1. No category

Module V
Laplace Transforms
Prepared by Reji P R
Assistant professor
SNGCE Kadayiruppu
Lapace Transforms
Let f(t) be a continuous function defined for all t≥0,then the Laplace transform of
f(t) is defined as L{f(t)} = ∫
f(t) dt, provided the integral exists.
Properties
1) L{f(t)+g(t)} = L{f(t)+g(t)}
2) L{c f(t)} = cL{f(t)},where c being a constant.
Laplace Transform of some elementary Functions:1) L (1) = ,s>0
Proof:L(1) = ∫
2) L(
)=
dt = {
}0
=
s>a
3) L(
4) L(sinat)=
5) L(cos at)=
6) L(
7) L(cosh at) =
8) L(sinh at)=
First Shifting theorem:If L{f(t)} = f(s) then L{
1) L(
sinbt)=
2) L(
cosbt)=
f(t)} = f(s-a)
3)
sinhbt)=
4)
coshbt)=
5)
)=
Find the Laplace transform of the following
1)
=
L{
}=
=
2) f(t) = ,when 0<t<T,f(t) = 1 ,when t>T
L{f(t)}= ∫
dt + ∫
=
Inverse Laplace Transforms
If L(f(t)) = f(s) then f(t) is called the inverse Laplace transform of f(s) and is
denoted by L-1 (f(s)) = f(t).
Some formulas:1) L-1 { } = 1
2) L-1 {
}=
3) L-1 { } =
4) L-1 {
if n is a positive integer.
=
5) L-1 {
} = sin at
6) L-1 {
} = sin hat
7)L-1 {
} = cos at
8) L-1 {
} = cos hat
9) L-1 {
}=
sin bt
10) L-1 {
}=
cos bt
11) L-1 {
}=
sinh bt
12) L-1 {
}=
cos h bt
Problems:1) Find the inverse Laplace transform of
=
=
L-1 (
=
{ }
=
+
{ }
=
2) Find the inverse Laplace transform of
=
+
ie, A = 2 ,B=-2,C=6,D=3
ie,L-1{
2L-1{
{
}
{
}
=2
3) Find
= s{
}=
L-1{
=
}=
coshhat+cosat)
Find the inverse Laplace transform of the following
1)
2)
LAPLACE TRANSFORMS OF DERIVATIVES
Theorem
If f ’(t) be continuous and L {f(t)} = f(s) , then L{f ’(t)} = s f(s) – f(0) , provided Lt [ e –st
f(t)] = 0
t→∞
Proof:∞
L{f ‘ (t)} = ∫ e –st f ‘ (t) dt
0
∞ ∞
= [ e –st f(t) ] - ∫ -s e –st f(t) dt
0 0
∞
= Lt [ e –st f(t)] - f(0) + s ∫ e –st f(t) dt
t→∞
0
= s f(s) – f(0)
Generalisation
If f(t) and its first (n-1) derivatives be continuous , then
L{ f(t)} = s n f(s) – s n-1 f(0) – s n-2 f ‘ (0) - …….- f n-1 (0)
In particular L{ f ’’(t)} = s2 f(s) – sf(0) – f ’(0)
L {f ’’’(t) } = s3f(s) – s2 f(0) – s f ‘(0) – f ‘’(0)
LAPLACE TRANSFORMS OF INTEGRALS
t
t
If L {f(t)} = f(s) , then L{ ∫ f (u) du } = 1 f(s) or L-1 {1 f(s) } = ∫ f (u) du
0
s
s
0
Proof:Ф ( t) = ∫ f(u) du , then Ф’(t) = f(t) and f(0) = 0
L { Ф’(t)} = s Ф( s) – Ф(0)
1)
=> sL{ f(t)} = s Ф(s)
=> f(s) = s L{Ф(t)}
=> L{Ф(t)} = 1 f(s)
t
s
t
ie, L{ ∫ f (u) du } = 1 f(s) or L-1 {1 f(s) } = ∫ f (u) du
0
s
s
0
Find the inverse Laplace transforms of i)
1
ii)
1
s3( s2 + a2)
s (s + 1 )3
i)
Since L -1 [ 1 / ( s2 + a2) ] = 1/a sin at
t
-1
L { 1
= ∫ 1/a sin au du = 1/a2 ( 1 – cos at)
s(s2+a2) 0
t
L-1 { 1
} = ∫1/a2 ( 1 – cos au) du = 1/a2 ( t – sin at )
s2(s2+a2 )
0
a
t
L-1 { 1
} = ∫ 1 /a2 ( u – sin au ) du = 1/a2 [ t2 + cos at – 1 ]
s3(s2+a2 ) 0
a
2
a2 a2
-1
–t 2
ii) since L [ 1
] =e t
3
(s + 1)
2!
t
-1
L [ 1
= 1 ∫ e – t t2 dt = 1 – e – t { 1 + t + t2 }
s ( s +1)3
2 0
2
MULTIPLICATION BY tn
If L {f(t)} = f(s), then L{ tn f(t) } = (-1)n dn [f(s)] , where n = 1,2,3 ….
dsn
Proof:- We prove the theorem by induction
t
L {f(t)} = f(s) => ∫ e – st f(t) dt = f(s)
0
Differentiating both sides with respect to s , we have
∞
d ∫ e – st f(t) dt = d[f(s)] or ∂ [e – st f(t)] dt = d[f(s)]
ds 0
ds
∂s
ds
∞
or ∫ -t e – st f(t) dt = d[f(s)] or ∂ e – st [ t f(t)] dt = - d[f(s)]
0
ds
∂s
ds
Or L [ t f(t) ] = - d[f(s)] Which confirms the truth of the theorem for n = 1
ds
Now assume the theorem to be true for n = m, so that
∞
m
m m
L{ t f(t) } = (-1) d [f(s)] Or ∫ e – st t m f(t) dt = (-1)m dm [f(s)]
dsm
0
dsm
Differentiating both sides w.r.to s, we have
∞
d ∫ e – st tm f(t) dt = (-1)mdm+1 [f(s)]
ds 0
ds m+1
∞
Or ∫ ∂ [e – st tm f(t) ]dt = (-1)mdm+1 [f(s)]
0 ∂s
ds m+1
∞
Or ∫ [- t e – st tm f(t) ]dt = (-1)mdm+1 [f(s)]
0
ds m+1
∞
Or ∫ e – st [ tm+1 f(t) ]dt = (-1)m+1d m+1 [f(s)]
0
ds m+1
Or L [tm+1 f(t) ] =
(-1)m+1d m+1 [f(s)]
ds m+1
Which shows that the theorem is true for n = m+1
Hence by mathematical induction the theorem is true for all n.
Cor. If L -1 { f (s)} = f(t) , then L-1 [ d n ] = (-1)n tn f(t)
dsn
In particular L -1 { d [ f(s)] } = (-t) f(t)
ds
DIVISION BY t
∞
If L {f(t)} = f(s), then L{1 f(t) } = ∫ f(s)ds, Provided the integral exists
t
s
Proof:∞
We have f(s) = ∫ e – st f(t) dt
0
Integrating both sides w.r. to s from s to ∞ , we have
∞
∞∞
∫f(s) ds = ∫ [ ∫ e – st f(t) dt ] ds
s
s 0
changing the order of integration on the right side , we have
∞
∞∞
∞
– st
∫f(s) ds = ∫ [ ∫ e
ds ]f(t) dt = ∫ e – st f(t) dt = L { 1 f(t)}
s
PROBLEMS
1) If L { t sin ωt }=
0
s
0
t
t
2ωs , evaluate L { 2 cos ωt – ωt sin ωt }
(ω 2 + s2)2
L{ f ’’(t)} = s2 f(s) – sf(0) – f ’(0)
=> L { 2 ωcos ωt – ω2t sin ωt }= s2 *
2ωs
- s.0-0 = 2ωs3
(ω 2 + s2)2
(ω 2 + s2)2
3
=> L { 2 cos ωt – ωt sin ωt } = 2s
(ω 2 + s2)2
2) Find the inverse Laplace transforms of
1 .
s3 (s2+a2)
1 ] = 1 sin at
s + a2
a
t
-1
so L [ 1
] = ∫ 1 sin au du = 1/a2 ( 1-cos at)
s(s2 + a2)
0 a
since L -1 [
2
so L -1 [
t
1
] = ∫ 1 ( 1-cos au) du = 1/a2 (t - sin at)
s2(s2 + a2) 0 a2
a
so L -1 [
1
] = ∫ 1 (u - sin au) du = 1/a2 (t2 + cos at - 1)
2
2
s ( s + a ) 0 a2
a
2
a2
a2
3
3) Find the Laplace transforms of t3 e – 3 t
Since L (e – 3 t ) = 1
s+3
L (t3 e – 3 t) = (-1)3 d3 { 1 } =
6.
ds3 s+3 ( s + 3)4
4) Find the Laplace transforms e – a t – e – bt
t
Since L { e – a t – e – bt } = 1 - 1
s +a s +b
∞
L { e – a t – e – bt } = ∫ [ 1 - 1
] ds
t
s s +a s +b
∞
= log[ s + a ] s
s+b
= log s+b
s +a
5) Find the Laplace transforms of t sin3t cos 2t
L (sin3t cos 2t) = ½ L ( sin 5t + sin t)
=½[ . 5 . +. 1.
]
s 2 + 52
s2 + 1
L (t sin3t cos 2t) = - d {½ [ . 5 . + . 1 .
ds
s 2 + 52 .. s2 + 1
=
]}
. 5s . + . s .
( s2 + 25 )2 (s2 +1)2
6) ) Find the Laplace transforms of e -
t
sint
t
. 1
.
( s+1)2 +1
∞
t
L (e - sint) = ∫ . 1 . ds = π
t
s ( s+1)2 +1
2
L(e - t sint) =
- tan -1 (s+1) = cot -1 (s+1)
CONVOLUTION THEOREM
If
t
L-1{f(s)}= f(t), L-1 { g(s)} = g(t) , then L-1{ f(s)g(s)} = ∫ f(u)g(t-u)du = f*g
0
-1
1) Apply Convolution theorem to evaluate L { . s . }
(s2+a2)2
Since L-1 {. s . } = cos at and L -1 { . 1
.} = 1/a sinat
s2 + a2
s2 + a2
By Convolution theorem , we have L-1 {.
s . }= L -1[ . s . . 1. }
s2 + a2
s2 + a2 s2 + a2
t
= ∫ cos a ( t – u)sin audu
0
a
= 1 t sin at
2a
APPLICATION TO DIFFERENTIAL EQUATIONS
1) Solve the equation d3y + 2 d2y – dy – 2y = 0 , dy = 2 , d2y = 2 at t = 0
dt3
dt2 dt
dt
dt2
The given equation is y’’’ + 2y’’ – y1 – 2y = 0
Taking the laplace transform on both sides , we get
[s3y – s2 y(0) – s y’(0) – y’’(0)] + 2[s2y – s y(0)- y’(0)] – [sy – y(0)]- 2y = 0
Using the given conditions y(0) = 1 , y’(0) = 2,y’’(0) = 2 we get
(s3+2s2-s-2)y –s2-2s-2-2s-4+1=0
(s3+2s2-s-2)y = s2+4s+5
y = s2+4s+5 = . s2+4s+5
.
(s3+2s2-s-2) ( s -1)(s +1)(s +2)
=. 5 .–1 +. 1 .
3(s -1) s+1 3(s+2)
Taking the inverse Laplace transform on both sides , we get
y = 1/3[ 5et + e -2t] – e-t
2) Solve the simultaneous equations
(D2 – 3)x – 4y = 0
(D2 +1)y +x = 0 for t > 0, given that x = y = dy = 0 and dx = 2 at t = 0
dt
dt
Taking the laplace transform of the given equations, we get
[s2x– sx(0) –x’(0) – 3x – 4y] =0
(s2 -3)x- 4y = 0 ……………..(1)
2
[x + s y – s y(0) – y’(0) +y = 0
x + (s2 + 1)y = 0……………..(2)
Solving (1) & (2) we get x = 2(s2 +1) = . 1
. +. 1 .
( s2 – 1) 2 (s – 1) 2 ( s + 1) 2
y = -. 2 . = - ½ [ . 1 - 1 .- 1 . + . . 1
]
2
2
2
2
(s – 1)
( s + 1) ( s – 1) ( s +1)
( s – 1)
Taking the inverse Laplace transform on both sides , we get
x = t et + t e –t = 2t [ et + e –t] = 2t cos ht
2
–t
t
t
t
y = -1/2 { e – e – t e + te } = et – e –t – t { et – e-t} =(1-t) sin ht
2
2
LAPLACE TRANSFORMS OF SOME OTHER USEFUL FUNCTIONS
(1) Unit Step Function ( Heaviside’s Unit Function )
The unit step function u(t-a) is defined as
u(t-a) = 0 for t < a
=1 for t ≥ a, where a ≥0
As a particular case ,
u (t) = 0 for t < 0
= 1 for t ≥ 0
Laplace transform of Unit Step function
∞
∞
∞
L[ u(t-a)] = ∫e –st u(t-a) dt = ∫ e –st f(t-a)u(t-a)dt = ∫ e –s(u+a)f(u)du, where u = t - a
0
0
0
∞
= e –as ∫ e-su f(u) du = e-as f(s)
0
1) Find the Laplace transforms of (t-1)2 u(t-1)
Comparing (t-1)2 u(t-1) with f(t-a)u(t-a),we have
a =1 and f(t) = t2
f(s) = L{f(t)} = 2/s3
L{(t-1)2 u(t-1)} = e –s f(s) = 2e-s/s3
2) Find the inverse Laplace transforms of e -2s
s -3
Let f(s) = . . 1 . , then f(t) = e 3t
s-3
L-1 { e -2s } = L-1 { e -2s f(s)} = f(t-2)u(t-2) = e 3(t-2)u(t-2)
Periodic functions
∫
If f(t) is a periodic function with period T ,then L{f(t)} =
f(t)dt
Proof:L{f(t)} = ∫
f(t)dt= ∫
f(t)dt +∫
f(t)dt+…….
Putting t=u,t=U+T,t=U+2T…..in the successive integrals.
L{f(t)}= ∫
f(u+T)du+…….
f(u)du +∫
Since f(u) = f(u+T)=f(u+2T)….we have
L{f(t)}= ∫
f(u)du +
= ( 1+
+
∫
=
∫
+….)∫
f(u)du+…….
f(u)du
f(t)dt
1)Find the Laplace Transform of the triangular wave function of period 2c
given by f(t) = t , 0<t<c and f(t) = 2c-t,c<t<2c.
L{f(t)} =
∫
f(t)dt =
{∫
.tdt+∫
(2c-t)dt
=
Home work
1)Find the Laplace transform of the saw toothed wave function of period T
,defined by f(t) = ,for 0<t<T
2) Find the Laplace transform of the square wave function of period a
defined as
f(t) = 1,when 0<t<a/2 and f(t) = -1 ,when a/2 < t<a.
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