Logarithms and Decibels
The Decibel
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Named for Alexander Graham Bell.
Originally used to measure power losses in
telephone lines.
A Bel is the common log of the ratio of two
power levels.
A decibel is one-tenth of a bel.
A Bel is not a unit of anything – but simply a
logarithmic ratio of two power levels.
Definition of a logarithm
A logarithm is an exponent, but is stated
differently.
 When we write: 53=125
 5 is the base, 3 is the exponent.
 This is exponential form.
 Logarithmic form is when we say: The log
(to the base 5) of 125 is 3.
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Notation
log form: log636 =2
 exponential form: 62=36
 Any expression in logarithmic form may
be indicated in exponential form.
 if any number b to the power of x = N:
bx=N, then the logarithm of N to the base
b = x. This is logarithmic form.
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Base 10
In computation the base 10 is used for
logarithms.
 It is so convenient and common that it is
not usually written as a subscript but is
understood if no base is shown.
 This is similar to scientific notation where
very large or small numbers are
expressed as X 10 to an exponent value.
 For example 93000000 = 93 x106
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Logarithms = Convenience
Comparing sounds at the threshold of
hearing to sounds at the threshold of pain
represents over a million fold difference in
pressure levels.
 The dB as a logarithmic measure of ratios
fits well with our perceived loudness of
sound intensity.
 Logarithms are used to help us condense
the huge range of SPL humans perceive
into a manageable scale.
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A Comparison of SPL’s
A 1 db change in level is barely noticeable
A 3db increase doubles MEASURED
power level, but is not perceived that way.
Range of Human Hearing
Formula for dB SPL measurements
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The formula for SPL:
20 log (p/p0)
P0 is the reference of 20 micropascals
(threshold of hearing).
P = the pressure level of the sound we are
comparing to the reference level.
Calculating dB SPL differences
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Any sound pressure can be expressed as dB
SPL by comparing the sound pressure to the 0
dB (threshold of hearing) reference point with
the 20 log formula.
For example, how many dB SPL is a sound that
is 50 μ Pa?
dB SPL = 20 log (50μPa/20μPa)
dB SPL = 20 log (2.5 Pa)
dB SPL = 20 (.39794)
dB SPL = 7.96
Using the Log formulas
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First, divide the numbers in parentheses.
Next, find the log of the result.
Multiply that number by 20.
We can express the difference between any 2
pressure levels as dB using the 20 log formula.
Using dBs to compare two SPL
levels
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Find the dB difference between 1000μPa and
100μPa.
dB = 20 log (1000μPa/100μPa)
dB = 20 log (10Pa)
dB = 20 (1)
dB difference = 20
dB SPL As a Function of
Distance
SPL changes with the square of distance,
meaning that....
 Doubling the distance results in a drop of
6 dB SPL.
 Halving the distance results in a 6 dB SPL
increase.
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dB PWL
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PWL or Lw (sound power level) is the total sound
power emitted by a source in all directions.
Like electrical power, PWL is measured in watts.
Formula: dB PWL = 10 log (W/W0) where W0 is one
picowatt (10-12 watt).
Rule of thumb: doubling sound pressure results in a
6 dB increase, whereas doubling the sound power
level results in a 3 dB increase.
Calculating dB PWL
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Any sound power level can be expressed in dB PWL by
comparing it to the 0 dB PWL reference point of 1pW.
How many dB PWL is 4pW?
dB PWL = 10 log (4pW/1pW)
dB PWL = 10 log (4)
dB PWL = 10 (.60206)
dB PWL = 6.02
We can express the difference between any two sound
power levels (including electrical power) by using the 10
log formula.
dB PWL differences
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What is the dB difference between a 100-watt and 350watt amplifier?
dB = 10 log (100/350)
dB = 10 log (.2857143)
dB = 10 (-0.544068)
dB = -5.44
The 100 watt amp is 5.44 dB less than the 350 watt
amp (or we could say the 350 watt amp is 5.44 dB
greater than the 100 watt amp).
The dB in Electronics
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dBs are used in audio electronics to express
differences in power levels and voltage levels.
In the early days of audio electronics all audio
equipment was designed to have a 600 ohm
output impedance.
The dBm is a dB standard from those times and
is not used for current audio equipment.
Today we have the dBu, which has replaced the
dBm.
The dB in Electronics
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Power: dBm (0 dBm = 1milliwatt into a 600 ohm
load), 0dBW =1 watt into a 600 ohm load.
dBm Power Formula: 10 log (p1/.001W)
dBW Power Formula: 10 log (p1/1W)
Voltage: dBu (0 dBu = .775 volts) not referenced to
any load - chosen for historical reasons which is the
voltage you get with 1mW in a 600 ohm load.
dBu Voltage Formula: 20 log (E1/.775V)
dBV Voltage Formula: 20 log (E1/1V)
Audio line level standards
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Today, in the United States, the professional
line level standard is +4 dBu.
+4 dBu audio gear generally uses balanced I/O.
-10 dBV is the standard today for consumer
audio gear.
-10 dBV audio gear generally uses unbalanced
I/O.
dB differences in power levels
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Any power level can be expressed as dBm or
dBW.
dBm and dBW both use the 10 log formula,
however dBm uses 1mW for the 0 dB reference
point; dBW uses 1W for the 0 dB reference
point.
Calculating dBm
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How many dBm is a signal that measures 4
mW?
dBm = 10 log (4mW/1mW)
dBm = 10 log (4)
dBm = 10 (.60206)
dBm = 6.02
Calculating dBW
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How many dBW is a signal that measures 5W?
dBW = 10 log (5W/1W)
dBW = 10 log (5)
dBW = 10 (.69897)
dBW = 6.9897
Calculating dB differences in
power levels
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We can express the difference between any 2
power levels as dB (no suffix) by using the 10 log
formula.
We simply write the 2 voltages as a ratio.
For example: What’s the dB difference between 10
watts and 15 watts?
dB = 10 log (10/15)
dB = 10 log (.66667)
dB = 10 (-0.1760913)
dB = -1.76, so 10W is 1.7609 dB less than 15W
dB differences in voltage levels
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Any voltage level can be expressed as dBu or
dBV.
dBu and dBV both use the 20 log formula,
however dBu uses .775V for the 0 dB reference
point; dBV uses 1V for the 0 dB reference point.
Calculating dBu
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How many dBu is a signal that measures 2
volts?
dBu = 20 log (2V/.775V)
dBu = 20 log (2.5806452)
dBu = 20 (.4117283)
dBu = 8.235
Calculating dBV
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How many dBV is a signal that measures 2
volts?
dBV = 20 log (2V/1V)
dBV = 20 log (2)
dBV = 20 (.30103)
dBV = 6.02
Calculating dB differences in
voltage
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We can express the difference between any 2
voltage levels as dB (no suffix) by using the 20 log
formula.
We simply write the 2 voltages as a ratio.
For example: What’s the dB difference between 5
volts and 10 volts?
dB = 20 log (5/10)
dB = 20 log (.5)
dB = 20 (-.30103)
dB = -6.02, so 5V is 6.02 less than 10V