Thursday, December 6, 2012

Decibel scale

Reference : https://ccrma.stanford.edu/~jos/st/Properties_DB_Scales.html
https://ccrma.stanford.edu/~jos/st/Decibels.html 

A decibel (abbreviated dB) is defined as one tenth of a bel.

The bel is an amplitude unit defined for sound as the log (base 10) of the intensity relative to some reference intensity, i.e., 

\begin{displaymath}
\mbox{Amplitude\_in\_bels} = \log_{10}\left(\frac{\mbox{Signal\_Intensity}}{\mbox{Reference\_Intensity}}\right)
\end{displaymath}     


The choice of reference intensity (or power) defines the particular choice of dB scale. Signal intensity, power, and energy are always proportional to the square of the signal amplitude. Thus, we can always translate these energy-related measures into squared amplitude:

\begin{displaymath}
\mbox{Amplitude\_in\_bels} =
\log_{10}\left(\frac{\mbox{Amp...
...ft\vert\mbox{Amplitude}_{\mbox{\small ref}}\right\vert}\right)
\end{displaymath}  

Since there are 10 decibels to a bel, we also have 
\begin{eqnarray*}
\mbox{Amplitude}_{\mbox{\small dB}} &=&
20\log_{10}\left(\fra...
...t(\frac{\mbox{Energy}}{\mbox{Energy}_{\mbox{\small ref}}}\right)
\end{eqnarray*} 
In every kind of dB, a factor of 10 in amplitude increase corresponds to a 20 dB boost (increase by 20 dB):

$\displaystyle 20\log_{10}\left(\frac{10 \cdot A}{A_{\mbox{\small ref}}}\right)
...
...)}_{\mbox{$20$\ dB}} + 20\log_{10}\left(\frac{A}{A_{\mbox{\small ref}}}\right)
$

A function $ f(x)$ which is proportional to $ 1/x$ is said to ``fall off'' (or ``roll off'') at the rate of $ 20$ dB per decade. That is, for every factor of $ 10$ in $ x$ (every ``decade''), the amplitude drops $ 20$ dB.   


Similarly, a factor of 2 in amplitude gain corresponds to a 6 dB boost:
$\displaystyle 20\log_{10}\left(\frac{2 \cdot A}{A_{\mbox{\small ref}}}\right)
=...
...2)}_{\mbox{$6$\ dB}}
+ 20\log_{10}\left(\frac{A}{A_{\mbox{\small ref}}}\right)
$
and

$\displaystyle 20\log_{10}(2) = 6.0205999\ldots \approx 6 \;$   dB$\displaystyle . \protect$ 
A function $ f(x)$ which is proportional to $ 1/x$ is said to fall off $ 6$ dB per octave. That is, for every factor of $ 2$ in $ x$ (every ``octave''), the amplitude drops close to $ 6$ dB. Thus, 6 dB per octave is the same thing as 20 dB per decade.  


A doubling of power corresponds to a 3 dB boost:
$\displaystyle 10\log_{10}\left(\frac{2 \cdot A^2}{A^2_{\mbox{\small ref}}}\righ...
...{\mbox{$3$\ dB}}
+ 10\log_{10}\left(\frac{A^2}{A^2_{\mbox{\small ref}}}\right)
$
and 

$\displaystyle 10\log_{10}(2) = 3.010\ldots \approx 3\;$dB$\displaystyle . \protect$ 


No comments:

Post a Comment