4
$\begingroup$

There is an analog system which includes the continuous-time linear equalizer (CTLE). With some .noise analysis the power-spectral density (PSD) of the noise in that system is provided. So let's not care about the details how PSD is provided, the only thing that matters is that I have access to the noise PSD of the circuit. The noise PSD is not necessarily white and it can have an arbitrary shape, but it dies out after a certain frequency (let's say most energy is contained within 100 GHz). By having the noise PSD, I integrate it over a large enough frequency range (in this example from 0 to 100 GHz) to get the (almost) total power of the noise, let's name it $\sigma^2$.

Now, in my MATLAB simulation I have a clean sampled signal $s(n \,\,\Delta t)$ with sampling frequency $f_s$, where $f_s = 1 / \Delta t$ (To add more details, $s(t)$ is a continuous-time signal that has a bandwidth of 30GHz and I am oversampling $s(t)$ by taking 32 samples per symbol time. So the sampling frequency in this example would be $ f_s $ = 32 x 30GHz = 960 GHz . Maybe these details don't matter).

I want to add discrete white Gaussian noise to $s(n \,\,\Delta t)$ such that this discrete Gaussian noise has the same power (i.e. variance) as the continuous (non-white) noise which was calculated to be $\sigma^2$. I know I can do something like $N = \sigma$ x randn( size(s) ) in MATLAB to generate noise with power $\sigma^2$, but my main question is that what is the dependency of the discrete noise in this example to the sampling rate $f_s$. If I change $f_s$ how the randn() function or the $\sigma$ should change.

Cause I see in some pages it is mentioned that:

"Given a continuous White Noise (Wide Sense) with variance $\sigma^2$ and you want sample it at rate of $f_s$ you should generate discrete noise samples with variance of $f_s \, \sigma^2$",

but I don't know what is the mathematical reasoning behind it and if the same applies in my case or not.

If you have any good read on this matter, that would be really appreciated.

$\endgroup$
4
  • $\begingroup$ I would refer to this answer dsp.stackexchange.com/a/8632/26081. Ideal sampling is filter with finite bandwidth $f_s$, then you generate samples with variance $f_s \sigma^2$. $\endgroup$
    – AlexTP
    May 31, 2018 at 12:07
  • $\begingroup$ I completely agree with @Royi that "It is important to have definitions straight." This really is about statistical communications, or communications within a context of added noise. We need to be careful about the definition of "white noise". You do well by giving it an overall limiting bandwidth that you say is 100 GHz. If you are replacing your noise with white noise over that same bandwidth, then that limiting frequency must be part of your mathematical description, if you are to understand a finite variance, $\sigma^2$ of the noise. That "100 GHz" must be part of it. $\endgroup$ Aug 7, 2023 at 1:43
  • $\begingroup$ So, at first stab, the equation @Royi put in, should be changed to: $$ \sigma_n^{2} = \frac{f_\mathrm{s}}{2 \times 100 \, \text{GHz}} \ \sigma^2 $$ $\endgroup$ Aug 7, 2023 at 1:47
  • $\begingroup$ Related: dsp.stackexchange.com/a/87654/21048 $\endgroup$ Aug 11, 2023 at 3:43

1 Answer 1

0
$\begingroup$

It is important to have definitions straight.
The RMS of a white noise, since it has zero mean, is its standard deviation.
So it is easier to talk on the Variance (The squared RMS or the MS).

As you wrote, the Variance of a sampled continuous white noise is given by:

$$ {\sigma}_{n}^{2} = {f}_{s} {\sigma}^{2} $$

Where $ {\sigma}_{n} $ is the RMS of the sampled noise and $ {R}_{xx} \left( t \right) = {\sigma}^{2} \delta \left( t \right) $ is the auto correlation of the White Noise Process.

As we can see, the noise RMS is higher the LPF bandwidth of the sampling system is. This is why it is advised to sample according to the data BW in order to accumulate the least noise as possible.

Remark 001: Just to make things more coherent. The actual noise variance is defined by the LPF of the sampling system. So if the LPF BW is given by $ B $ (Assuming ideal LPF) then the noise variance is given by $ {\sigma}_{n}^{2} = B {\sigma}^{2} $ where $ B $ is the total area below the LPF graph (Both sides).

Remark 002: White noise, in the context of Signal Processing, is usually defined only by its Auto Correlation function. Mathematically there are deeper models to define it. Yet it is out of the scope of this question.

$\endgroup$
52
  • $\begingroup$ Can you be more mathematically specific about your answer. Basically what you propose does not make sense energy-wise. Cause higher sampling frequency means more frequency separation and hence wider bandwidth for the white noise. So I don't get why the RMS should remain the same. I tried to explain the problem in other words in the following post. Please read it and let me know if you have any explanation on that. dsp.stackexchange.com/questions/49574/… $\endgroup$
    – shampar
    Jul 19, 2018 at 19:52
  • $\begingroup$ It does makes sense. Imagine ideal LPF. Then the RMS of the Noise going through the analog chain (LPF + Sampling) is exactly the White Noise Level multiplied by Bandwidth. $\endgroup$
    – Royi
    Apr 16, 2021 at 9:26
  • $\begingroup$ @shampar, Could you please mark my answer? $\endgroup$
    – Royi
    Jun 30, 2022 at 5:46
  • 1
    $\begingroup$ Hay, I just saw this, 5 years later. Royi, continuous-time white noise (ya know the thing we represented with $\frac{\eta}{2}$ as the level over all frequencies, positive and negative) has infinite bandwidth, infinite power, infinite variance, and infinite standard deviation. In continuous time, white noise only makes sense in a model when that $\frac{\eta}{2}$ is multiplied by some bandwidth $B$ in both the positive and negative frequencies (leaving a noise variance of $\eta B$). But, of course, sampled signals (including noise signals) have finite bandwidth $\frac{f_\mathrm{s}}{2}$. $\endgroup$ Aug 5, 2023 at 4:07
  • $\begingroup$ @robertbristow-johnson, The bandwidth is $ {f}_{s} $. It is spanned over $ \left[ -\frac{ {f}_{s} }{2}, \frac{ {f}_{s} }{2} \right] $. So the above holds. To make it more accurate, we actually need to take into account the BW of the low pass filter. So there is an implicit assumption the pre sampling low pass filter matches exactly the sampling frequency. This assumption is common, so no issues in there. $\endgroup$
    – Royi
    Aug 5, 2023 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.