Why factor of 20 is multiplied while converting the ratio of two powers/voltages to dB value?
SdB=20 Log (Smag)
where:
SdB = value of the magnitude, in dB
Smag = value of the magnitude, as a number
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1\$\begingroup\$ You've not even got the right facts in the question. I suggest you do some double-checking on this formula. \$\endgroup\$– Andy akaSep 29, 2020 at 11:33
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\$\begingroup\$ Corrected my question.. Waiting for your valuable reply.. \$\endgroup\$– AnantheshSep 29, 2020 at 12:16
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\$\begingroup\$ Your question is still incorrect. Read it. \$\endgroup\$– Andy akaSep 29, 2020 at 12:26
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\$\begingroup\$ It's times ten because you want deciBel and not Bel. times 2 because the log of x^2 is twice the log of x. So when you have voltages, it's only times 10; when you have powers, that are basically voltage^2 you have times 2 times 10. \$\endgroup\$– Sredni VashtarSep 29, 2020 at 14:20
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\$\begingroup\$ Thanks @SredniVashtar. Now I understood the logical reason behind it. \$\endgroup\$– AnantheshOct 1, 2020 at 6:51
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2 Answers
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20 is used for voltage ratio. SdB = 20.log(Vout/Vin)
10 used used for power ratio. SdB = 10.log(Pout/Pin)
They are different by a factor of 2, as power = k.voltage2
Why don't we use smaller, or other, numbers, as the multiplier?
I think it was felt in the days when the ratio that log(Pout/Pin) was starting to be used, that this ratio, known as the bel (symbol B), was too large to be convenient. Hence it was multiplied by 10 to get the decibel, or dB. It's no more complicated than that. See wikipedia for more details and history.
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The delightful fact about the 10 and the 20 is that a value, say 87dB, becomes the correct ratio for current and voltage and power IF you have equal SOURCE and LOAD resistors.
answered Sep 29, 2020 at 14:06