The matched filter for burst detection

• Let us consider a signal $s(t)$ in presence of noise $n(t)$. The available information is the sum
  
  $$x(t)=s(t)+n(t)$$
  
  
  $x(t)$ is the measurement at the output of the low noise amplifier and $n(t)$ is a random process with known properties. Let us start by applying to $x(t)$ a linear filter which must be such to maximize the signal to noise ratio SNR at a given time $t_o$ (we emphasize the fact that we search the signal at a given time $t_o$).

  Indicating with $w(t)$ the impulse response of the filter (to be determined) and with $y_s(t)=s(t)*w(t)$ and $y_n(t)=n(t)*w(t)$ respectively the convolutions of the signal and of the noise we have
  

  $$SNR=\frac{\vert y_s(t_o)\vert^2}{E[\vert y_n(t_o)\vert^2]}$$
  
  

  The expectation of the noise after the filter, indicating with $N(\omega)$ the power spectrum of the noise $n(t)$ is
  

  $$E[\vert y_n(t_o)\vert^2]=\frac{1}{2\pi}\int_{-\infty}^{\infty}N(\omega)\vert W(\omega)\vert ^2d\omega$$
  
  
  where $W(\omega)$ is the Fourier transform of the unknown $w(t)$.

  At $t=t_o$ the output due to $s(t)$ with Fourier transform $S(\omega)$ is given by
  

  $$y_s(t_o)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S(\omega)W(\omega) e^{j\omega t_o} d\omega$$
  
  

  We apply now the Schwartz' inequality to the integral and using the identity
  

  $$S(\omega)W(\omega)=\frac{S(\omega)}{\sqrt{N(\omega)}} W(\omega)\sqrt{N(\omega)}$$
  
  
  we obtain
  
  $$SNR \leq \frac{1}{2 \pi}\int_{-\infty}^{\infty} \frac{\vert S(\omega)\vert^2}{N(\omega)}df$$
  
  
  It can be shown that the equal sign holds if and only when
  
  $$W(\omega)=constant\frac{S(\omega)^*}{N(\omega)}e^{-j\omega t_o}$$
  
  
  Applying this optimum filter to the data we obtain the maximum SNR
  
  $$SNR =\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{\vert S(\omega)\vert^2}{N(\omega)}d\omega$$
  
  
  where $S(\omega)$ and $N(\omega)$ as already specified are, respectively, the Fourier transform of the signal and the power spectrum of the noise at the end of the electronic chain where the measurement $x(t)$ is taken.

  Let us apply the above result to the case of measurements $x(t)$ done at the end of a chain of two filters with transfer functions $W_a$ (representing the bar) and $W_e$ (representing the electronics).

  Be $S_{uu}$ the white spectrum of the brownian noise entering the bar and $S_{ee}$ the white spectrum of the electronics noise. The total noise power spectrum is
  

  $$N(\omega)=S_{uu}\vert W_a\vert^2\vert W_e\vert^2+S_{ee}\vert W_e\vert^2$$
  
  
  The Fourier transform of the signal is
  
  $$S(\omega)=S_g(\omega)W_a W_e.$$
  
  
  where $S_g$ is the Fourier transform of the GW signal at the bar entrance.

  The optimum filter will have the transfer function
  

  $$W(\omega)=\frac{S_g^* e^{-j\omega t_o}}{S_{uu}} \frac{1}{W_a W_e} \frac{1}{1+\frac{\Gamma}{\vert W_a\vert^2}}$$
  
  
  where
  
  $$\Gamma=\frac{S_{ee}}{S_{uu}}$$
  
  

  We obtain
  

  $$SNR=\frac{1}{2\pi S_{uu}}\int_{-\infty}^{\infty} \frac{\vert S_g(\omega)\vert^2 d\omega}{1+\frac{\Gamma}{\vert W_a\vert^2}}$$
  
  

  We apply now the above result to a delta GW with Fourier transform $S_g$ independent on $\omega$. The integral can be easily solved if we consider that near the resonance $\pm \omega_o$ we can approximate

  
  

  $$W_a=\frac{\beta_1}{\beta_1+i\omega}$$
  
  
  We estimate now signal and noise just after the first transfer function, before the electronic wide-band noise. For the signal due to a delta excitation we have
  
  $$V(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\beta_1}{\be... ...omega}S_g d\omega =S_g\beta_1e^{-\beta_1 t}=V_s e^{-\beta_1 t}$$
  
  
  where $V_s$ is the maximum signal. For the thermal noise we have
  
  $$V^2_{nb}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\beta_1^2} {\beta_1^2+\omega^2}S_{uu} d\omega=\frac{S_{uu}\beta_1}{2}$$
  
  
  where $V_{nb}^2=\frac{\alpha^2 k T_e}{m \omega_o^2}$ is the mean square narrow-band noise .

  Introducing the signal energy $E_s=\frac{1}{2}m\omega_o^2 (\frac{V_s}{\alpha})^2$ we calculate (with the variable $y=\frac{\omega}{\beta_1}$)
  

  $$SNR= \frac{S_g^2\beta_1}{2 \pi S_{uu}}\int_{-\infty}^{\infty... ..._s^2}{4V_{nb}^2 \sqrt{\Gamma}}= \frac{E_s}{2kT_e\sqrt{\Gamma}}$$
  
  

  The effective noise temperature is
  

  $$T_{eff}=2T_e\sqrt{\Gamma}$$