Log base of decibel scale4/13/2024 You can find functions to carry out most of the above calculations without needing the formulae or a calculator on our System Calculations page. Yet using decibels we find that the increase is still the same: 6dB (10 × log 4 = 10 × 0.6 = 6). However, we have quadrupled the power generated. Using decibels we find that doubling the output voltage is an increase of 6dB (20 × log 2 = 20 × 0.3 = 6). If we double the output voltage (40 Volts) we find that the power generated is now (40^2)/4 Watts = 1,600/4 watts = 400 Watts. From this, we can calculate the power generated into a 4Ω load as (20^2)/4 Watts = 400/4 Watts = 100 Watts. Where power and pressure combine, the values remain consistent.įor example, an amplifier produces an output of 20 Volts. Multiply your reference value by 10^(n/20).įor example, if your microphone produces 2.6 millivolts, an increase of 60 decibels will produce 2.6 × 10^(60/20) millivolts = 2.6 × 10^3 millivolts = 2.6 × 1,000 millivolts = 2,600 millivolts = 2.6 volts.At 20 decibels below full output, your 100 watt system is only running at 1 watt! Negative values work in exactly the same way, so that if your system is rated at 100 watts, a reduction of 20 decibels (−20 dB) will produce 100 × 10^−(20/10) watts = 100 × 10^−2 watts = 100 × 0.01 watts = 1 watt. Multiply your reference value by 10^(n/10).įor example, if your system is rated at 100 watts, a 20 decibel increase represents 100 × 10^(20/10) watts = 100 × 10^2 watts = 100 × 100 watts = 10,000 watts!. to work out what value is n decibels larger or smaller than a reference value): This may be a measured value, or may be a common standard reference point (e.g. ‡ The first value is your reference value.
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