No!
Voltage and SPL are 20 log measures (so 1/2 is -6dB and 1/4 is -12dB)
Power, energy, Intensity are 10 log measures (so 1/2 is -3dB and 1/4 is -6dB)
thanks tarrogan and andrewT.
Ive just realised that db ratio isnt really necessary unless i need say 5db reduction, since i just need to halve it, the absolute ratio is all i need. Gosh i feel a fool...
Re reading my post i spotted my error, voltage is 20log. So, to get a halving of acoustic output, i need an absolute ratio of 0.5? Thats correct, isnt it?
Ive just realised that db ratio isnt really necessary unless i need say 5db reduction, since i just need to halve it, the absolute ratio is all i need. Gosh i feel a fool...
Re reading my post i spotted my error, voltage is 20log. So, to get a halving of acoustic output, i need an absolute ratio of 0.5? Thats correct, isnt it?
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Gents
dB is a power ratio normally with a reference to a given power.
Say we put 4mw into a cable (or a filter) and the output is 1 mw
? dB = 20 Log ( Power in / Power out )
? dB = 20 Log ( 1/4 )
? dB = -20 Log ( 4 )
? dB = -20 * 0.3
? dB = - 6
So our circuit loss is 6 dB
If we are measuring voltage, power = volts^2/R: volts squared divided by resistance.
?dB = 10 Log volts in^2/volts out^2
? dB = 20 Log volts in/volts out
Don
dB is a power ratio normally with a reference to a given power.
Say we put 4mw into a cable (or a filter) and the output is 1 mw
? dB = 20 Log ( Power in / Power out )
? dB = 20 Log ( 1/4 )
? dB = -20 Log ( 4 )
? dB = -20 * 0.3
? dB = - 6
So our circuit loss is 6 dB
If we are measuring voltage, power = volts^2/R: volts squared divided by resistance.
?dB = 10 Log volts in^2/volts out^2
? dB = 20 Log volts in/volts out
Don
I don't fully understand your post.
But there are some gross errors.
decibel ratio = 10*log[p1/p2]
decibel ratio = 20*log[v1/v2]
all the rest is gobbledegook except
this is correct
this one is also correct but why would you ever use it? It is far simpler to use 20times rather than 10times the log[x].
Oh, your first statement was almost right:
dB decibel is a ratio. don't add voltage or power or current or anything else. It is a ratio.
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Hi,
Very confused.
8R in series with 8R reduces power and acoustic output by 6dB.
It reduces the voltage level expressed as power and dB by 6dB.
Looking at purely voltage (not dB) the voltage level is halfed.
Power is V[squared]/R, 20log converts the ratio to power.
(doubllng a log squares the value).
You can't talk about pure voltage ratios in dB, like half = 3dB,
its just wrong ,10log V2/V1 does not = dB, and is not used.
rgds, sreten.
you cant talk of voltage ratios and express them in dB? Op amp gain is expressed in such a way, so i find that remark a little perplexing. The question was written whilst enebriated somewhat. I guess that i shouldve asked if an absolute voltage ratio will also result in an equal acoustic ratio. 3db is half power, 6db half amplitude, If i now recall correctly. I shouldve asked if spl is a amplitude or power type unit.
Hi,
All dB's are power - related amplitude ratios are 0.707 for -3dB, 0.5 for -6dB.
A x10 voltage gain op-amp stage has 20 (i.e. 100) dB gain.
You can't talk about x10 in dB, as its x100 in dB.
You can of course just talk about voltage gain, e.g. typically 20 to 30
for a power amplifier and it some respects its a lot simpler to most.
(Nobody talks about power amplifier gain in dB much .... )
Same with the gain of preamp stages, but here dB totally
dominates in expressing the gain, not voltage ratios.
rgds, sreten.
10log is used only for comparing power, e.g. 10W vs 1W = +10dB.
All dB's are power - related amplitude ratios are 0.707 for -3dB, 0.5 for -6dB.
A x10 voltage gain op-amp stage has 20 (i.e. 100) dB gain.
You can't talk about x10 in dB, as its x100 in dB.
You can of course just talk about voltage gain, e.g. typically 20 to 30
for a power amplifier and it some respects its a lot simpler to most.
(Nobody talks about power amplifier gain in dB much .... )
Same with the gain of preamp stages, but here dB totally
dominates in expressing the gain, not voltage ratios.
rgds, sreten.
10log is used only for comparing power, e.g. 10W vs 1W = +10dB.
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ah ha. I see where i was mistaken. Thanks.
So, to attenuate 3db, my potential divider of Resistor and driver should have an absolute ratio of root2/2? The op amp thing confuses me still, but nevermind. Reading Capel's book for a minute revealed where my 3db half power misconception originated. Its half power in the sense that 2 drivers crossing at -3db sum to 0db. Sometimes im so daft. Thanks again.
So, to attenuate 3db, my potential divider of Resistor and driver should have an absolute ratio of root2/2? The op amp thing confuses me still, but nevermind. Reading Capel's book for a minute revealed where my 3db half power misconception originated. Its half power in the sense that 2 drivers crossing at -3db sum to 0db. Sometimes im so daft. Thanks again.
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there are times in my life when i regret asking questions when im half cut. I just wanted to confirm that a voltage ratio would translate to the same acoustic ratio. I have to admit that to my knowledge a crossover does 'cross' at -6db, even though in my confusion i said the contrary. Im at a loss to explain where the -3db half power bit comes from.
3dB half power is not a misconception, it is true. if you have something with 100Watts (of anything, electrical or acoustical power or whatever you like) and you reduce it to 50Watts (of anything) you have 3dB less.
Thing is just that in order to get half the power into a given impedance, you need to reduce the voltage not to half its original value, but to 0.707 times its original value.
p (pressure), v (velocity), V (voltage)... all use 20log(out/in).
I (intensity), P (power)... all use 10log(out/in).
reason for this is simple, it's because the lower values are proportional to the second power of the upper values (P=V^2/R, I=p*v, P_ac=p*v*A...)
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