Click here to purchase the entire book in PDF format. dBu Let's think back to the 1mW into 600$\Omega$ situation. What will be the voltage required to generate 1mW in a 600$\Omega$ resistor? $$P = \frac{V^{2}}{R}$$ (3.37) $$\textrm{therefore}$$ (3.38) $$V ^{2} = P * R$$ (3.39) $$V = \sqrt {P * R}$$ (3.40) $$= \sqrt {1 \textrm{ mW} * 600 \Omega}$$ (3.41) $$= \sqrt {0.001 \textrm{ W} * 600 \Omega}$$ (3.42) $$= \sqrt {0.6}$$ (3.43) $$= 0.774596669 \textrm{ V}$$ (3.44) Therefore, the voltage required to generate the reference power was about 0.775 Vrms. Nowadays, we don't use the 600$\Omega$ impedance anymore, but the rounded-off value of 0.775 Vrms was kept as a standard reference. So, if you use 0.775 Vrms as your reference voltage in the equation like this: $$\textrm{Voltage (in dBu)} = 20 \log \left ( \frac{\textrm{Voltage 1} }{ 0.775 \textrm{ Vrms}} \right )$$ (3.45) your unit of measure is called dBu. Where Voltage1 is measured in Vrms. (It used to be called dBv, but people kept mixing up dBv with dBV and that couldn't continue, so they changed the dBv to dBu instead. You'll still see dBv occasionally - it is exactly the same as dBu... just different names for the same thing.) Remember - we're still measuring pressure so it's a 20 instead of a 10, and, like the dBV measurement, there is no specified impedance.