The decibel unit was originally defined by Bell Labs as a standard way to relate power losses in circuits and gain in amplifiers. It has since been expanded into many engineering branches, especially acoustics. A decibel relates the power or intensity of a physical quantity as a ratio to a reference level or to another quantity. The decibel is useful because a large range of values are managed with a small range of decibel numbers. These ratios can also be expressed as a percent in order to give an indication of the scale of change in power with a certain change in decibels.
Decibel measurements of different types are usually denoted with a suffix, to indicate the reference unit, or the scale being measured. For instance, dBu measures voltages compared to 0.775 Volts RMS. Other scales are:
dBA , a Sound Pressure measurement that is weighted for human ear sensitivity;
dBm or dBmW, the power relative to one milliwatt.
Amplifier gain usually has the input power as the reference voltage, and is usually noted as just dB, since there is no standardized reference in this case.
The computation of the decibel level depends on the type of physical quantity being measured. If you are measuring power levels, such as acoustic energy or light intensity, then decibel levels (LdB) are proportional to the logarithm (base 10) of the ratio of the power (P) to a reference level (Pref). The decibel in this case is defined as:
LdB = 10 log(P/Pref) : Note that the logarithm is multiplied by 10 for the answer in dB.
When measuring field amplitude such as sound or voltage levels, then the power is measured proportional to the square of the amplitude. So the decibel increase is then the logarithm of the ratio of the square of the amplitude (A) to the reference level (Aref). Most uses of decibel in everyday terms fall into this category.
Ldb = 10 log(A^2 / Aref^2)
Since log(A^2) = 2 log(A), this simplifies to:
Ldb = 20 log(A/Aref)
All decibel measurements must have a reference level. If sound pressure levels from a speaker are being measured, then the reference is usually the limit of human sound sensitivity, stated as a sound pressure level of 20 micro-pascals (0.02mPa). A sound with this level has a measurement of 0 dB. A sound with twice this level has a dB measurement of:
20 log ( 0.04 / 0.02 ) = 20 log 2 = 6.0 dB
If you are measuring sound intensity, that is all the power available from a sound source, including reflected and transmitted sound, then the dB increase is:
10 log (0.04 / 0.02) = 3.0 dB
This also the amount of power needed by the amplifier if the speakers have a linear response. An increase of power by a factor of 4 gives a 6 dB increase, an an increase by a factor of 10 gives a 10 dB increase.
Compute the percent increase from the dB power increase by first solving the decibel formula for the ratio of the powers.
L = 10 log ( P / Pref), L is measured in dB
L /10 = log (P / Pref)
P/Pref = 10^(L/10)
The percent change would then be (P-Pref) (100%) / Pref = 10^(L/10). If the value of P is very much larger than Pref, Then this simplifies to approximately:
percent change = 100% * 10^(L/10); with L in dB.
Compute the percent increase from the dB amplitude increase by first solving the decibel formula for the ratio of the powers.
L = 20 log ( A / Aref), L is measured in dB
L /20 = log (A / Aref)
A/Aref = 10^(L/20)
The percent change would then be (A-Aref) (100%) / Aref = 10^(L/20). Once again, as is typical, the value of A is very much larger than Aref, Then this simplifies to approximately:
percent change = 100% * 10^(L/20); with L in dB.
So a change in Voltage Amplitude of 6 dBu would be a change of:
100% * 10^(6/20) = 100% * 1.995 = 199.5%, usually written as 200%
A change in sound pressure of -3.0 dbA would be:
100% * 10^(-3/20) = 100% * 0.7079 = 70.8% decrease in sound pressure.
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