From Creighton and Anderson Chapter 7 on Gravitational Wave data analysis:
Please explain where the delta t and the limit came from in this derivation.
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Marcus Müller
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Jyothis Chandran
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2I really don't think the right equality in $(7.22)$ makes much sense; $\lim\limits_{\Delta! t\to 0}2\sigma^2\Delta! t\equiv0$ for all finite $\sigma$. – Marcus Müller Jan 28 '19 at 20:46
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But this result is needed in that text to derive the optimal matched filter used in GW data analysis, which they got right...this can be found in section 7.1.2 in the same text...are you sure that this is wrong. – Jyothis Chandran Jan 30 '19 at 09:39
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yes, I'm sure. That's the definition of $\lim\limits_{\Delta t\to 0}\Delta t$. – Marcus Müller Jan 30 '19 at 09:56
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See this answer for a somewhat different viewpoint. – Dilip Sarwate Feb 17 '21 at 22:37
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Marcus is correct, so this means something else is wrong. The left equality simply shows that, for continuous time, the unilateral PSD, $S_x(f)$, is two times the Fourier transform of the autocorrelation function, $R_x(\tau)$. See, e.g., A.B. Carlson, Introduction to Communication Systems, 2nd Ed., McGraw-Hill, NY ©1975, Chapter 2, equation 18a. Or see pretty much any other book on Fourier transforms.
In discrete time, with consecutive samples spaced by $\Delta t$, the unilateral PSD for white noise is $2\sigma^2\Delta t$. Small values of $\Delta t$ are usually desirable, since the sampling frequency is the reciprocal of $\Delta t$ and the Nyquist frequency is half the sampling frequency. But $\Delta t$ cannot go to zero, as Marcus commented.
What happens if you just delete the limit stuff, on the right hand side of equation 7.22, and see what happens in the textbook's subsequent matched filter treatment? For a matched filter, the maximum signal-to-noise ratio is simply twice the pulse energy divided by the unilateral (white noise) PSD. (Same reference as above, Chapter 4, equation 12a.) So it looks like $2\sigma^2\Delta t$ will be used in your textbook. Hope this helps a bit.
Ed V
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