I am in 7th grade, and recently conducted an experiment testing the dB levels different headphones gave off at a given distance from a microphone connected to a computer program spectrum analyzer (the microphone portraying the distance the headphones would be from the ear canal). My results were that the Canalphones peak dB level was -13.7 dB, the earbuds peak dB level was -20 dB, the Supra-Aural headphones peak dB level was -19.5, and the Circumaural headphones had a dB level of -23.5.
My problem is that when I started researching, many of the web pages used positive decibels in saying what was too loud and what decimel levels significantly damaged a person's ablility to hear. Is there any way to convert the negative decibels into positive decibels, or are there statistics on negative decibels too? Also, why are there positive and negative decibels?
I don't know if this will help, but on my frequency charts, the highest decibel level it displayed was 0 dB.
Thank you in advance.
Need help with positive and negative dB
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Re: Need help with positive and negative dB
Hi,
I think your questions will be answered by this Wikipedia entry:
http://en.wikipedia.org/wiki/Decibel
Best Regards,
Barrett Tomlinson
I think your questions will be answered by this Wikipedia entry:
http://en.wikipedia.org/wiki/Decibel
Best Regards,
Barrett Tomlinson
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Re: Need help with positive and negative dB
The hearing loss and pain data are typically published in dBm which uses a 0 dBm reference corresponding to 1 milliwatt of audio power measured in the ear canal.
The negative sign on a dB (logrithmic) comparison means that the mesurement was less than the reference where a positive sign means it was more than the reference. So the only way to convert a negative dB indication is to change the reference. For example -2 dBm would be +1 dBu.
Now back to what your test setup and measurement results and what they mean. Headphones are "transducers". They convert electrical signals (electrical waves) into acustic signals (sound waves). If all of the headphones you tested had the same impedance and you utilized the same electrical signal and signal strength to test with, about all you can say is that in your test setup, the Canalphones unit was 5.8 dB more efficient than Supra-Aural model, which was 0.5 dB more efficient than the earbuds model, which was 3.5 dB more efficient than the Circumaural model.
Unless you have a calibrated microphone and spectrum analyzer you don't know what a 0 dB reading from the spectrum analyzer really means in terms of actual power. Calibrating sound measurement equipment requires some calibrated transducers sound booths and knowledge of how to utilize them for calibration purposes.
The negative sign on a dB (logrithmic) comparison means that the mesurement was less than the reference where a positive sign means it was more than the reference. So the only way to convert a negative dB indication is to change the reference. For example -2 dBm would be +1 dBu.
Now back to what your test setup and measurement results and what they mean. Headphones are "transducers". They convert electrical signals (electrical waves) into acustic signals (sound waves). If all of the headphones you tested had the same impedance and you utilized the same electrical signal and signal strength to test with, about all you can say is that in your test setup, the Canalphones unit was 5.8 dB more efficient than Supra-Aural model, which was 0.5 dB more efficient than the earbuds model, which was 3.5 dB more efficient than the Circumaural model.
Unless you have a calibrated microphone and spectrum analyzer you don't know what a 0 dB reading from the spectrum analyzer really means in terms of actual power. Calibrating sound measurement equipment requires some calibrated transducers sound booths and knowledge of how to utilize them for calibration purposes.
-Craig
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Re: Need help with positive and negative dB
Thank you. So, I would have to change my reference point in order to make the comparison? Are the negative decibels dbU or dbM, or just dB?
Secondly, I came across the equation 20log10(A1/A2). A1 being the measurement, and A2 being the reference point. Could I use this to change my results?
Secondly, I came across the equation 20log10(A1/A2). A1 being the measurement, and A2 being the reference point. Could I use this to change my results?
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Re: Need help with positive and negative dB
You really need to look up the specifications on the spectrum analyzer and measurement circuitry used in your computer system to be able to answer part of this question. My guess is that if they were designed to work together and you didn't do any calibration step to establish what 0 dB meant, it probably is based on 0 dBm (e.g. a 1 milliwatt electrical signal). Without knowing how your microphone behaves at different frequencies, you can't make any assumptions on what the power of the electrical signal output is for the power of a given sound wave. Absent any additional information, what you have is just relative dB that allows you to compare different sound sources without any absolute calibration to scientific units.Are the negative decibels dbU or dbM, or just dB?
Maybe. The process of calibrating a test setup involves having a known. In this case, you need to have a known sound source that produces a known amount of sound power. Without having a calibrated sound source or a calibrated sound measuring device, you won't have a reference that is tracable to scientific units.Secondly, I came across the equation 20log10(A1/A2). A1 being the measurement, and A2 being the reference point. Could I use this to change my results?
-Craig