The unit is the Bel. By convention, the Bel has not been seen as convenient,
so most applications use the metric system to divide it by 10 using the
designator deci. So the correct capitalization is "dB."
DB, db or Db are all incorrect. (It is kind of like MHz; lots of folks
miscapitalize that one, too. My favorite is mHz. If someone asks me how
long to make a 7 mHz antenna, I give them the correct answer -- and then
answer the question I think they intended to ask. (<smiley>)
As an aside, if one were to say that a signal were to increase by 100 dB, one could also say that it had increased by 10 Bels. Few do, and fewer yet would instantly understand it.
The term dB expresses a ratio. It is common to see a third letter after the dB, indicating some standard things that are used for a reference for that ratio. Common ones in RF electronics are dBm (dB relative to a milliwatt), dBW (watt), dBc (carrier), dBi (dB relative to an isotropic antenna), dBd (dB relative to a half-wave dipole in free space). Once the reference has been established, an actual value is being expressed, not a ratio.
As ratios, dB can be added and subrtacted. If you say that you have
a power of 1 watt and you amplify it by 10 dB, put it into an antenna with
10 dB' of gain and a
feedline loss of 3 dB, you can sum the gains and losses and determine
by how many dB you have increased that signal. In this case, + 10 dB +
10 dB - 3 dB, or a total of +17 dB. You can apply that addition to a dB
expression that is not a ratio, but only once. So, if we were to start
with 2 watts of power (3 dBW), we could do:
+ 3 dBW
+ 10 dB amplifier gain
+ 10 dBi
- 3 dB
----------
+20 dBW effective isotropically radiated power (EIRP). In this case,
one can mix the "W" and "i" because the forumula for EIRP includes power
and antenna gain. If the amplifier gain were not specified as +10 dB, but
only its output power, you could not add the initial 3 dBW and the amplifier
output power of 13 dBW, because the forumula for EIRP doesn't care about
the input and output power of the amplifier. It really only cares about
the output power fed to the feed line, so you could either use the input
power and amplifier gain, as shown above, or could replace the +3 dBW and
10 dB gain with +13 dBW, the output of the amplifier.
One cannot, as we saw earlier, add up the dB when dB is used directly in a power expression, such as dBW or dBm. To add those powers, one would have to convert the dBm or dBW to power, add those powers. If one wanted the power expressed with a decibel ratio, one could convert the result to dBm or dBW.
(A quick caution: many software programs, especially Basic, do NOT use
log10 for its log calculations, but rather log to the base e, a different
beast altogether.
Some Basics have both a loge and log10 function, but many do not. I
discovered a trick to use that will give log10 no matter what log the Basic
actually uses. If N is the number you want to manipulate, you can use log(N)
/ log(10) and, no matter what log the Basic program is using, you will
end up with log10 results.)
Okay, so after that digression, here are the basics for manipulating dB:
For power, use:
dB = 10 * log10(power ratio)
So, if power increases to 4 watts from 1 watts, one would have
dB = 10 * log10(4 / 1) = 6.02 dB
If the resistance is the same for two voltages, dB can be calculated as:
dB = 20 * log10(voltage ratio) or
dB = 10 * log10(voltage ratio ^ 2)
Note: ^ is the Basic designator for "raised to the power of."
Now, this would be a bit ugly in ascii, but even if the impedances are not the same, you can use a dB formula to compare the two levels, by squaring the voltages and dividing by the appropriate impedance.
For V1 and R1 and V2 and R2:
dB = 10 * log10 ((V1^2/R1) / (V2^2/R2))
It is even easier to convert the other way. If, for example, one has a power of 23 dBW, one can use the formula:
Power (watts) = 10 ^ (dBW/10)
In this case, the answer is 199.5 watts.
It is also common to see dB referenced to voltages, such as dBV or dBuV, etc. In that case, use the formula:
Voltage (units) = 10 ^ (dBunits / 20)
So, if a voltage were expressed as 12 dBuV, one could calculate
Voltage uV = 10 ^ (12 / 20 ) = 3.98 uV
It is early in the AM, so if I made any goofs in the above, please do
correct them. I would hate my
morning haze to give anyone bad information. :-)
73,
Ed, W1RFI