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Table of Content
 1 Inverse Laplace Transform Formula: 2 How to Find Inverse Laplace Transform? 3 Inverse Laplace Table: 4 What is the importance of the inverse Laplace transform? 5 What is the inverse of Laplace’s constant? 6 What does inverse Fourier transform can do?
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An online inverse Laplace transform calculator will convert the complex function F(s) into a simple function f(t) in the real-time domain. The inverse Laplace calculator has several properties that make it useful for analyzing linear dynamical systems. So, start learning to understand how to find the inverse Laplace transform with the help of an example and inverse Laplace table.

## What is Inverse Laplace Transform?

In mathematics, the inverse Laplace transform is the opposite method, starting from F(s) of the complex variable s, and then returning it to the real variable function f(t). Ideally, we want to simplify the F(s) of the complex variable to the point where we make the comparison for the formula from an inverse Laplace transform table.

If the Laplace inverse formula corresponding to each part of F(s) found on the table, then the inverse Laplace Transform calculator can be transformed. Let us start over from the perspective of the real variable t and obtain the function of f(t).

However, an Online Laplace Transform Calculator provides the transformation of the real variable function to the complex variable.

### Inverse Laplace Transform Formula:

The inverse Laplace transform of the function F(s) is a real function f(t), which is piecewise continuous and exponentially restricted. Its properties are:

$$L {f}(s) = L {f(t)} (s) = F(s)$$

It can be proved that if the function F(s) has the inverse Laplace transform f(t), then f(t) is uniquely determined (considering that the function is divided only by a set of distinguishing points, use the same Null Lebesgue metric).

If given the two Laplace transforms G (s) and F (s), then

$$L^{−1} {xF(s) + y G(s)} = x L^{−1} {F(s)} + y L^−1{G(s)}$$

With any constants x and y.

### How to Find Inverse Laplace Transform?

There are many inverse Laplace transform examples available for determining the inverse transform.

Example 1:

Find the inverse transform:

$$F(s) = 21 / s − 1 /(s − 17) + 15 (s − 33)$$

Solution:

As can be seen from the denominator of the first term, it is just a constant. The correct numerator of this term is “1”. If we use the inverse Laplace Transform Calculator, then we will only consider factor 21 before the inverse transformation. Therefore, a = 17 is a numerator which exactly what it needs to be. The third term also seems to be exponential, but this time a = 33, we need to factor the 15 before performing the inverse transformation.

More details than what we usually enter,

$$F(s) = 21/s − 1 / (s − 17) + 15 (s − 33)$$
$$f(t) = 21(1) − e^{17t} + 15 (e^{33t})$$

$$= 21 − e^{17t} + 15 e^{33t}$$

Example 2:

Determine the inverse Laplace transform:
$$F(s) = 9s / (s^2 + 36) +3 / (s^2 + 36)$$

Solution:

The numerators in the inverse Laplace table tell us about which denominator we really have. In this case, we need to multiply the numerator by 9. The second term has only one constant in the numerator. Therefore, we need to multiply/divide by 6 for inverse Laplace:

$$F(s) = 9s / (s^2 + (6)^2) + 7 (6 / 6) / s^2 + (6)^2$$

$$= 9s / (s^2 + (6)^2) + (7 / 6) 6 / s^2 + (6)^2$$

By taking the inverse transform provides,

$$f(t) = 9cos (6t) + 7 / 6sin (6t)$$

However, you can get the same results by substituting these values in the inverse Laplace Transform Calculator.

Example 3:

Finding the inverse Laplace transform of

$$F(s) = 3s − 13 / s^2 + 11$$

The denominator indicates whether it is sine or cosine. However, the numerator does not match any counter in the table. Cosine needs only one s in the numerator, and at the most multiplicative constant, while sine needs only one constant in the numerator without s. We got both in the numerator.

However, this is easy to solve. We will break the conversion into two different parts and then perform the reverse conversion.

$$F(s) = 3s / s^2 + 11 − 13 (\sqrt {11} / \sqrt {11}) / s^2 + 11$$

$$f(t) = 3cos (\sqrt{11t}) − 13 / \sqrt{11t} sin( \sqrt{11t})$$

### Inverse Laplace Table:

Here’s a table with factor in denominators for making all these calculations handy.

 Factor in denominator Term in partial fraction decomposition Ax + b A / ax + b (ax + b)^k A_1 / ax + b + A_2(ax + b)^2 + ⋯ + Ak / (ax + b)k Ax^2 + bx + c Ax + B / ax^2 + bx + c (ax^2 + bx + c)^k A_1x + B1 / ax^2 + bx + c + A_2x + B_2 / (ax^2 + bx + c)^2+ . . . + A_kx + B_k / (ax^2 + bx + c)^k

#### Inverse Laplace Transform Example with Partial Fractions Decomposition:

Example:

Use the inverse Laplace to find f(t).

$$F(s) =​ s + 19​​ / s^​2​​ − 3s − 10​​$$

Solution:

Simplify F(s) so that we can identify the inverse Laplace transform formula for each part of F(s) from the Laplace inverse table. However, F(s) is too complicated to fit with any formula in the table. Therefore, we will try to simplify them by considering the denominator.

$$F(s)= ​s + 19 / ​(s + 2) (s − 5)$$

Now try a partial fractions decomposition so:

​$$s + 19 / (s + 2) (s − 5)​​ = A / ​s + 2 + B / ​s − 5$$

​Multiplying both sides by the denominator we get:

$$S + 19 = A (s − 5) + B (s + 2)$$

If we set s = -2, s = −2,

If we set s = 5, this will remove A and allow us to solve for B.

$$5 + 19 = A (5 − 5) + B(5 + 2)$$

$$24 = A (0) + B (7)$$

$$24 = 7 B$$

$$B=\frac{24}{7}​​$$

If we solve for A where s = -5, then inverse laplace transform calculator will remove the B:

$$−2 + 19 = A (−2 − 5) + B (−2 + 2)$$

$$17 = A(-7) + B (0)$$

$$17 = -7 A$$

$$A = -17 / 7$$

Substituting the values for A and B back into the fractions decomposition provides:

$$F(s) =​​​ −​5 / 19 ( 1 / s + 2) ​​​​+ 12 / 5 (1/ ​s – 5)$$

Now, factoring the constants out of every numerator, the inverse Laplace calculator uses the function as:

$$F(s) = −​5 / 19 ​​(​1 / s + 2​​1​​) + ​​​12 / ​7 ​(​1 / s − 5​​​)​​​$$

We can observe that both values in the parentheses resemble the inverse Laplace transform formula

$$e​^{at} ​​= ​1 / s – a$$

We need to change the 1/(s+2) so that it exactly matches the inverse Laplace formula:​

$$F(s)=− 5 /19 [1 / ​s − (−2)​​]+​12 / 7​​(1 / ​s−5​​)$$

Then, identify the $$a_1 = -2 and a_2 = 5$$ ,

Laplace inverse calculator use formula for the equation as

$$f(t) = −​5 / 19 ​(e^{​−2t}​​) +​ 12 / 19 ​​(e​^{5t}​​)$$

$$f(t)=\frac{12} {19}e^{5t}-\frac {5} {19} e^{-2t}$$

However, an online Riemann Sum Calculator helps you to approximate the definite integral and sample points of midpoints, right and left endpoints using finite sum.

## How Inverse Laplace Transform Calculator Works?

An online inverse Laplace calculator allows you to transform a complex Function F(s) into a simple real function f(t) by following these instructions:

### Input:

• Enter a complex function F(s) and see the equation preview in Laplace form.
• Hit the calculate button to see the results.

### Output:

• The Laplace inverse calculator transforms the given equation into a simple form.
• You can transform many equations with this Laplace step function calculator numerous times quickly without any cost.

## FAQ:

### What is the importance of the inverse Laplace transform?

The Laplace transform provides some useful techniques to determine certain types of differential equations when the initial conditions are met, especially when the original value is zero.

### What is the inverse of Laplace’s constant?

The sum of the two separate terms is the reciprocal of the sum of the inverse transformation of each term, the latter being considered separately. The Laplace transform is a constant multiplied by a function with an inverse constant.

### What does inverse Fourier transform can do?

The Fourier transform is used to transform the signal from the time domain to the frequency domain, and the inverse Fourier transform is used to transform the signal from the frequency domain back to the time domain. This is achieved by using the inverse fast Fourier transform IFFT.

## Conclusion:

Use this online inverse Laplace transform calculator, which solves differential equations by transforming one form F(s) into another simple form f(t). The inverse Laplace transform can more easily solve the problems in technical applications in real life and makes the differential equations simple to solve.

## Reference:

From the source of Wikipedia: Inverse Laplace transform, Mellin’s inverse formula, Post’s inversion formula.

From the source of Paul’s online notes: Inverse Laplace Transform, Factor in the denominator, Term in partial fraction decomposition.

From the source of Science Direct: Linear Systems Analysis, inverse transform, Additive property, First shift theorem, The Convolution Theorem.