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Discussion: Theoretical signal-to-noise ratio for the stochastic background


In order to reliably detect a stochastic background of gravitational radiation, we will need to be able to say (with a certain level of confidence) that an observed positive mean value for the cross-correlation signal measurements is not the result of detector noise alone, but rather is the result of an incident stochastic background. This leads us natually to consider the signal-to-noise ratio, since the larger its value, the more confident we will be in saying that the observed mean value of our measurements is a valid estimate of the true mean value of the stochastic background signal. Thus, an interesting question to ask in regard to stochastic background searches is: ``What is the theroretically predicted signal-to-noise ratio after a total observation time $T$, for a given pair of detectors, and for a given strength of the stochastic background?'' In this section, we derive the mathematical equations that we need to answer this question. Numerical results will be calculated by example programs in Secs. [*] and [*].

To answer the above question, we will need to evaluate both the mean value

\mu:=\langle S\rangle
\end{displaymath} (11.19.253)

and the variance
\sigma^2:=\langle S^2\rangle-\langle S\rangle^2
\end{displaymath} (11.19.254)

of the stochastic background cross-correlation signal $S$. The signal-to-noise ratio SNR is then given by
{\rm SNR}:={\mu\over\sigma}\ .
\end{displaymath} (11.19.255)

As described in Sec. [*], if the magnitude of the noise intrinsic to the detectors is much larger than the magnitude of the signal due to the stochastic background, then

$\displaystyle \mu$ $\textstyle =$ $\displaystyle T\ {3H_0^2\over 20 \pi^2}\ \int_{-\infty}^\infty df\gamma(\vert f\vert)\vert f\vert^{-3}\Omega_{\rm gw}(\vert f\vert)\tilde Q(f)$ (11.19.256)
$\displaystyle \sigma^2$ $\textstyle \approx$ $\displaystyle {T\over 4}\int_{-\infty}^\infty df\ P_1(\vert f\vert) P_2(\vert f\vert)
\vert\tilde Q(f)\vert^2\ ,$ (11.19.257)

where $\tilde Q(f)$ is an arbitrary filter function. The choice
\tilde Q(f):=\lambda{\gamma(f)\Omega_{\rm gw}(f)\over f^3 P_1(f) P_2(f)}
\end{displaymath} (11.19.258)

maximizes the signal-to-noise ratio ([*]). It is the optimal filter for stochastic background searches. As also described in Sec. [*], if the stochastic background has a constant frequency spectrum

\Omega_{\rm gw}(f)=\left\{
\Omega_0 & \qua...
... f_{\rm high}\0 & \quad {\rm otherwise,}

it is convenient to choose the normalization constant $\lambda$ so that
\mu=\Omega_0\ T\ .
\end{displaymath} (11.19.259)

For such a $\lambda$,
\sigma^2\approx{T\over 2}\left({10\pi^2\over 3 H_0^2}\right)...
df\ {\gamma^2(f)\over f^6 P_1(f) P_2(f)}\right]^{-1}\ ,
\end{displaymath} (11.19.260)

which leads to the squared signal-to-noise ratio
({\rm SNR})^2=T\ \Omega_0^2\ {9 H_0^4\over 50\pi^4}
...w}}^{f_{\rm high}} df\ {\gamma^2(f)\over f^6 P_1(f) P_2(f)}\ .
\end{displaymath} (11.19.261)

This is equation (3.33) in Ref. [36].

But suppose that we do not assume that the noise intrinsic to the detectors is much larger in magnitude than that of the stochastic background. Then Eq. ([*]) for $\sigma^2$ needs to be modified to take into account the non-negligible contributions to the variance brought in by the stochastic background signal. (Equation ([*]) for $\mu$ is unaffected.) This change in $\sigma^2$ implies that Eq. ([*]) for $\tilde Q(f)$ is no longer optimal. But to simplify matters, we will leave $\tilde Q(f)$ as is. Although such a $\tilde Q(f)$ no longer maximizes the signal-to-noise ratio, it at least has the nice property that, for a stochastic background having a constant frequency spectrum, the normalization constant $\lambda$ can be chosen so that $\tilde Q(f)$ is independent of $\Omega_0$. The expression for the actual optimal filter function, on the other hand, would depend on $\Omega_0$.

So keeping Eq. ([*]) for $\tilde Q(f)$, let us consider a stochastic background having a constant frequency spectrum as described above. Then we can still choose $\lambda$ so that

\mu=\Omega_0\ T\ ,
\end{displaymath} (11.19.262)

(the same $\lambda$ as before works), but now
$\displaystyle {
\sigma^2={T\over 2}\left[
\int_{f_{\rm low}}^{f_{\rm high}} df\...
\int_{f_{\rm low}}^{f_{\rm high}}
df\ {\gamma^2(f)\over f^6 P_1(f) P_2(f)}}$
    $\displaystyle +\Omega_0\left({10\pi^2\over 3 H_0^2}\right)
\int_{f_{\rm low}}^{...
...\int_{f_{\rm low}}^{f_{\rm high}} df{\gamma^2(f)\over f^9 P_1(f) P^2_2(f)}\ $  
    $\displaystyle +\Omega_0^2
\int_{f_{\rm low}}^{f_{\rm high}} df{\gamma^2(f)\over f^{12} P^2_1(f) P^2_2(f)}
\left(1+\gamma^2(f)\right)\Bigg\}\ .$ (11.19.263)

The new squared signal-to-noise ratio is $\Omega_0^2\ T^2$ divided by the above expression for $\sigma^2$.

Note the three additional terms that contribute to the variance $\sigma^2$. Roughly speaking, they can be thought of as two ``signal+noise'' cross-terms and one ``pure signal'' variance term. These are the terms proportional to $\Omega_0$ and $\Omega_0^2$, respectively. When $\Omega_0$ is small, the above expression for $\sigma^2$ reduces to the pure noise variance term ([*]). This is what we expect to be the case in practice. But for the question that we posed at the beginning of the section, where no assumption is made about the relative strength of the stochastic background and detector noise signals, the more complicated expression ([*]) for $\sigma^2$ should be used. The function calculate_var(), which is defined in the following section, calculates the variance using this equation.

next up previous contents
Next: Function: calculate_var() Up: GRASP Routines: Stochastic background Previous: Example: optimal_filter program   Contents
Bruce Allen 2000-11-19