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Based on this post, let me assume that I have a input source of 94 dB SPL / 1 Pa that a microphone is picking sounds from.

The microphone has a sensitivity of -46dBV/Pa , this gives 0.005012 V RMS/ Pa.

Let us assume the ADC has unity preamp gain and no additional gain. Let us also assume an ideal case where there is no degradation in signal due to noise before ADC.

Now, I'm guessing the signal measured at ADC would be the same as input, which is 0.005012 Vrms.

20×log 0.005012/005012 = 0 dB

so the dB SPL will be (-46) + 0 = -46 + 94 = 48 dB SPL.

Why is the output not 94 dB SPL which is what I thought the output should be because we are inputting a 1 Pa sound?

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    \$\begingroup\$ The microphone converts the sound pressure level (SPL) signal into a voltage signal. So after that it doesn't make sense to talk about dB SPL. You could convert it back into SPL using an amplifier and a speaker if you want. \$\endgroup\$ – mkeith Feb 19 at 10:13
  • \$\begingroup\$ @mkeith I'm clear now with the explanation that 'justme' provided. But just to take you up on your answer,even if I do not have an amplifier, can i not work the dB SPL value with the measured voltage referenced to the input voltage ? \$\endgroup\$ – whoknowsmerida Feb 20 at 1:24
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You can't add different units such as dB(V/Pa), dB(1/1) and dB(SPL) together and expect a result in dB(SPL).

You are correct that 94 dB(SPL) is 1 Pa. Feeding that 1 Pa to microphone with sensitivity of -46 dB (V/Pa) does give you -46 dB (V) or 5 millivolts (0.005V). But then I don' know what you are expecting to happen next.

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  • \$\begingroup\$ I am aware that I can't add different units such as dB(V/Pa), dB(1/1) and dB(SPL) together. But if you read the post linked in my OP, with the resultant dB V value, you can obtain/convert the Voltage to dB SPL. I'll copy paste a snippet of that post below " If we do an example calculation for the measurement of 2.5v (assuming a unity gain for the amplifier) we get 20×log 2.5/0.005012=53.96dB so the SPL will be (-46) + 53.96 = 7.95 + 94 = 101.95 Db SPL " Using the same logic, with the 0.005 mV obtained, i should be able to convert that into a dB SPL value, shouldn't I? \$\endgroup\$ – whoknowsmerida Feb 19 at 7:10
  • \$\begingroup\$ No that is wrong. If calculated with 2.5V, that is 500 times the 0.005 Volts, which is +54dB more. So -46dB(V) + 54dB is 8dB(V) of mic voltage which does equal 2.5V. And since voltage is 500 times, also means there are 500 times the pressure of 1 Pa as well. 94dB(SPL) + 54dB = 148dB(SPL) which does equal 500 Pa. Therefore, as 0.005V referenced to 0.005V is 0dB, mic voltage is still -46dB referenced to 1V with 1 Pa signal, and 1 Pa signal is 94dB(SPL). \$\endgroup\$ – Justme Feb 19 at 10:16
  • \$\begingroup\$ I'm clear now. But to clarify, is the Math in that earlier referenced post wrong ? \$\endgroup\$ – whoknowsmerida Feb 20 at 1:17

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