Question

Find \(\mathcal{L}\\{f(t)\\}\) for each of the functions \(f\) defined. $$ f(t)=\left\\{\begin{array}{ll} |\sin t|, & 0 \leq t<\pi \\ f(t+\pi)=f(t), & \text { for all } t \geq 0 \end{array}\right. $$

Short answer

The Laplace Transform of the given function \(f(t)\) is \(\mathcal{L}\\{f(t)\\} = \frac{1}{(1-e^{-s\pi})(s^2+1)}\left(\int_{0}^{\pi}\sin(t)e^{-st}dt - \int_{\pi}^{\infty}\sin(t)e^{-st}dt\right)\).

Step-by-step solution

Calculate the Laplace Transform for the given interval

Recall the definition of the Laplace Transform for a function \(g(t)\): \[ \mathcal{L}\\{g(t)\\} = \int_{0}^{\infty} g(t)e^{-st} dt \newline \] Now, let's consider the function \(\vert \sin t \vert\) and find its Laplace Transform: \[ \mathcal{L}\\{|\sin t|\\} = \int_{0}^{\infty} |\sin t|e^{-st} dt \newline \] Since the function \( |\sin t| \) is defined for \(0 \le t < \pi\), we have: \[ \mathcal{L}\\{|\sin t|\\} = \int_{0}^{\pi} |\sin t|e^{-st} dt \]

Apply the s-shift property to the function

We can simplify the Laplace Transform of \(\vert \sin t \vert\) by applying the s-shifting theorem: \begin{align*} \mathcal{L}\\{|\sin t|\\} &= \int_{0}^{\pi}\sin(t)e^{-st}dt - \int_{\pi}^{\infty}\sin(t)e^{-st}dt \\ &= \frac{1}{s^2+1}\int_{0}^{\pi}\sin(t)e^{-st}dt - \frac{1}{s^2+1}\int_{\pi}^{\infty}\sin(t)e^{-st}dt \end{align*}

Apply the periodic property of the Laplace Transform

Since \(f\) is periodic with period \(\pi\), we can use the following formula for its Laplace Transform: \[ \mathcal{L}\\{f(t)\\} = \frac{1}{1-e^{-s\pi}}\mathcal{L}\\{|\sin t|\\} \] Substituting the Laplace Transform of \(|\sin t|\) from Step 2, we have: \[ \mathcal{L}\\{f(t)\\} = \frac{1}{1-e^{-s\pi}}\left(\frac{1}{s^2+1}\int_{0}^{\pi}\sin(t)e^{-st}dt - \frac{1}{s^2+1}\int_{\pi}^{\infty}\sin(t)e^{-st}dt\right) \]

Simplify the expression

We can now simplify the expression and write the Laplace Transform of the function \(f(t)\): \[ \mathcal{L}\\{f(t)\\} = \frac{1}{(1-e^{-s\pi})(s^2+1)}\left(\int_{0}^{\pi}\sin(t)e^{-st}dt - \int_{\pi}^{\infty}\sin(t)e^{-st}dt\right) \] So this is the Laplace Transform of the given function \(f(t)\).

Key concepts

Periodic Functions

In mathematics, a periodic function is a function that repeats its values in regular intervals or periods. The simplest example of a periodic function is the sine function, which repeats every \(2\pi\). In our exercise, the function \(f(t)\) is given by \(|\sin t|\) for \(0 \leq t < \pi\) and is extended to a periodic function with period \(\pi\). This means that its values repeat every \(\pi\) units.Periodic functions are important because they allow us to predict behavior over time. For example:

• Signals in electronics often repeat, making periodic functions useful for analysis.
• Periodic functions help model cyclic phenomena such as sound waves or seasonal patterns.

When applying mathematical analysis tools like the Laplace Transform, knowing that a function is periodic can simplify calculations. This is because the periodicity can often be leveraged to reduce the complexity of the functions being analyzed.

S-shift Property

The s-shift property is a valuable tool in working with Laplace Transforms. This property helps simplify computations by shifting the variable \(s\) in the exponent during integral evaluations. The general form of the s-shift property states that \( \mathcal{L}\{e^{at}f(t)\} = F(s-a) \), where \(F(s)\) is the Laplace Transform of \(f(t)\). In the context of the current problem, we start with the function \(|\sin t|\) and evaluate its Laplace Transform. The s-shift comes into play when computing the integral \(\int_{0}^{\pi}\sin(t)e^{-st}dt\), allowing us to extract terms involving \(s\) to simplify calculations:

• We first evaluate on a standard interval \([0, \pi]\) using parts simplified by the s-shift.
• The shift helps adjust for periodic extensions in a useful way.

Understanding this property is key in handling transforms of functions involving exponential terms.

Integral Transform

An integral transform is a mathematical operation that takes a given function and transforms it into a new function. The Laplace Transform is a specific kind of integral transform often used in complex and real analysis for solving differential equations.The general form of the Laplace Transform is as follows:\[\mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t)e^{-st} dt\]This operation translates a function of time \(f(t)\) into a function of a complex variable \(s\), simplifying the problem of solving differential equations by converting them into algebraic equations.Key points to understand about integral transforms include:

• They help shift problems from the time domain to the frequency domain.
• They are useful in analyzing systems through their behavior over frequency ranges.

In the exercise, the Laplace Transform offers a convenient way to handle the periodic function \(|\sin t|\), involving complex components that arise during periodic extensions.

Mathematical Analysis

Mathematical analysis involves techniques used to understand the behavior and properties of functions. It uses formal mathematics to explore functions' limits, derivatives, integrals, and convergence. The Laplace Transform is an advanced topic within mathematical analysis that allows for systematic approaches to time-dependent problems.Steps involved in the mathematical analysis of the given problem include:

• Recognizing the periodic nature of \(f(t)\) and how it influences the domain of integration.
• Utilizing the s-shift property to handle complex exponential terms—a method confirming the transform leads to effective simplifications.
• Applying periodic properties to ultimately find the transform of a function \(f(t)\) that repeats, using formula extensions.

In summary, mathematical analysis provides the tools to delve deeper into the function's properties and streamline solutions, as demonstrated in finding the Laplace Transform of periodic functions.