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## Comparison of signal detectability for single-phase and two-phase searches

The previous Section described optimal filtering searches in two cases - looking for:
• A signal of known phase, proportional to , and
• A signal of unknown phase, which is some linear combinations of and .
With the choice of filter normalizations made previously, the expected signal produced by a source would be the same for both searches, but the expected (noise) was higher in the two-phase case. One might wonder if this reduced SNR means that a two-phase search reduces ones ability to identify signals. The answer turns out to be not significantly".

The reason for this is that the distribution of signal values produced by detector noise alone in the single- and two-phase cases are quite different. In order to answer the question: what is the smallest signal detectable" we need to fix a false alarm rate. For a given time-duration of data, this is equivalent to fixing a false alarm probability. Let us assume that this probability has been fixed to be a small value , and compare the single- and two-phase searches.

In the single-phase case, in the absence of a source, the values of the signal () are Gaussian random variables with a mean-squared value of 1. Hence the threshold determined by the false alarm rate must be set so that there is probability of falling outside the range . This means that

 (6.15.88)

The solution to this equation is the threshold as a function of the false alarm probability:
 (6.15.89)

Thus, for example, to obtain a false alarm probability of we need to set a threshold . In this case, our minimum detectable signal has amplitude .

In the two-phase case, the probability distribution of the signal in the absence of a source is different, because in this case the signal () is described by the probability distribution of a random variable , where and and are independent random Gaussian variables with unit rms. Here, and are the real and imaginary parts of the signal ( in the absence of a source. Their probability distribution is:

 (6.15.90) (6.15.91)

In the final line, we have integrated over the irrelevant angular variable . So in the two-phase case, as before, the threshold value of the signal is set by requiring that the false alarm probability be :
 (6.15.92)

The solution here is that the threshold is . For example, to obtain a false alarm probability of we need to set a threshold . In this case, our minimum detectable (0-phase) signal has amplitude , which is only slightly higher than in the single-phase case.

It is not a coincidence that for a given false alarm rate, the amplitude of the minimum detectable signals are almost the same. Although the expected value of the single-phase signal in the absence of a source is smaller than the expected value of the two-phase in the absence of a source, the tails of the two probability distributions are almost identical. For the same false alarm probability the thresholds in the two instances are related by

 (6.15.93)

But for thresholds of reasonable size (small ) both integrals are dominated by the region just to the right of , and in this neighborhood the integrands differ by a small factor of approximately . Since varies exponentially with the threshold, there is a logarithmically small difference between the thresholds and .

For a fixed false alarm probability, we can write the the two-phase threshold as a function of the one-phase threshold :

 (6.15.94)

The plot of this relationship in Figure shows clearly that once the thresholds are reasonably large, they are very nearly equal.

Next: Function: correlate() Up: GRASP Routines: Gravitational Radiation Previous: Wiener (optimal) filtering   Contents
Bruce Allen 2000-11-19