Recent content by arcTomato

@Twigg P.S. I was able to understand it somehow. First of all, as for units, they should be the same for both. (I figured it out by thinking that each Fourier transform, A and W, has the same units as the units before the Fourier transform.) Also, I think the ##|A(\omega) * W(\omega)|^{2}##...

@rude man Thank you. My understanding is that the sign of exp in the Fourier transform must be opposite to the sign on the shoulder in the forward and inverse transforms, but it depends on the definition of which should have which sign.

@Twigg Thank you. I didn't realize that the units were different. I will check it myself. It's going to take me a while to figure it out, but your advice has been very helpful. If I get any more tips myself, I'll post them in this thread. I wish you all the best with your exams!

@Twigg I am grateful for your thoughts on the issues I am facing. I am still aware that it is generally defined as the former (##P(\omega)=|A(\omega)|^{2} *|W(\omega)|^{2}##). I am thinking about under what circumstances the two coincide.

@Twigg Thank you. I see. If you could give me a detailed explanation, that would be very helpful.

I'm sorry for my poor English @Charles Link . I can observe x-rays with an x-ray observer and get the light curve itself as data. The goal is to Fourier transform the light curve, or time series data, and get the power spectrum. My question was about the defining equation for this.

Hello PF. I am thinking about the power spectrum when observing X-rays. We are trying to obtain the power spectrum by applying a window function ##w(t)## to a light curve ##a(t)## and then Fourier transforming it. I have seen the following definition of power spectrum ##P(\omega)##. Suppose...

I see. So you mean that the effect of the cross term is visible at ## j=0## and nearby, but becomes invisible at higher frequencies. The discussion was very easy to understand. Thank you once again, and sorry for my bad English.

No, it's because I didn't explain it well enough. I'm sorry. About ##\left\langle a_{j, b g}\right\rangle=0##, this was my misunderstand. So, is there a term that remains in the cross-term? I mean, ##2\left\langle a_{j, \mathrm{bg}}\right\rangle \left\langle a_{j...

It was the Fourier transform of the X-ray count rate. In other words, it is the one that appears in the definition of the power spectrum, and ##j## is the wave number. ##P_{j}=\left|a_{j}\right|^{2}##

I see, even if there is background, as long as it is uncorrelated with the signal, the cross term disappears. I rewrote the previous equation in my own way. ##a_{j, \text { signal }}=\left\langle a_{j, \text { signal }}\right\rangle+\delta a_{j, \text { signal }}##, and ##a_{j, \text { bg...

Thank you for explaining exactly the part I was wondering about.I would like to ask further questions if you don't mind. I am now thinking about the analysis of data containing Poisson noise, which is the sum of the time variability of a signal and a almost constant value of background noise...

I thought that if we Fourier transformed the counts of the sum of the signal from the source and the Poisson noise, and obtained the power spectrum, we would get the following, ##P_{j}=P_{j, \text { signal }}+P_{j, \text { noise }}+\text { cross terms }## but I found the following description...

@vanhees71 Thank you for your very detailed answer. It made a lot of sense to me. It's very helpful.

Sorry for the late reply. I am thinking about that very situation. I see. Thank you for the detailed explanation. However, this part bothered me a little. From which equation exactly do you get that the electric field is equal to the magnetic field?