So guys i am trying to make a sound sensing meter which gives me output in decibels but the problem is can't figure out which sensor to buy any help would be welcome.
Yeah but the one i got is giving readings in volts and there is nothing given in its datasheet about conversion to dB. Any idea exactly which one will have that available?
So i get one off the shelf and then calibrate it to different readings? but what d use as a base line how do i get that exact reading? Like if my baseline is 80 dB how do i get that 80 dB precise sound?
The decibel is a ratio of two measurements,
Generally with electronics or sound the decibel is a ratio of two power measurements,
Google it.
I suspect you will need a 'proper' decibel meter to calibrate it.
Thanks man. So what do i do get the voltage reading and just graph it? All I am suppose to do is analyze some data? and can you help me with another how to make live graphs from reading i get off an arduino.
clueless1994:
Thanks man. So what do i do get the voltage reading and just graph it? All I am suppose to do is analyze some data? and can you help me with another how to make live graphs from reading i get off an arduino.
Each microphone and it's amplifier will have different characteristics.
Graphing the output is an excellent idea.
You will need some method of calibrating the graph results though.
Live graphs from arduino output,- you can make something like a spreadsheet output or csv delimited file from that.
There is software that can generate graphs from that , but you may have to pay for it.
Do you have a specification for the microphone sensitivity - ?
0dBA iis -94dB from 1 Pascal. ( 1 newton/square metre)
Mike sensitivity can be specified eg in dB referenced to 0dbA==1volt
Here are a couple of exam - like questions!
ezample : suppose we have a mike whose sensitivity is quoted as - 43dB. What output in dbV would it give when exposed to a sound pressure level of 75dBA ? Express that in volts.
75dbA above -94dBa is - 19dbA and the mike's output would be -19 + -43 or -62dBV
the voltage is therefore 10^(-62/20) = 7.94 x 10^4 volts.
A milke can also be specified in mV/pascal.
ezample : suppose we have a mike whose sensitivity is quoted as 22dB mv/pascal . What output in dbV would it give when exposed to a sound pressure level of 75bBA ? Express that in volts.
as above the mike's output would be -19 + 22 = +3dB above 1 mV (-60dBV) ie -57dBV or
10(-57/20) = 1.41 x 10^-3 volts
Note that both these values are tiny compared to the 0-5v an arduino requires to do a meaningful a-d conversion, and you will need an amplifier of perhaps 60dB ( 1000x) gain to get it to a sensible level...........
Note also that professional meters have a filter with a specific frequency response eg 'A' weighting - which attempts to approximate the sensitivity of the ear - if you want to do this you'll have to do some filtering too....
All the above refers to any single frequency. The microphone will have a senstivity which various with frequency, and in practise this has to be taken into account as well.
if you need help with either of the above, please repost.
I designed pro audio for a while, and have dragged these sums out from ancient memories!
hope this helps.
regards
Allan
Quite accurate, but can be easier.
You have to know what output the microphone makes when having 94 dBSPL, which is exactly 1 Pa.
This is marked as sensitivity on the microphones datasheet.
Quick correction:
This is not the 0 dBA, but will be a reference point in our case.
Let's go through an example!
Let the microphone output 24 mV/Pa
This means that if something is 94 dBSPL loud, the microphone will output 24 mV.
Let's say you read the voltage at a lower sound level, and it writes you 0.12 mV.
Since we know that the microphone outputs 24 mV at 94 dBSPL, then how many dBSPL equals to the 0.12 mV?
It can be calculated by the two voltage like this:
24 mV/0.12 mV
This will give us the number 200, which means that the sound pressure is 200 times less than the 94 dBSPL.
But we would love to have dB's, so we have to convert that ratio to decibels, using this formula:
dB = 20*log (ratio)
This 200 ratio will give us 46 dB.
From now on, all we have to do is subtract the 46 dB (since it is 200 times lower volume) from the reference point 94 dB
94 dB - 46 dB = 48 dBSPL
You got what this volume is.
But be aware. Microphones hear better than us, they hear every frequency almost perfectly, not like us. If you want to measure how a human would hear, you have to apply on these readings a filter, which reduces the dBSPL even further, based on what frequency, AND loudness the sound is. Look up the word "weighted curves" in Google, you'll find the curve "A", mentioned above, but you'll also find curve "B" and "C", which has to be applied at different sound levels.
Source: I'm a sound engineer
A bit of history . I worked for Neve back in the 70's, and designed , among other things, the concept and detail design for the remote mike preamps on the Montserrat series of consoles - still in use!
In those days signal levels were always quoted in dBm.
I thought this was nonsense, as 0dBm is a power of 1mW. It would give 0.775V but only when terminated in 600 ohms .
But our console output stages had an impedance of about 50 ohms, the line input impedance was 10k ohms, and the mike inputs 1.2k ohms. Where's the 600?
In fact we needed a voltage standard rather than a power one - and so I wrote 0dBu = 0.775V rms on all the drawings I made for customers , and referenced everything to that.
The 'u' is for 'units' - whatever they might be.
The convention seems to have stuck.
regards
Allan
clueless1994:
So i get one off the shelf and then calibrate it to different readings? but what d use as a base line how do i get that exact reading? Like if my baseline is 80 dB how do i get that 80 dB precise sound?
This is the EXACT question many municipalities have asked after passing regulations that specify how loud certain sounds may be in a neighborhood, or how loud the exhaust note can be from an automobile. I think in all cases they have rescinded the rule because of the cost and time required to calibrate their sound measuring system. Like radar speed measurements, they must prove calibration before taking a measurement to court.
They lost too many cases to make the rule enforceable.
Paul