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## Logarithms and Decibels

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**The Decibel**• 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.**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.**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**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.**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.**Formula for dB SPL measurements**• 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**• 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**• 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**• 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.**dB PWL**• 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**• 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**• What is the dB difference between a 100-watt and 350-watt 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**• 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**• 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**• 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**• 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**• 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**• 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**• 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**• 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**• 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**• 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**• 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