klieberschnitzel
Electrical
- Mar 10, 2010
- 6
I'm trying to justify my empirical finding with an intuitive and mathematical explanation.
I am receiving an LFM centered at 100kHz with fmin=95kHz, fmax=105kHz. My receiver is a matched filter with a pre-defined threshold at the output that decides there is a detection if the threshold is exceeded.
Before the matched filter, at my receiver is a BPF centered at 100kHz with variable BW. Initially, I thought to maximize the SNR going to the matched filter by minimizing the noise being received, so I made the BW of my BPF slightly larger than the BW of my LFM, i.e. 1.75 times the BW of my LFM or 17.5kHz. I then started performing False Alarm tests to determine what threshold I needed to achieve a false alarm rate of 1detect/day.
I then re-ran the false alarm tests with a BW of the BPF at 50kHz. The threshold for the 1 detect/day was substantially smaller.
My question is why? Intuitively I can explain it as the reduction of the BW of the BPF colors the noise to within the BW of what the matched filter is detecting.
But how do I explain it in terms of Kay's probability of detection equation:
Pd = Q(Q^-1(Pfa)-sqrt(d^2))
d^2 is the deflection coefficient
d^2 = intergral of the signal power spectrum divided by the PSD of the noise, i.e. the signal to noise ratio.
Thanks,
klieberschnitzel
I am receiving an LFM centered at 100kHz with fmin=95kHz, fmax=105kHz. My receiver is a matched filter with a pre-defined threshold at the output that decides there is a detection if the threshold is exceeded.
Before the matched filter, at my receiver is a BPF centered at 100kHz with variable BW. Initially, I thought to maximize the SNR going to the matched filter by minimizing the noise being received, so I made the BW of my BPF slightly larger than the BW of my LFM, i.e. 1.75 times the BW of my LFM or 17.5kHz. I then started performing False Alarm tests to determine what threshold I needed to achieve a false alarm rate of 1detect/day.
I then re-ran the false alarm tests with a BW of the BPF at 50kHz. The threshold for the 1 detect/day was substantially smaller.
My question is why? Intuitively I can explain it as the reduction of the BW of the BPF colors the noise to within the BW of what the matched filter is detecting.
But how do I explain it in terms of Kay's probability of detection equation:
Pd = Q(Q^-1(Pfa)-sqrt(d^2))
d^2 is the deflection coefficient
d^2 = intergral of the signal power spectrum divided by the PSD of the noise, i.e. the signal to noise ratio.
Thanks,
klieberschnitzel