Rectangular function: Difference between revisions

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==History==

The ''rect'' function has been introduced 1953 by [[Philip Woodward|Woodward]]<ref>{{Cite journal |last=Klauder |first=John R |title=The Theory and Design of Chirp Radars |pages=745–808 |journal=Bell System Technical Journal |year=1960 |volume=39 |issue=4 |doi=10.1002/j.1538-7305.1960.tb03942.x |url=https://ieeexplore.ieee.org/document/6773600 |url-access=subscription }}</ref> in "Probability and Information Theory, with Applications to Radar"<ref>{{Cite book |last=Woodward |first=Philipp M |title=Probability and Information Theory, with Applications to Radar |publisher=Pergamon Press |pages=29 |year=1953 }}</ref> as an ideal [[Window function#Rectangular window|cutout operator]], together with the [[Sinc function|''sinc'' function]]<ref>{{Cite book |last=Higgins |first=John Rowland |title=Sampling Theory in Fourier and Signal Analysis: Foundations |pages=4 |publisher=Oxford University Press Inc. |year=1996 |isbn=0198596995 }}</ref><ref>{{Cite book |last=Zayed |first=Ahmed I |title=Handbook of Function and Generalized Function Transformations |pages=507 |publisher=CRC Press |year=1996 |isbn=9780849380761 }}</ref> as an ideal [[Whittaker–Shannon interpolation formula|interpolation operator]], and their counter operations which are [[Sampling (signal processing)|sampling]] ([[Dirac comb#Dirac-comb identity|''comb'' operator]]) and [[Periodic summation|replicating]] ([[Dirac comb#Dirac-comb identity|''rep'' operator]]), respectively.

==Relation to the boxcar function==

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where <math>H(x)</math> is the [[Heaviside step function]]; the function is centered at <math>X</math> and has duration <math>Y</math>, from <math>X-Y/2</math> to <math>X+Y/2.</math>

==Fourier transform of the rectangular function==

[[File:Sinc_function_(normalized).svg|thumb|400px|right|Plot of normalized <math>\mathrmoperatorname{sinc}(x)</math> function (i.e. <math>\mathrmoperatorname{sinc}(\pi x)</math>) with its spectral frequency components.]]

The [[Fourier transform#Tables of important Fourier transforms|unitary Fourier transforms]] of the rectangular function are<ref name="wolfram"/>

<math display="block">\int_{-\infty}^\infty \mathrmoperatorname{rect}(t)\cdot e^{-i 2\pi f t} \, dt

=\frac{\sin(\pi f)}{\pi f} = \mathrmoperatorname{sinc}{(\pi f) =\operatorname{sinc}_\pi(f),</math>

using ordinary frequency {{mvar|f}}, where [[sinc function|<math>\mathrmoperatorname{sinc}_\pi</math>]] is the normalized form<ref>Wolfram MathWorld, https://mathworld.wolfram.com/SincFunction.html</ref> of the [[sinc function]] and

<math display="block">\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \mathrmoperatorname{rect}(t)\cdot e^{-i \omega t} \, dt

=\frac{1}{\sqrt{2\pi}}\cdot \frac{\mathrm{sin}\left(\omega/2 \right)}{\omega/2}

=\frac{1}{\sqrt{2\pi}} \mathrmcdot \operatorname{sinc}\left(\omega/2 \right),

</math>

using angular frequency <math>\omega</math>, where [[sinc function|<math>\mathrmoperatorname{sinc}</math>]] is the unnormalized form of the [[sinc function]].

For <math>\operatorname{rect} (x/a)</math>, its Fourier transform is<math display="block">\int_{-\infty}^\infty \operatorname{rect}\left(\frac{t}{a}\right)\cdot e^{-i 2\pi f t} \, dt

Note that as long as the definition of the pulse function is only motivated by its behavior in the time-___domain experience, there is no reason to believe that the oscillatory interpretation (i.e. the Fourier transform function) should be intuitive, or directly understood by humans. However, some aspects of the theoretical result may be understood intuitively, as finiteness in time ___domain corresponds to an infinite frequency response. (Vice versa, a finite Fourier transform will correspond to infinite time ___domain response.)

=a \frac{\sin(\pi af)}{\pi af} = a\ \operatorname{sinc}_\pi{(a f)}.</math>

==Relation to the triangular function==

We can define the [[triangular function]] as the [[convolution]] of two rectangular functions:

<math display=block>\mathrmoperatorname{tri(t/T)} = \mathrmoperatorname{rect(2t/T)} * \mathrmoperatorname{rect(2t/T)}.\,</math>

==Use in probability==

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The pulse function may also be expressed as a limit of a [[rational function]]:

<math display="block">\Pi(t) = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1}.</math>

===Demonstration of validity===

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It follows that:

<math display="block">\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \frac{1}{0+1} = 1, |t|<\tfrac{1}{2}.</math>

Second, we consider the case where <math display="inline">|t|>\frac{1}{2}.</math> Notice that the term <math display="inline">(2t)^{2n}</math> is always positive for integer <math>n.</math> However, <math>2t>1</math> and hence <math display="inline">(2t)^{2n}</math> grows very large for large <math>n.</math>

It follows that:

<math display="block">\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \frac{1}{+\infty+1} = 0, |t|>\tfrac{1}{2}.</math>

Third, we consider the case where <math display="inline">|t| = \frac{1}{2}.</math> We may simply substitute in our equation:

<math display="block">\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{1^{2n}+1} = \frac{1}{1+1} = \tfrac{1}{2}.</math>

We see that it satisfies the definition of the pulse function. Therefore,

<math display="block">\mathrmoperatorname{rect}(t) = \Pi(t) = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \begin{cases}

0 & \mbox{if } |t| > \frac{1}{2} \\

\frac{1}{2} & \mbox{if } |t| = \frac{1}{2} \\

1 & \mbox{if } |t| < \frac{1}{2}. \\

\end{cases}</math>

== Dirac delta function ==

The rectangle function can be used to represent the [[Dirac delta function]] <math>\delta (x)</math>.<ref name=":0">{{Cite book |last1=Khare |first1=Kedar |title=Fourier Optics and Computational Imaging |last2=Butola |first2=Mansi |last3=Rajora |first3=Sunaina |publisher=Springer |year=2023 |isbn=978-3-031-18353-9 |edition=2nd |pages=15–16 |chapter=Chapter 2.4 Sampling by Averaging, Distributions and Delta Function |doi=10.1007/978-3-031-18353-9}}</ref> Specifically,<math display="block">\delta (x) = \lim_{a \to 0} \frac{1}{a}\operatorname{rect}\left(\frac{x}{a}\right).</math>For a function <math>g(x)</math>, its average over the width ''<math>a</math>'' around 0 in the function ___domain is calculated as,

<math display="block">g_{avg}(0) = \frac{1}{a} \int\limits_{- \infty}^{\infty} dx\ g(x) \operatorname{rect}\left(\frac{x}{a}\right).</math>

To obtain <math>g(0)</math>, the following limit is applied,

<math display="block">g(0) = \lim_{a \to 0} \frac{1}{a} \int\limits_{- \infty}^{\infty} dx\ g(x) \operatorname{rect}\left(\frac{x}{a}\right)</math>

and this can be written in terms of the Dirac delta function as,

<math display="block">g(0) = \int\limits_{- \infty}^{\infty} dx\ g(x) \delta (x).</math>The Fourier transform of the Dirac delta function <math>\delta (t)</math> is

<math display="block">\delta (f)

= \int_{-\infty}^\infty \delta (t) \cdot e^{-i 2\pi f t} \, dt

= \lim_{a \to 0} \frac{1}{a} \int_{-\infty}^\infty \operatorname{rect}\left(\frac{t}{a}\right)\cdot e^{-i 2\pi f t} \, dt

= \lim_{a \to 0} \operatorname{sinc}{(a f)}.</math>

where the [[sinc function]] here is the normalized sinc function. Because the first zero of the sinc function is at <math>f = 1 / a</math> and <math>a</math> goes to infinity, the Fourier transform of <math>\delta (t)</math> is

<math display="block">\delta (f) = 1,</math>

means that the frequency spectrum of the Dirac delta function is infinitely broad. As a pulse is shorten in time, it is larger in spectrum.

==See also==

*[[Fourier transform]]

*[[Square wave (waveform)|Square wave]]

*[[Step function]]

*[[Top-hat filter]]

*[[Boxcar function]]

==References==