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I'm reading the Wikipedia article on spectral density. It is said:

Sometimes one encounters an amplitude spectral density (ASD), which is the square root of the PSD; the ASD of a voltage signal has units of V Hz−1/2.[6] This is useful when the shape of the spectrum is rather constant, since variations in the ASD will then be proportional to variations in the signal's voltage level itself. But it is mathematically preferred to use the PSD, since only in that case is the area under the curve meaningful in terms of actual power over all frequency or over a specified bandwidth.

Could somebody elaborate what this means? Why is it that only in the case of power spectral density (PSD) the area under the curve is meaningful?

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3 Answers 3

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The units are the key, so let's consider an example. Suppose we are measuring a voltage. Thus, the unit of the $PSD$ is $V^2/ Hz$, and the integral over the frequency range $$ \int_{f_1}^{f_2}PSD\; df $$ has the unit $V^2$. Hence, the result of the integration has the unit $[signal]^2$. In signal processing the square of the signal is called "power" -- note that this differs from the definition we use in physics ($\rm{Watt} \ne \rm{Voltage}^2$). Also note that the square root of this integral is the RMS value of the signal within the frequency range $[f_1, f_2]$.

In contrast, if we integrate the ASD across the frequency range $$ \int_{f_1}^{f_2}ASD\; df $$ the result has the unit $V \cdot \sqrt{Hz}$. The unit of this result does not have a simple interpretation or relation to the measured signal. However, I would not call it a meaningless result.

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  • $\begingroup$ Thank you for the answer, but I don't see how we got the units for in the integral of ASD. Why is the ASD defined as the square root of PSD, why don't we define it as V/Hz, and then the integral unit would be just V? $\endgroup$
    – S. Rotos
    Commented Feb 18, 2021 at 9:18
  • $\begingroup$ The short answer to the question "why is $ASD = \sqrt{PSD}$?" is: The properties of the absolute value are not very "nice". The long version has deep roots in mathematics (i.e. statistics) which go way beyond the original question. However, if we accept that we use the $ASD$ as defined above, the unit of the integral does not allow a straight forward interpretation. Do you agree? $\endgroup$
    – Semoi
    Commented Feb 18, 2021 at 17:08
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Both are useful but sometimes one is more practical than the other. For example, most electronic noisy components are specified with their ASD be it an opamp, D2A, or an oscillator when it comes to their output level. But ASD by itself is meaningless unless one specifies the impedance of the noisy source. One can claim that because $(PSD) = 4 \Re\{Z\} \cdot (ASD)^2 $ power spectral density represents the (maximum) available noise power that can be delivered to a load if matched, and hence by definition it is independent of termination but this independence is an illusion for one must know what $Z$ is to terminate it for maximum noise power.

Of course, ASD may be represented as either voltage or current noise; which one is preferred depends on the output impedance: if the source is a voltage source, i.e., has small output impedance, then ASD is measured in $\rm{V/\sqrt{Hz}}$, contrarywise if the impedance is high then ASD is more conveniently measured as current noise in units of $\rm{A/\sqrt{Hz}}$

When it comes to phase noise you only see PSD and never ASD because there is no natural voltage (current) scale on which one should measure angle. In fact, let the signal be immersed in band pass noise $$Acos(\omega_c t) + n_X cos(\omega_c t) - n_Y sin(\omega_c t) \approx Acos\left(\omega_c t +\frac{n_Y}{A}\right)$$ for low level noise $|\frac{Var\{n_X\}}{A^2}|=|\frac{Var\{n_Y\}}{A^2}|<< 1$. Phase noise is then the ratio of noise voltage to signal amplitude $\phi = \frac{\{n_Y\}}{A}$, and as such it is natural to characterize it with its power spectral density that here is a mean square of the phase error per unit frequency width at a particular frequency *deviation* measured from the carrier frequency $\omega_c$.

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S.Rotos, I think the reason why we don't simply define the amplitude spectral density with the unit of V/Hz is that the level of the resulting amplitude spectral density curve would become dependent on the frequency resolution of the FFT used in obtaining the amplitude spectral density.

If we defined:

amplitude spectral density ~ |FFT| / N / df

(where N is the number of data points in FFT, and df is the frequency resolution of the FFT.)

then two analysts who use two different time-history sample lengths will report two different curves for the same overall random signal.

So instead we define:

amplitude spectral density ~ |FFT| / N / sqrt(df)

whose overall level is independant of the the choice of df and N (as long as they are fine enough).

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