I'm reading the datasheet of a DAC (see page 6), and the "Output Noise Spectral Density" has units V / Hz^(1/2).

The Wikipedia article on noise spectral density however states that the noise spectral density has unit Watt / Hz.

Those two units are not equivalent. Why is there a discrepancy in the units?

  • \$\begingroup\$ see also this question. (possible duplicate) \$\endgroup\$ – stevenvh Aug 10 '12 at 14:48

The \$\mathrm{W/Hz}\$ may be a bit confusing as it looks like it refers to a single frequency. But that's just the dimension, it actually refers to a bandwidth, which is also expressed in Hz: maximum frequency - minimum frequency. So it's the power over a given bandwidth.

If you divide power by the load's resistance you get voltage squared. So for a given load you can express noise power as

\$ \mathrm{\dfrac{V^2}{BW}}\$

where \$\mathrm{BW}\$ is bandwidth. If you want to know the voltage you take the square root:

\$ \mathrm{\sqrt{\dfrac{V^2}{BW}} = \dfrac{V}{\sqrt{BW}}}\$

which indeed has the dimension of \$\mathrm{V/\sqrt{Hz}}\$.


Both units show you the same thing, just in a slightly different form. From the perspective of a DAC, the noise voltage is what matters since the DAC is outputting specific voltages. However, in a communication system, the noise power is what matters, so in that case people prefer to use Watts.

Also remember that Watts and Volts are related by a square. P=V^2/R, we can ignore the R here since its just a scaling factor. So inversely if you take the square-root of Watts you will end up with volts.

  • \$\begingroup\$ Ok, so that explains that Watt versus Volts discrepancy. What about the square root? \$\endgroup\$ – Randomblue Aug 10 '12 at 14:30

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