Fourier Transform
functionh:\mathbb{R}\to\mathbb{C}h:R→CAbout\hat{h} =
F\left[h\right]:\mathbb{R}\to\mathbb{C}h^=F[h]:R→Ca function called
\displaystyle \hat{h}(t) = F\left[h\right]\left(t\right)
\equiv \int_{-\infty} ^{\infty} e^{-2 \pi itx}
h\left(x\right) \mathrm{d}x \quad \left(i = \sqrt{1}\right)h^(t)=F[h](t)≡∫− ∞∞And−2πitxh(x)dx(i=− 1)
, and transform the above\hat{h}, F[h]h^,F[h]functionhhIt is defined as the Fourier
transform of[One][2] It can be defined similarly in higher dimensions. Fourier transform\int e^{itx} h\left(x\right) \mathrm{d}x∫And−itxh(x)dxSome mathematicians define it as[3] In
this case, the relationship between the derivative of the function and the Fourier transform is
neat (F\left[h'\right](t)=itF\left[h\right](t)F[h'](t)=i t F[h](t)) is mainly used in partial
differential equations. In physics, normal timettThe angular frequency is the Fourier domain
of\omegaωuse , locationxxwave number into the Fourier domain oftotouse
The above integral is an arbitrary functionhhis not well defined
(e.g.h(x)=1h(x)=Oneif ). However, in most cases, the integral expression is interpreted as
improper Riemann integral,h(x)e^{-\epsilon x^2}h(x)e−ϵx2After calculating the Fourier
transform of\ epsilon \ to 0ϵ→0It is solved by sending the extreme. In any case, it is
necessary to strictly define the Fourier transform in order to solve such a small problem.
Please refer to the following.
1.1. in Schwarz space[edit]
schwarz space\mathcal{S}Sis a space of functions that are infinitely differentiable and
whose derivatives are rapidly decreasing. In this case, the Fourier transform as an integral
expression is well defined. especially,F:\mathcal{S}\to \mathcal{S}F:S→Sis a one-toone correspondence, and allh,g\in
\mathcal{S}h,g∈SAbout
\displaystyle \int_{-\infty}^{\infty} h(x)\bar{g}(x) \mathrm{d}x = \int_{\infty}^{\infty} F[h](t) \overline{F[g]}(t) \mathrm{d}t∫− ∞∞h(x)gˉ
(x)dx=∫− ∞∞F[h](t)F[g](t)dt
satisfies Plancherel's theorem.
1.2. in L 1 space[edit]
First, you need to know about the Lebesgue space. (see comments)[4] L^{1}IOneThe
Fourier transform as an integral expression for a function is well defined, but the transformed
function isL^{1}IOnemay not go into, in factF:L^{1} \to L^{\infty}F:IOne→I∞Only Im
can know. Unlike in Schwarz space,F:L^{1} \to L^{\infty}F:IOne→I∞is not a
transcription function.[5]
1.3. in L 2 space[edit]
with this nature\mathcal{S}\subset
L^{2}S⊂I2by the compactness ofF:\mathcal{S}\to
\mathcal{S}F:S→SIsF:L^{2}\to L^{2}F:I2→I2It has a unique extension to the causal
operator. because of thisF:L^{2}\to L^{2}F:I2→I2is well defined
1.4. in a controlled distribution space.[edit]
\mathcal{S}'S'space is schwarz space\mathcal{S}SAs a dual space of , it has a weak-*
phase.\mathcal{S}'S'The elements of is called the control distribution. control distributionh
\in \mathcal{S}'h∈S'Fourier transform ofF[h] \in \mathcal{S}'F[h]∈S'is naturally
defined as follows by Plancherel's theorem.
This Fourier transform is very similar to the Laplace transform . right abovetttoisistry
putting The derivative of a function is multiplied with a variable by Fourier transform, and the
product is transferred to convolution. Therefore, the Laplace transform solution of the
differential equation can be corrected with the Fourier transform solution as it is. However,
what is even better than the Laplace transform is that the inverse transform is very easy. No,
it's just an inverse transformation! strictly speakingF^{2} h\left(t\right) =
F\left[F\left[h\right]\right]\left(t\right) = h\left(t\right)F2h(t)=F[F[h]](t)=h(−t)is established.[6]