stylistic:style_transfer, linguistic:paraphrasing
Dirac delta function

The rectangle function can be used to represent the Dirac delta function δ (x) {\displaystyle \delta (x)}. 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.