The decibel unit does not merit the extensive bewilderment which it
attracts. Electrical engineers who need to understand what a
*decibel* is, and how to use it, need not panic - it really is fully
comprehensible.

- What's in a name? - how the decibel got its.
- The oddball bel
- Two power levels, one unit
- Re. your P
_{r} - Examples 1: simultaneous measurement
- Examples 2: measurement at a point
- Examples 3: absolute references
- Dealing with voltages: Alex and Jim agree
- The rationale Very interesting, but ...
- Why the deciBewilderment?

See also ...

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The name decibel honours the inventor of the telephone: Alexander Graham
Bell, in recognition of his work in acoustics. *Sound intensity*
levels are frequently specified using decibels. Quasimodo's famous exclamation
"The bels! The bels!" was, in fact, a reference to the sound intensity.

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As a unit of measurement the decibel is peculiar in two ways. Firstly the
'deci' prefix (meaning 0.1) is, unlike its smaller relations milli and
micro, almost unused, except when a volume is given in 'cubic decimetres'.
Just be clear that if you have n decibels (dB) that means you have n/10
*bels* (B).

Secondly, if the metre is a unit of length, the kelvin is a unit of temperature and the henry is a unit of inductance then what is the decibel a a unit of? For the purposes of exposition we'll invent a quantity called lorp (logarithm of a ratio of powers) and give it a symbol ð.

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Lorp, the quantity which decibels express, is calculated like this:

Equation DBO

where P_{c} is a *compared power level* in watts and P_{r} is a
*reference power level*, also in watts. So lorp is the logarithm
of a ratio, and we can use it to compare power levels. An understanding
of this equation is essential; the rest is detail. So, for example, if
you wish to compare a power of 42 watts with a reference of 3 watts then

ð = 10 log_{10}(42/3) = 11.5 dB

Equation DBA

Decibels are so easy they're boring.

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What is the reference power level? The answer is 'any level which is both
convenient and made clear to everyone'. A particular ref level will
usually fall into one of the three types in the list below. If this
confuses you then try clicking the asterisks to get an example for each
type. You often know what P_{r} is by the context in which the lorp is specified, but sometimes the units are changed to
'dBx' where the x is some letter or symbol indicating what P_{r} is.

- * A power in your circuit at some point
which is separate from
the point at which P
_{c}is measured but which is present at the same time as P_{c}. - * A power in your circuit at the same
point at which P
_{c}is measured but which is present at a different time or under different circuit conditions or with a different measurement method. - * A value of power which is fixed in absolute terms.

You are here comparing power in your circuit at some point which is
separate from the point at which P_{c} is measured but which is
present at the same time as P_{c}. Suppose that you have an
amplifier into which you feed 3 mW and which delivers 42 mW from its
output. Then P_{r} = 3 mW and P_{c} = 42 mW and you find
the amplifier's lorp , its gain in decibels, like this:

ð = 10
log_{10}(42/3) = 11.5 dB

Equation DBB

If you have a '5 dB attenuator' then this means that:

10 log_{10}(P_{input}/P_{output}) = 5 dB

Equation DBC

Knowing that

log(a/b) = log(a) - log(b)

Equation DBC

you can transpose a and b to prove

log(b/a) = -log(a/b)

Equation DBD

so that, for the same attenuator, if we treat it as the amplifier
and put P_{r} = P_{input}

10 log_{10}(P_{output}/P_{input}) = -5 dB

Equation DBE

In other words its 'gain' is -5 dB.

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You are comparing a power in your circuit at the same point at which
P_{c} is measured but which is present at a different time or under
different circuit conditions or with a different measurement method.

Suppose that you are designing a radio communications system and need to
select an antenna for the transmitter. If your receiver gives an
indication of received power then you can note what that is for one type of
antenna driven by the transmitter and call that P_{r}. You switch
to a second antenna and call the received power P_{c}. You can
then say that antenna 2 has a gain *over* antenna 1 of

Equation DBF

If the first antenna happened to be a dipole then you would say
that "the gain of antenna 2 is ð_{2} dBd";
the suffix d indicating that the comparison is with a dipole.
If the gain is given as 'dBi' then the comparisson is with
an antenna that radiates energy equally in all directions (an
'isotropic' antenna).

Later in the design of your communications system you become interested
in the purity of the signal. You use a spectrum analyzer to look at the
received power at the fundamental frequency of the transmitter and call
that P_{r}. Next you look at the power at twice this frequency,
the 'second harmonic', and call that P_{c}. The lorp in this case is denoted by dBc, the suffix
indicating that you have taken the power of the fundamental or
*carrier* frequency as your reference.

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Here P_{r} is established by convention.

In acoustics, for sound intensity levels P_{r} is sometimes
10^{-2} W m^{-2}. This is about as loud as a pneumatic
drill two metres away. More usually, P_{r} is 10^{-12} W
m^{-2}, which makes the drill 100 decibels. On this scale
0dB corresponds approximately to a whispered conversation. This
illustrates how wide the range over which our hearing works and why
a logarithmic scale is useful.

In electronics, dBm uses 1 milliwatt for P_{r} and
dBW uses 1 watt. For example, if in a 50 ohm system you know a signal
is ð dBm then you convert to voltage using -

V_{rms} = √0.05 × 10^{(ð/20)}

Equation DBG

and back the other way with -

ð = 10 log(20 V_{rms}^{2})

Equation DBH

So -12 dBm is equivalent to 0.0562 volts rms. Clearly, with all absolute
references for P_{r} you can deduce the absolute value of
P_{c} for any value of ð

P_{c} = P_{r} × 10^{(ð /10)} watts

Equation DBI

Audio engineers like to use dBu, where 0 dBu is 0.775 volts RMS (1 mW
into 600 ohms, see the next section).

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As it is often easier to make voltage measurements than power measurements you may wonder if lorp can be calculated from the former.

V_{c}/V_{r} | P_{c}/P_{r} | lorp (dB) |
---|---|---|

1000 | 1e6 | 60 |

500 | 2.5e5 | 53.98 |

316.2 | 1e5 | 50 |

200 | 4e4 | 46.02 |

100 | 1e4 | 40 |

50 | 2500 | 33.98 |

31.62 | 1000 | 30 |

20 | 400 | 26.02 |

10 | 100 | 20 |

5 | 25 | 13.98 |

3.162 | 10 | 10 |

2 | 4 | 6.021 |

1.414 | 2 | 3.010 |

1 | 1 | 0 |

0.7071 | 0.5 | -3.010 |

0.5 | 0.25 | -6.021 |

0.3162 | 0.1 | -10 |

0.2 | 0.04 | -13.98 |

0.1 | 0.01 | -20 |

0.05 | 2.5e-3 | -26.02 |

3.162e-2 | 1e-3 | -30 |

2e-2 | 4e-4 | -33.98 |

1e-2 | 1e-4 | -40 |

5e-3 | 2.5e-5 | -46.02 |

3.162e-3 | 1e-5 | -50 |

2e-3 | 4e-6 | -53.98 |

1e-3 | 1e-6 | -60 |

The answer is yes - provided that you are careful about your load resistance. Power is related to voltage and resistance by

P = V^{2}/R watts

Equation DBJ

We can use this to convert the standard equation for lorp into

For decibel measurements at a point R_{c} will usually be
equal to R_{r} (typically 50 ohms). We can then cancel the impedance
from the equation

ð =
10 log_{10}(V_{c}^{2} / V_{r}^{2})

ð =
10 log_{10}((V_{c} / V_{r})^{2})

ð =
10 * 2 log_{10}(V_{c} / V_{r})

Equation DBL

Remember, though, that this equation is only only applicable when the
impedance fed by
V_{r} is the same as for V_{c}.

Look at the decibel equivalents for a power change of 2.0 and for 0.5.
If you double or halve the power then this gives very close to 3 dBs.
This is only coincidence, but it explains the frequent references you
see to 'the -3 dB point': it really means half the power.

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OK, so now you know how to use decibels, but perhaps you don't see the point of doing so. After all, if you can do nothing with them except express a ratio then why not say "my output power is 14 times my input power" rather than "my amplifier gives 11.5 dB". Well, clearly it's quicker to do the latter. It's also convenient to label graph scales in dB. In Figure DSC you might not agree that the decibel scale is more informative but it surely gives a different slant (ha ha) to the response.

Mostly, though, it is the logarithmic nature of the unit that has ensured it's survival. Communication systems, as with acoustics, tend to operate over a large dynamic range and decibels cope with that. Acoustically, the human ear perceives sound levels in a logarithmic fashion; so you 'feel' that an 18 dBu sound is about three times as loud as a 6 dBu one, even though the power ratio is nearly 16 times.

The use of decibels is also partly for historical reasons. When a
communications engineer had, let's say, an 11.5 dB amplifier followed by
a 5 db attenuator he could quickly calculate that the overall gain was
11.5 - 5 = 6.5 db. This works because adding logarithms of ratios
corresponds to multiplying the straight ratios: 14 * 0.316 = 4.42 (this
was the rationale behind 'log tables'). With a hand held calculator
it's as quick to do the multiplication anyway.

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If you can understand logarithms then you can understand decibels. So why do they cause confusion? One problem is that statements like "The output power is 15dBm" are very common. The trouble with this is that power is measured in watts, which have dimensions of joules per second, whereas decibels are dimensionless. Watts are not really decibels of any variety. Life is too short, however, to be pedantic by saying "Ten times the logarithm of the ratio of the output power to 1mW is 15dB". Just make a mental note of the pitfall, say that "The output is 15dBm", and get on with the job.

More damaging is equation DBL. Engineers having a faint knowledge of this eventually start asking questions like "Does a '5dB attenuator' attenuate the voltage by 5dB or the power by 5dB?" The best answer to that is probably "Yes". If you have gone through all of this page then you will understand that such an attenuator produces a lorp of -5dB. If you want to know the straight ratio of the voltage after the attenuator to that before it then set ð = -5 and use

V_{output}/V_{input} = 10^{(ð /20)} = 0.5623

Equation DBM

and if you want the straight power ratio use

P_{output}/P_{input} = 10^{(ð /10)} = 0.3162

Equation DBN

The final way to go wrong is to attempt to use type 1
reference levels when the circuit impedance is not the same for both
V_{c} and V_{r}. Rather, in fact, as
with figure DSC. This page goes no further into such
murky and uncharted waters.

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E-mail:
R.Clarke@surrey.ac.uk

Last revised: 2016 March 10^{th}.