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I read a paper reference at http://arxiv.org/pdf/1101.1764.pdf that if we average a set $V=\{V(t_0,\nu_0), V({t_1,\nu_1),..., V(t_n,\nu_n)}\}$; with $V(t_i,\nu_i)=e^{i\sigma(t_i,\nu_i)}$ then we can approximate this average as:

$<V>\simeq sinc(\frac{\Delta \phi}{2})sinc(\frac{\Delta \theta }{2})V(t_c,\nu_c)$

$\\$where $c=(n+1)/2$ is the mid point,

$\Delta \phi=arg V(t_n,\nu_c)-argV(t_0,\nu_c)$,

$\Delta \theta=arg V(t_c,\nu_n)-argV(t_c,\nu_0)$. where arg denotes the complex argument or complex angle.

Please can some one help me to demonstrate this?

Also, if I weight this set with a window $W(t,\nu)$, what will be this approximation?

From what I understood, averaging meant weighting the data with a normalize top-hat ("box") window, and the Fourier transform of a top-hat window is a sinc. But I am not so sure that my terminology works.

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  • $\begingroup$ It has to depend on what $\sigma(t_i,\nu_i)$ is. For instance, if $\sigma = 2\pi f t + 0 \nu$, where $f >> 0$, then one would expect the DC value (which is just the average), to be near zero. However, I don't see anything to prevent either $\Delta\phi$ nor $\Delta \theta$ from being equal to zero, which means that $<V> \approx V(t_c,\nu_c)$. Perhaps I am missing something. $\endgroup$ – AnonSubmitter85 Oct 31 '14 at 20:10
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    $\begingroup$ Thank you. I did some errors and I duplicate the question, I have to delete the first one. In additional from what I understood $\int_{0}^{x_0}e^{ix}dx\simeq sinc(\frac{x_0}{2})e^{\frac{ix_0}{2}}$ if the phase $x_0$ is kept small enough. Please advice $\endgroup$ – armando Oct 31 '14 at 20:55

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