Presentation on theme: "Logarithms and Decibels. The Decibel Named for Alexander Graham Bell. Originally used to measure power losses in telephone lines. A Bel is the common."— Presentation transcript:
Logarithms and Decibels
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: 5 3 =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: log 6 36 =2 exponential form: 6 2 =36 Any expression in logarithmic form may be indicated in exponential form. if any number b to the power of x = N: b x =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 x10 6
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.
Range of Human Hearing
Formula for dB SPL measurements The formula for SPL: 20 log (p/p 0 ) P 0 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 L w (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/W 0 ) where W 0 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 (p 1 /.001W) dBW Power Formula: 10 log (p 1 /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 (E 1 /.775V) dBV Voltage Formula: 20 log (E 1 /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