 # Inverse Laplace Transform

## Laplace Transform and Inverse Laplace Transform

The Inverse Laplace Transform can be described as the transformation into a function of time. In the Laplace inverse formula  F(s) is the Transform of F(t) while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). Therefore, we can write this Inverse Laplace transform formula as follows:

f(t) = L⁻¹{F}(t) = $\frac{1}{2\pi i} \lim_{T\rightarrow \infty} \oint_{\gamma - iT}^{\gamma + iT} e^{st} F(s) ds$

If the integrable functions differ on the Lebesgue measure then the integrable functions can have the same Laplace transform. Therefore, there is an inverse transform on the very range of transform. The inverse of a complex function F(s) to generate a real-valued function f(t) is an inverse Laplace transformation of the function. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. This function is therefore an exponentially restricted real function.

## Inverse Laplace Transform Table

 Function y(a) Transform Y(b) b 1 $\frac{1}{b}$ b>0 a $\frac{1}{b^{2}}$ b>0 Aⁱ , i = integer $\frac{i!}{s(i + 1)}$ b>0 exp (ta), where t = constant $\frac{1}{(b - t)}$ b>t cos (sa), s= constant $\frac{b}{b^{2} + s^{2}}$ b>0 Sin (sa), s = constant $\frac{t}{b^{2} + s^{2}}$ b>0 exp(ta)cos(sa) $\frac{b - t}{(b - t)^{2} + s^{2}}$ b>t exp(ta)sin(sa) $\frac{s}{(b - t)^{2} + s^{2}}$ b>t

### Inverse Laplace Transform Theorems

Theorem 1: When a and b are constant,

L⁻¹ {a f(s) + b g(s)} = a L⁻¹ {f(s)} + b L⁻¹{g(s)}

Theorem 2: L⁻¹ {f(s)} = $e^{-at} L^{-1}$ {f(s - a)}

### Inverse Laplace Transform Examples

Example 1)  Compute the inverse Laplace transform of Y (s) = $\frac{2}{3−5s}$.

Solution 1) Adjust it as follows:

Y(s) = $\frac{2}{3 - 5s} = \frac{-2}{5}. \frac{1}{s - \frac{3}{5}}$

Thus, by linearity,

Y(t) = $L^{-1}[\frac{-2}{5}. \frac{1}{s - \frac{3}{5}}]$

= $\frac{-2}{5} L^{-1}[\frac{1}{s - \frac{3}{5}}]$

= $\frac{-2}{5} e^{(\frac{3}{5})t}$

Example 2) Compute the inverse Laplace transform of Y (s) = $\frac{5s}{s^{2} + 9}$

Solution 2) Adjust it as follows:

Y (s) = $\frac{5s}{s^{2} + 9} = 5. \frac{s}{s^{2} + 9}$

Thus, by linearity,

y(t) = $L^{-1} [5. \frac{s}{s^{2} + 9}]$

= $5 L^{-1} [\frac{s}{s^{2} + 9}]$

= 5 Cos 3t

Example 3) Compute the inverse Laplace transform of Y (s) = $\frac{2}{3s^{4}}$.

Solution 3)  Adjust it as follows:

Y (s) = $\frac{2}{3s^{4}} = \frac{1}{9} . \frac{3!}{s^{4}}$

Thus, by linearity,

y(t) = $L^{-1} [ \frac{1}{9}. \frac{3!}{s^{4}}]$

= $\frac{1}{9} L^{-1} [\frac{3!}{s^{4}}]$

= $\frac{1}{9}t^{3}$

Example 4) Compute the inverse Laplace transform of Y (s) = $\frac{3s + 2}{s^{2} + 25}$.

Solution 4)  Adjust it as follows:

Y (s) = $\frac{3s + 2}{s^{2} + 25}$

= $\frac{3s}{s^{2} + 25}$ + $\frac{2}{s^{2} + 25}$

= $3. \frac{s}{s^{2} + 25} + \frac{2}{5} . \frac{5}{s^{2} + 25}$

Thus,

$L^{-1}[3. \frac{s}{s^{2} + 25} + \frac{2}{5} . \frac{5}{s^{2} + 25}]$

= $3 L^{-1} [\frac{s}{s^{2} + 25}] + \frac{2}{5} L^{-1} [\frac{5}{s^{2} + 25}]$

= 3 Cos 5t + $\frac{2}{5}$ Sin 5t.

Example 5) Compute the inverse Laplace transform of Y (s) = $\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}$

Solution 5) Adjust it as follows:

Y (s) = $\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}$

= $\frac{1}{-4} . \frac{1}{s - \frac{3}{4}} + \frac{3}{s^{2} + 49} - \frac{2s}{s^{2} + 49}$

= $\frac{1}{-4} . \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{7}{s^{2} + 49} -2. \frac{s}{s^{2} + 49}$

Thus,

y(t) = $L^{-1} [\frac{-1}{4}. \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{7}{s^{2} + 49} -2. \frac{s}{s^{2} + 49}]$

= $-\frac{1}{4} L^{-1} [\frac{1}{s - \frac{3}{4}}] + \frac{3}{7} L^{-1}[\frac{7}{s^{2} + 49}] -2 L^{-1} [\frac{s}{s^{2} + 49}]$

= $-\frac{1}{4} e^{(\frac{3}{4})t} + \frac{3}{7} sin 7t - 2 cos 7t$

Example 6)  Compute the inverse Laplace transform of Y (s) = $\frac{5}{(s + 2)^{3}}$

Solution 6) The transform pair is:

$t \Leftrightarrow \frac{2}{s^{3}}$

According to the proposition,

$e^{-2t}t^{2} \Leftrightarrow \frac{2}{(s + 2)^{3}}$

Therefore,

y(t) = $L^{-1} [\frac{5}{(s + 2)^{3}}]$

= $L^{-1} [\frac{5}{2} . \frac{2}{(s + 2)^{3}}]$

= $\frac{5}{2} L^{-1} [\frac{2}{(s + 2)^{3}}]$

= $\frac{5}{2} e^{-2t}t^{2}$

Example 7) Compute the inverse Laplace transform of Y (s) = $\frac{4(s - 1)}{(s - 1)^{2} + 4}$

Solution 7) The transform pair is:

$cos 2t \Leftrightarrow \frac{s}{s^{2} + 4}$

According to the proposition,

$e^{t} cos 2t \Leftrightarrow \frac{s - 1}{(s - 1)^{2} + 4}$

Hence,

y(t) = $L^{-1} [\frac{4(s - 1)}{(s - 1)^{2} + 4}]$

= $4 L^{-1} [\frac{s - 1}{(s - 1)^{2} + 4}]$

= $4e^{t} cos 2t$

Question 1) What is the Inverse Laplace Transform of 1?

Answer 1) First we have to discuss the unit impulse function :-

In mechanics, the idea of a large force acting for a short time occurs frequently.

Therefore, to deal with such similar ideas, we use the unit impulse function which is also called Dirac delta function. We can define the unit impulse function by the limiting form of it.

δₑ(t - a) = 1/e , a ≤ t ≤ a + e

Otherwise, 0.

If ϵ → 0, the height of the strip will increase indefinitely and the width will decrease in such a manner that its area is always unity.

Thus the unit impulse function δ(t - a) can be defined as

δ(t - a) = ∞  where t= a

Solving it, our end result would be L⁻¹ = δ(t).

Question 2) What is the Main Purpose or Application of Inverse Laplace Transform?

Answer 2) The Inverse Laplace Transform can be described as the transformation into a function of time. In the Laplace inverse formula  F(s) is the Transform of F(t) while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). Therefore, Inverse Laplace can basically convert any variable domain back to the time domain or any basic domain for example, from frequency domain back to the time domain. These properties allow them to be utilized for solving and analyzing linear dynamical systems and optimisation purposes. Both Laplace and inverse laplace transforms can be used to solve differential equations in an extremely easy way.