/*! 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}

91Ó°ÊÓ

Find the fourier transform off(x)=e−x2/(2σ2). Hint: Complete the square in the xterms in the exponent and make the change of variabley=x+σ2iα .Use tables or computer to evaluate the definite integral.

Short Answer

Expert verified

The fourier transform of f(x)=e−x2/(2σ2) isg(α)=σ2πe−σ2α2/2.

Step by step solution

01

Given Information.

The given function isf(x)=e−x2/(2σ2).

02

Definition of fourier transform

The Fourier transform is a mathematical technique for expressing a function as the summation of sines and cosines functions.

03

Step 3: To find the fourier transform of the given function

The fourier transform is given below.

g(α)=12π∫−∞∞e−x2/(2σ2)e−iαxdx

Put u=xσ2

Differentiate with respect to x.

du=dxσ2

g(α)=12π∫−∞∞e−u2−iα2σu2σdug(α)=σπ2∫−∞∞e−u2−iα2σudug(α)=σπ2∫−∞∞e−(u+iασ/(2))2−σ2α2/2du

Simplify further

g(α)=σπ2e−σ2α2/2∫−∞∞e−(u+iασ/(2))2d(u+iσα/2)

Putu+iσα/2=η

g(α)=σπ2e−σ2α2/2∫−∞∞e−η2dη

But the integral ∫−∞∞e−η2dηis Euler-poisson integral and its value isπ.

g(α)=σ2πe−σ2α2/2

The fourier transform of f(x)=e−x2/(2σ2)isg(α)=σ2πe−σ2α2/2.

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 91Ó°ÊÓ!

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

See all solutions

Recommended explanations on Physics Textbooks

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.