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Natural Science Forum / Physics / Acoustics / September 2006novice: loudness of a wav
Hi, If i multiply each sample in my wav file by 0.1, will my wav file sound half as loud? Thanks, Aine >Hi, > [quoted text clipped - 4 lines] > >Aine No, you need to use something more like 0.3 for that (roughly 10dB, 0.1 would give you 20dB) d  SignaturePearce Consulting http://www.pearce.uk.com > Hi, > [quoted text clipped - 4 lines] > > Aine Nope, it's a power thing, not a voltage level thing, and multiplying each sample's level by .1 would alter the final voltage level of the amp driving the speakers. You'd need to calculate for -10dB power levels. In any event the 10dB thing is both subjective and relative and varies by the frequency of the tone as well as the initial loudness. In some cases a difference of 6dB will result in an apparent doubling or halving of the perceived loudness. > Nope, it's a power thing, not a voltage level thing, and multiplying each > sample's level by .1 would alter the final voltage level of the amp driving > the speakers. You'd need to calculate for -10dB power levels. Nonsense! -10dB is -10dB, period, and you are obviously clueless about the subject matter about which you pretend to be knowledgeable. Perhaps you should stick to posting your drivel alt.music.home-studio where you can more easily foist your ignorance as knowledge, becasue it won't fly here. >> Nope, it's a power thing, not a voltage level thing, and multiplying each >>sample's level by .1 would alter the final voltage level of the amp driving [quoted text clipped - 5 lines] > where you can more easily foist your ignorance as knowledge, becasue it > won't fly here. Not clear at all. When multiplying a wave file, I interpret that the sound pressure level created by the sound card varies directly as the wav amplitude. If it's power, nobody here ever declared it to be. In the predominance of such occasions, the amplitude expressed applies to the voltage or the current applied to the PC loudspeaker the user's earphone. In those cases, the sound pressure created is in proportion to that voltage or current. In that case, "multiplication by 0.1" causes the amplitude to be diminished to 1/10th and hence the sound power emitted to be 1/100dth, which is -20 dB, making the sound to appear to be 1/4 (half of half). BTW, does you byline mean that you are purely imaginary? Angelo Campanella > >> Nope, it's a power thing, not a voltage level thing, and multiplying each > >>sample's level by .1 would alter the final voltage level of the amp driving [quoted text clipped - 20 lines] > diminished to 1/10th and hence the sound power emitted to be 1/100dth, > which is -20 dB, making the sound to appear to be 1/4 (half of half). I agree with you. My point was that half loudness at mid frequencies corresponds approximately to a decrease of 10dB in the signal, and that it's a 10dB decrease irrespective of whether you are talking about voltage, current or power. If you are talking about either voltage or current, it's a decrease in a factor of 1/sart(10). If you are talking about power, it's a decrease by a factor of 10. > BTW, does you byline mean that you are purely imaginary? It simply reflects the fact that I am in transition from being a disembodied spirit, to being purely imaginary to being totally anonymous. > Not clear at all. > [quoted text clipped - 14 lines] > > Angelo Campanella Since a speaker will be reproducing the sound in the wave file,and a power amp will be driving the speaker, my understanding is that in order to provide a 6 dB increase in SPL, it will require a 6 dB increase in amplifier power, is this not correct? Since P= E2/R or I2R, a 6dB increase in voltage or current would result in a 16 fold increase in amplifier power. The case of an earphone is considerably different from the case on a speaker in open air, since the earphone is firing into a very small totally enclosed space and the speaker is firinf into a large open space. > > Not clear at all. > > [quoted text clipped - 23 lines] > speaker in open air, since the earphone is firing into a very small totally > enclosed space and the speaker is firinf into a large open space. You are undisputedly "The Ultimate Techncial Ignoramus." Every time you open your mouth (post) you reveal even more of the unbouned depth of your technical ignorance. > Since a speaker will be reproducing the sound in the wave file,and a power > amp will be driving the speaker, my understanding is that in order to [quoted text clipped - 4 lines] > speaker in open air, since the earphone is firing into a very small totally > enclosed space and the speaker is firinf into a large open space. Somebody give Mike a crash course... better make that a comprehensive course... on decibel math. Angelo Campanella |
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