Question

Why a zero watt cannot as chose as zero decibel level ??

Answer 1

Answer:

Explanation:

The decibel (dB) is a logarithmic unit used to measure sound level. It is also widely used in electronics, signals and communication. The dB is a logarithmic way of describing a ratio. The ratio may be power, sound pressure, voltage or intensity or several other things. Later on we relate dB to the phon and to the sone, which measures loudness. But first, to get a taste for logarithmic expressions, let's look at some numbers. (If you have forgotten, go to What is a logarithm?)

For instance, suppose we have two loudspeakers, the first playing a sound with power P1, and another playing a louder version of the same sound with power P2, but everything else (how far away, frequency) kept the same.

Using the decibel unit, the difference in sound level, between the two is defined to be

10 log (P2/P1) dB        where the log is to base 10.

If the second produces twice as much power than the first, the difference in dB is

10 log (P2/P1) = 10 log 2 = 3 dB     (to a good approximation).

This is shown on the graph, which plots 10 log (P2/P1) against P2/P1. To continue the example, if the second had 10 times the power of the first, the difference in dB would be

10 log (P2/P1) = 10 log 10 = 10 dB.

If the second had a million times the power of the first, the difference in dB would be

10 log (P2/P1) = 10 log 1,000,000 = 60 dB.

This example shows a feature of decibel scales that is useful in discussing sound: they can describe very big ratios using numbers of modest size. But note that the decibel describes a ratio: so far we have not said what power either of the speakers radiates, only the ratio of powers. (Note also the factor 10 in the definition, which puts the 'deci' in decibel: level difference in bels (named for Alexander Graham Bell) is just log (P2/P1).)

Sound pressure, sound level and dB. Sound is usually measured with microphones and they respond proportionally to the sound pressure, p. Now the power in a sound wave, all else equal, goes as the square of the pressure. (Similarly, electrical power in a resistor goes as the square of the voltage.) The log of x2 is just 2 log x, so this introduces a factor of 2 when we convert pressure ratios to decibels. The difference in sound pressure level between two sounds with p1 and p2 is therefore:

20 log (p2/p1) dB   =  10 log (p22/p12) dB   = 10 log (P2/P1) dB       (throughout, the log is to base 10).

What happens when you halve the sound power? The log of 2 is 0.3010, so the log of 1/2 is -0.3, to a good approximation. So, if you halve the power, you reduce the power and the sound level by 3 dB. Halve it again (down to 1/4 of the original power) and you reduce the level by another 3 dB. If you keep on halving the power, you have these ratios.

Answer 2

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Prior to the invention of electronic calculators, multiplying and dividing long numbers was painful. Yet that was something engineers (including audio ones) had to do often.

Prior to the invention of electronic calculators, multiplying and dividing long numbers was painful. Yet that was something engineers (including audio ones) had to do often.Logarithms were invented in the 17th century to simplify calculations.

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➡️Answer by @crazygirl169

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