Question 458326: 1)If my meter is showing -8dBu what is the voltage on the meter?
dBu=20xlog10(v/vref)0.775
-8x20log(10/0.775
=0.31
2)If you move the fader to +24 dbu what will be the voltage?
12.3
3)If I can measure 5 volts across my channel strip, what will be the read out of dBu, 5 volts equals how many dBu?
dBV = 20 x log10(V/VREF) where VREF is 1V
16.19 dBu
i have the answers i just dnt know if i am using the appropraite formulas
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! With certain assumptions, the equation:
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is the correct equation to use for this problem. (For simplicity, I am using D to represent the decibels, V to represent the output or measured voltage, and I to represent the input or reference voltage.)
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For your first problem, D = -8 and you apparently meant that I, the reference voltage, is 0.775 volts. Substituting these values into the equation results in:
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Get rid of the multiplier of 20 on the right side by dividing both sides of the equation by 20:
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Divide out the left side:
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Transform this from a log equation by raising the base of the log (which is 10) to the exponent consisting of the left side of this equation and you have:
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Use a calculator to raise 10 to the power of -0.4 and you get 0.39810717. (If your calculator doesn't handle negative exponents, raise 10 to the power +0.4 to get 2.511886432 and then divide the result into 1.) Substituting the result for the left side, the equation becomes:
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Solve for V by multiplying both sides of the equation by 0.775 to get:
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which reduces to:
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and this rounds to the answer you gave of:
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 volts
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Your second problem is identical to the first with the single exception that the value of D is changed to +24. The process is:
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and this rounds to
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 volts
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The last problem is a little different and the answer is not the same as you indicated. The change is that this time you have V = 5 volts, I = 1 volt, and you are trying to solve for D. Use the same formula with these new values substituted:
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Use your calculator to find that  and substitute this value to get:
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Do the multiplication on the right side:
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Which rounds to:
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 decibels
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and this is different from your answer of 16.19. Check both your work and mine to see why we have this difference.
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Hope this helps.
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If you are into electronics, make sure you understand that most meters are calibrated with the assumption that the waveform being used is sinusoidal and also that the load is purely resistance. If you change the waveform so that its rms (root mean square) value changes or if the load has a significant reactive component, the meter does not adjust for the differences and its reading will not be correct. If you're not into electronics, just ignore this last set of comments. They don't affect the problems that you presented above.
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