Rate my filtering

I'm trying to get my understanding in order. The monopoles have the LP and HP LR4 applied, so are –6dB at crossover, and sum to 0dB. The power response at crossover is –3dB by virtue of the magnitude responses both being –6dB (i.e., Power = sqrt(0.5^2+0.5^2) = 0.7071 = –3dB). If I'm getting it correct, the directivity is referenced back to the power computed on the basis of the "on-axis" SPL, which is 0dB. The two unfiltered monopoles each have a directivity index DI = 0dB (as they are omnidirectional). So, at the crossover frequency, do we have DI = 0dB or DI = −3dB? I can't see which answer is correct...
 
Exactly. I would use the listening axis, whatever that is chosen to be.
So, in my admittedly simplified theoretical example, the combined source comprised of the two coincident monopoles is itself omnidirectional. Thus, any axis is the same as any other. Hence, the DI = 10log10((Pressureon-axis)^2/Powertotal) = 10log10((1^2/(0.5^2+0.5^2) = 10log10(1/0.5) = +3.0dB. Does that look correct?
 
Last edited:
This is one of those times it can help to take a step back because simply put, directivity index is the power response normalised to an axis. Effectively it is the direct level compared to the reverberent.

I could suggest areas of doubt by saying the power remains the same while DI and response changes with reference angle, or that the response pressure has no directivity.

When you see it from the point of view that the power is averaged over nominal omnidirectionality anyway, in other words the power is counted without even considering where it came from, it's actually possible to end up with a negative DI.

Edit.. This was written before your post edit. I'd have to dig out my text books to verify what you have shown, but it looks right.
 
Last edited:
  • Like
Reactions: witwald
Directivity index is the 10log function of the sound intensity on a certain point at a fixed distance (usually far field) divided by the averaged 4π sound intensity of the source at that same distance. So yes, you can find negative Di’s. Certainly ‘off axis’…