The power spectrum $\texttt{PS}$ of a (white noise) signal depends on the frequency resolution, $df=f_s/N$, where $f_s$ is the sample rate ($\texttt{Hz}$) and $N$ is the sample size.

The power spectral density $$\texttt{PSD} = \texttt{PS}/(df \cdot \texttt{noise power window bandwidth})$$ effectively removes the dependency on sample size $N$.

What I don't understand is what happens to the dependency on sampling rate. It seems that $$\texttt{PSD} = \texttt{PS}\cdot N/(f_s \cdot \texttt{noise power window bandwidth})$$ This still clearly depends on $f_s$. Then, how can say, an amplifier have a certain level of input-referred votlage noise, quoted as $\texttt{PSD}$ in $\texttt{V/rtHz}$?


1 Answer 1


If this is with regards to a measurement of sampled white noise with an FFT (let’s assume without further windowing where the resolution bandwidth is simply $f_s/N$), where there was proper consideration to pre-sampling gain and anti-alias filtering to measure properly the noise, then I think the confusion is how $PS$ changes with regards to $N$ and $f_s$. If I understand the question, $PS$ is the measured spectrum as the total power within each bin, and therefore the power spectral density would be this measurement divided by the resolution bandwidth of each bin, as:

$$(RBW)= f_s/N$$

$$(PSD) = (PS)/ (RBW)$$

So we get

$$(PSD) = (PS) N/f_s$$

However what is missing from this final formula above is how PS changes if either $N$ or $f_s$ changes:

$$(PS) = (PSD) f_s/N$$

So the formula and logic results in a trivial $(PSD)= (PSD)$ which is indeed the case, neither $N$ nor $f_s$ will effect the power spectral density of the properly sampled signal.

Here is an example that may help understand why the sampling rate will not change the power spectral density of the measured signal:

If we double the sampling rate but keep N the same we will have doubled the resolution BW of each bin and therefore the noise in each bin will have gone up by 3 dB. But since the resolution BW doubled there is no change in power spectral density.

In the last formula $(PSD)=(PS) N/f_s$ with respect to the above example, $PS$ went up by 2, and $f_s$ went up by 2, $N$ is unchanged so $PSD$ is unchanged.

So in conclusion the power spectral density of the actual waveform sampled is independent of the sampling rate (or the total number of samples we may use in an FFT) as long as we had sufficient gain and properly filtered to eliminate aliasing. However to be clear, this is not the case for the power spectral density if limited by the quantization noise floor.

Power spectral density of quantization noise due to sampling is entirely dependent on the sampling rate, and the type of sampler. For example with most A/D converters the quantization noise is white and well estimated as having a total noise power from DC to the sampling rate as 6.02 dB/bit + 1.76 dB below that power level where a full scale sine wave would clip. The noise density is this noise divided by the sampling rate (to be in units of W/Hz for example as a noise density), so if we increase the sampling rate up to the allowable conversion rate for a particular converter, the total noise power will not have changed but the noise density will decrease (in practical application, as we increase the sampling rate the total noise power will typically increase due to other factors). With Sigma Delta converters the noise is shaped and no longer white, but we will still have a total noise power over a given sampling rate that will spread out over more frequencies if the sampling rate is increased.

In contrast, if the noise is dominated by the signal itself that we are sampling, then nothing about the sampling rate will change that noise density. So the noise density of the actual waveform is independent of the sampling rate. This is assuming we sample properly, in that we low pass (or bandpass) filter to avoid aliasing from other Nyquist bins, which would increase the measurement of the noise, and noise density, beyond the reality which is in the signal itself.

For further details and considerations on A/D conversion and related noise, this post may also be of related interest.

  • $\begingroup$ Thanks. However, I don't fully understand. Also, the noise density is not the noise divided by the sampling rate fs, it is the noise divided by the frequency resolution bandwidth df. Also, I'm talking about a white noise signal, like thermal noise from a resistor--not quantization noise due to sampling. Also, it seems like your second paragraph contradicts your first: are you saying that if the noise is from the signal itself, then no, the sampling rate will not change the noise density? Can you please explain that further? Thanks a lot! $\endgroup$
    – Nick
    Jan 12, 2023 at 16:03
  • $\begingroup$ White noise from a resistor is not at all affected by the sampling rate unless you sample in a way that causes aliasing (you must low pass filter so that any noise above half the sampling rate, for real signals, does not alias). Regarding your first comment; the noise density frequency is quantization noise IS the total noise divided by the sampling rate. You might be thinking of what you get from an FFT since you mention resolution BW (that is not what I am talking about): if the total quantization noise is 10 mW and that noise is spread across 10 Hz then the noise density is 1mW/Hz. $\endgroup$ Jan 13, 2023 at 0:45
  • $\begingroup$ This is with reference to the formula I gave: a converter with an ENOB of 10 bits at a given sampling rate would have a total noise power (when integrated from -fs/2 to +fs/2) of -61.8 dB below the power level of a full scale sine wave right at clipping. This noise if white (and usually is except when we sample a sine wave at an integer multiple of its rate) will be evenly distributed across that bandwidth, so the noise density would be that total power level divided by that sampling rate $\endgroup$ Jan 13, 2023 at 0:50
  • $\begingroup$ (Or a spectrum analyzer; with your RBW comment): either way that is a measurement of what the noise density is but doesn’t set the noise density; correct measurement does not change what’s there. If we don’t amplify that resistor noise sufficiently before we sample, then the noise we measure is the quantization noise (and affected by the sampling rate)- if we do amplify enough (and filter properly) then the noise we measure is the amplified thermal noise plus noise figure and is not affected by the sampling rate. Hope that helps clear it up. $\endgroup$ Jan 13, 2023 at 1:03
  • $\begingroup$ @Nick I think I see what your confusion was; see update at start of my answer $\endgroup$ Jan 13, 2023 at 1:28

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