[Fourier transform and its application study notes] 27. Gauviffli Leaf Transformation, review

This lesson is mainly about Fourier transform calculation, because the Gauviffli leaf transformation has multiple variables, multiple integrals, so in the calculation will have greater difficulties. However, some functions will have a relatively simple calculation method, the following analysis of two such functions.

Separable functions

A high-dimensional Fourier transform with a class of functions can be obtained by calculating a series of low Viffoury leaf transformations, which are called separable functions. (There ' an important class of functions for which can compute a higher-dimensional transform by computing a series of Lower-dimensional transforms. These is separate functions.)

Example One

Two-dimensional rectangular function $\pi (x_1,x_2) $

$\pi (x_1,x_2) =\begin{cases}
1 & \text{,} |x_1|<\frac{1}{2}\ \& \ |x_2|<\frac{1}{2} \ \
0 & \text{,} otherwise
\end{cases}$

In addition, the function can also be written as a product of two one-dimensional rectangular functions

$\pi (x_1,x_2) = \pi (x_1) \pi (x_2) $

Its Fourier transform is

$\begin{align*}
\mathcal{f}f (\xi_1,\xi_2)
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2\pi I (x_1\xi_1+x_2\xi_2)}\Pi (x_1,x_2) dx_1dx_2\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2\pi ix_1\xi_1}e^{-2\pi ix_2\xi_2}\Pi (x_1) \Pi (x_2) DX_1DX _2\\
&=\left (\int_{-\infty}^{\infty}e^{-2\pi ix_1\xi_1}\pi (x_1) dx_1 \right) \left (\int_{-\infty}^{\infty}e^{-2\pi Ix_2\xi_2}\pi (x_2) dx_2 \right) \ \
&=\mathcal{f}\pi (\xi_1) \mathcal{f}\pi (\xi_2) \ \
&= (sinc\xi_1) (sinc\xi_2)
\end{align*}$

In general, if a high-dimensional function can be written as a product of a low-dimensional function, the Fourier transform of the high-dimensional function can also be written as the product of these low-dimensional Fourier transforms.

$f (x_1,x_2,..., x_n) = F_1 (x_1) f_2 (x_2) ... f_n (x_n) $

$\rightarrow \mathcal{f}f (\xi_1,\xi_2,..., \xi_n) = \mathcal{f}f_1 (\xi_1) \mathcal{f}f_2 (\xi_2) ... \mathcal{F}f_n (\xi_n) $

Example Two

Ivigos function

$g (x_1,x_2) = E^{-\pi (x_1^2+x_2^2)}$

It can be divided into two of a Gaussian function of the product

$g (x_1,x_2) = E^{-\pi x_1^2}e^{-\pi X_2^2} = g_1 (x_1) g_2 (x_2) $

Its Fourier transform is

$\begin{align*}
\mathcal{f}g (\xi_1,\xi_2)
&=\mathcal{f}g_1 (\xi_1) \mathcal{f}g_2 (x_2) \ \
&=e^{-\pi\xi_1^2}e^{-\pi\xi_2^2}\\
&=e^{-\pi (\xi_1^2+\xi_2^2)}
\end{align*}$

(two-dimensional) the Fourier transform of the Gaussian function itself.

Radial functions (Radial function)

The Ivigos function is an example of a radial function, which is circular symmetric. When we introduce polar coordinates, we have

$\left.\begin{matrix}
R=\sqrt{x_1^2+x_2^2} \ \
\theta=arctan\frac{x_2}{x_1}
\end{matrix} \quad \right| \quad
\left.\begin{matrix}
X_1=rcos\theta \ \
X_2=rsin\theta
\end{matrix} \right.$

Then the Ivigos function can become

$g (x_1,x_2) = E^{-\pi (x_1^2+x_2^2)} = E^{-\pi r^2}$

It relies only on $r$ and not on individual $x_1,x_2$, which is the definition of radial functions. The radial function relies only on the distance $r$ to a point of origin.

The radial function has one feature: the Fourier transform of the radial function is still a radial function.

The certification process is as follows:

Converting Fourier transform in Cartesian coordinate system into polar coordinate system form

The Fourier transform in Cartesian coordinates has the following form

$\displaystyle{\mathcal{f}f (\xi_1,\xi_2) =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2\pi I (x_1\xi_1+x_2\ xi_2)}f (x_1,x_2) dx_1dx_2}$

Variable for polar coordinate conversion

$\begin{matrix}
X_1=rcos\theta &\qquad \xi_1=\rho cos\varphi \ \
X_2=rsin\theta &\qquad \xi_2=\rho Sin\varphi
\end{matrix}$

Assuming that $f$ is a radial function, then $f (x_1,x_2) = f (f) $, $dx _1dx_2$ is converted to $rdrd\theta$ by polar coordinates

The inner product of the variable in the complex exponent becomes

$\begin{align*}
X_1\xi_1+x_2\xi_2
&=rcos\theta \rho Cos\varphi+rsin\theta \rho sin\varphi\\
&=r\rho (Cos\theta cos\varphi+sin\theta sin\varphi) \ \
&=r\rho cos (\theta-\varphi)
\end{align*}$

The above variables are transformed into the Fourier transform of the Cartesian coordinate system, which has

$\begin{align*}
& \quad \int_0^{\infty}\int_0^{2\pi}e^{-2\pi ir\rho cos (\theta-\varphi)}f (R) rdrd\theta\\
&=\int_0^{\infty}\left (\int_0^{2\pi}e^{-2\pi ir\rho cos (\theta-\varphi)}d\theta \right) F (r) rdr\\
&=\int_0^{\infty}\left (\int_0^{2\pi}e^{-2\pi Ir\rho Cos\theta}d\theta \right) F (r) rdr \ \
&\quad (cos\ is\ a\ periodic\ function\ of\ 2\pi\, Shift\ won ' t\ effect\ its\ Integral)
\end{align*}$

We define the integral in parentheses as a function

$\displaystyle{J_0 (a) = \frac{1}{2\pi}\int_0^{2\pi}e^{-iacos\theta}d\theta}$

$J _0 (a) $ is referred to as Class 0 Bessel function (0 ' th order Bessel function of the first kind), Bessel has other classes, other order functions, which often appear in the scenario where the radial function appears.

As a result, the two-dimensional Fourier transform of the radial function is eventually transformed through the polar coordinate system

$\displaystyle{\mathcal{f}f (\rho) =2\pi\int_{0}^{\infty}f (R) j_0 (2\pi r\rho) rdr}$

It is a function that relies only on $\rho$, and $\varphi$ has been eliminated in front of the $\theta$ points. This transformation is called the 0-order Henkel transformation (0 ' th order Henkel transformation). This also proves that the Fourier transform of the radial function is still a radial function.

Convolution theorem of Gauviffli leaf transformation

As with the convolution theorem for one-dimensional Fourier transforms, they are equally applicable in high-dimensional

Vector form

$\displaystyle{(F*g) (\underline{x}) =\int_{\mathbb{f}^n}f (\underline{x}-\underline{y}) g (\underline{y}) d\ Underline{y}}$

Two-dimensional component form

$\displaystyle{(f*g) (x_1,x_2) =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f (x_1-y_1,x_2-y_2) g (y_1,y_2) Dy_1dy _2}$

When analyzing a Viffoury leaf transform convolution, we do not recommend analyzing convolution in the time domain, but rather the convolution as the product of the frequency domain. Similarly, it is not recommended here to analyze convolution in airspace, as more variables lead to more complex thinking, fortunately, the convolution on the Fourier transform formula is still used in the high-dimensional, i.e.

$\MATHCAL{F} (F*G) =\mathcal{f}f\mathcal{f}g$

$\MATHCAL{F} (FG) =\MATHCAL{F}F * \mathcal{f}g$

[Fourier transform and its application study notes] 27. Gauviffli Leaf Transformation, review