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When computing power spectral density via Fourier transform and Parseval's theorem, one gets units of $V^2/Hz$ (e.g. Wiki). As a popular next step, the following unit conversion is widely used: $V^2/Hz \to dB/Hz$, and it's done by taking the famous $10.\log10(\cdot)$.

Question: while $\mathrm{dB}$ is a log of a ratio of two $\mathrm{V}^2$'s (second $\mathrm{V}^2$ is a fictive one) and I get where $\mathrm{dB}$ comes from, I'm struggling to understand why $1/\mathrm{Hz}$ remains unchanged after taking the logarithm?

Alex

P.S. Some obscure explanation is in the last paragraph here (and I'm afraid the word 'loosely' conceals all the real stuff here):

Since the argument of the logarithm is in units of $\mathrm{Hz}^{-1}$, this spectral measure can loosely be said to be in units of `dB/Hz'

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  • $\begingroup$ i have trouble with the notion of dB/Hz in the first place. with V$^2$/Hz, we know that when we double a thin bandwidth, the power (or square of voltage) will also double. but the dB will not. $\endgroup$ – robert bristow-johnson Apr 22 '17 at 22:33
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Say you have a measurement in unit $U$ (V$^2$) per unit $u$ (Hz). I do claim (although not in public) that the logarithm is a kind of unitless transformation, see for instance What is the logarithm of a kilometer? Is it a dimensionless number? In other words, it is somehow $0$-homogeneous, while a power is $2$-homogeneous: multiplying by a constant yields a constant to some power.

First, the logarithm is kind of homogeneous to a $0$-th power: it is the integral of $1/x$ (power $-1$). The derivative of an affine function is a constant. Derivatives decrease powers by one unit, integrals increase them of the same amount. The logarithm somehow stands with constants between $x^{-1}$ and $x^{1}$. Additionally, the coefficient of variation of $x^p$ when $p\to 0$ is about:

$$ \frac{x^p-x^0}{p-0}\approx \log x\,.$$

Second, a dB scale is often relative: to some signal measurement, to some "absolute" value. The $\log$ argument is better unitless: something in U divided by some unit $1$ U, called U$_0$. Thus, computing $\log(\mathrm{U}/\mathrm{U}_0)$ could be considered.

So converting U/Hz into dB$_\text{U}$/Hz, to me, only affects the numerator.

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    $\begingroup$ it is a meaningless unit of measure. even when referenced to a specific voltage. $\endgroup$ – robert bristow-johnson Apr 22 '17 at 22:34
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    $\begingroup$ I see your point! Thanks for your answer! $\endgroup$ – agronskiy Apr 23 '17 at 13:52
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Simply because the intention of converting to a dB scale, here, is only to scale the power density, and not the bandwidth.

So, that's simply by definition.

Look at it this way: if you have a PSD plot with frequency on the x-axis and power density on y, you'd usually only want to scale the y-axis logarithmically.

You can, of course, also scale the frequency axis, but then, you get a different thing.

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