Laplace Transform

Laplace transform is named in honor of the great French mathematician, Pierre Simon De Laplace (1749-1827). Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. The best way to convert differential equations into algebraic equations is the use of Laplace transform. In this section, students get step-by-step explanation to every concept. Students will find it extremely easy to understand differential equations and how to go about solving them.

Laplace Transform Definition

For t ≥ 0, let f(t) be given and assume the function satisfy certain conditions to be stated later on.

The Laplace transform of f(t), that it is denoted by

f(t) or F(s)

is defined by the equation

F(s)= \(\int_0^\infty f(t)\ e^{-st} dt \)

whenever the improper integral converges.

Laplace Transform of Differential Equation

The Laplace transform is a well established mathematical technique for solving a differential equation. Many mathematical problems are solved using transformations. The idea is to transform the problem into another problem that is easier to solve. On the other side, inverse transform is helpful to calculate the solution to the given problem.

For better understanding, let us solve a first order differential equation with the help of laplace transformation, Consider y’- 2y = e3x and y(0) = -5. Find the value of L(y).

First step of the equation can be solved with the help of the linearity equation:

L(y’ – 2y] = L(e3x)

L(y’) – L(2y) = 1/(s-3)   

(because L(eax) = 1/(s-a))

L(y’) – 2s(y) = 1/(s-3)

sL(y) – y(0) – 2L(y) = 1/(s-3)       

(Using Linearity property of the Laplace transform)

L(y)(s-2) + 5 = 1/(s-3) (Use value of y(0) ie -5 (given))

L(y)(s-2) = 1/(s-3) – 5

L(y) = (-5s+16)/(s-2)(s-3) …..(1)

here (-5s+16)/(s-2)(s-3) can be written as -6/s-2 + 1/(s-3) using partial fraction method

(1) implies L(y) = -6/(s-2) + 1/(s-3)

L(y) = -6e2x + e3x

Properties of Laplace Transform

Some of the properties of Laplace transforms are:

If \(\,f_1 (t) \stackrel{\mathrm{L.T}}{\longleftrightarrow} F_1(s)\) and

\(\,f_2 (t) \stackrel{\mathrm{L.T}}{\longleftrightarrow} F_2(s)\), then

Linearity Property \(\,A\ f_1 (t) + B\ f_2(t) \stackrel{\mathrm{L.T}}{\longleftrightarrow} AF_1(s)+BF_2(s)\)
Frequency Shifting Property \(\,e^{s_0t}f(t)) \stackrel{\mathrm{L.T}}{\longleftrightarrow}F(s-s_0)\)
nth Derivative \(\,\frac{\mathrm{d^nf(t)} }{\mathrm{dt^n}} \stackrel{\mathrm{L.T}}{\longleftrightarrow}s^nF(s)-\sum_{i=1}^{n} s^{n-i}f^{i-1}(0^-)\)
Integration \(\,\int_{0}^{t}f(\lambda)d\lambda \stackrel{\mathrm{L.T}}{\longleftrightarrow}\frac{1}{s} F(s)\)
Multiplication by Time T f(t) \(\stackrel{\mathrm{L.T}}{\longleftrightarrow}\) (-dF(s)/ds)
Complex Shift f(t) e^(-at) \(\stackrel{\mathrm{L.T}}{\longleftrightarrow}\) F(s+a)
Time Reversal Property f(-t) \(\stackrel{\mathrm{L.T}}{\longleftrightarrow}\) F(-s)
Time Scaling f(t/a) \(\stackrel{\mathrm{L.T}}{\longleftrightarrow}\) aF(as)

Laplace Transform Table

Sl No. f(t) L(f(t) = F(s) Sl No. f(t) L(f(t) = F(s)
1 1 1/s 11 e(at) 1/(s-a)
2 tn at t = 1,2,3,…   n!/s(n+1) 12 tp, at p>-1 Γ(p+1)/s(p+1)
3 sqrt(t)   sqrt(pi)/2s(3/2) 13 t(n-1/2) at n = 1,2,.. (1.3.5…(2n-1)\(\sqrt{\pi}\))/(2^n s(n+1/2)
4 sin(at) a/(s2+a2) 14 cos(at) s/(s2+a2)
5 t sin(at) 2as/(s2+a2)2 15 t cos(at) (s2-a2)/(s2+a2)2
6 sin(at+b) (s sin(b)+ a cos(b)/(s2+a2) 16 cos(at+b) (s cos(b)-a sin(b)/(s2+a2)
7 sinh(at) a/(s2-a2) 17 cosh(at) s/(s2-a2)
8 e(at)sin(bt) b/((s-a)2+b2) 18 e(at)cos(bt) (s-a)/((s-a)2+b2)
9 e(ct)f(t) F(s-c) 19 tnf(t) at n = 1,2,3.. (-1)n Fn s
10 f'(t) sF(s) – f(0) 20 f”(t) s2F(s) – sf(0)-f'(0)

Laplace Transform Examples

Below examples are based on some important elementary functions of Laplace transform.

Laplace transform

Laplace Transform

Read More:

Laplace Transform Formulas Inverse Laplace Transform
Laplace Transform Differential Equations Laplace Transform Table

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Practise This Question

[51425×559+(12+34)] is equal to: