[Fourier transform and its application study notes] nine. Continuation of the convolution discussion

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

The application of convolution in filtering

Turbidity (turbidity) research is about measuring the clarity of water. The approximate method is to place the light sensor in the deep water area, and then measure the dim degree of the light, and the measured value will change over time.

(as there is no real data, the following is a rough simulation of water turbidity data using Mathematica)

Can see the signal is mainly concentrated in the low frequency, we need to remove the glitch, that is, the high-frequency removal, low-pass filtering in the frequency domain (lower pass Filtering)

The waveform after filtering is as follows

Frequency domain operation: $\pi_{2\nu_c} F (s) $; the time domain operation is convolution: $2\nu_c sinc (2\nu_c t) *f (t) $.

Filtering concepts

filtering (Filtering) is usually equivalent to convolution, which is implemented by filters.

A filter is a system in which the input variable function (signal) is convolution with a fixed function (signal). This fixed signal is called the impulse response (impulse response).

$g \quad = \quad F \qquad * \qquad h$
$\qquad output \qquad input \qquad impulse \ response$

Convolution is a representation of time domain, in general, the frequency domain operation is much simpler than the time domain, because the frequency domain simply performs the multiplication operation.

$G (s) = F (s) H (s) $

$H (s) $ is called the transfer function (transfer functions), when designing the filter is usually designed with the appropriate transfer function $h (s) $.

Below are the more commonly used filters.

Low pass filter, often used for image compression.

High-pass filter (High pass filter), commonly used for edge detection (detection edge)

Band pass filter (band pass filter)

The meaning of convolution

The professor thinks that it is only necessary to understand the multiplication of functions from the frequency domain, and there is no need to go to the figurative convolution in the time domain. (I think it was equally idiotic to try to visualize convolution. I think the visualize convolution, if there is a-a, is-to-think in terms of multiplying in the frequency domain.)

Properties of convolution

Generally $f*g$ are generally smoother than individual $f$ and $g$.

For example, the $\pi$ of a rectangular function is discontinuous, and the convolution of two $\pi$ functions is a trigonometric $\lambda$, which is continuous.

$\eta (\PI * \pi) = (\eta \pi) (\eta \pi) = sinc^2 = \eta \lambda$

Fourier derivative theorem

After the original function is differential, its Fourier transform is equal to the Fourier transform of its original function multiplied by the $2\PI is$

$\eta (f ') (s) = 2\pi is (\eta f) (s) $

The certification process is as follows:

Fourier inverse transformations are:

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

To differentiate it,

$\begin{align*}
\frac{\partial f}{\partial T}
&= \int_{-\infty}^{\infty}f (s) (2\pi ise^{2\pi ist}) DS \ \
&= \int_{-\infty}^{\infty} (2\pi IsF (s)) E^{2\pi ist}ds \ \
\end{align*}$

Then there is the relationship between $f ' $ and $2\pi IsF (s) $ for the Fourier transform

$f ' \ \leftrightarrow \ 2\pi IsF (s) $

Spread out there

$\eta (F^n) (s) = (2\pi is) ^n (\eta f) (s) $

Thermal equation on an infinitely long column

The $U (x,t) $ represents the time $t$, the temperature on the position $x$.

The known initial temperature is $u (x,0) = f (x) $, and the thermal equation is $u_t = \frac{1}{2}u_{xx}$.

The solution for $U (x,t) $ is as follows:

The position variable is $x$ Fourier transform, assuming that the result of the transformation is $u (s,t) $.

Fourier transform to the left of the equation of heat,

$\begin{align*}
\eta (u_t)
&= \int_{-\infty}^{\infty}e^{-2\pi ISX} \frac{\partial}{\partial t}u (x,t) dx \ \
&= \frac{\partial}{\partial T}\int_{-\infty}^{\infty} e^{-2\pi isx}u (x,t) dx \ \
&= \frac{\partial}{\partial t}u (s,t)
\end{align*}$

Fourier transform to the right of the equation of heat,

$\eta (\frac{1}{2}u_{xx}) = \frac{1}{2} (2\pi is) ^2u (s,t) =–2\pi ^2s^2u (s,t) $

that there

$\frac{\partial}{\partial t}u (s,t) =–2\pi^2s^2u (s,t) $

To find partial differential equations, it is

$U (s,t) = U (s,0) e^{-2\pi^2s^2t}$

$U (s,0) = \displaystyle{\int_{-\infty}^{\infty}u (x,0) e^{-2\pi isx}dx=\int_{-\infty}^{\infty}f (x) e^{-2\pi ISX}DX = F (s)}$

Substituting the result of $u (s,0) $ into $u (s,t) $,

$U (s,t) = F (s) e^{-2\pi ^2s^2t}$

Convert to convolution format

$e ^{-2pi ^2s^2t} = \eta (\frac{1}{\sqrt{2\pi t}}e^{\frac{x^2}{2t}}) $

$\begin{align*}
U (s,t)
&= F (s) e^{-2\pi ^2s^2t}\\
&= (\eta F) (\eta (\frac{1}{\sqrt{2\pi t}}e^{\frac{x^2}{2t})) \ \
&= \eta (f* \frac{1}{\sqrt{2\pi t}}e^{\frac{x^2}{2t}})
\end{align*}$

$U (X,t) = f (x) * \frac{1}{\sqrt{2\pi t}}e^{\frac{x^2}{2t}}$

[Fourier transform and its application study notes] nine. Continuation of the convolution discussion