/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The rectangular function is ofte... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 11

The rectangular function is often defined as $$\operatorname{rect}\left|\frac{x-x_{0}}{a}\right|=\left\\{\begin{array}{ll}0, & \left|\left(x-x_{0}\right) / a\right|>\frac{1}{2} \\ \frac{1}{2}, & \left|\left(x-x_{0}\right) / a\right|=\frac{1}{2} \\\1, & \left|\left(x-x_{0}\right) / a\right|<\frac{1}{2}\end{array}\right.$$ where it is set equal to \(\frac{1}{2}\) at the discontinuities (Fig. P.11.11). Determine the Fourier transform of $$f(x)=\operatorname{rect}\left|\frac{x-x_{0}}{a}\right|$$ Notice that this is just a rectangular pulse, like that in Fig. \(11.1 b\) shifted a distance \(x_{0}\) from the origin.

Short Answer

Expert verified
The Fourier transform is \( F(k) = a \cdot \text{sinc}(\pi k a) \cdot e^{-2\pi i k x_0} \).

Step by step solution

01

Understand the Rectangular Function

The rectangular function \( \operatorname{rect}\left|\frac{x-x_{0}}{a}\right| \) is defined in pieces depending on the value of \( \left|\frac{x-x_{0}}{a}\right| \). It represents a pulse centered at \( x_0 \) with width \( a \).
02

Identify the Fourier Transform Formula

The Fourier transform of a function \( f(x) \) is given by the formula:\[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} \, dx \]Here, we need to apply this formula to the rectangular function \( f(x) = \operatorname{rect}\left|\frac{x-x_{0}}{a}\right| \).
03

Set Up the Integration Limits for the Rectangular Function

The rectangular function \( \operatorname{rect}\left|\frac{x-x_{0}}{a}\right| \) is 1 for \( |x-x_0| < \frac{a}{2} \). Therefore, the integration limits will be from \( x_0 - \frac{a}{2} \) to \( x_0 + \frac{a}{2} \).
04

Substitute into the Fourier Transform Formula

Since \( f(x) = 1 \) between \( x_0 - \frac{a}{2} \) and \( x_0 + \frac{a}{2} \), the Fourier transform becomes:\[ F(k) = \int_{x_0 - \frac{a}{2}}^{x_0 + \frac{a}{2}} e^{-2\pi i k x} \, dx \]
05

Compute the Integral

Integrate \( e^{-2\pi i k x} \) with respect to \( x \):\[ \int e^{-2\pi i k x} \, dx = \frac{e^{-2\pi i k x}}{-2\pi i k} \]Evaluate this antiderivative from \( x_0 - \frac{a}{2} \) to \( x_0 + \frac{a}{2} \).
06

Evaluate the Definite Integral

Evaluating the integral from the previous step results in:\[ F(k) = \left[ \frac{e^{-2\pi i k x}}{-2\pi i k} \right]_{x_0 - \frac{a}{2}}^{x_0 + \frac{a}{2}} \]= \( \frac{1}{-2\pi i k} \left( e^{-2\pi i k (x_0 + \frac{a}{2})} - e^{-2\pi i k (x_0 - \frac{a}{2})} \right) \)
07

Simplify the Expression

Use Euler's formula to express the terms as sines:\[ e^{-2\pi i k (x_0 + \frac{a}{2})} - e^{-2\pi i k (x_0 - \frac{a}{2})} = -2i \sin(\pi k a) e^{-2\pi i k x_0} \]Thus, the Fourier Transform becomes:\[ F(k) = \frac{a \sin(\pi k a)}{\pi k a} e^{-2\pi i k x_0} \]
08

Finalize the Fourier Transform Result

Recognize that \( \frac{\sin(\pi k a)}{\pi k a} \) is the sinc function, often written as \( \text{sinc}(\pi k a) \). Hence, the Fourier Transform simplifies to:\[ F(k) = a \cdot \text{sinc}(\pi k a) \cdot e^{-2\pi i k x_0} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Rectangular Function
The rectangular function, often notated as \( \operatorname{rect}\left| \frac{x-x_{0}}{a} \right| \), is a simple piecewise function that resembles a pulse or flat-top waveform.
This function is defined by three different values based on distinct domains:
  • It equals 0 when \( \left|\frac{x-x_{0}}{a}\right| > \frac{1}{2} \), denoting that outside this range, the pulse does not exist.
  • It equals 1 where \( \left|\frac{x-x_{0}}{a}\right| < \frac{1}{2} \), capturing the pulse's full height.
  • At the boundaries where \( \left|\frac{x-x_{0}}{a}\right| = \frac{1}{2} \), it is conventionally set to \( \frac{1}{2} \) to handle the discontinuities.
This function is centered at \( x_0 \) and possesses a width of \( a \), where \( a \) affects how wide the "base" of the pulse is when plotted on a graph.
Sinc Function
The sinc function plays a pivotal role in signal processing and Fourier analysis. The term "sinc" commonly refers to a normalized version defined as \( \text{sinc}(x) = \frac{\sin(x)}{x} \).
In the context of Fourier transforms, particularly for the rectangular function, it emerges naturally from the computations.
Specifically, the sinc function appears in the formula:
  • \( \text{sinc}(\pi k a) \) becomes \( \frac{\sin(\pi k a)}{\pi k a} \).
This describes how the Fourier transform of a rectangular pulse reveals a sinc-shaped spectrum in the frequency domain. The parameter \( a \) reflects the duration of the pulse, impacting the spacing of the zeros in the sinc function.
Fourier Analysis
Fourier analysis is a method used to decompose functions, signals, or shapes into their constituent frequencies. It's like taking a song and breaking it down into individual musical notes.
This approach helps us understand and manipulate these frequencies independently.
The Fourier Transform is a vital tool in this analysis:
  • It translates a time-domain signal (like our rectangular function) into a frequency-domain representation.
  • In practical terms, this means turning a time-based signal into a spectrum that shows its frequency components.
For instance, transforming a rectangular function via Fourier provides insights into its frequency response, represented as a combination of sinusoidal functions.
Integration
Integration is a fundamental concept in calculus, used here for transforming functions. The process of integrating helps find the area under the curve of a function, which is crucial in determining the Fourier transform.
For the rectangular function, integration focuses on evaluating:
  • The integral of the function \( \int e^{-2\pi i k x} \, dx \) over the limits indicated by the pulse, from \( x_0 - \frac{a}{2} \) to \( x_0 + \frac{a}{2} \).
This calculation eventually leads to finding the frequency response of a signal, and simplification of complex expressions using exponential terms into sines and cosines.
Pulse Function
A pulse function describes a signal that remains constant over a defined period and zero elsewhere. In this case, it's represented by the rectangular function - a basic pulse shape.
  • Pulses are fundamental in digital communication, appearing in bits of data transmitted over networks or signals in switch circuits.
  • Its simplicity makes it an ideal candidate for theoretical analysis, including Fourier transforms, to understand how signals might behave in complex systems.
The pulse's width \( a \) dictates its duration, and shifting the pulse modifies its phase or starting point in applications like signal processing or control systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the Fourier transform of $$f(x)=\left\\{\begin{array}{ll} \sin ^{2} k_{p} \cdot x & |x|L\end{array}\right.$$ Make a sketch of it.

Determine the Fourier transform of $$f(t)=\left\\{\begin{array}{ll} \cos ^{2} \omega_{p} t & |t|T\end{array}\right.$$ Make a sketch of \(F(\omega),\) then sketch its limiting form as \(T \rightarrow \pm \infty.\)

Show that \(\mathscr{F}\\{1\\}=2 \pi \delta(k).\)

Suppose we have a single slit along the \(y\) -direction of width \(b\) where the aperture function is constant across it at a value of \(\mathscr{A}_{0}.\) What is the diffracted field if we now apodize the slit with a cosine function amplitude mask? In other words, we cause the aperture function to go from \(\mathscr{A}_{0}\) at the center to 0 at \(\pm b / 2\) via a cosinusoidal dropoff.

Prove analytically that the convolution of any function \(f(x)\) with a delta function, \(\delta(x),\) generates the original function \(f(X) .\) You might make use of the fact that \(\delta(x)\) is even.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Thermodynamics

Read Explanation

Medical Physics

Read Explanation

Kinematics Physics

Read Explanation

Nuclear Physics

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.