On Mon, Apr 01, 2024 at 01:50:32PM +1300, Al Grant wrote: > I have a block of code in my wildlife tracker that detects high/low beeps > in a frequency.
It's not clear what the signal (blue trace in your png) actually represents and what exactly you are trying to do. - Detect any signal that exceeds the noise level by some margin ? In that case S/N ratio makes sense. - Detect a signal at a particular frequency ? In that case S/N isn't relevant, S/No (signal to noise density ratio) is the relevant metric. So until you provide a bit more detail, just a few comments. I assume the signal (blue trace) is the absolute value or square of something - it must be nonnegative for what you do to make sense. > samples = signal.convolve(samples, [1]*189, 'same')/189 So you are convolving with a rectangular pulse in the time domain, which becomes a sinc in the frequency domain. This is not a very good choice. If you are convolving anyway (there may be simpler solutions), better use a shape that transforms to a good lowpass filter. This would be a truncated or windowed sinc in the time domain. > 189 is the number of expected samples What do you mean by 'expected' ? Why 189 ? > 1. Beep length calculations are off because of the extended > length of "high samples" Look at the points in your plot where the orange line intersects the blue pulse. This happens at half the peak value of the orange trace. A threshold at that level will give you a good estimate of the pulse lenght. > 2. SNR calculations are considerably higher (and I am not sure > which SNR is "correct") Which may be because S/N isn't the relevant parameter (see above). You need to provide more details about what comes before (how you get the blue signal). Ciao, -- FA
