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.}
Alternative definitions of the function define rect ( ± 1 2 ) {\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.
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)}
where H ( x ) {\displaystyle H(x)} is the Heaviside step function; the function is centered at X {\displaystyle X} and has duration Y {\displaystyle Y} , from X − Y / 2 {\displaystyle X-Y/2} to X + Y / 2. {\displaystyle X+Y/2.}
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),} using ordinary frequency f, where sinc π {\displaystyle \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),} using angular frequency ω {\displaystyle \omega } , where sinc {\displaystyle \operatorname {sinc} } is the unnormalized form of the sinc function.
For rect ( x / a ) {\displaystyle \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)}.}
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)} .\,}
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.} The characteristic function is
φ ( k ) = sin ( k / 2 ) k / 2 , {\displaystyle \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}},}
where sinh ( t ) {\displaystyle \sinh(t)} is the hyperbolic sine function.
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}}.}
First, we consider the case where | t | < 1 2 . {\textstyle |t|<{\frac {1}{2}}.} Notice that the term ( 2 t ) 2 n {\textstyle (2t)^{2n}} is always positive for integer n . {\displaystyle n.} However, 2 t < 1 {\displaystyle 2t<1} and hence ( 2 t ) 2 n {\textstyle (2t)^{2n}} approaches zero for large n . {\displaystyle 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}}.}
Second, we consider the case where | t | > 1 2 . {\textstyle |t|>{\frac {1}{2}}.} Notice that the term ( 2 t ) 2 n {\textstyle (2t)^{2n}} is always positive for integer n . {\displaystyle n.} However, 2 t > 1 {\displaystyle 2t>1} and hence ( 2 t ) 2 n {\textstyle (2t)^{2n}} grows very large for large n . {\displaystyle 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}}.}
Third, we consider the case where | t | = 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}}.}
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}}}
The rectangle function can be used to represent the Dirac delta function δ ( x ) {\displaystyle \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).} For a function g ( x ) {\displaystyle g(x)} , its average over the width a {\displaystyle 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).} To obtain g ( 0 ) {\displaystyle 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)} 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).} The Fourier transform of the Dirac delta function δ ( t ) {\displaystyle \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)}.} 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} and a {\displaystyle a} goes to infinity, the Fourier transform of δ ( t ) {\displaystyle \delta (t)} is
δ ( f ) = 1 , {\displaystyle \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.