Question

Use the result of Problem 28 to find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{aligned} 1, & 0 \leq t<1 \\\\-1, & 1 \leq t<2 \end{aligned}\right. $$ $$ f(t+2)=f(t) $$

Short answer

Question: Find the Laplace transform of the given piecewise-defined function: $$ f(t) = \begin{cases} 1, & 0 \leq t < 1 \\ -1, & 1 \leq t < 2 \end{cases} $$ where the function is periodic with period 2. Answer: The Laplace transform of the given function is $$ F(s) = \frac{1}{s} \left(1 - 2e^{-s} + e^{-2s} \right). $$

Step-by-step solution

Write down the definition of the Laplace transform

Recall that the Laplace transform of a function f(t) is defined as $$ F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty}e^{-st}f(t)dt. $$ We will apply this definition to our given piecewise function.

Separate the function into its two pieces

Since the given function is defined differently for \(0 \leq t < 1\) and \(1 \leq t < 2\), we will separate the function into these two parts and find the Laplace transform of each part individually: $$ F(s) = \int_0^{1} e^{-st} \cdot 1 dt + \int_1^{2} e^{-st} \cdot (-1) dt. $$

Calculate the integrals

Now we evaluate the two integrals separately. First, the integral of \(e^{-st}\) with respect to \(t\): $$ \int e^{-st} dt = -\frac{1}{s}e^{-st} + C. $$ Plugging this back into the previous equation, we get $$ F(s) = -\frac{1}{s}e^{-s(1)} - \left(-\frac{1}{s}e^{-s(0)}\right) - \left[-\frac{1}{s}e^{-s(2)} - \left(-\frac{1}{s}e^{-s(1)}\right)\right]. $$

Simplify the expression

Now, we will simplify the expression for \(F(s)\): $$ F(s) = -\frac{1}{s}e^{-s} - \left(-\frac{1}{s}\right) - \left[-\frac{1}{s}e^{-2s} - \left(-\frac{1}{s}e^{-s}\right)\right]. $$ Simplifying further, $$ F(s) = \frac{1}{s} \left(1 - e^{-s} + e^{-2s} - e^{-s} \right). $$

Apply the periodic condition

Now, we will make use of the information that the function is periodic with a period of 2, which means that: $$ f(t+2) = f(t). $$ This allows us to represent the Laplace transform more compactly: $$ F(s) = \frac{1}{s} \left(1 - 2e^{-s} + e^{-2s} \right). $$ Finally, we have found the Laplace transform of the given function: $$ F(s) = \frac{1}{s} \left(1 - 2e^{-s} + e^{-2s} \right). $$

Key concepts

Piecewise Functions

When studying mathematics, one often encounters piecewise functions, which are defined by different expressions over various intervals. In the context of the Laplace Transform, piecewise functions require particular attention because each segment must be transformed separately.

The example function provides a perfect illustration of how a piecewise function can be handled. In this case, the function switches from 1 to -1 at the point t=1. To find the Laplace Transform, we consider each piece in the interval it's defined. We integrate the corresponding function segment within its specific interval and then aggregate the results.

Understanding piecewise functions is crucial for analyzing complex signals in engineering and physics, as real-world phenomena often behave differently under various conditions. By mastering the approach to these functions, one gains the versatility to tackle a wide range of problems where standard functions fall short.

Periodic Functions

Another significant concept in mathematics and engineering is that of periodic functions. A periodic function is one that repeats its values at regular intervals, known as the period. The function from the exercise demonstrates periodicity with a period of 2. This repetition simplifies many operations, including the Laplace Transform.

For periodic functions, once the Laplace Transform for one period is determined, the entire transform can be constructed due to the function's repetitive nature. This characteristic is particularly useful when analyzing signals in electronics or understanding harmonic motions in mechanics. By recognizing periodicity, one can streamline many complex calculations to a simpler analysis of a single period, saving time and reducing the potential for errors.

Integral Transforms

Lastly, the core of the exercise lies within the realm of integral transforms. Integral transforms, such as the Laplace Transform, are powerful mathematical tools used to convert functions from the time domain to the complex frequency domain. This alter-ego of the function opens doors to easier manipulation and analysis, particularly for differential equations and system responses in engineering.

In the case of the Laplace Transform, we apply an integral operation to the function over the entire range of time, using the kernel e^-st. It is vital for students to understand that integral transforms can greatly simplify the process of solving complex equations, making them a cornerstone in the fields of control theory, signal processing, and more.