Laplace Transform of Periodic Function

Example

To find the Laplace transform of given periodic function.

Solution: 

Calculation of the first period is given by:

f1(t) = t*[u(t) - u(t-1)]
period=2
Taking Laplace on both the sides,
L{f1(t)} = L{t*[u(t) - u(t-1)]}
           = L{t*u(t)} - L{t*u(t-1)}
Now,
t⋅u(t−1) = (t−1)⋅u(t−1)+u(t−1)
Then,
L{t⋅u(t)}−L{t⋅u(t−1)}
            = L{t⋅u(t)}− L{(t−1)⋅u(t−1)+u(t−1)}
            = L{t⋅u(t)}− L{(t−1)⋅u(t−1)}− L{u(t−1)}
            = (1-e-s-se-s) / s2
Hence, the Laplace transform of above function is given by
L{f(t)} = L{f1(t)} * (1/1 - e-sp)
L{f(t)} = {(1-e-s-se-s) / s2} * { (1/1 - e-sp)}
          = (1-e-s-se-s) / s2(1/1 - e-sp)

Frequently Asked Questions

1. What is a Laplace transform?

A Laplace transform is an integral transform of a derivative function with a real variable ‘t’ which can be used to convert it into a complex function with a variable ‘s’. For t ≥ 0, let f(t) be given and assume the function satisfies certain conditions to be stated later on.

The Laplace transform of f(t), that is denoted by L{f(t)} or F(s) is defined by the Laplace transform formula:

2. Why do we use Laplace transform?

A Laplace transform is used generally for solving linear ordinary differential equations which involve the use of integral transforms. Differential equations can also be solved using a Laplace transform. 

It is widely accepted in many fields. A Laplace transform reduces linear differential equations (LDE) to an algebraic equation (relates calculus and algebra), which may then be solved using basic algebraic identities. This has 

a lot of applications in physics, electrical engineering, optics, control engineering, mathematics, signal processing, etc.

Key Takeaways

Congratulations on finishing the blog!! After reading this blog, you will grasp the concept of the Laplace Transform.

Recommended Readings:

• Initial Value Theorem

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