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PASHKOV 8th December 2012 08:54 PM

Please explain the words meaning (a couple obscure sentences from docs)
 
Here I see the next text in LEAP documentation:

Ground plane measurements provide the nearest ideal to an anechoic measurement without gating thus allowing full range response to be obtained. Turning the box upside down permits the critical high frequency tweeter to be closer to the boundary.

Can someone make me understand it? Especially what is 'gating' and 'boundary' in the context, and what is the general meaning of this text.

Thanks in advance.

Speedskater 8th December 2012 09:49 PM

This is what I think it means:
By "ground plane" they place the speaker on Mother Earth or at least a very large flat surface.
By "gating" they mean a time window where the program starts accepting data after a certain time lapse then stops after another time period.

dmills 8th December 2012 09:57 PM

Take a speaker outside in a large open air car park.... The concrete is your boundary, now place the measurement microphone diaphram within a small fraction of a wavelength of this boundary and make your measurements.

Because the microphone is acoustically close to the boundary, reflections from this surface do not modify the frequency response, and the gain from the boundary effect is trivially calculated.

Gating is a time domain windowing tequnique used to eliminate the reflected energy when measuring a transducer in a reverberent environment, you measure only the energy contained in a short window after applying the stimulus thereby eliminating the reflected signal that arrives later. Most of the better network analysers can be set up to do gated measurements, and any good acoustics measuring package sould be trivially programmable to do them.

The limitation on gated measurements is that you still need an acoustic environment large enough that you can get the data you need before the reflections arrive, easy at high frequency (as long as driver Q is sane) rather harder at 30Hz when looking at the time domain behaviour of a high Q transducer.

Regards, Dan.

PASHKOV 9th December 2012 07:54 PM

Speedskater, dmills,

thanks for your answers. Now I understand it.


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