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Chapter 4: Problem 2

Find the Laplace transform of $$ f(t)= \begin{cases}\sin t, & 0 \leq t \leq 2 \pi, \\ 0, & 2 \pi

Short Answer

Expert verified
\( \mathcal{L}\{f(t)\} = \frac{1 - e^{-2\pi s}}{s^2 + 1} \).

Step by step solution

01

Identify the Function

The given function is a piecewise function defined as \( f(t) = \sin t \) for \( 0 \leq t \leq 2\pi \) and \( f(t) = 0 \) for \( t > 2\pi \). This means that the function has a discontinuity at \( t = 2\pi \) where it drops from \( \sin t \) to 0.
02

Setup the Laplace Transform

The Laplace transform of a piecewise function is calculated using the integral from 0 to \( \infty \). In this case, for \( 0 \leq t \leq 2\pi \), the integral of \( \sin t \) should be considered. Thus, we write:\[\mathcal{L}\{f(t)\} = \int_{0}^{2\pi} e^{-st} \sin t \, dt\].
03

Use the Formula for Laplace Transform of Sine

The Laplace transform of \( \sin(at) \) is given by \( \frac{a}{s^2 + a^2} \). For \( \sin t \), \( a = 1 \), so we start with:\[\int_{0}^{2\pi} e^{-st} \sin t \, dt = \frac{1}{s^2 + 1} \left(1 - e^{-2\pi s}(\cos(2\pi) + s \sin(2\pi))\right)\]. Since \( \cos(2\pi) = 1 \) and \( \sin(2\pi) = 0 \), this simplifies further.
04

Simplify the Expression

Recognizing \( \cos(2\pi) = 1 \) and \( \sin(2\pi) = 0 \), we simplify the expression from Step 3:\[\int_{0}^{2\pi} e^{-st} \sin t \, dt = \frac{1}{s^2 + 1} (1 - e^{-2\pi s})\].
05

Final Expression of Laplace Transform

Thus, the Laplace transform of the given function \( f(t) \) is:\[\mathcal{L}\{f(t)\} = \frac{1 - e^{-2\pi s}}{s^2 + 1}\], which is the final solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise Functions
A piecewise function is a type of function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. Here's what makes them special:
  • They allow different rules to be applied to different parts of their domain.
  • Are usually represented using braces which group the different parts of the function.
  • Often used to address real-world scenarios where a situation changes at certain points.
For example, in this exercise, \( f(t) \) switches from \( \sin t \) when \( 0 \leq t \leq 2\pi \) to 0 for \( t > 2\pi \). This portrays a real-life event where a process might work continuously for a time and then stop. The flexibility of a piecewise structure allows mathematical models to accurately reflect such situations.
Integrals
Integrals are a fundamental concept in calculus, often used to calculate areas under curves. In the context of Laplace transforms, integrals play a vital role to convert a function of time into a function of a complex variable. Here's why they're important:
  • The definite integral \( \int_{a}^{b} f(t) \, dt \) gives the area under the curve from \( t = a \) to \( t = b \).
  • In Laplace transforms, we use the integral \( \int_{0}^{\infty} e^{-st} f(t) \, dt \) to transform the function.
  • This process captures the essence of the function over its entire range.
In this exercise, the definite integral \( \int_{0}^{2\pi} e^{-st} \sin t \, dt \) is used to handle the piecewise nature of the function, since after \( 2\pi \), the function reduces to zero.
Sine Functions
Sine functions are trigonometric functions that display periodic behavior, repeating every \( 2\pi \). They play a prominent role in oscillation and wave studies. Here are some key insights:
  • Defined mathematically as \( \sin(t) \).
  • They oscillate between -1 and 1.
  • In the Laplace domain, \( \sin(at) \) transforms into \( \frac{a}{s^2 + a^2} \).
In this problem, \( f(t) = \sin t \) is transformed considering \( a = 1 \). This results in the integral being evaluated to include exponential decay terms derived from the Laplace transform, capturing the sine wave behavior in the time domain and representing it in the frequency domain.
Discontinuity
Discontinuity in a function occurs where a sudden jump or break happens in the graph of the function. It can make calculations slightly tricky, as special attention is needed at these points:
  • A function is continuous if you can draw it without lifting your pen.
  • Discontinuities are points where this is not possible.
  • They might signify changes in state or behavior of phenomena being modeled.
In the original exercise, the piecewise function becomes discontinuous at \( t = 2\pi \) where \( f(t) \) suddenly drops from \( \sin t \) to 0. This kind of jump is seamlessly handled by the Laplace transform approach, as it employs integrals that inherently capture these shifts鈥攈elping you mathematically model and understand sudden changes within processes.

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Most popular questions from this chapter

Calculate the Laplace transform of \(f(t)=\frac{\sin t}{t}\). (Hint: First compute \(\left.\frac{d}{d s} \mathcal{L}[f](s)_{+}\right)\)

Prove, for every \(p, q>-1, \quad \int_{0}^{1} x^{p}\left(\ln \frac{1}{x}\right)^{q} d x=\frac{\Gamma(q+1)}{(p+1)^{q+1}}\)

Let \(f\) and \(g\) be continuously differentiable and such that \(f, g, f^{\prime}\), and \(g^{\prime}\) are absolutely integrable on the interval \([0, \infty)\), and \(f(0)=g(0)=0\). Prove that $$ (f * g)^{\prime}=f^{\prime} * g=f * g^{\prime} . $$

Prove that if \(f:[0, \infty) \rightarrow \mathbb{C}\) is piecewise continuous and \(p\)-periodic \((p>0)\), then $$ \mathcal{L}[f](s)=\frac{1}{1-e^{-p s}} \int_{0}^{p} e^{-s t} f(t) d t, \quad s>0 . $$

For each of the following differential equations with associated initial conditions, find a solution on the interval \([0, \infty)\). (a) \(y^{\prime \prime}(t)+y(t)=g(t), \quad y(0)=0, \quad y^{\prime}(0)=0\), where $$ g(t)= \begin{cases}t, & 0 \leq t<1 \\ 1, & 1 \leq t\end{cases} $$ (b) \(y^{\prime \prime}(t)+2 y^{\prime}(t)-3 y(t)=f(t), \quad y(0)=1, \quad y^{\prime}(0)=0\) where $$ f(t)= \begin{cases}0, & 0 \leq t \leq 2 \pi, \\ \sin t, & t>2 \pi .\end{cases} $$ (c) \(y^{\prime \prime}(t)+2 y^{\prime}(t)+3 y(t)=\delta_{\pi}(t)+\sin t, \quad y(0)=0, \quad y^{\prime}(0)=1\). (d) \(y^{\prime \prime}(t)+y(t)=\delta_{\pi}(t) \cos t, \quad y(0)=0, \quad y^{\prime}(0)=1\). (e) \(y^{\prime \prime \prime}(t)-y^{\prime \prime}(t)+4 y^{\prime}(t)-4 y(t)=68 e^{t} \sin 2 t\), \(y(0)=1, \quad y^{\prime}(0)=-19, \quad y^{\prime \prime}(0)=-37 .\) (f) \(y^{\prime \prime}(t)+9 y(t)=f_{c}(t), \quad y(0)=a, \quad y^{\prime}(0)=b\), where for \(c>0\), $$ f_{c}(t)= \begin{cases}0, & 0 \leq t \leq c \\ t-c, & c
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