Rectangular function (original) (raw)

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Function whose graph is 0, then 1, then 0 again, in an almost-everywhere continuous way

Rectangular function with a = 1

The rectangular function (also known as the rectangle function, rect function, Pi function, Heaviside Pi function,[1] gate function, unit pulse, or the normalized boxcar function) is defined as[2]

rect ⁡ ( t a ) = Π ( t a ) = { 0 , if | t | > a 2 1 2 , if | t | = a 2 1 , if | t | < a 2 . {\displaystyle \operatorname {rect} \left({\frac {t}{a}}\right)=\Pi \left({\frac {t}{a}}\right)=\left\{{\begin{array}{rl}0,&{\text{if }}|t|>{\frac {a}{2}}\\{\frac {1}{2}},&{\text{if }}|t|={\frac {a}{2}}\\1,&{\text{if }}|t|<{\frac {a}{2}}.\end{array}}\right.} $\operatorname {rect} \left({\frac {t}{a}}\right)=\Pi \left({\frac {t}{a}}\right)=\left\{{\begin{array}{rl}0,&{\text{if }}|t|>{\frac {a}{2}}\\{\frac {1}{2}},&{\text{if }}|t|={\frac {a}{2}}\\1,&{\text{if }}|t|<{\frac {a}{2}}.\end{array}}\right.$

Alternative definitions of the function define rect ⁡ ( ± 1 2 ) {\textstyle \operatorname {rect} \left(\pm {\frac {1}{2}}\right)} ${\textstyle \operatorname {rect} \left(\pm {\frac {1}{2}}\right)}$ to be 0,[3] 1,[4][5] or undefined.

Its periodic version is called a rectangular wave.

The rect function has been introduced 1953 by Woodward [6] in "Probability and Information Theory, with Applications to Radar"[7] as an ideal cutout operator, together with the sinc function [8][9] as an ideal interpolation operator, and their counter operations which are sampling (comb operator) and replicating (rep operator), respectively.

Relation to the boxcar function

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The rectangular function is a special case of the more general boxcar function:

rect ⁡ ( t − X Y ) = H ( t − ( X − Y / 2 ) ) − H ( t − ( X + Y / 2 ) ) = H ( t − X + Y / 2 ) − H ( t − X − Y / 2 ) {\displaystyle \operatorname {rect} \left({\frac {t-X}{Y}}\right)=H(t-(X-Y/2))-H(t-(X+Y/2))=H(t-X+Y/2)-H(t-X-Y/2)} $\operatorname {rect} \left({\frac {t-X}{Y}}\right)=H(t-(X-Y/2))-H(t-(X+Y/2))=H(t-X+Y/2)-H(t-X-Y/2)$

where H ( x ) {\displaystyle H(x)} $H(x)$ is the Heaviside step function; the function is centered at X {\displaystyle X} $X$ and has duration Y {\displaystyle Y} $Y$ , from X − Y / 2 {\displaystyle X-Y/2} $X-Y/2$ to X + Y / 2. {\displaystyle X+Y/2.} $X+Y/2.$

Fourier transform of the rectangular function

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Plot of normalized sinc ⁡ ( x ) {\displaystyle \operatorname {sinc} (x)} $\operatorname {sinc} (x)$ function (i.e. sinc ⁡ ( π x ) {\displaystyle \operatorname {sinc} (\pi x)} $\operatorname {sinc} (\pi x)$ ) with its spectral frequency components.

The unitary Fourier transforms of the rectangular function are[2] ∫ − ∞ ∞ rect ⁡ ( t ) ⋅ e − i 2 π f t d t = sin ⁡ ( π f ) π f = sinc π ⁡ ( f ) , {\displaystyle \int _{-\infty }^{\infty }\operatorname {rect} (t)\cdot e^{-i2\pi ft}\,dt={\frac {\sin(\pi f)}{\pi f}}=\operatorname {sinc} _{\pi }(f),} $\int _{-\infty }^{\infty }\operatorname {rect} (t)\cdot e^{-i2\pi ft}\,dt={\frac {\sin(\pi f)}{\pi f}}=\operatorname {sinc} _{\pi }(f),$ using ordinary frequency f, where sinc π {\displaystyle \operatorname {sinc} _{\pi }} $\operatorname {sinc} _{\pi }$ is the normalized form[10] of the sinc function and 1 2 π ∫ − ∞ ∞ rect ⁡ ( t ) ⋅ e − i ω t d t = 1 2 π ⋅ sin ⁡ ( ω / 2 ) ω / 2 = 1 2 π ⋅ sinc ⁡ ( ω / 2 ) , {\displaystyle {\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\operatorname {rect} (t)\cdot e^{-i\omega t}\,dt={\frac {1}{\sqrt {2\pi }}}\cdot {\frac {\sin \left(\omega /2\right)}{\omega /2}}={\frac {1}{\sqrt {2\pi }}}\cdot \operatorname {sinc} \left(\omega /2\right),} ${\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\operatorname {rect} (t)\cdot e^{-i\omega t}\,dt={\frac {1}{\sqrt {2\pi }}}\cdot {\frac {\sin \left(\omega /2\right)}{\omega /2}}={\frac {1}{\sqrt {2\pi }}}\cdot \operatorname {sinc} \left(\omega /2\right),$ using angular frequency ω {\displaystyle \omega } $\omega$ , where sinc {\displaystyle \operatorname {sinc} } $\operatorname {sinc}$ is the unnormalized form of the sinc function.

For rect ⁡ ( x / a ) {\displaystyle \operatorname {rect} (x/a)} $\operatorname {rect} (x/a)$ , its Fourier transform is ∫ − ∞ ∞ rect ⁡ ( t a ) ⋅ e − i 2 π f t d t = a sin ⁡ ( π a f ) π a f = a sinc π ⁡ ( a f ) . {\displaystyle \int _{-\infty }^{\infty }\operatorname {rect} \left({\frac {t}{a}}\right)\cdot e^{-i2\pi ft}\,dt=a{\frac {\sin(\pi af)}{\pi af}}=a\ \operatorname {sinc} _{\pi }{(af)}.} $\int _{-\infty }^{\infty }\operatorname {rect} \left({\frac {t}{a}}\right)\cdot e^{-i2\pi ft}\,dt=a{\frac {\sin(\pi af)}{\pi af}}=a\ \operatorname {sinc} _{\pi }{(af)}.$

Relation to the triangular function

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

t r i ( t / T ) = r e c t ( 2 t / T ) ∗ r e c t ( 2 t / T ) . {\displaystyle \operatorname {tri(t/T)} =\operatorname {rect(2t/T)} *\operatorname {rect(2t/T)} .\,} $\operatorname {tri(t/T)} =\operatorname {rect(2t/T)} *\operatorname {rect(2t/T)} .\,$

Viewing the rectangular function as a probability density function, it is a special case of the continuous uniform distribution with a = − 1 / 2 , b = 1 / 2. {\displaystyle a=-1/2,b=1/2.} $a=-1/2,b=1/2.$ The characteristic function is

φ ( k ) = sin ⁡ ( k / 2 ) k / 2 , {\displaystyle \varphi (k)={\frac {\sin(k/2)}{k/2}},} $\varphi (k)={\frac {\sin(k/2)}{k/2}},$

and its moment-generating function is

M ( k ) = sinh ⁡ ( k / 2 ) k / 2 , {\displaystyle M(k)={\frac {\sinh(k/2)}{k/2}},} $M(k)={\frac {\sinh(k/2)}{k/2}},$

where sinh ⁡ ( t ) {\displaystyle \sinh(t)} $\sinh(t)$ is the hyperbolic sine function.

Rational approximation

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

Π ( t ) = lim n → ∞ , n ∈ ( Z ) 1 ( 2 t ) 2 n + 1 . {\displaystyle \Pi (t)=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}.} $\Pi (t)=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}.$

Demonstration of validity

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First, we consider the case where | t | < 1 2 . {\textstyle |t|<{\frac {1}{2}}.} ${\textstyle |t|<{\frac {1}{2}}.}$ Notice that the term ( 2 t ) 2 n {\textstyle (2t)^{2n}} ${\textstyle (2t)^{2n}}$ is always positive for integer n . {\displaystyle n.} $n.$ However, 2 t < 1 {\displaystyle 2t<1} $2t<1$ and hence ( 2 t ) 2 n {\textstyle (2t)^{2n}} ${\textstyle (2t)^{2n}}$ approaches zero for large n . {\displaystyle n.} $n.$

It follows that: lim n → ∞ , n ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = 1 0 + 1 = 1 , | t | < 1 2 . {\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{0+1}}=1,|t|<{\tfrac {1}{2}}.} $\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{0+1}}=1,|t|<{\tfrac {1}{2}}.$

Second, we consider the case where | t | > 1 2 . {\textstyle |t|>{\frac {1}{2}}.} ${\textstyle |t|>{\frac {1}{2}}.}$ Notice that the term ( 2 t ) 2 n {\textstyle (2t)^{2n}} ${\textstyle (2t)^{2n}}$ is always positive for integer n . {\displaystyle n.} $n.$ However, 2 t > 1 {\displaystyle 2t>1} $2t>1$ and hence ( 2 t ) 2 n {\textstyle (2t)^{2n}} ${\textstyle (2t)^{2n}}$ grows very large for large n . {\displaystyle n.} $n.$

It follows that: lim n → ∞ , n ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = 1 + ∞ + 1 = 0 , | t | > 1 2 . {\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{+\infty +1}}=0,|t|>{\tfrac {1}{2}}.} $\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{+\infty +1}}=0,|t|>{\tfrac {1}{2}}.$

Third, we consider the case where | t | = 1 2 . {\textstyle |t|={\frac {1}{2}}.} ${\textstyle |t|={\frac {1}{2}}.}$ We may simply substitute in our equation:

lim n → ∞ , n ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = lim n → ∞ , n ∈ ( Z ) 1 1 2 n + 1 = 1 1 + 1 = 1 2 . {\displaystyle \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}}.} $\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{1^{2n}+1}}={\frac {1}{1+1}}={\tfrac {1}{2}}.$

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

rect ⁡ ( t ) = Π ( t ) = lim n → ∞ , n ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = { 0 if | t | > 1 2 1 2 if | t | = 1 2 1 if | t | < 1 2 . {\displaystyle \operatorname {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}}} $\operatorname {rect} (t)=\Pi (t)=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\begin{cases}0&{\mbox{if }}|t|>{\frac {1}{2}}\\{\frac {1}{2}}&{\mbox{if }}|t|={\frac {1}{2}}\\1&{\mbox{if }}|t|<{\frac {1}{2}}.\\\end{cases}}$

Dirac delta function

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The rectangle function can be used to represent the Dirac delta function δ ( x ) {\displaystyle \delta (x)} $\delta (x)$ .[11] Specifically, δ ( x ) = lim a → 0 1 a rect ⁡ ( x a ) . {\displaystyle \delta (x)=\lim _{a\to 0}{\frac {1}{a}}\operatorname {rect} \left({\frac {x}{a}}\right).} $\delta (x)=\lim _{a\to 0}{\frac {1}{a}}\operatorname {rect} \left({\frac {x}{a}}\right).$ For a function g ( x ) {\displaystyle g(x)} $g(x)$ , its average over the width a {\displaystyle a} $a$ around 0 in the function domain is calculated as,

g a v g ( 0 ) = 1 a ∫ − ∞ ∞ d x g ( x ) rect ⁡ ( x a ) . {\displaystyle g_{avg}(0)={\frac {1}{a}}\int \limits _{-\infty }^{\infty }dx\ g(x)\operatorname {rect} \left({\frac {x}{a}}\right).} $g_{avg}(0)={\frac {1}{a}}\int \limits _{-\infty }^{\infty }dx\ g(x)\operatorname {rect} \left({\frac {x}{a}}\right).$ To obtain g ( 0 ) {\displaystyle g(0)} $g(0)$ , the following limit is applied,

g ( 0 ) = lim a → 0 1 a ∫ − ∞ ∞ d x g ( x ) rect ⁡ ( x a ) {\displaystyle g(0)=\lim _{a\to 0}{\frac {1}{a}}\int \limits _{-\infty }^{\infty }dx\ g(x)\operatorname {rect} \left({\frac {x}{a}}\right)} $g(0)=\lim _{a\to 0}{\frac {1}{a}}\int \limits _{-\infty }^{\infty }dx\ g(x)\operatorname {rect} \left({\frac {x}{a}}\right)$ and this can be written in terms of the Dirac delta function as, g ( 0 ) = ∫ − ∞ ∞ d x g ( x ) δ ( x ) . {\displaystyle g(0)=\int \limits _{-\infty }^{\infty }dx\ g(x)\delta (x).} $g(0)=\int \limits _{-\infty }^{\infty }dx\ g(x)\delta (x).$ The Fourier transform of the Dirac delta function δ ( t ) {\displaystyle \delta (t)} $\delta (t)$ is

δ ( f ) = ∫ − ∞ ∞ δ ( t ) ⋅ e − i 2 π f t d t = lim a → 0 1 a ∫ − ∞ ∞ rect ⁡ ( t a ) ⋅ e − i 2 π f t d t = lim a → 0 sinc ⁡ ( a f ) . {\displaystyle \delta (f)=\int _{-\infty }^{\infty }\delta (t)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}{\frac {1}{a}}\int _{-\infty }^{\infty }\operatorname {rect} \left({\frac {t}{a}}\right)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}\operatorname {sinc} {(af)}.} $\delta (f)=\int _{-\infty }^{\infty }\delta (t)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}{\frac {1}{a}}\int _{-\infty }^{\infty }\operatorname {rect} \left({\frac {t}{a}}\right)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}\operatorname {sinc} {(af)}.$ where the sinc function here is the normalized sinc function. Because the first zero of the sinc function is at f = 1 / a {\displaystyle f=1/a} $f=1/a$ and a {\displaystyle a} $a$ goes to infinity, the Fourier transform of δ ( t ) {\displaystyle \delta (t)} $\delta (t)$ is

δ ( f ) = 1 , {\displaystyle \delta (f)=1,} $\delta (f)=1,$ 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.

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