Wavelet Analysis or MRA

March 21, 2017

GoLuckyRyan Math, theory , , , , , , 2 Comments

Although the Fourier transform is a very powerful tool for data analysis, it has some limit due to lack of time information. From physics point of view, any time-data should live in time-frequency space. Since the Fourier transform has very narrow frequency resolution, according to  uncertainty principle, the time resolution will be very large, therefore, no time information can be given by Fourier transform.

Usually, such limitation would not be a problem. However, when analysis musics, long term performance of a device, or seismic survey, time information is very crucial.

To over come this difficulty, there a short-time Fourier transform (STFT) was developed. The idea is the applied a time-window (a piecewise uniform function, or Gaussian) on the data first, then FT. By applying the time-window on difference time of the data (or shifting the window), we can get the time information. However, since the frequency range of the time-window  always covers the low frequency, this means the high frequency  signal is hard to extract.

To improve the STFT, the time-window can be scaled (usually by 2). When the time window is shrink by factor of 2, the frequency range is expanded by factor of 2. If we can subtract the frequency ranges for the time-window and the shrink-time-window, the high frequency range is isolated.

To be more clear, let say the time-window function be

\phi_{[0,1)}(t) = 1 , 0 \leq t < 1

its FT is

\hat{\phi}(\omega) = sinc(\pi \omega)

Lets also define a dilation operator

Df(t) = \sqrt{2} f(2t)

the factor \sqrt{2} is for normalization.

The FT of D\phi(t) has smaller frequency range, like the following graph.

Capture.PNG

We can subtract the orange and blue curve to get the green curve. Then FT back the green curve to get the high-frequency time-window.

We can see that, we can repeat the dilation, or anti-dilation infinite time. Because of this, we can drop the FT basis Exp(-2\pi i t \omega), only use the low-pass time-window to see the low-frequency behaviour of the data, and use the high-pass time-window to see the high-frequency behaviour of the data. Now, we stepped into the Multi-resolution analysis (MRA).

In MRA, the low-pass time-window is called scaling function \phi(t) , and the high-pass time-window is called wavelet \psi(t).

Since the scaling function is craetd by dilation, it has the property

\phi(t) = \sum_{k} g_{0}(k) \phi(2t-k)

where k is integer. This means the vector space span by {\phi(t-k)}_{k}=V_0 is a subspace of the dilated space DV_0 =V_1. The dilation can be go one forever, so that the whole frequency domain will be covered by V_{\infty}.

Also, the space span by the wavelet, {\psi(t-k)}=W_0, is also a subspace of V_1. Thus, we can imagine the structure of MRA is:

Capture.PNG

Therefore, any function f(t) can also be expressed into the wavelet spaces. i.e.

f(t) = \sum_{j,k} w_{j,k} 2^{j/2}\psi(2^j t - k)

where j, k are integers.

I know this introduction is very rough, but it gives a relatively smooth transition from FT to WT (wavelet transform), when compare to the available material on the web.

 

 

 

 

decay time constant and line width

December 20, 2010

GoLuckyRyan advance, Basic, theory, Working on , , , , , , , , Leave a comment

the spectrum of energy always has a peak and a line width.

the reason for the line width is, this is decay.

i give 2 explanations, once is from classical point of view and i skipped the explanation for the imaginary part. so, i am not fully understand. the 2nd explanation is look better, but it is from QM. however, there is one hide question for that explanation is, why the imaginary energy is negative?

the simplest understanding of the relation is using fourier transform. (i think)

Fourier transform is changing the time-frame into the frequency frame. i.e, i have a wave, propagating with frequency w. we can see a wave shape when plot with time. and we only see a line, when we plot with frequency, since there is only 1 single frequency. however, for a general wave, it is composite of many different frequencies, using fourier transform can tell us which frequency are involved. And energy is proportional to frequency.

when the particle or state under decay. the function is like

f(t) = Exp(-R t) Exp ( i \omega_0 t)

where the R is decay constant, and ω0 is the wave frequency.

after fourier transform, assume there is nothing for t < 0

F(t) = \frac {1} { R + i ( \omega_0 - \omega )}

the real part is

Re(F(t)) = \frac {R} { R^2 + ( \omega_0 - \omega )^2}

which is a Lorentzian shape and have Full-Width-Half-Maximum (FWHM) is 2R. it comes from the cosine part of the fourier transform. thus, the real part.

and the imaginary part is

Im(F(t)) = \frac {\omega_0 - \omega}{R^2 + ( \omega_0 -\omega )^2 }

the imaginary part is corresponding to the since part, so, we can neglect it. (how exactly why we can neglect it? )

Thus, we can see, if there is no decay, R → 0, thus, there is no line width.

therefore, we can see the line width in atomic transition, say, 2p to 1s. but there are many other mechanism to the line width, like Doppler broadening, or power broadening. So, Decay will product line width, but not every line width is from decay.

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another view of this relation is from the quantum mechanics.

the solution of Schroedinger equation is

\Psi (x,t) = \phi(x) Exp \left( - i \frac {E}{\hbar} t \right)

so, the probability conserved with time, i.e.:

|\Psi(x,t)|^2 = |\Psi (x,0)|^2

if we assume the energy has small imaginary part

E = E_0 - \frac {i} {2} R \hbar

( why the imaginary energy is nagative?)

|\Psi(x,t)|^2 = |\Psi (x,0)|^2 Exp ( - R t)

that make the wavefunction be :

\Psi (x,t) = \phi (x) Exp( - i \frac {E}{\hbar} t ) Exp( - \frac {R}{2} t )

what is the meaning of the imaginary energy?

the wave function is on time-domain, but what is “physical”, or observable is in Energy -domain. so, we want Psi[x,E] rather then Psi[x,t], the way to do the transform is by fourier transform.

and after the transform, the probability of finding particle at energy E is given by

|\Psi(x,E)|^2 = \frac {Const.}{R^2 +(\omega_0 - \omega )^2}

which give out the line width in energy.

and the relation between the FWHM(line width) and the decay time is

mean life time ≥ hbar / FWHM

which once again verify the uncertainty principle.

Method

December 16, 2010

GoLuckyRyan Basic, no math , , , , Leave a comment

As we know the world of nuclear physics is so small. ordinary method is not applicable to “see” this world.

we have no choice but just bombard the nucleus with electrons, protons, neutrons, etc… if we are Alice, who become much bigger then our earth and touch the moon easily. How do we understand human world? we simply pick a human, hitting on each other, see what is going on, what is the result. If we want to know how elevator work, we put a human, let him ride on it and see the result. but sometime, we will accidentally, put a car in elevator.

so, Most nuclear experiment is SCATTERING EXPERIMENT.

and the machine to conduce this kind of experiment is called Accelerator.

there is a famous quote i forget where it come from:

nuclear (particle) physics is like we figure out how a watch work by broken it and see the fragments.

The picture is really like this. we shoot particles into nucleus, and see what was knot out. how the incident particle changed. we can extract the energy change, the scatter angle, the polarization. basically are these 3 things. and using these 3 data, we construct the world of nucleus. Is it amazing???

There are some fundamental limits of the scattering experiment.

1) since the size of the nucleus is very small, the chance of hitting it is very small for 1 particle to hit another particle. thus, we use many particles hit many particles. but even doing so, only a tiny fraction of reaction takes place. most of them just pass by, say a hello. Thus, we have to create a high density particle beam, and target.

2) the particle should be moving very fast, almost same as speed of light, in order to carry enough energy to go inside the nucleus. because there is a barrier form the forces. the nucleus is something like a fortress, walled by forces. (sound like a star wars movie) another reason is, the larger the energy, the shortest distance we can probe. According to De Broglie, every particle can be treated as wave with wavelength is inversely proportional to the momentum. Thus, a faster particle has larger momentum and shorter wavelength. so, can see a smaller world. [ the De Broglie’s wavelength has some debt on weather it is a physical wave or probability amplitude, this was solved by Dirac and proved by experiment that, it means both. for more info, see discussion on “interpolation on quantum wave function” ]

This is a general property of wave. a Radio wave can easy pass though us because we are small compare to the wavelength, which is about 5 to 10 meters long. but red light can “see” us, or we can block red light, because we are much bigger. the idea is, if we want to see the detail, you have to use a smaller ruler.

3) since we are using the building block to hit another building block in investigation. they are similar size. imagine a scenario that you want to measure to speed of a car, and you use another car to hit on it, and see the bounded back car to find out the original speed. you can see, the cars hit each other and changed the original speed, and there is no way to accurate to measure to speed! same things happen in nuclear physics. the scatter particle will change to state of target, that create an uncertainty. This was formulated by Heisenberg and now called the Heisenberg’s Uncertainty Principle. which state that

change of position X change of momentum  ≥ Planck’s constant