# why the deconvolved response and impulse response in soundcheck has unit of Pa/s?

I am recently try to figure out the time selective response algorithm of soundcheck. I found:

1. It uses a logarithmic sweep sine as a stimulus, x(t) for example, with unit of Volt.
2. The excitation signal is sent to the sound card to achieve D / A conversion, and then sent to the speaker through the power amplifier. While the speaker plays the sound, the microphone collects the sound to obtain the signal y (t), the unit is Pa. Assuming the sensitivity of the microphone is s0 V/Pa, we can convert the unit of y(t) to Volt( same as x(t)), for example y1(t)=y(t)[Pa]*s0[V/Pa].
3. According to Farina's paper "Advancements in impulse response measurements by sine sweeps", firstly I calculate the inverse filter f(t)=Am*fliplr(x(t)), where Am is an amplitude modulation. Then I perform the deconvolution:

deconv=conv(y(t),f(t)), it should be unit-less.

4. However the de-convolved response in soundcheck has a unit of Pa/s, and is much smaller than the results I calculated, even though they have exactly the same shape.

Here below are some pictures: can anybody give me some tips, thanks!

• Not having used soundcheck before, but having had quite a bit of experience with inverse convolution, I would say initially that the result could be due to the gain structure of the sensing microphone. Everything else looks as per expectations. – Mark Mar 6 at 11:16
• The thing that I do observe, which gives me cause for concern, is the considerable pre-ring in the resulting impulse response which should not be there. I suspect this is something to do with your filtering. I have not read Angelo's paper yet, so will do so, but I am unsure as to why there is a need to associate Amplitude Modulation with the filter and why you can't just proceed straight to a deconvolution. Will read the paper and hopefully will have a better understanding of what he is trying to do there. – Mark Mar 6 at 23:29