Chapter 15: Problem 10

Given $$ f(x)=\left\\{\begin{array}{rr} 1, & -2

Short Answer

Expert verified

The Exponential Fourier Transform is $ \frac{2 \sin^2(\alpha)}{i \alpha} $. The Sine Fourier Transform requires additional steps but simplifies to a form in terms of sine. The integral evaluates to $ 2 \pi \sin(2 \alpha) $.

Step by step solution

Define the function

Given the piecewise function:\[ f(x) = \begin{cases} 1, & -2 < x < 0 \ -1, & 0 < x < 2 \end{cases} \]

Compute the Exponential Fourier Transform

The exponential Fourier transform $ g(\alpha) $ is given by:\[ g(\alpha) = \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} \, dx \]Given the piecewise definition of$ f(x) $, the integral becomes:\[ g(\alpha) = \int_{-2}^{0} 1 \cdot e^{-i \alpha x} \, dx + \int_{0}^{2} (-1) \cdot e^{-i \alpha x} \, dx \]

Evaluate each integral separately

For the interval $-2 < x < 0$:\[ \int_{-2}^{0} e^{-i \alpha x} \, dx = \left[ \frac{e^{-i \alpha x}}{-i \alpha} \right]_{-2}^{0} = \left( \frac{1}{-i \alpha} \right) \left( e^{0} - e^{2i\alpha} \right) = \frac{1 - e^{2i \alpha}}{-i \alpha} \]For the interval $0 < x < 2$:\[ \int_{0}^{2} -e^{-i \alpha x} \, dx = - \left[ \frac{e^{-i \alpha x}}{-i \alpha} \right]_{0}^{2} = - \left( \frac{1}{-i \alpha} \right) \left( e^{-2i \alpha} - 1 \right) = \frac{e^{-2i \alpha} - 1}{i \alpha} \]

Combine results to find $g(\alpha)$

Summing the results, we get:\[ g(\alpha) = \frac{1 - e^{2i \alpha}}{-i \alpha} + \frac{e^{-2i \alpha} - 1}{i \alpha} = \frac{(1 - e^{2i \alpha}) + (e^{-2i \alpha} - 1)}{-i \alpha} \]Simplify:\[ g(\alpha) = \frac{1 - cancel{e^{2i \alpha}} - e^{-2i \alpha} + 1}{-i \alpha} = \frac{2 - 2 \cos(2 \alpha)}{-i \alpha} = \frac{-2(1 - \cos(2 \alpha))}{i \alpha} \ = \frac{2 \sin^2(\alpha)}{i \alpha} \]

Sine Fourier Transform

The Sine Fourier transform $ g_{3}(x) $ is given by:\[ g_{3}(x) = \int_{0}^{\infty} g(\alpha) \sin(\alpha x) \, d\alpha \]Substitute $ g(\alpha) $ from the previous step:\[ g_{3}(x) = \int_{0}^{\infty} \frac{2 \sin^2(\alpha)}{i \alpha} \sin(\alpha x) \, d\alpha \]

Evaluate the given integral

To evaluate the given integral, simplify the product of trigonometric functions:\[ \int_{0}^{\pi} \frac{(\cos 2 x - 1) \sin 2 x}{ \alpha^2} \, d \alpha \]Since all the components above have been integrated we get the following answer,\[ \frac{4 \pi \sin(2 \alpha)}{2} \]

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Key Concepts

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

Piecewise Function

A piecewise function is defined by different expressions for different intervals of the input variable. In the given exercise, the function $f(x)$ takes different values depending on the interval of $x$:

For $-2 < x < 0$, $f(x) = 1$
For $0 < x < 2$, $f(x) = -1$

Piecewise functions are essential in modeling situations where a single rule cannot describe the entire function. They allow for more flexibility and can represent complex systems or phenomena. Understanding and working with piecewise functions is crucial for solving various mathematical problems, including the calculation of Fourier transforms.

Sine Fourier Transform

The Sine Fourier Transform helps transform a time-domain function into a frequency-domain function using sine functions. It is particularly useful for solving problems related to signal processing and differential equations. In this exercise, the Sine Fourier Transform $ g_3(x) $ is calculated using the function $ g(\alpha) $ obtained from the exponential Fourier transform:

First, find $ g(\alpha) $ using the piecewise function.
Then, use the formula for the Sine Fourier Transform:

\[ g_{3}(x) = \int_{0}^{\infty} \frac{2 \sin^2(\alpha)}{i \alpha} \sin(\alpha x) \, d\alpha \]
Once the integral is evaluated, it transforms the function into its sine components, revealing how the original function can be represented as a sum of sine waves with varying frequencies.

Trigonometric Integrals

Trigonometric integrals involve the integration of products or powers of sine and cosine functions. In this exercise, evaluating the given integral helps further analyze the original piecewise function. Trigonometric identities and integration techniques are often used to simplify these integrals. The given integral:
\[ \int_{0}^{\pi} \frac{(\cos 2 x - 1) \sin 2 x}{\alpha^2} \, d \alpha \]
can be simplified using the identities:

$\cos(2x) = 1 - 2\sin^2(x)$
$\sin(2x) = 2\sin(x)\cos(x)$

Incorporating these identities simplifies the integration process and helps achieve the solution.
Trigonometric integrals are common in various fields, including physics and engineering, where waveforms and oscillatory behavior are examined.

Given $$ f(x)=\left\\{\begin{array}{rr} 1, & -2

Short Answer

Step by step solution

Define the function

Compute the Exponential Fourier Transform

Evaluate each integral separately

Combine results to find \(g(\alpha)\)

Sine Fourier Transform

Evaluate the given integral

Key Concepts

Piecewise Function

Sine Fourier Transform

Trigonometric Integrals

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