# Laplace transform of the multiplication of two functions

• Debdut
In summary, the Laplace transform of the multiplication of two functions can be calculated by taking the convolution of their individual Laplace transforms. This is known as the convolution theorem and is a powerful tool for solving complex differential equations involving multiplication of functions. By using the properties of the Laplace transform and the convolution theorem, the multiplication of two functions can be simplified and solved without the need for complicated integration techniques. This makes the Laplace transform an essential tool in mathematical and engineering applications.
Debdut
I have two functions ##\phi(t)=\cos(\omega t)## and ##f(t)=u(t)−u(t−k)## with ##f(t)=f(t+T)##, ##u(t)## is the unit step function.
The problem is to find Laplace transform of ##\phi(t) \cdot f(t)##.

I have tried convolution in frequency domain, but unable to solve it because of gamma functions. Also a doubt is arising about the limits of the convolution. Laplace transform of multiplication of simple functions and convolution of their individual transforms are not matching when taking limits like ##0 \rightarrow s## or ##s \rightarrow \infty## or ##0 \rightarrow \infty##.

I have also thought about integration by parts of ##\phi(t)f(t)e^{−st}## as its limits are known: ##0 \rightarrow \infty##. But value of a periodic function at infinity is undefined.

I am stuck, please help...

<Moderator's note: Member has been warned not to remove template.>

Last edited by a moderator:
Debdut said:
I have two functions ##\phi(t)=\cos(\omega t)## and ##f(t)=u(t)−u(t−k)## with ##f(t)=f(t+T)##, ##u(t)## is the unit step function.
The problem is to find Laplace transform of ##\phi(t) \cdot f(t)##.

I have tried convolution in frequency domain, but unable to solve it because of gamma functions. Also a doubt is arising about the limits of the convolution. Laplace transform of multiplication of simple functions and convolution of their individual transforms are not matching when taking limits like ##0 \rightarrow s## or ##s \rightarrow \infty## or ##0 \rightarrow \infty##.

I have also thought about integration by parts of ##\phi(t)f(t)e^{−st}## as its limits are known: ##0 \rightarrow \infty##. But value of a periodic function at infinity is undefined.

I am stuck, please help...

<Moderator's note: Member has been warned not to remove template.>

Assuming that ##T > k## the problem is do-able. Here is one way to do it.

Your transform has the form
$$\sum_{n=0}^{\infty} \int_0^k e^{-s(t+nT)} \cos(wt+nT)) \, dt$$
Expand the ##\cos(wt +n w T)## and then do the integrations.

Last edited:
Thank you very much sir.
I must say that I don't fully understand the above expression. The only formulae that I have got are from this site:
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
Could you please offer some explanation or redirect me to any reference where I can learn it more?

Debdut said:
Thank you very much sir.
I must say that I don't fully understand the above expression. The only formulae that I have got are from this site:
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
Could you please offer some explanation or redirect me to any reference where I can learn it more?

Just apply the definition of the Laplace transform: ##\int_0^{\infty} e^{-st} F(t) \, dt##. Now use the fact that ##F(t)## is the product of a periodic (period ##T##) "rectangular" function of width ##k## and ##\cos(wt)##. Express the integrals over ##T < t < 2T, 2T < t < 3T, \ldots## as integrals over ##0 < t < T## with shifted values of ##t##.

Reading about Laplace transforms won't help much; just go back to the start and apply the definition.

Debdut
OK, thank you again.

## What is a Laplace transform?

The Laplace transform is a mathematical operation that transforms a time-domain function into a complex frequency-domain function. It is commonly used in engineering and physics to solve differential equations and analyze systems.

## How is the Laplace transform of the multiplication of two functions calculated?

The Laplace transform of the multiplication of two functions is calculated by taking the Laplace transform of each individual function, multiplying them together, and then taking the inverse Laplace transform of the resulting expression.

## What are the properties of the Laplace transform of the multiplication of two functions?

The Laplace transform of the multiplication of two functions follows the same properties as the individual Laplace transforms, such as linearity, time shifting, and frequency shifting. Additionally, the Laplace transform of the multiplication of two functions can be written as the convolution of their individual Laplace transforms.

## Why is the Laplace transform of the multiplication of two functions useful?

The Laplace transform of the multiplication of two functions allows us to easily solve differential equations and analyze complex systems in the frequency domain. It also provides a simpler and more efficient way to solve problems involving convolutions.

## Are there any limitations to using the Laplace transform of the multiplication of two functions?

One limitation of the Laplace transform of the multiplication of two functions is that it can only be applied to functions that satisfy certain conditions, such as having finite integrals and being of exponential order. Additionally, the inverse Laplace transform may not exist for certain functions.

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