Heaviside Function: What is H(-x)? Fourier Transform

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Discussion Overview

The discussion revolves around the Heaviside function, specifically the evaluation of H(-x) and its implications for Fourier transforms. Participants explore the properties of the Heaviside function and how it relates to Fourier transforms in various contexts.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant questions the value of H(-x), suggesting it is 1 if x < 0 and 0 if x ≥ 0.
  • Another participant clarifies that by definition, H(x) is 0 if x < 0 and 1 if x > 0, leading to the conclusion that H(-x) is 0 if x > 0 and 1 if x < 0. They also propose a relationship: H(-x) = 1 - H(x).
  • A participant presents a Fourier transform result as f(w) = 1/(1-iw) without further context.
  • Another participant inquires whether the Fourier transform of the expression H(x)exp(-2x) + H(-x)exp(x) can be expressed as f(w) = 1/(2+iw) + 1/(1-iw).

Areas of Agreement / Disagreement

Participants express differing views on the evaluation of H(-x) and its relationship to H(x). The discussion remains unresolved regarding the Fourier transform of the combined expression.

Contextual Notes

There are assumptions about the definitions of the Heaviside function and its properties that are not explicitly stated. The mathematical steps leading to the Fourier transform results are not fully detailed.

squenshl
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What is H(-x)? Is it 1 if x < 0, 0 if x [tex]\geq[/tex] 0. If so what is the Fourier transform of H(-x)exp(x)?
 
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By definition, H(x) is 0 if x < 0, 1 if x > 0, so setting x -> -x gives H(-x) is 0 if x > 0, 1 if x < 1. By inspection, you should be able to relate H(-x) to H(x), by H(-x) = 1 - H(x), which will help you with your Fourier transform.
 
Sweet.
My Fourier transform is f(w) = 1/(1-iw)
 
So is the Fourier transform of H(x)exp(-2x) + H(-x)exp(x)
f(w) = 1/(2+iw) + 1/(1-iw)?
 

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