[Fourier transform and its application study notes] six. Thermal equation discussion

This is my study notes, the course for NetEase Open Class Stanford University Open Class: Fourier transform and its application.

As mentioned in the previous lesson, in Fourier analysis of aperiodic functions, there is

$C _k = \displaystyle{\frac{1}{t}\int_{-\frac{t}{2}}^{\frac{t}{2}}f (T) e^{-2\pi I\frac{k}{t}t}dt}$

$f (t) = \displaystyle{\sum_{k=-\infty}^{\infty}c_ke^{2\pi i\frac{k}{t}t}}$

We hope that $t\to \infty$ will get the results we want: Fourier transforms are suitable for aperiodic functions. But the results proved that this is not feasible, and finally concluded: for any $c_k$, there are $c_k \leqslant \frac{m}{t}$, when $t \to \infty \, C_k \to 0$. $C _k$ is inversely proportional to $t$.

According to this relationship, can we introduce $t$ to $c_k$ side?

New Symbol $\eta$

Make

$\displaystyle{\eta f (\frac{k}{t}) =c_k \times T = \int_{-\frac{t}{2}}^{\frac{t}{2}}e^{-2\pi i \frac{k}{T}t}f (t) DT}$

That is,

$f (t) = \displaystyle{\sum_{k=-\infty}^{\infty}\eta f (\frac{k}{t}) e^{2\pi i\frac{k}{t}t} \frac{1}{t}}$

Now make $t \to \infty$, then $\frac{k}{t}$ 's value range is $ (\frac{k=-\infty}{t\to \infty},\frac{k=+\infty}{t\to \infty}) $, which is $ (-\infty, +\infty) $. The value interval is $\frac{1}{t} \to 0$, tending to continuous variables. Now use continuous variable $s$ to represent $\frac{k}{t}$:

$s = \frac{k}{t} \, –\infty < s < \infty$

$\eta f (s) = \displaystyle{\int_{-\infty}^{\infty}e^{-2\pi ist}f (t) DT}$

Since $\frac{k}{t}$ is replaced with a continuous variable $s$, the polynomial of the Fourier series is replaced with integral, where $\frac{1}{t}$ is $\bigtriangleup S $, which is $ds$

$f (t) = \displaystyle{\int_{-\infty}^{\infty}\eta f (s) e^{2\pi Ist}ds}$

Conclusion (definition)

If the period of $f (t) $ is defined in the entire real number field, i.e. $-\infty < T < \infty$, then its

Fourier transform:

$\eta f (s) = \displaystyle{\int_{-\infty}^{\infty} e^{-2\pi ist}f (t) DT}$

Fourier inverse transformation:

$f (t) = \displaystyle{\int_{-\infty}^{infty}e^{2\pi IS0} \eta F (s) DS}$

You can also write the following form:

$\eta f (s) = \displaystyle{\int_{-\infty}^{\infty} e^{-2\pi ist}f (t) DT}$

$\eta^{-1}g (t) = \displaystyle{\int_{-\infty}^{infty}e^{2\pi ist}g (s) DS}$

The symbol $\eta$ represents the Fourier transform, and the $\eta^{-1}$ represents the inverse Fourier transform.

The Fourier transform function is decomposed into continuous complex exponent, and the Fourier inverse transform combines these successive complex exponents into the original function.

0-point value

$\eta F (0) = \displaystyle{\int_{-\infty}^{\infty} e^{-2\pi i0t}f (t) dt = \int_{-\infty}^{\infty}f (t) DT} $

$\eta^{-1}g (0) = \displaystyle{\int_{-\infty}^{infty}e^{2\pi IS0} g (s) ds = \int_{-\infty}^{infty}g (s) DS}$

Fourier Transform Example 1. Rectangle function

$\PI (t) =
\left\{\begin{matrix}
1 & \left| T \right| < \frac{1}{2}\\
0 & \left| T \right| \geqslant \frac{1}{2}
\end{matrix}\right.$

The Fourier transform is as follows:

$\begin{align*}
\eta \pi (s)
&= \displaystyle{\int_{-\infty}^{\infty}e^{-2\pi ist}\pi (t) dt} \ \
&= \displaystyle{\int_{-\frac{1}{2}}^{\frac{1}{2}}e^{-2\pi Ist}dt} \ \
&= \left. -\frac{1}{2\pi is}e^{-2\pi ist} \right |_{\frac{1}{2}}^{\frac{1}{2}} \ \
&=-\frac{1}{2\pi is}e^{-2\pi Is\frac{1}{2}}-(-\frac{1}{2\pi Is}e^{-2\pi is (-\frac{1}{2})}) \ \
&=-\frac{1}{2\pi Is}e^{-\pi is} + \frac{1}{2\pi is}e^{\pi is} \ \
&= \frac{1}{\pi S} (\frac{e^{\pi is}-e^{-\pi is}}{2i}) \ \
&= \frac{1}{\pi S} (\frac{cos (\pi s) +isin (\pi s)-cos (-\pi s)-Isin (-\pi s)}{2i}) \ \
&= \frac{1}{\pi S} (\frac{2isin (\pi s)}{2i}) \ \
&= \frac{sin (\pi s)}{\pi s} \ \
&= sinc \ S
\end{align*}$

This function is called the $sinc$ function

2. Triangle function

$\lambda (t) =
\left\{\begin{matrix}
1-\left|t\right| & \left|t\right|<1 \ \
0 & \left|t\right| \geqslant 1
\end{matrix}\right.$

The Fourier transform is as follows:

\begin{align*}\eta\lambda (s)
&= \displaystyle{\int_{-\infty}^{\infty}e^{-2\pi ist}\lambda (t) dt}\\
&=\displaystyle{\int_{-1}^{0}e^{-2\pi ist} (1+t) dt + \int_{0}^{1}e^{-2\pi ist} (1-t) dt}\\
&=\left (\left. ( 1+t) (-\frac{2\pi is}{1}e^{-2\pi ist}) \right|_{-1}^0-\displaystyle{\int_{-1}^{0}-\frac{1}{2\pi is}e^{-2\pi Ist}dt}\ right) +\left (\left. ( 1-T) (-\frac{2\pi is}{1}e^{-2\pi ist}) \right|_{0}^1-\displaystyle{\int_{0}^{1}-\frac{1}{2\pi is}e^{-2\pi Ist}dt}\ right) \ \
&=\left (-\frac{1}{2\pi is}-\left. \frac{1}{4\pi^2i^2s^2}e^{-2\pi ist}\right|_{-1}^{0}\right) +\left (\frac{ 1}{2\pi Is}+\left. \frac{1}{4\pi^2i^2s^2}e^{-2\pi ist}\right|_{0}^{1}\right) \ \
&=-\left (\frac{1}{-4\pi^2s^2}-\frac{1}{-4\pi^2s^2}e^{2\pi is}\right) +\left (\frac{1}{-4\pi^2s^2}e^{-2\pi is}-\frac{1}{-4\pi^2s^2}\right) \ \
&=\frac{-2+cos (2\pi s) +isin (2\pi s) +cos ( -2\pi s) +isin ( -2\pi s)}{-4\pi^2s^2}\\
&=\frac{-2+2cos (2\pi s)}{-4\pi^2s^2}\\
&=\frac{-4cos^2 (\pi s)}{-4\pi^2s^2}\\
&=\frac{cos^2 (\pi s)} {(\pi s) ^2}\\
&=sinc^2s
\end{align*}

[Fourier transform and its application study notes] six. Thermal equation discussion