The name ‘Laplace Transform’ was kept in honor of the great mathematician from France, Pierre Simon De Laplace. Moreover, the Laplace transform converts one signal into another conferring to the fixed set of rules or equations. However, the best method to change the differential equations into algebraic equations is using the Laplace transformation.
Formula
The Laplace transform is the essential makeover of the given derivative function. Moreover, it comes with a real variable (t) for converting into complex function with variable (s). For ‘t’ ≥ 0, let ‘f(t)’ be given and assume the function fulfills certain conditions to be stated later.
Further, the Laplace transform of ‘f(t)’, denoted by ‘f(t)’ or ‘F(s)’ is definable with the equation:
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The Laplace transform is referred to as the one-sided Laplace transform sometimes. In addition, there is a 2 sided type where the integral goes from ‘−∞’ to ‘∞’.
Properties of the Laplace Transform
If, f1 (t) ⟷ F1 (s) and [note: ‘⟷’ implies the Laplace Transform].
f2 (t) ⟷ F2 (s), then:
| Linearity Property | A f1(t) + B f2(t) ⟷ A F1(s) + B F2(s) |
| Frequency Shifting Property | es0t f(t)) ⟷ F(s – s0) |
| Nth Derivative Property | (dn f(t)/ dtn) ⟷ sn F(s) − n∑i = 1 sn − i fi − 1 (0−) |
| Integration | t∫0 f(λ) dλ ⟷ 1⁄s F(s) |
| Multiplication by Time | T f(t) ⟷ (−d F(s)⁄ds) |
| Complex Shift Property | f(t) e−at ⟷ F(s + a) |
| Time Reversal Property | f (-t) ⟷ F(-s) |
| Time Scaling Property | f (t⁄a) ⟷ a F(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 | √(t) | √π/2s(3/2) | 13 | t(n-1/2) at n = 1,2,.. | (1.3.5…(2n-1)√π)/(2ns(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 of Differential Equation
The Laplace transform is a deep-rooted mathematical system for solving the differential equations. Therefore, there are so many mathematical problems that are solved with the help of the transformations. However, the idea is to convert the problem into another problem which is much easier for solving. On the other hand, the inverse transform is useful for calculating the solution to any given problem.
For a brief understanding, we will have to solve a first-order differential equation using the Laplace transformation:
Consider y’- 2y = e3x and ‘y(0) = -5’.
Find out the value of ‘L(y)’.
Step 1 of the equation can be solved using 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 ‘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) is writable as -6/s-2 + 1/(s-3) using the partial fraction technique.
(1) implies L(y) = -6/(s-2) + 1/(s-3)
L(y) = -6e2x + e3x.
Solved Example for You
Question: Confirm that any bounded function ‘f(t)’ for ‘t ≥ 0’ is exponentially bounded.
Solution: Since ‘f(t)’ is bounded for ‘t ≥ 0’, there is a positive constant ‘M’ such that:
|f(t)| ≤ M for all t ≥ 0, But this is similar as (1) with ‘a = 0’ and ‘C = 0’.
Therefore, ‘f(t)’ is exponentially bounded.
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