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26 Feb, 2026
Laplace transform is an integral transform used in mathematics and engineering to convert a function of time f(t) into a function of a complex variable s, denoted as F(s), where s =
Ο
+
ΞΉ
Ο
\sigma+\iota\omega
.
Let us assume
f
(
t
)
f(t)
is a function, be it a real or complex function of the variable
t
>
0
t>0
, where
t
t
is time. Then, the Laplace transform
F
(
s
)
F(s)
of
f
(
t
)
f(t)
is the complex function defined for
s
β
C
s\in \Complex
, given by:
F
(
s
)
=
L
{
f
(
t
)
}
=
β«
0
β
e
β
s
t
f
(
t
)
β
d
t
F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt
where s = \sigma + i\omega.
Standard Notation:
If a function of
t
t
is indicated as f(t),
f
(
t
)
,
g
(
t
)
,
f(t),g(t),
or
y
(
t
)
y(t)
, then their respective Laplace transforms are represented by
F
(
s
)
,
G
(
s
)
F(s), G(s)
and
Y
(
s
)
Y(s)
. Besides the notation
F
(
s
)
F(s)
, we can also use
L
{
f
(
t
)
}
\mathcal{L}\{f(t)\}
or
L
{
f
}
(
s
)
\mathcal{L}\{f\}(s)
.
Laplace transforms of some elementary functions:
Function
f
(
t
)
f(t)
Laplace Transform
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\} =F(s)
1
1
s
;
β
s
>
0
\frac{1}{s};\,s>0
t
t
1
s
2
;
β
s
>
0
\frac{1}{s^2};\,s>0
t
n
n
=
0
,
1
,
2
,
.
.
.
t^n\\n = 0,1,2,...
n
!
s
n
+
1
;
β
s
>
0
\frac{n!}{s^n+1};\,s>0
e
a
t
e^{at}
1
s
β
a
;
β
s
>
a
\frac{1}{s-a};\,s>a
sin
β‘
a
t
\sin at
a
s
2
+
a
2
;
β
s
>
0
\frac{a}{s^2+a^2};\,s>0
cos
β‘
a
t
\cos at
s
s
2
+
a
2
;
β
s
>
0
\frac{s}{s^2+a^2};\,s>0
sinh
β‘
a
t
\sinh at
a
s
2
β
a
2
;
β
s
>
β£
a
β£
\frac{a}{s^2-a^2};\,s>|a|
cosh
β‘
a
t
\cosh at
s
s
2
β
a
2
;
β
s
>
β£
a
β£
\frac{s}{s^2-a^2};\,s>|a|
Existence of the Laplace Transform
Here are some definitions before delving into the sufficient conditions for the existence of the Laplace transform:
Sectional Continuity:
A function is said to be
sectionally or piecewise continuous
in an interval
t
1
β€
t
β€
t
2
t_1\le t\le t_2
if that interval can be subdivided into a finite number of subintervals, in each of which the function is continuous and has finite left- and right-hand limits.
Functions of Exponential Order
: If real constants
k
>
0
k>0
and
Ξ³
\gamma
exist such that for all
t
>
N
t>N
, the function
f
(
t
)
f(t)
satisfies the condition
β£
f
(
t
)
β£
β€
k
e
Ξ³
t
|f(t)|\le ke^{\gamma t}
, then
f
(
t
)
f(t)
is said to be of exponential order
Ξ³
\bm\gamma
.
Sufficient Condition for Existence of Laplace Transform:
If
f
(
t
)
f(t)
is sectionally continuous in every finite interval
0
β€
t
β€
N
0\le t\le N
and of exponential order
Ξ³
\bm\gamma
for
t
>
N
\bm{t>N}
, then its Laplace transform
F
(
s
)
F(s)
exists for all
s
>
Ξ³
\bm{s>\gamma}
.
Important Properties of Laplace Transformation:
Linearity
Shifting
Change of Scale
Laplace Transforms of Derivatives
Laplace Transforms of Integrals
Multiplication by
t
n
t^n
Division by t
Laplace Transform of a Periodic function
Behavior of F(s) as s
β
β
\to\infty
Initial value theorem
Final value theorem
Convolution theorem for Laplace transform
Linearity
If
c
1
c_1
and
c
2
c_2
are two constants, while
f
1
(
t
)
f_1(t)
and
f
2
(
t
)
f_2(t)
are functions, and
F
1
(
s
)
F_1(s)
and
F
2
(
s
)
F_2(s)
are their respective Laplace transforms, then:
L
{
c
1
f
1
(
t
)
+
c
2
f
2
(
t
)
}
=
c
1
L
{
f
1
(
t
)
}
+
c
2
L
{
f
2
(
t
)
}
=
c
1
F
1
(
s
)
+
c
2
F
2
(
s
)
\begin{aligned}\mathcal{L}\{c_1f_1(t) +c_2f_2(t)\} &= c_1\mathcal{L}\{f_1(t)\}+c_2\mathcal{L}\{f_2(t)\}\\&=c_1F_1(s) +c_2F_2(s)\end{aligned}
Shifting
It constitutes of two properties
First shifting property
: If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\}=F(s)
, then
L
{
e
a
t
f
(
t
)
}
=
F
(
s
β
a
)
\mathcal{L}\{e^{at}f(t)\}=F(s-a)
.
Second Shifting property
: If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\}=F(s)
and
g
(
t
)
=
{
f
(
t
β
a
)
;
t
>
a
0
;
t
<
a
g(t)=\begin{cases}
f(t-a)\quad&; {t>a}\\
0 \quad&;{t<a}
\end{cases}
, then
L
{
g
(
t
)
}
=
e
β
a
s
F
(
s
)
\mathcal{L}\{g(t)\}=e^{-as}F(s)
.
Change of Scale
If f(t) is a function and F(s) is its Laplace transform, and c is a constant, then
L
{
f
(
a
t
)
}
=
1
a
F
(
s
a
)
\mathcal{L}\{f(at)\}=\frac{1}{a}F(\frac{s}{a})
.
Laplace Transform of Derivatives
If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\}=F(s)
, then
L
{
f
β²
(
t
)
}
=
s
F
(
s
)
β
f
(
0
)
\mathcal{L}\{f'(t)\}=sF(s)-f(0)
, where f(t) is continuous for
0
β€
t
β€
N
0\le t\le N
and of exponential order
Ξ³
\gamma
, and its derivative
f
β²
(
t
)
f'(t)
is sectionally continuous for
0
β€
t
β€
N
0\le t\le N
.
If f(t) fails
to be
continuous
at t=0 but
lim
β‘
t
β
0
f
(
t
)
=
f
(
0
+
)
\lim\limits_{t\to0}f(t)=f(0^+)
, then
L
{
f
β²
(
t
)
}
=
s
F
(
s
)
β
f
(
0
+
)
\mathcal{L}\{f'(t)\}=sF(s)-f(0^+)
.
If f(t) fails
to be
continuous
at t=a, then
L
{
f
β²
(
t
)
}
=
s
F
(
s
)
β
f
(
0
)
β
e
β
a
s
{
f
(
a
+
)
β
f
(
a
β
)
}
\mathcal{L}\{f'(t)\}=sF(s)-f(0)-e^{-as}\{f(a^+)-f(a^-)\}
If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\}=F(s)
, then
L
{
f
(
n
)
(
t
)
}
=
s
n
F
(
s
)
β
s
n
β
1
f
(
0
)
β
s
n
β
2
f
β²
(
0
)
β
.
.
.
β
f
(
n
β
1
)
(
0
)
\mathcal{L}\{f^{(n)}(t)\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-...-f^{(n-1)}(0)
, where
f
(
t
)
,
β
f
β²
(
t
)
,
β
f
β²
β²
(
t
)
,
β
.
.
.
β
,
f
(
n
β
1
)
(
t
)
f(t),\,f'(t),\,f''(t),\,...\,,f^{(n-1)}(t)
are continuous for
0
β€
t
β€
N
0\le t\le N
and of exponential order for
t
>
N
t>N
while
f
(
n
)
(
t
)
f^{(n)}(t)
is sectionally continuous for
0
β€
t
β€
N
0\le t\le N
.
Laplace Transform of Integrals:
If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\} =F(s)
, then
L
{
β«
0
t
f
(
v
)
d
v
}
=
F
(
s
)
s
\mathcal{L}\{\int\limits_0^t f(v)dv\}=\frac{F(s)}{s}
.
Multiplication by
t
n
t^n
If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\} = F(s)
, then
L
{
t
n
f
(
t
)
}
=
(
β
1
)
n
d
n
d
s
n
F
(
s
)
\mathcal{L}\{t^nf(t)\} = (-1)^n \frac{d^n}{ds^n}F(s)
, where
d
n
d
s
n
\frac{d^n}{ds^n}
denotes the n-th derivative.
Division by
t
t
If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\} = F(s)
, then
L
{
f
(
t
)
t
}
=
β«
s
β
f
(
v
)
d
v
\mathcal{L}\{\frac{f(t)}{t}\} = \int\limits_{s}^{\infty}f(v)dv
.
Laplace Transform of Periodic functions
Let a function f(t) be periodic with period T>0, such that
f
(
t
+
T
)
=
f
(
t
)
f(t+T)=f(t)
. Then
L
{
f
(
t
)
}
=
β«
0
T
e
β
s
t
f
(
t
)
d
t
1
β
e
β
s
T
\mathcal{L}\{f(t)\}=\frac{\int\limits_0^T e^{-st}f(t)dt}{1-e^{-sT}}
.
Behavior of F(s) as s
β
β
\to\infty
If
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\}=F(s)
, then
lim
β‘
s
β
β
F
(
s
)
=
0
\lim\limits_{s\to\infty}F(s)=0
.
Initial Value Theorem
Let f(t) be a sectional continuous with Laplace transform F(s). Then
lim
β‘
s
β
β
s
F
(
s
)
=
f
(
0
+
)
\lim\limits_{s\to\infty}sF(s) = f(0^+)
, where the limit
s
β
β
s\to\infty
has to be taken in such a way that the real part of s,
Re
s \to\infty as well.
Final Value Theorem
Let f(t) be a sectional continuous with Laplace transform F(s). When
f
(
β
)
=
lim
β‘
t
β
β
f
(
t
)
f(\infty) =\lim\limits_{t\to\infty}f(t)
exists, then
lim
β‘
s
β
0
s
F
(
s
)
=
f
(
β
)
\lim\limits_{s\to0}sF(s) = f(\infty)
, where the limit
s
β
0
s\to0
has to be taken in such a way that the real part of s,
Re(
s) \to0 as well.
Convolution Theorem for Laplace Transform
Let f(t) and g(t) be piecewise continuous and of exponential order
Ξ³
\gamma
with their Laplace transforms
F
(
s
)
=
L
{
f
(
t
)
}
F(s)=\mathcal{L}\{f(t)\}
and
G
(
s
)
=
L
{
g
(
t
)
}
G(s)=\mathcal{L}\{g(t)\}
, respectively. Then
L
{
f
β
g
}
\mathcal{L}\{f*g\}
exists for Re(s)>
Ξ³
\gamma
and
L
{
f
β
g
}
=
F
(
s
)
β
G
(
s
)
\mathcal{L}\{f*g\}=F(s)\cdot G(s)
.
Inverse Laplace Transformation:
If the Laplace transform of a function f(t) is F(s), i.e., if
L
{
f
(
t
)
}
=
F
(
s
)
\mathcal{L}\{f(t)\}=F(s)
, then f(t) is called the inverse Laplace transform of F(s), and we write it symbolically as
f
(
t
)
=
L
β
1
{
F
(
s
)
}
f(t)=\mathcal{L}^{-1}\{F(s)\}
where
L
β
1
\mathcal{L^{-1}}
is called the
inverse Laplace transformation
operator.
Example: Find the inverse Laplace transform of F(s)=
1
s
+
3
\frac{1}{s+3}
.
We know that
L
{
e
a
t
}
=
1
s
β
a
\mathcal{L}\{e^{at}\}=\frac{1}{s-a}
, s>a. Then,
L
β
1
{
1
s
β
a
}
=
e
a
t
\mathcal{L^{-1}}\{\frac{1}{s-a}\}=e^{at}
. Similarly,
L
β
1
{
1
s
+
3
}
=
e
β
3
t
\mathcal{L^{-1}}\{\frac{1}{s+3}\}=e^{-3t}
.
Bilateral Laplace Transform
The bilateral Laplace transform involves the values of a function for both t<0 and t
β₯
\ge
0. This means that the bilateral Laplace transform is well-suited for non-causal signals and functions. If f(t) is the function and F(s) is its bilateral Laplace transform, then F(s) is given by:
F
(
s
)
=
β«
β
β
β
f
(
t
)
e
β
s
t
d
t
F(s) = \int_{-\infty}^{\infty} f(t) e^{-st}dt
Applications of Laplace Transform
Various applications of the Laplace transform include:
The Laplace transform can be used to find the transfer function of linear time-invariant continuous-time systems: In linear time-invariant continuous-time systems (LTI systems), h(t) is the impulse response. The system function, or transfer function, H(s), of the LTI system is the Laplace transform of h(t).
Laplace transform can be used to solve
differential equation
problems, including initial value problems. In an initial value problem, the solution to a differential equation is determined by the initial conditions of the system, such as the initial values of the function and its derivatives.
Let H(s) be the transfer function of a causal LTI system described by the following differential equation:
a
n
d
n
y
(
t
)
d
t
n
+
a
n
β
1
d
n
β
1
y
(
t
)
d
t
n
β
1
+
β―
+
a
1
d
y
(
t
)
d
t
+
a
0
y
(
t
)
=
b
m
d
m
x
(
t
)
d
t
m
+
b
m
β
1
d
m
β
1
x
(
t
)
d
t
m
β
1
+
β―
+
b
1
d
x
(
t
)
d
t
+
b
0
x
(
t
)
a_n \frac{d^n y(t)}{dt^n} + a_{n-1} \frac{d^{n-1} y(t)}{dt^{n-1}} + \cdots + a_1 \frac{dy(t)}{dt} + a_0 y(t) = b_m \frac{d^m x(t)}{dt^m} + b_{m-1} \frac{d^{m-1} x(t)}{dt^{m-1}} + \cdots + b_1 \frac{dx(t)}{dt} + b_0 x(t)
satisfying the following condition of initial rest,
y
(
0
)
=
y
β²
(
0
)
=
y
β²
β²
(
0
)
=
.
.
.
=
y
m
β
2
(
0
)
=
y
m
β
1
(
0
)
=
0
y(0)=y'(0)=y''(0)=...=y^{m-2}(0)=y^{m-1}(0)=0
Then the system is stable if and only if the poles of H(s) lie in the left half-plane Re(s)<0.
Laplace transform can be applied to analyze electrical circuits, simplifying the process of solving circuits with capacitors, inductors, and resistors by converting the time-domain equations into s-domain equations.
Laplace transform is used in
probability theory
to find the distribution of sums of random variables and to solve problems related to stochastic processes. For example: Transforming Probability Density Functions (PDFs). The Laplace transform can be used to transform the probability density function (PDF) of a random variable. For a non-negative
random variable
X with PDF
f
X
(
x
)
f_X(x)
, the Laplace transform is
L
{
f
X
(
x
)
}
=
F
(
s
)
=
β«
0
β
e
β
s
x
f
X
(
x
)
β
d
x
\mathcal{L}\{f_X(x)\} = F(s) = \int_0^\infty e^{-sx} f_X(x) \, dx
.
Related Articles:
Laplace Transform Practice Questions
Inverse Laplace Transform
Fourier Transform |
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# Laplace Transform
Last Updated : 26 Feb, 2026
Laplace transform is an integral transform used in mathematics and engineering to convert a function of time f(t) into a function of a complex variable s, denoted as F(s), where s = Ο \+ ΞΉ Ο \\sigma+\\iota\\omega Ο\+ΞΉΟ.
Let us assume f ( t ) f(t) f(t) is a function, be it a real or complex function of the variable t \> 0 t\>0 t\>0, where t t t is time. Then, the Laplace transform F ( s ) F(s) F(s) of f ( t ) f(t) f(t) is the complex function defined for s β C s\\in \\Complex sβC, given by:
> F ( s ) \= L { f ( t ) } \= β« 0 β e β s t f ( t ) d t F(s) = \\mathcal{L}\\{f(t)\\} = \\int\_{0}^{\\infty} e^{-st} f(t) \\, dt F(s)\=L{f(t)}\=β«0ββeβstf(t)dt
where s = \\sigma + i\\omega.
### Standard Notation:
If a function of t t t is indicated as f(t), f ( t ) , g ( t ) , f(t),g(t), f(t),g(t), or y ( t ) y(t) y(t), then their respective Laplace transforms are represented by F ( s ) , G ( s ) F(s), G(s) F(s),G(s) and Y ( s ) Y(s) Y(s). Besides the notation F ( s ) F(s) F(s), we can also use L { f ( t ) } \\mathcal{L}\\{f(t)\\} L{f(t)} or L { f } ( s ) \\mathcal{L}\\{f\\}(s) L{f}(s).
### Laplace transforms of some elementary functions:
| Function f ( t ) f(t) f(t) | Laplace Transform L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} =F(s) L{f(t)}\=F(s) |
|---|---|
| 1 | 1 s ; s \> 0 \\frac{1}{s};\\,s\>0 s1β;s\>0 |
| t t t | 1 s 2 ; s \> 0 \\frac{1}{s^2};\\,s\>0 s21β;s\>0 |
| t n n \= 0 , 1 , 2 , . . . t^n\\\\n = 0,1,2,... tnn\=0,1,2,... | n \! s n \+ 1 ; s \> 0 \\frac{n!}{s^n+1};\\,s\>0 sn\+1n\!β;s\>0 |
| e a t e^{at} eat | 1 s β a ; s \> a \\frac{1}{s-a};\\,s\>a sβa1β;s\>a |
| sin β‘ a t \\sin at sinat | a s 2 \+ a 2 ; s \> 0 \\frac{a}{s^2+a^2};\\,s\>0 s2\+a2aβ;s\>0 |
| cos β‘ a t \\cos at cosat | s s 2 \+ a 2 ; s \> 0 \\frac{s}{s^2+a^2};\\,s\>0 s2\+a2sβ;s\>0 |
| sinh β‘ a t \\sinh at sinhat | a s 2 β a 2 ; s \> β£ a β£ \\frac{a}{s^2-a^2};\\,s\>\|a\| s2βa2aβ;s\>β£aβ£ |
| cosh β‘ a t \\cosh at coshat | s s 2 β a 2 ; s \> β£ a β£ \\frac{s}{s^2-a^2};\\,s\>\|a\| s2βa2sβ;s\>β£aβ£ |
### Existence of the Laplace Transform
Here are some definitions before delving into the sufficient conditions for the existence of the Laplace transform:
- ****Sectional Continuity:**** A function is said to be ****sectionally or piecewise continuous**** in an interval
t
1
β€
t
β€
t
2
t\_1\\le t\\le t\_2
t1ββ€tβ€t2β
if that interval can be subdivided into a finite number of subintervals, in each of which the function is continuous and has finite left- and right-hand limits.
- ****Functions of Exponential Order****: If real constants
k
\>
0
k\>0
k\>0
and
Ξ³
\\gamma
Ξ³
exist such that for all
t
\>
N
t\>N
t\>N
, the function
f
(
t
)
f(t)
f(t)
satisfies the condition
β£
f
(
t
)
β£
β€
k
e
Ξ³
t
\|f(t)\|\\le ke^{\\gamma t}
β£f(t)β£β€keΞ³t
, then
f
(
t
)
f(t)
f(t)
is said to be of exponential order
Ξ³
\\bm\\gamma
Ξ³
.
#### Sufficient Condition for Existence of Laplace Transform:
If f ( t ) f(t) f(t) is sectionally continuous in every finite interval 0 β€ t β€ N 0\\le t\\le N 0β€tβ€N and of exponential order Ξ³ \\bm\\gamma Ξ³ for t \> N \\bm{t\>N} t\>N, then its Laplace transform F ( s ) F(s) F(s) exists for all s \> Ξ³ \\bm{s\>\\gamma} s\>Ξ³.
## Important Properties of Laplace Transformation:
- Linearity
- Shifting
- Change of Scale
- Laplace Transforms of Derivatives
- Laplace Transforms of Integrals
- Multiplication by
t
n
t^n
tn
- Division by t
- Laplace Transform of a Periodic function
- Behavior of F(s) as s
β
β
\\to\\infty
ββ
- Initial value theorem
- Final value theorem
- Convolution theorem for Laplace transform
### Linearity
If c 1 c\_1 c1β and c 2 c\_2 c2β are two constants, while f 1 ( t ) f\_1(t) f1β(t) and f 2 ( t ) f\_2(t) f2β(t) are functions, and F 1 ( s ) F\_1(s) F1β(s) and F 2 ( s ) F\_2(s) F2β(s) are their respective Laplace transforms, then:
> L { c 1 f 1 ( t ) \+ c 2 f 2 ( t ) } \= c 1 L { f 1 ( t ) } \+ c 2 L { f 2 ( t ) } \= c 1 F 1 ( s ) \+ c 2 F 2 ( s ) \\begin{aligned}\\mathcal{L}\\{c\_1f\_1(t) +c\_2f\_2(t)\\} &= c\_1\\mathcal{L}\\{f\_1(t)\\}+c\_2\\mathcal{L}\\{f\_2(t)\\}\\\\&=c\_1F\_1(s) +c\_2F\_2(s)\\end{aligned} L{c1βf1β(t)\+c2βf2β(t)}β\=c1βL{f1β(t)}\+c2βL{f2β(t)}\=c1βF1β(s)\+c2βF2β(s)β
### Shifting
It constitutes of two properties
1. ****First shifting property****: If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
L{f(t)}\=F(s)
, then
L
{
e
a
t
f
(
t
)
}
\=
F
(
s
β
a
)
\\mathcal{L}\\{e^{at}f(t)\\}=F(s-a)
L{eatf(t)}\=F(sβa)
.
2. ****Second Shifting property****: If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
L{f(t)}\=F(s)
and
g
(
t
)
\=
{
f
(
t
β
a
)
;
t
\>
a
0
;
t
\<
a
g(t)=\\begin{cases} f(t-a)\\quad&; {t\>a}\\\\ 0 \\quad&;{t\<a} \\end{cases}
g(t)\={f(tβa)0β;t\>a;t\<aβ
, then
L
{
g
(
t
)
}
\=
e
β
a
s
F
(
s
)
\\mathcal{L}\\{g(t)\\}=e^{-as}F(s)
L{g(t)}\=eβasF(s)
.
### Change of Scale
If f(t) is a function and F(s) is its Laplace transform, and c is a constant, then L { f ( a t ) } \= 1 a F ( s a ) \\mathcal{L}\\{f(at)\\}=\\frac{1}{a}F(\\frac{s}{a}) L{f(at)}\=a1βF(asβ).
### Laplace Transform of Derivatives
1. If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
L{f(t)}\=F(s)
, then
L
{
f
β²
(
t
)
}
\=
s
F
(
s
)
β
f
(
0
)
\\mathcal{L}\\{f'(t)\\}=sF(s)-f(0)
L{fβ²(t)}\=sF(s)βf(0)
, where f(t) is continuous for
0
β€
t
β€
N
0\\le t\\le N
0β€tβ€N
and of exponential order
Ξ³
\\gamma
Ξ³
, and its derivative
f
β²
(
t
)
f'(t)
fβ²(t)
is sectionally continuous for
0
β€
t
β€
N
0\\le t\\le N
0β€tβ€N
.
2. ****If f(t) fails**** to be ****continuous**** at t=0 but
lim
β‘
t
β
0
f
(
t
)
\=
f
(
0
\+
)
\\lim\\limits\_{t\\to0}f(t)=f(0^+)
tβ0limβf(t)\=f(0\+)
, then
L
{
f
β²
(
t
)
}
\=
s
F
(
s
)
β
f
(
0
\+
)
\\mathcal{L}\\{f'(t)\\}=sF(s)-f(0^+)
L{fβ²(t)}\=sF(s)βf(0\+)
.
3. ****If f(t) fails**** to be ****continuous**** at t=a, then
L
{
f
β²
(
t
)
}
\=
s
F
(
s
)
β
f
(
0
)
β
e
β
a
s
{
f
(
a
\+
)
β
f
(
a
β
)
}
\\mathcal{L}\\{f'(t)\\}=sF(s)-f(0)-e^{-as}\\{f(a^+)-f(a^-)\\}
L{fβ²(t)}\=sF(s)βf(0)βeβas{f(a\+)βf(aβ)}
4. If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
L{f(t)}\=F(s)
, then
L
{
f
(
n
)
(
t
)
}
\=
s
n
F
(
s
)
β
s
n
β
1
f
(
0
)
β
s
n
β
2
f
β²
(
0
)
β
.
.
.
β
f
(
n
β
1
)
(
0
)
\\mathcal{L}\\{f^{(n)}(t)\\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-...-f^{(n-1)}(0)
L{f(n)(t)}\=snF(s)βsnβ1f(0)βsnβ2fβ²(0)β...βf(nβ1)(0)
, where
f
(
t
)
,
f
β²
(
t
)
,
f
β²
β²
(
t
)
,
.
.
.
,
f
(
n
β
1
)
(
t
)
f(t),\\,f'(t),\\,f''(t),\\,...\\,,f^{(n-1)}(t)
f(t),fβ²(t),fβ²β²(t),...,f(nβ1)(t)
are continuous for
0
β€
t
β€
N
0\\le t\\le N
0β€tβ€N
and of exponential order for
t
\>
N
t\>N
t\>N
while
f
(
n
)
(
t
)
f^{(n)}(t)
f(n)(t)
is sectionally continuous for
0
β€
t
β€
N
0\\le t\\le N
0β€tβ€N
.
### Laplace Transform of Integrals:
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} =F(s) L{f(t)}\=F(s), then L { β« 0 t f ( v ) d v } \= F ( s ) s \\mathcal{L}\\{\\int\\limits\_0^t f(v)dv\\}=\\frac{F(s)}{s} L{0β«tβf(v)dv}\=sF(s)β.
### Multiplication by t n t^n tn
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} = F(s) L{f(t)}\=F(s), then L { t n f ( t ) } \= ( β 1 ) n d n d s n F ( s ) \\mathcal{L}\\{t^nf(t)\\} = (-1)^n \\frac{d^n}{ds^n}F(s) L{tnf(t)}\=(β1)ndsndnβF(s), where d n d s n \\frac{d^n}{ds^n} dsndnβ denotes the n-th derivative.
### Division by t t t
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} = F(s) L{f(t)}\=F(s), then L { f ( t ) t } \= β« s β f ( v ) d v \\mathcal{L}\\{\\frac{f(t)}{t}\\} = \\int\\limits\_{s}^{\\infty}f(v)dv L{tf(t)β}\=sβ«ββf(v)dv.
### Laplace Transform of Periodic functions
Let a function f(t) be periodic with period T\>0, such that f ( t \+ T ) \= f ( t ) f(t+T)=f(t) f(t\+T)\=f(t). Then L { f ( t ) } \= β« 0 T e β s t f ( t ) d t 1 β e β s T \\mathcal{L}\\{f(t)\\}=\\frac{\\int\\limits\_0^T e^{-st}f(t)dt}{1-e^{-sT}} L{f(t)}\=1βeβsT0β«Tβeβstf(t)dtβ.
### Behavior of F(s) as sβ β \\to\\infty ββ
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\}=F(s) L{f(t)}\=F(s), then lim β‘ s β β F ( s ) \= 0 \\lim\\limits\_{s\\to\\infty}F(s)=0 sββlimβF(s)\=0.
### Initial Value Theorem
Let f(t) be a sectional continuous with Laplace transform F(s). Then lim β‘ s β β s F ( s ) \= f ( 0 \+ ) \\lim\\limits\_{s\\to\\infty}sF(s) = f(0^+) sββlimβsF(s)\=f(0\+), where the limit s β β s\\to\\infty sββ has to be taken in such a way that the real part of s, ****Re**** s \\to\\infty as well.
### Final Value Theorem
Let f(t) be a sectional continuous with Laplace transform F(s). When f ( β ) \= lim β‘ t β β f ( t ) f(\\infty) =\\lim\\limits\_{t\\to\\infty}f(t) f(β)\=tββlimβf(t) exists, then lim β‘ s β 0 s F ( s ) \= f ( β ) \\lim\\limits\_{s\\to0}sF(s) = f(\\infty) sβ0limβsF(s)\=f(β), where the limit s β 0 s\\to0 sβ0 has to be taken in such a way that the real part of s, ****Re(****s) \\to0 as well.
### Convolution Theorem for Laplace Transform
Let f(t) and g(t) be piecewise continuous and of exponential order Ξ³ \\gamma Ξ³ with their Laplace transforms F ( s ) \= L { f ( t ) } F(s)=\\mathcal{L}\\{f(t)\\} F(s)\=L{f(t)} and G ( s ) \= L { g ( t ) } G(s)=\\mathcal{L}\\{g(t)\\} G(s)\=L{g(t)}, respectively. Then L { f β g } \\mathcal{L}\\{f\*g\\} L{fβg} exists for Re(s)\>Ξ³ \\gamma Ξ³ and L { f β g } \= F ( s ) β
G ( s ) \\mathcal{L}\\{f\*g\\}=F(s)\\cdot G(s) L{fβg}\=F(s)β
G(s).
## Inverse Laplace Transformation:
If the Laplace transform of a function f(t) is F(s), i.e., if L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\}=F(s) L{f(t)}\=F(s), then f(t) is called the inverse Laplace transform of F(s), and we write it symbolically as f ( t ) \= L β 1 { F ( s ) } f(t)=\\mathcal{L}^{-1}\\{F(s)\\} f(t)\=Lβ1{F(s)}where L β 1 \\mathcal{L^{-1}} Lβ1 is called the [inverse Laplace transformation](https://www.geeksforgeeks.org/electronics-engineering/inverse-laplace-transform/) operator.
- Example: Find the inverse Laplace transform of F(s)=
1
s
\+
3
\\frac{1}{s+3}
s\+31β
.
We know that
L
{
e
a
t
}
\=
1
s
β
a
\\mathcal{L}\\{e^{at}\\}=\\frac{1}{s-a}
L{eat}\=sβa1β
, s\>a. Then,
L
β
1
{
1
s
β
a
}
\=
e
a
t
\\mathcal{L^{-1}}\\{\\frac{1}{s-a}\\}=e^{at}
Lβ1{sβa1β}\=eat
. Similarly,
L
β
1
{
1
s
\+
3
}
\=
e
β
3
t
\\mathcal{L^{-1}}\\{\\frac{1}{s+3}\\}=e^{-3t}
Lβ1{s\+31β}\=eβ3t
.
## Bilateral Laplace Transform
The bilateral Laplace transform involves the values of a function for both t\<0 and tβ₯ \\ge β₯0\. This means that the bilateral Laplace transform is well-suited for non-causal signals and functions. If f(t) is the function and F(s) is its bilateral Laplace transform, then F(s) is given by:
> F ( s ) \= β« β β β f ( t ) e β s t d t F(s) = \\int\_{-\\infty}^{\\infty} f(t) e^{-st}dt F(s)\=β«ββββf(t)eβstdt
## Applications of Laplace Transform
Various applications of the Laplace transform include:
- The Laplace transform can be used to find the transfer function of linear time-invariant continuous-time systems: In linear time-invariant continuous-time systems (LTI systems), h(t) is the impulse response. The system function, or transfer function, H(s), of the LTI system is the Laplace transform of h(t).
- Laplace transform can be used to solve [differential equation](https://www.geeksforgeeks.org/maths/differential-equations/) problems, including initial value problems. In an initial value problem, the solution to a differential equation is determined by the initial conditions of the system, such as the initial values of the function and its derivatives.
- Let H(s) be the transfer function of a causal LTI system described by the following differential equation:
a
n
d
n
y
(
t
)
d
t
n
\+
a
n
β
1
d
n
β
1
y
(
t
)
d
t
n
β
1
\+
β―
\+
a
1
d
y
(
t
)
d
t
\+
a
0
y
(
t
)
\=
b
m
d
m
x
(
t
)
d
t
m
\+
b
m
β
1
d
m
β
1
x
(
t
)
d
t
m
β
1
\+
β―
\+
b
1
d
x
(
t
)
d
t
\+
b
0
x
(
t
)
a\_n \\frac{d^n y(t)}{dt^n} + a\_{n-1} \\frac{d^{n-1} y(t)}{dt^{n-1}} + \\cdots + a\_1 \\frac{dy(t)}{dt} + a\_0 y(t) = b\_m \\frac{d^m x(t)}{dt^m} + b\_{m-1} \\frac{d^{m-1} x(t)}{dt^{m-1}} + \\cdots + b\_1 \\frac{dx(t)}{dt} + b\_0 x(t)
anβdtndny(t)β\+anβ1βdtnβ1dnβ1y(t)β\+β―\+a1βdtdy(t)β\+a0βy(t)\=bmβdtmdmx(t)β\+bmβ1βdtmβ1dmβ1x(t)β\+β―\+b1βdtdx(t)β\+b0βx(t)
satisfying the following condition of initial rest,
y
(
0
)
\=
y
β²
(
0
)
\=
y
β²
β²
(
0
)
\=
.
.
.
\=
y
m
β
2
(
0
)
\=
y
m
β
1
(
0
)
\=
0
y(0)=y'(0)=y''(0)=...=y^{m-2}(0)=y^{m-1}(0)=0
y(0)\=yβ²(0)\=yβ²β²(0)\=...\=ymβ2(0)\=ymβ1(0)\=0
Then the system is stable if and only if the poles of H(s) lie in the left half-plane Re(s)\<0.
- Laplace transform can be applied to analyze electrical circuits, simplifying the process of solving circuits with capacitors, inductors, and resistors by converting the time-domain equations into s-domain equations.
- Laplace transform is used in [probability theory](https://www.geeksforgeeks.org/maths/probability-theory/) to find the distribution of sums of random variables and to solve problems related to stochastic processes. For example: Transforming Probability Density Functions (PDFs). The Laplace transform can be used to transform the probability density function (PDF) of a random variable. For a non-negative [random variable](https://www.geeksforgeeks.org/engineering-mathematics/random-variable/) X with PDF
f
X
(
x
)
f\_X(x)
fXβ(x)
, the Laplace transform is
L
{
f
X
(
x
)
}
\=
F
(
s
)
\=
β«
0
β
e
β
s
x
f
X
(
x
)
d
x
\\mathcal{L}\\{f\_X(x)\\} = F(s) = \\int\_0^\\infty e^{-sx} f\_X(x) \\, dx
L{fXβ(x)}\=F(s)\=β«0ββeβsxfXβ(x)dx
.
### Related Articles:
> - [Laplace Transform Practice Questions](https://www.geeksforgeeks.org/maths/laplace-transforms-practice-problems/#:~:text=What%20is%20the%20Laplace%20transform%3F,the%20solution%20of%20differential%20equations.)
> - [Inverse Laplace Transform](https://www.geeksforgeeks.org/dsa/solving-initial-value-2nd-order-differential-equation-problem-using-laplace-transform-in-matlab/)
> - [Fourier Transform](https://www.geeksforgeeks.org/maths/fourier-transform/)
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| Readable Markdown | Last Updated : 26 Feb, 2026
Laplace transform is an integral transform used in mathematics and engineering to convert a function of time f(t) into a function of a complex variable s, denoted as F(s), where s = Ο \+ ΞΉ Ο \\sigma+\\iota\\omega.
Let us assume f ( t ) f(t) is a function, be it a real or complex function of the variable t \> 0 t\>0, where t t is time. Then, the Laplace transform F ( s ) F(s) of f ( t ) f(t) is the complex function defined for s β C s\\in \\Complex, given by:
> F ( s ) \= L { f ( t ) } \= β« 0 β e β s t f ( t ) d t F(s) = \\mathcal{L}\\{f(t)\\} = \\int\_{0}^{\\infty} e^{-st} f(t) \\, dt
where s = \\sigma + i\\omega.
### Standard Notation:
If a function of t t is indicated as f(t), f ( t ) , g ( t ) , f(t),g(t), or y ( t ) y(t), then their respective Laplace transforms are represented by F ( s ) , G ( s ) F(s), G(s) and Y ( s ) Y(s). Besides the notation F ( s ) F(s), we can also use L { f ( t ) } \\mathcal{L}\\{f(t)\\} or L { f } ( s ) \\mathcal{L}\\{f\\}(s).
### Laplace transforms of some elementary functions:
| Function f ( t ) f(t) | Laplace Transform L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} =F(s) |
|---|---|
| 1 | 1 s ; s \> 0 \\frac{1}{s};\\,s\>0 |
| t t | 1 s 2 ; s \> 0 \\frac{1}{s^2};\\,s\>0 |
| t n n \= 0 , 1 , 2 , . . . t^n\\\\n = 0,1,2,... | n \! s n \+ 1 ; s \> 0 \\frac{n!}{s^n+1};\\,s\>0 |
| e a t e^{at} | 1 s β a ; s \> a \\frac{1}{s-a};\\,s\>a |
| sin β‘ a t \\sin at | a s 2 \+ a 2 ; s \> 0 \\frac{a}{s^2+a^2};\\,s\>0 |
| cos β‘ a t \\cos at | s s 2 \+ a 2 ; s \> 0 \\frac{s}{s^2+a^2};\\,s\>0 |
| sinh β‘ a t \\sinh at | a s 2 β a 2 ; s \> β£ a β£ \\frac{a}{s^2-a^2};\\,s\>\|a\| |
| cosh β‘ a t \\cosh at | s s 2 β a 2 ; s \> β£ a β£ \\frac{s}{s^2-a^2};\\,s\>\|a\| |
### Existence of the Laplace Transform
Here are some definitions before delving into the sufficient conditions for the existence of the Laplace transform:
- ****Sectional Continuity:**** A function is said to be ****sectionally or piecewise continuous**** in an interval
t
1
β€
t
β€
t
2
t\_1\\le t\\le t\_2
if that interval can be subdivided into a finite number of subintervals, in each of which the function is continuous and has finite left- and right-hand limits.
- ****Functions of Exponential Order****: If real constants
k
\>
0
k\>0
and
Ξ³
\\gamma
exist such that for all
t
\>
N
t\>N
, the function
f
(
t
)
f(t)
satisfies the condition
β£
f
(
t
)
β£
β€
k
e
Ξ³
t
\|f(t)\|\\le ke^{\\gamma t}
, then
f
(
t
)
f(t)
is said to be of exponential order
Ξ³
\\bm\\gamma
.
#### Sufficient Condition for Existence of Laplace Transform:
If f ( t ) f(t) is sectionally continuous in every finite interval 0 β€ t β€ N 0\\le t\\le N and of exponential order Ξ³ \\bm\\gamma for t \> N \\bm{t\>N}, then its Laplace transform F ( s ) F(s) exists for all s \> Ξ³ \\bm{s\>\\gamma}.
## Important Properties of Laplace Transformation:
- Linearity
- Shifting
- Change of Scale
- Laplace Transforms of Derivatives
- Laplace Transforms of Integrals
- Multiplication by
t
n
t^n
- Division by t
- Laplace Transform of a Periodic function
- Behavior of F(s) as s
β
β
\\to\\infty
- Initial value theorem
- Final value theorem
- Convolution theorem for Laplace transform
### Linearity
If c 1 c\_1 and c 2 c\_2 are two constants, while f 1 ( t ) f\_1(t) and f 2 ( t ) f\_2(t) are functions, and F 1 ( s ) F\_1(s) and F 2 ( s ) F\_2(s) are their respective Laplace transforms, then:
> L { c 1 f 1 ( t ) \+ c 2 f 2 ( t ) } \= c 1 L { f 1 ( t ) } \+ c 2 L { f 2 ( t ) } \= c 1 F 1 ( s ) \+ c 2 F 2 ( s ) \\begin{aligned}\\mathcal{L}\\{c\_1f\_1(t) +c\_2f\_2(t)\\} &= c\_1\\mathcal{L}\\{f\_1(t)\\}+c\_2\\mathcal{L}\\{f\_2(t)\\}\\\\&=c\_1F\_1(s) +c\_2F\_2(s)\\end{aligned}
### Shifting
It constitutes of two properties
1. ****First shifting property****: If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
, then
L
{
e
a
t
f
(
t
)
}
\=
F
(
s
β
a
)
\\mathcal{L}\\{e^{at}f(t)\\}=F(s-a)
.
2. ****Second Shifting property****: If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
and
g
(
t
)
\=
{
f
(
t
β
a
)
;
t
\>
a
0
;
t
\<
a
g(t)=\\begin{cases} f(t-a)\\quad&; {t\>a}\\\\ 0 \\quad&;{t\<a} \\end{cases}
, then
L
{
g
(
t
)
}
\=
e
β
a
s
F
(
s
)
\\mathcal{L}\\{g(t)\\}=e^{-as}F(s)
.
### Change of Scale
If f(t) is a function and F(s) is its Laplace transform, and c is a constant, then L { f ( a t ) } \= 1 a F ( s a ) \\mathcal{L}\\{f(at)\\}=\\frac{1}{a}F(\\frac{s}{a}).
### Laplace Transform of Derivatives
1. If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
, then
L
{
f
β²
(
t
)
}
\=
s
F
(
s
)
β
f
(
0
)
\\mathcal{L}\\{f'(t)\\}=sF(s)-f(0)
, where f(t) is continuous for
0
β€
t
β€
N
0\\le t\\le N
and of exponential order
Ξ³
\\gamma
, and its derivative
f
β²
(
t
)
f'(t)
is sectionally continuous for
0
β€
t
β€
N
0\\le t\\le N
.
2. ****If f(t) fails**** to be ****continuous**** at t=0 but
lim
β‘
t
β
0
f
(
t
)
\=
f
(
0
\+
)
\\lim\\limits\_{t\\to0}f(t)=f(0^+)
, then
L
{
f
β²
(
t
)
}
\=
s
F
(
s
)
β
f
(
0
\+
)
\\mathcal{L}\\{f'(t)\\}=sF(s)-f(0^+)
.
3. ****If f(t) fails**** to be ****continuous**** at t=a, then
L
{
f
β²
(
t
)
}
\=
s
F
(
s
)
β
f
(
0
)
β
e
β
a
s
{
f
(
a
\+
)
β
f
(
a
β
)
}
\\mathcal{L}\\{f'(t)\\}=sF(s)-f(0)-e^{-as}\\{f(a^+)-f(a^-)\\}
4. If
L
{
f
(
t
)
}
\=
F
(
s
)
\\mathcal{L}\\{f(t)\\}=F(s)
, then
L
{
f
(
n
)
(
t
)
}
\=
s
n
F
(
s
)
β
s
n
β
1
f
(
0
)
β
s
n
β
2
f
β²
(
0
)
β
.
.
.
β
f
(
n
β
1
)
(
0
)
\\mathcal{L}\\{f^{(n)}(t)\\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-...-f^{(n-1)}(0)
, where
f
(
t
)
,
f
β²
(
t
)
,
f
β²
β²
(
t
)
,
.
.
.
,
f
(
n
β
1
)
(
t
)
f(t),\\,f'(t),\\,f''(t),\\,...\\,,f^{(n-1)}(t)
are continuous for
0
β€
t
β€
N
0\\le t\\le N
and of exponential order for
t
\>
N
t\>N
while
f
(
n
)
(
t
)
f^{(n)}(t)
is sectionally continuous for
0
β€
t
β€
N
0\\le t\\le N
.
### Laplace Transform of Integrals:
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} =F(s), then L { β« 0 t f ( v ) d v } \= F ( s ) s \\mathcal{L}\\{\\int\\limits\_0^t f(v)dv\\}=\\frac{F(s)}{s}.
### Multiplication by t n t^n
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} = F(s), then L { t n f ( t ) } \= ( β 1 ) n d n d s n F ( s ) \\mathcal{L}\\{t^nf(t)\\} = (-1)^n \\frac{d^n}{ds^n}F(s), where d n d s n \\frac{d^n}{ds^n} denotes the n-th derivative.
### Division by t t
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\} = F(s), then L { f ( t ) t } \= β« s β f ( v ) d v \\mathcal{L}\\{\\frac{f(t)}{t}\\} = \\int\\limits\_{s}^{\\infty}f(v)dv.
### Laplace Transform of Periodic functions
Let a function f(t) be periodic with period T\>0, such that f ( t \+ T ) \= f ( t ) f(t+T)=f(t). Then L { f ( t ) } \= β« 0 T e β s t f ( t ) d t 1 β e β s T \\mathcal{L}\\{f(t)\\}=\\frac{\\int\\limits\_0^T e^{-st}f(t)dt}{1-e^{-sT}}.
### Behavior of F(s) as sβ β \\to\\infty
If L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\}=F(s), then lim β‘ s β β F ( s ) \= 0 \\lim\\limits\_{s\\to\\infty}F(s)=0.
### Initial Value Theorem
Let f(t) be a sectional continuous with Laplace transform F(s). Then lim β‘ s β β s F ( s ) \= f ( 0 \+ ) \\lim\\limits\_{s\\to\\infty}sF(s) = f(0^+), where the limit s β β s\\to\\infty has to be taken in such a way that the real part of s, ****Re**** s \\to\\infty as well.
### Final Value Theorem
Let f(t) be a sectional continuous with Laplace transform F(s). When f ( β ) \= lim β‘ t β β f ( t ) f(\\infty) =\\lim\\limits\_{t\\to\\infty}f(t) exists, then lim β‘ s β 0 s F ( s ) \= f ( β ) \\lim\\limits\_{s\\to0}sF(s) = f(\\infty), where the limit s β 0 s\\to0 has to be taken in such a way that the real part of s, ****Re(****s) \\to0 as well.
### Convolution Theorem for Laplace Transform
Let f(t) and g(t) be piecewise continuous and of exponential order Ξ³ \\gamma with their Laplace transforms F ( s ) \= L { f ( t ) } F(s)=\\mathcal{L}\\{f(t)\\} and G ( s ) \= L { g ( t ) } G(s)=\\mathcal{L}\\{g(t)\\}, respectively. Then L { f β g } \\mathcal{L}\\{f\*g\\} exists for Re(s)\>Ξ³ \\gamma and L { f β g } \= F ( s ) β
G ( s ) \\mathcal{L}\\{f\*g\\}=F(s)\\cdot G(s).
## Inverse Laplace Transformation:
If the Laplace transform of a function f(t) is F(s), i.e., if L { f ( t ) } \= F ( s ) \\mathcal{L}\\{f(t)\\}=F(s), then f(t) is called the inverse Laplace transform of F(s), and we write it symbolically as f ( t ) \= L β 1 { F ( s ) } f(t)=\\mathcal{L}^{-1}\\{F(s)\\}where L β 1 \\mathcal{L^{-1}} is called the [inverse Laplace transformation](https://www.geeksforgeeks.org/electronics-engineering/inverse-laplace-transform/) operator.
- Example: Find the inverse Laplace transform of F(s)=
1
s
\+
3
\\frac{1}{s+3}
.
We know that
L
{
e
a
t
}
\=
1
s
β
a
\\mathcal{L}\\{e^{at}\\}=\\frac{1}{s-a}
, s\>a. Then,
L
β
1
{
1
s
β
a
}
\=
e
a
t
\\mathcal{L^{-1}}\\{\\frac{1}{s-a}\\}=e^{at}
. Similarly,
L
β
1
{
1
s
\+
3
}
\=
e
β
3
t
\\mathcal{L^{-1}}\\{\\frac{1}{s+3}\\}=e^{-3t}
.
## Bilateral Laplace Transform
The bilateral Laplace transform involves the values of a function for both t\<0 and tβ₯ \\ge0\. This means that the bilateral Laplace transform is well-suited for non-causal signals and functions. If f(t) is the function and F(s) is its bilateral Laplace transform, then F(s) is given by:
> F ( s ) \= β« β β β f ( t ) e β s t d t F(s) = \\int\_{-\\infty}^{\\infty} f(t) e^{-st}dt
## Applications of Laplace Transform
Various applications of the Laplace transform include:
- The Laplace transform can be used to find the transfer function of linear time-invariant continuous-time systems: In linear time-invariant continuous-time systems (LTI systems), h(t) is the impulse response. The system function, or transfer function, H(s), of the LTI system is the Laplace transform of h(t).
- Laplace transform can be used to solve [differential equation](https://www.geeksforgeeks.org/maths/differential-equations/) problems, including initial value problems. In an initial value problem, the solution to a differential equation is determined by the initial conditions of the system, such as the initial values of the function and its derivatives.
- Let H(s) be the transfer function of a causal LTI system described by the following differential equation:
a
n
d
n
y
(
t
)
d
t
n
\+
a
n
β
1
d
n
β
1
y
(
t
)
d
t
n
β
1
\+
β―
\+
a
1
d
y
(
t
)
d
t
\+
a
0
y
(
t
)
\=
b
m
d
m
x
(
t
)
d
t
m
\+
b
m
β
1
d
m
β
1
x
(
t
)
d
t
m
β
1
\+
β―
\+
b
1
d
x
(
t
)
d
t
\+
b
0
x
(
t
)
a\_n \\frac{d^n y(t)}{dt^n} + a\_{n-1} \\frac{d^{n-1} y(t)}{dt^{n-1}} + \\cdots + a\_1 \\frac{dy(t)}{dt} + a\_0 y(t) = b\_m \\frac{d^m x(t)}{dt^m} + b\_{m-1} \\frac{d^{m-1} x(t)}{dt^{m-1}} + \\cdots + b\_1 \\frac{dx(t)}{dt} + b\_0 x(t)
satisfying the following condition of initial rest,
y
(
0
)
\=
y
β²
(
0
)
\=
y
β²
β²
(
0
)
\=
.
.
.
\=
y
m
β
2
(
0
)
\=
y
m
β
1
(
0
)
\=
0
y(0)=y'(0)=y''(0)=...=y^{m-2}(0)=y^{m-1}(0)=0
Then the system is stable if and only if the poles of H(s) lie in the left half-plane Re(s)\<0.
- Laplace transform can be applied to analyze electrical circuits, simplifying the process of solving circuits with capacitors, inductors, and resistors by converting the time-domain equations into s-domain equations.
- Laplace transform is used in [probability theory](https://www.geeksforgeeks.org/maths/probability-theory/) to find the distribution of sums of random variables and to solve problems related to stochastic processes. For example: Transforming Probability Density Functions (PDFs). The Laplace transform can be used to transform the probability density function (PDF) of a random variable. For a non-negative [random variable](https://www.geeksforgeeks.org/engineering-mathematics/random-variable/) X with PDF
f
X
(
x
)
f\_X(x)
, the Laplace transform is
L
{
f
X
(
x
)
}
\=
F
(
s
)
\=
β«
0
β
e
β
s
x
f
X
(
x
)
d
x
\\mathcal{L}\\{f\_X(x)\\} = F(s) = \\int\_0^\\infty e^{-sx} f\_X(x) \\, dx
.
### Related Articles:
> - [Laplace Transform Practice Questions](https://www.geeksforgeeks.org/maths/laplace-transforms-practice-problems/#:~:text=What%20is%20the%20Laplace%20transform%3F,the%20solution%20of%20differential%20equations.)
> - [Inverse Laplace Transform](https://www.geeksforgeeks.org/dsa/solving-initial-value-2nd-order-differential-equation-problem-using-laplace-transform-in-matlab/)
> - [Fourier Transform](https://www.geeksforgeeks.org/maths/fourier-transform/) |
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