How to Find the Laplace Transform of ft t - 1, t ≥ 1 and t - 1, 0 ≤ t ≤ 1

The Laplace transform is a crucial tool in solving differential equations and analyzing linear time-invariant systems. In this article, we will explore the process of finding the Laplace transform of the piecewise-defined function:

1. Simplifying the Function

Consider the function defined in two intervals:

For 0 ≤ t ≤ 1:

ft 1 - t

For t ≥ 1:

ft t - 1

We can combine these intervals to write the function ft(t) as:

ft(t) {1 - t, if 0 ≤ t ≤ 1; t - 1, if t ≥ 1}

2. Expressing ft(t) using Unit Step Functions

Unit step functions, denoted as u(t), can simplify the expression. Specifically, we use the Heaviside step function u(t-a) which is 0 for t and 1 for t ≥ a. Using this function, we can rewrite ft(t) as:

ft(t) 1 - (t - 1)u(t - 1) (t - 1)u(t - 1)

This simplifies to:

ft(t) 2 - 2u(t - 1) (t - 1)u(t - 1)

3. Calculating the Laplace Transform

The Laplace transform of a function f(t) is defined as:

mathcal{L}{f(t)} F(s) int_0^∞e^{-st}f(t)dt

Given the function ft(t) in terms of unit step functions, we can calculate the Laplace transform using linearity and the known transforms of unit step functions:

mathcal{L}{2 - 2u(t - 1) (t - 1)u(t - 1)}

Breaking it down:

mathcal{L}{2 - 2u(t - 1)} 2/s - 2e^{-s}/s

mathcal{L}{(t - 1)u(t - 1)} de^{-s}/ds e^{-s}/s^2 - e^{-2s}/s^2

Combining the results:

mathcal{L}{ft(t)} 2/s - 2e^{-s}/s e^{-s}/s^2 - e^{-2s}/s^2

Final Result

The Laplace transform of ft(t) 2 - 2u(t - 1) (t - 1)u(t - 1) is:

mathcal{L}{ft(t)} 2/s - 2e^{-s}/s e^{-s}/s^2 - e^{-2s}/s^2

References and Further Reading

For a deeper understanding of Laplace transforms and their applications in various fields, including engineering and physics, you may refer to:

Stakgold, I. (2000). . Wiley. Debnath, L., Bhatta, D. (2007). . Chapman Hall/CRC. Boyce, W. E., DiPrima, R. C. (2012). . Wiley.