + Sl

following equations may be used for the evaluation of coefficients of re-

The
peated roots.
Klr = (s + Sl)r

— + •I(s)

— + sly •I(s)
I(r-2) - 2! ds 2

(s + SIY •I(s)
11 ¯ (r —1)! ds
(iii)Partial Fraction Expansion When two roots of Q(s) are of Complex
Conjugate Pair :
If two roots of Q(s) = 0, which form a complex conjugate pair, then

¯ (s + 01+ jo.)) (s +

Where KI = (s + a + j(D) s =-(a +jo)

and is the complex conjugate ofK1. An expression of the type shown above
is necessary for each pair complex conjugate roots.
of
Severalexampleswill illustrate the partial fraction expansion and the evalu-
ation t
of R s.
Example5.1.Solvethe Differential Equation

Solution; Taking
the Laplace transform,
- x'(0+) + 3sX(s) - 3x(0+) + 2X(s) = O
(82 + 38 +
= sx(0+) + x'(0+) +
+ 38+ = 2s+3 [By putting initial conditions]
or 28+3 28+3
82 +38 +2
 86 CJDCircuits and Systems

Hence, all roots of denominator polynomial are

fraction expansion, simple. Then
by
28+3

Where

and

The result of the partial fraction expansion is thus,

2s+3 1 1

Therefore, the solution of given differential equation is

1

or x(t) = e-t + e-2t

1
Example 5.2. Find i(t) , ifl(s) =

1
Solution :
Kl + 1<21 22

1 1
Where

21 ds

¯ds s(s+2) s2(s +2) 2

The complete expansion is
1 1 1
28 (s +1)2 20+2)
Therefore,
1
te- + —e- A
2
8 2 +58+9
Example 5.3,
88 +582 +128+8 ; find i(t).
Solution ; I(s) 8 2 +53+9
8 3 +58 2 + 128 + 8 (8 + +48 +8)
 Circuit Analygig by Laplace Transform 87

t/ +91

or s+l (s+2+j2) (s +2-j2)

s2 +58 +9

1
-(2+j2)=
j4
1
j4
Therefore,the complete expansion is
1 1
1 j4 j4

-t 1 -(2+j2)t 1 -(2-j2)t
Hence, i(t) = e e e
j4 j4
-t I _ej2t
j4
or i(t) = e-t ——e-
j4
sin2t A

S +5.9+9 1 1
Alternatively, I(s) =

1 1 2
s +1 2 (s +2) 2

Therefore, i(t) = = e-t + Le-2t sin2tA

as E-l
= e -at sin ot
(s + + 0 2

5.3TRANSFORMEDcmCUIT CONTONENTS REPRESENTATION

5-3.1.Independent Sources
The sourcesv(t) and i(t) may be represented by their transformations, namely
V(s)and I(s) respectively as
shown in figure 5.1.

't I(s)

(a)
F ig, 5.1, Representation of (a) voltage source (b) current source.
 88 Circuits and Systems

5.3.2.Resistance Parameter
By Ohm's law, the v—irelationship for a resistor in t-domain is
VR(t) = RiR(t)
In the complex-frequencydomain (s-domain), above equation becomes
VR(s) = RJR(s)
From above two equations, we iR(t)
observe that the representation of a
resistor in t-domain and s-domain
VR(t) R
are one and the same as shown in VR(s)
figure 5.2.
5.3.3.Inductance Parameter Fig. 5.2 Representation ofa resistor.
The v-i relationship for an
inductor is
VL(t)= LdiL(t)
dt

iL(t) = VL(t)d(t) +iL(0+)

0+
The Laplace transforms are
VL(s)= -

1 iL(0+)
or IL(s) - —
s
From above equations, we get the transformed circuit representationforthe
inductor as shown in figure 5.3.
IL(s)
IL(s)
iL(t)

VL(8) 1 iL(0
VL(s)

LiL(o+)

Fig. 5.3 Representationof an inductor.

5.3.4.Capacitance Parameter
For a capacitor, the v—irelationship
is
dv (t)
ic(t) =
dt
or vc(t) = —
The corresponding Laplace
Transform are
1/8) = c vc(s)-c

or vc(8) = 1
From above equations, we tb
get the transformed
capacitor as shown in figure 5.4. circuit representati0n for
 Circuit Analysis by Laplace Transform 89

Ir(g) lc(g)

it(t) 1

Vc(s) vc(6)
vt(t)
cvc(o+)
+ vc(0+)

Fig. 5.4 Representation ofa capacitor.

Example5.4.Consider the differential equation
d 2 y(t) + 3dy(t)
+ 2y(t) = 5 U(t)
WhereU(t)is unit-step function. The initial conditions are y(O+)= —1 and
4(0+) = 2. Determine '(t) for t 0.
Solution: Taking Laplace transform on both sides of given differential equation :
5
s2Y(s) - sy(0+) - + 3sY(s) - 3y(0+) + 2Y(s) = -
s
Substituting the values of initial conditions and solving for Y(s),we get
s 2Y(s) + s -2 + 3sY(s) + 3 + 2Y(s) = —5
s
2 2
or
s s +3s+2

Expanded by partial fraction expansion,

2

-4+2+5 3
- (-2)(-1)
=i
Therefore, 5 5 3
2s 8+1 2(s+2)
Hence,taking the inverse Laplace transform, we get the complete solution as
5
2
Example5.5.Consider
Qand the
L IH excited by circuit withR 4 s
a 48V dec. source as shown 40
in figure5.5 (a).
Assume the initial current
throughthe inductor 48V
transformdetermine is 3A.Using the Laplace i(t)
draw the current i (t); t O.Also
the s.domain
representation of the circuit.
Fig. 5.5(a)
90 Circuits and Systems

Solution : Applying KVL,

di(t) = 48
dt
Taking Laplace transform
48
RI(s)+ - =

-31 = —
48
41(s)+
3s+48
or
Applying the partial fraction expansion, we get
3S+48 Kl
s
3s+48
where Kl = = 12
4
and =
(3s +48)

12 9
Then, I(s) =
s Fig. 5.50)
or i(t) = e-1[ I(s)] = 12 - 9e-4t A
And the s-domain representation is shown
in figure 5.5(b).
s 20
Example 5.6.Consider a series R-L•Ccircuit
with the capacitor initially charged to volt-
age of 1 V as indicated in figure 5.6 (a). Find
the expression for i(t). Also draw the s-do-
main representation of the circuit.
Solution : The differential equation for the Fig. 5.6(a)
current i(t) is
di(t)
L + Ri(t) + BJi(t) dt+ vc(0+) = O
dt
and the correspondingtransform equation is

i(0+))+ RI(s) + —I(s) +

8

The parameters have been specified

onc = —F, R = 2Q, L = IH, and
Vc

—IV(with the given polarity), The

initial current i(0+) = O,because initially
behavesas a open circuit,The transform equation I(s) then becomes
21 (8) + —I (8) 1
8 8

or
8 +28+2
 Circuit Analysis by Laplace Transform t)CJ 91

the square, 2
2
Completing
or, 8 8

1
=
Therefore,i(t) e-t sint A
i(t) is
or representation Fig. 5.60)
The s—domain
5.6(b).
shown in figure shown in figure 5.7 (a) with
th e R-C parallel circuit as 10 A. Determine the voltage
Consider
Esgmple 5.7.
4F excited by d.c. current source of
andC
R capacitor by applying Lapla representation of the
acrossthe the capacitor as 2V.Also draw the *domain
across
circuit.
Applying KCL,
Solution:
dv(t) C=4F
= 10 t IOA v(t) R=O.5Q
R dt
transform
TakingLaplace
10
Fig. 5.7(a)

10
2V(s)+ - 2] =
8s+10
or VT(s) = s(4s+2)

the partial fraction expansion,

Applying
% +2.5 Kl
¯
s(s+0.5) S s+0.5

and

or
5 -3
s s+0.5
Therefore, v(t) =
= 5 - 3e-0•5t V
Thes-domain representation of
thecircuitis shown
in figure 5.7
Fig. 5.7(b)
Example 5.8.In the network
shownin
isclosedatfigure 5.8, the switch S
t 0. With the
parametervalues network 100
expression shown, find the + s
for
thenetwork il (t) and i2 (t), if 100v 100 100
is unenergized be- i2(t)
tore the switch ii(t)
is closed.
solution :
equations Applying RVL, Loop Fig. 5.8.
are

dh(t)
dt = 100
or
dh(t) ...(i)
+ 20i1(t) - 10i2(t) = 100