Laplace Transform Calculator

Table of Content

Laplace transform calculator with steps

The Laplace transform calculator is used to convert the real variable function to a complex-valued function. This Laplace calculator provides the step-by-step solution of the given function.

By using our Laplace integral calculator, you can also get the differentiation and integration of the complex-valued function.

How does the Laplace transformation calculator work?

Follow the below steps to transform a real-valued function.

• Enter the function into the input box.

• Use the keypad icon for entering the mathematics symbols.

• Press the calculate button to get the result.

• Click the show more button to view the solution with steps.

• To enter another input hit the reset button.

What is Laplace transformation?

Laplace transform is a method to convert the given function into some other function of s. It is an improper integral from zero to infinity of e to the minus st times f of t with respect to t. The notation of Laplace transform is an L-like symbol used to transform one function into another.

\(L\left\{f\left(t\right)\right\}=F\left(s\right)\)

Laplace transform converts the given real-valued function into a complex-valued function by integrating the function.

The formula for Laplace Transform

The formula used for the transformation of the function is given below.

\( F\left(s\right)=L\left\{f\left(t\right)\right\}=\int _0^{\infty \:}e^{-st}f\left(t\right)dt\)

f(t) is the function of t
s is the frequency parameter of complex number.

How to find the Laplace transform of a function?

Following are some examples solved by our Laplace solver.

Example 1

Find the Laplace transform of \(f\left(t\right)=e^t+sin\left(t\right)\).

Solution

Step 1: Write the given functions equal to f(t) & g(t).

\(f\left(t\right)=e^t\)

\(g\left(t\right)=sin\left(t\right)\)

Step 2: Now Apply the Laplace notation.

\(L\left\{f\left(t\right)+g\left(t\right)\right\}=L\left\{e^t+sin\left(t\right)\right\}\)

Step 3: Now apply the linearity property of Laplace.

\(L\left\{f\left(t\right)+g\left(t\right)\right\}=L\left\{e^t\right\}+L\left\{sin\left(t\right)\right\}\)

Step 4: Now transform the functions by using the Laplace table.

\(L\left\{e^t\right\}=\frac{1}{s-1}\)

\(L\left\{sin\left(t\right)\right\}=\frac{1}{s^2+1}\)

Step 5: Put the values.

\(L\left\{e^{t\:}+sin\left(t\right)\right\}=\frac{1}{s-1}+\frac{1}{s^2+1}\)

Example 2

Find the Laplace transform of \( e^{-2t}\sin ^2\left(t\right)\).

Solution

Step 1: Apply the notation of Laplace .

\(L\left\{e^{-2t}\:sin\:^2\left(t\right)\right\}\)

Step 2: Use a sine identity and put it into the given function.

We know that:

\(\sin ^2\left(x\right)=\frac{1}{2}-\cos \left(2x\right)\frac{1}{2}\)

So,

\(L\left\{e^{-2t}\left(\frac{1}{2}-\cos \:\:\left(2t\right)\frac{1}{2}\right)\right\}\)

Step 3: Now apply the linearity property of Laplace.

\(L\left\{a\cdot \:\:f\left(t\right)+b\cdot \:\:g\left(t\right)\right\}=a\cdot \:L\left\{f\left(t\right)\right\}+b\cdot \:L\left\{g\left(t\right)\right\}\)

\(L\left\{e^{-2t}\left(\frac{1}{2}\cos \:\left(2t\right)\frac{1}{2}\right)\right\}=\frac{1}{2}L\left\{e^{-2t}\right\}-\frac{1}{2}L\left\{e^{-2t}\cos \:\:\left(2t\right)\right\}\)

Step 4: Use the Laplace table to get the result.

\(L\left\{e^{-2t}\right\}=\frac{1}{s+2}\)

\(L\left\{e^{-2t}cos\left(2t\right)\right\}=\frac{s+2}{\left(s+2\right)^2+4}\)

Step 5: Put the values.

\(L\left\{e^{-2t}\left(\frac{1}{2}-\cos \:\:\left(2t\right)\frac{1}{2}\right)\right\}=\frac{1}{2}\cdot \:\frac{1}{s+2}-\frac{1}{2}\cdot \:\frac{s+2}{\left(s+2\right)^2+4}\)

\(L\left\{e^{-2t}\left(\frac{1}{2}-\cos \:\:\left(2t\right)\frac{1}{2}\right)\right\}=\frac{2}{\left(s+2\right)\left(s^2+4s+8\right)}\)

Table of Laplace Transform

This Laplace Transform calculator with steps follows the below table to transform the functions.

\(f\left(t\right)\)	\(F\left(s\right)=L\left\{f\left(t\right)\right\}\)
1	\(\frac{1}{s}\)
\(e^{at}\)	\(\frac{1}{s-a}\)
\(t^n,n=1,2,3,...\)	\(\frac{n}{s^{n+1}}\)
\(t^p, p>-1\)	\(\frac{Γ\left(p+1\right)}{s^{p+1}}\)
\(\sqrt{t}\)	\(\frac{\sqrt{\pi \:}}{2s^{\frac{3}{2}}}\)
\(t^{n-\frac{1}{2}},n=1,2,3,...\ \)	\(\frac{1⋅3⋅5⋯\left(2n−1\right)\sqrt{\pi }}{2^ns^{n+\frac{1}{2}}}\)
\(sin\left(at\right)\)	\(\frac{a}{s^2+a^2}\)
\(cos\left(at\right)\)	\(\frac{s}{s^2+a^2}\)
\(tsin\left(at\right)\)	\(\frac{2as}{\left(s^2+a^2\right)^2}\)
\(tcos\left(at\right)\)	\(\frac{s^2-a^2}{\left(s^2+a^2\right)^2}\)
\(sin\left(at\right)−atcos\left(at\right)\)	\(\frac{2a^3}{\left(s^2+a^2\right)^2}\)
\(sin\left(at\right)+atcos\left(at\right)\)	\(\frac{2as^2}{\left(s^2+a^2\right)^2}\)
\(cos\left(at\right)−atsin\left(at\right)\)	\(\frac{s\left(s^2-a^2\right)}{\left(s^2+a^2\right)^2}\)
\(cos\left(at\right)+atsin\left(at\right)\)	\(\frac{s\left(s^2+)3a^2\right)}{\left(s^2+a^2\right)^2}\)
\(sin\left(at+b\right)\)	\(\frac{ssin\left(b\right)+acos\left(b\right)}{s^2+a^2}\)
\(cos\left(at+b\right)\)	\(\frac{scos\left(b\right)−asin\left(b\right)}{s^2+a^2}\)
\(sinh\left(at\right)\)	\(\frac{a}{s^2-a^2}\)
\(cosh\left(at\right)\)	\(\frac{s}{s^2-a^2}\)
\(e^{at}sin\left(bt\right)\)	\(\frac{b}{\left(s-a\right)^2+b^2}\)
\(e^{at}cos\left(bt\right)\)	\(\frac{s-a}{\left(s-a\right)^2+b^2}\)
\(e^{at}sinh\left(bt\right)\)	\(\frac{b}{\left(s-a\right)^2-b^2}\)
\(e^{at}cosh\left(bt\right)\)	\(\frac{s-a}{\left(s-a\right)^2-b^2}\)
\(t^ne^{at},\:n=1,2,3,...\)	\(\frac{n!}{\left(s-a\right)^{n+1}}\)
\(f\left(ct\right)\)	\(\frac{1}{s}F\left(\frac{s}{c}\right)\)
\(uc\left(t\right)=u\left(t−c\right)\)	\(e^{-cs}\) /s
\(δ\left(t−c\right)\)	\(e^{-cs}\)
\(u_c\left(t\right)f\left(t−c\right)\)	\(e^{-cs}F\left(s\right)\)
\(u_c\left(t\right)g\left(t\right)\)	\(e^{-cs}L\left\{g\left(t+c\right)\right\}\)
\(e^{ct}f\left(t\right)\)	\(F\left(s−c\right)\)
\(t^nf\left(t\right),n=1,2,3,...\ \)	\(\left(-1\right)^nF^{\left(n\right)}\left(s\right)\)
\(\frac{1}{t}f\left(t\right)\)	\(\int _s^{\infty }F\left(u\right)du\)
\(\int _0^tf\left(v\right)dv\)	F(s)/s
\(\int _0^tf\left(t−τ\right)g\left(τ\right)dτ\)	\(F\left(s\right)G\left(s\right)\)
\(f\left(t+T\right)=f\left(t\right)\)	\(\frac{\int _0^Te^{-st}f\left(t\right)dt\:}{1-e^{-sT}}\)
\(f'\left(t\right)\)	\(sF\left(s\right)−f\left(0\right)\)
\(f''\left(t\right)\)	\(s^2F\left(s\right)-s f\left(0\right)−f′\left(0\right)\)