# Why do we use dB to represent the difference between two voltages?

For example, the passband of a LC resonant circuit is the differences of frequency at +3db and -3db.

Why do we prefer dB?

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the "3 dB point" is actually 10⋅log10(1/2) = -3.0102999566398... dB. It's chosen because 1/2 power is exactly where the asymptotes meet if you plot it on a log-log plot (I believe). –  endolith May 29 '11 at 18:16
dB doesn't represent the difference, but rather the ratio. It is another way to write percentage. For an attenuator, "power reduced to 50%" and "power reduced by 3dB" mean the same thing, but put two attenuators in series and 3dB + 3dB is easier to computer than 50% * 50%. –  markrages Jun 2 '11 at 22:33
Neper (Np) is pretty common too, especially in RF engineering. Neper's are like dB's, though based on ln(value) instead of 20.log(value). –  jippie Oct 18 '12 at 19:16

Many processes in nature are either of logarithmic nature (like human senses) or have a great dynamic range.

Describing them on a logarithmic scale and expressing differences in dB has several advantages:

• often the absolute difference doesn't matter, but the ratio (that's what dB is used for) does (e.g. signal-to-noise ratio)
• smaller numbers can be used
• there's an approximately linear relation between measurement and perceived sensation
• chained attenuations or amplifications can be expressed by addition instead of multiplication (easier to calculate in the head)

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dB is useful since it is a relative expression. +/-3dB is a doubling or halving of power.

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In many cases, voltage ratios are expressed in terms of dB rather than absolute numbers because there are many relationships which end up being linear when expressed in terms of dB. It is simpler, for example, to say that an N-stage low-pass filter will attenuate frequencies above the cutoff by $(6 \times N) \frac{dB}{octave}$ than it is to say that it will attenuate frequencies above the cutoff by a ratio of $({\frac{f_c}{f}})^N$.

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Right idea, but mixing two concepts. First, dB is a ratio relative to some reference, not an absolute value. Second, as you point out, it's a logarithmic representation of that ratio rather than a linear one. –  Chris Stratton May 30 '11 at 18:18
@Chris Stratton: By "absolute number" I didn't mean an absolute quantity, but rather a "bare" number without a dB suffix, as distinct from one with such a suffix. I should also have mentioned that it's easier to compare things that attenuate by 40, 50, 60, and 120dB than things which scale a signal by 0.01, 0.0033, and 0.001, and 0.000001. –  supercat May 31 '11 at 16:21
the word you want is 'linear' not 'absolute' –  Chris Stratton May 31 '11 at 16:31