I'm trying to wrap my head around power spectral density on a conceptual level, but I am having some difficulty. Suppose I have a communication system where I am receiving and sampling white Gaussian noise, which has a uniform PSD. Increasing the bandwidth of my receiver results in greater noise power across all frequency components. I can follow the math of how this occurs: integrate the PSD across more frequencies and you will increase the total power, which is uniform for white noise. However, how does it conceptually make sense that letting in higher frequency power components directly increases the power of the low frequency components?

EDIT: Some commentators have pointed out that my question was worded poorly. I apologize. I will try to clarify what I mean by turning it into a concrete thought experiment. Suppose I have a LPF with bandwidth $B_1$. Then, the variance of the DC component in the DFT domain after sampling WGN will be some $\sigma_1^2$. Now, compare that with a wider LPF with bandwidth $B_2$, where $B_2 > B_1$. Then the DC component variance would be greater, $\sigma_2^2 > \sigma_1^2$. Now, what if instead I implemented this second LPF filter a different way, with a parallelized structure: the sum of 2 different filter paths. The first filter path is the LPF with bandwidth $B_1$ and the second is a bandpass filter with passband $[B_1, B_2]$. This is followed by sampling. This is an equivalent structure to the LPF with bandwidth $B_2$, so the output statistics should be the same. However, notice that the second path is DC blocking, so it will never have a DC component. This should mean that the variance of the DC of the sum of the paths is equal to just the variance of LPF path, $\sigma_1^2$. But clearly, despite this being an equivalent filter structure, the output statistics are not the same. Why?

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    $\begingroup$ I’m having trouble wrapping my head around your question $\endgroup$ – user28715 Jun 16 '18 at 15:32
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    $\begingroup$ Duplicated dsp.stackexchange.com/questions/49555/… ? $\endgroup$ – AlexTP Jun 16 '18 at 17:30
  • $\begingroup$ @StanleyPawlukiewicz please see my edit, hopefully it clarifies my question. $\endgroup$ – Probably Jun 17 '18 at 23:41
  • $\begingroup$ @AlexTP please see my edit, it is not a duplicate of the linked question. $\endgroup$ – Probably Jun 17 '18 at 23:41
  • $\begingroup$ A puzzle it is. The observed power of kdiscrete time system would depend on sample rate . There is more sample to sample correlation the nearer in time they are sampled. This is just a guess. Probably not a very good one $\endgroup$ – user28715 Jun 18 '18 at 0:59

The spectrum allows you giving another complementary look on the data.
It is the other side of the same coin.

In Frequency Domain we are looking on the energy of each spectral component of our data.
Each component is independent of other component in the meaning each component is orthogonal to any other component.

In the case of White Noise, the PSD is the Fourier transform of the Auto Correlation function.
The model is Continuous White Noise is one which has infinite power.
Its power becomes finite once it is sampled by a sampling chain which has Low Pass Filter (Which the component which limits the incoming power) and AD.

Now, just like LPF will limit the power of any signal which has bandwidth which is higher the LPF bandwidth it does so to White Noise as well.

Namely, the Constant Level of the PSD is a property of the White Noise alone (In case of Gaussian White Noise it is the $ \sigma $ parameter) while the power of the sampled noise is derived by the level of the PSD and its bandwidth (Which is a property of the LPF).

There is no high frequencies which changes the values of low frequencies.
The power gets bigger because of the hidden assumption that when you sample with higher rate you use LPF with higher bandwidth which lets more frequencies in and hence more power.


After giving it more thought, I believe I've come up with an explanation. The issue was the mapping of continuous-time power to discrete-time power. The gain in the DFT domain after sampling is actually directly governed by the sampling rate, $F_s$, not by the LPF cutoff (the cutoff just prevents any excess gain that would arise from aliasing). Recall that when sampling, the amplitude of the continuous-time spectrum actually gets scaled by $F_s$.

For example, if I passed the noise input through a LPF with cutoff $B_1$, followed by a sampling rate of $F_{s1}=2B_1$, I would have variance $\sigma_1^2=N_0B_1$ across all frequency components. Now if I kept the same LPF but doubled the sampling rate to $F_{s2}=4B_1$, then the variance of my passed frequency components will each double to $2\sigma_1^2$. Of course, the stopband frequencies would have zero variance, so the overall average power would be the same. Notice that the increase of my lower frequency components did not come from increasing the LPF cutoff to encompass higher frequencies, but rather from increasing my sampling rate.

  • $\begingroup$ Even though you are using some unclear notations (for me), it seems that this is what I was trying to say in my comments, isn't it? $\endgroup$ – AlexTP Jun 19 '18 at 9:46
  • $\begingroup$ @AlexTP No, you (and all the links you gave) were focusing on $|H(f)^2|$ as opposed to the act of sampling itself (I realize real world samplers cannot be divorced from some sort of system response to go along with it, but this is from a theoretical mathematical perspective). It is a subtle distinction, but a distinction nonetheless. See my answer where I talk about keeping $|H(f)^2|$ fixed while changing the sampling rate. Also, I disagree with the notation being unclear, as it is pretty standard. But thanks for your attention to my question, it is much appreciated! $\endgroup$ – Probably Jun 19 '18 at 15:02

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