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White Gaussian noise has constant power spectral density $N_0/2$. I know that the area under the power spectral density curve between two points gives the power of the signal between these two points.

  1. If I want to know the power of a certain frequency in the signal (not in a range of frequencies), can we say that the power of each frequency in the signal is exactly $N_0/2$?

  2. The total of power of additive white Gaussian noise is infinity, what does this mean? Is it reasonable to assume that the noise added to the signal have an infinite power?

Gilles
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Noha
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3 Answers3

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If I want to know the power of a certain frequency in the signal (not in a range of frequencies), can we say that the power of each frequency in the signal is exactly No/2?

No, BUT: You mean the right thing, you just say it wrongly:

The Power Spectral Density is constant – a single frequency doesn't have any power; it has a "power per bandwidth"! To arrive at a power, you need to integrate the density over a non-zero mass of frequencies.

(This is kind of an important distinction to make – only infinitely long periodic signals, e.g. sine waves, have power at a single frequency; everything else has a "power distributed over frequencies".)

The total of power of additive white Gaussian noise is infinity, what does this mean? Is it reasonable to assume that the noise added to the signal have an infinite power?

Yes. Notice that you're never dealing with a truly white Gaussian noise in continuous-time systems (luckily for the universe, I might add); it's always approximately white for some bandwidth. Everything else is physically impossible – but rarely matters. Example: the thermal noise you can measure over a resistor is the classical example of white Gaussian noise in systems. However, it's not really white – the power density decreases at very high frequencies. But that's totally irrelevant to your observation – your measurement doesn't go into the terahertzes.

In time-discrete systems, things look different: for a sampled time-continuous stochastic signal (noise) to be white, it's sufficient that the original time-continuous signal had a constant PSD over a bandwidth. So, there's no physical problem in the time-continuous world. Since a discrete signal is just a sequence of numbers, there's no concern for "physicality" anyways.

Marcus Müller
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  • The power spectral density of AWGN is constant at No/2 or No? What I read is that it is constant at No/2, while the definition of No is the amount of power per unit bandwidth watt/Hz. Why No is divided by two? If we don't divide by two, then the value of the constant PSD is No (Watt/Hz), which when multiplied by the total bandwidth will give the total power. – Noha Dec 08 '20 at 12:05
  • I need to understand the following: why we can't define power for a single frequency component, unless there is a sinusoid at that frequency? infinitely long periodic signals, e.g. sine waves, have power at a single frequency, but any signal consists of a range of frequencies in the form of sinusoidal signals. Each sinusoid has a certain duration in the signal. Infinitely long sinusoidal signals have power, and also finite sinusoidal signals have power. – Noha Dec 08 '20 at 12:49
  • $N_0$ vs $N_0/2$: complex or real signals; depends on your definition of bandwidth. So, watch out for how the texts define bandwidth. – Marcus Müller Dec 08 '20 at 13:30
  • We can define power at a single frequency. It's 0 for white noise. It's because an integral over a single point of a bounded function is always 0, see my comments under Mark's answer. – Marcus Müller Dec 08 '20 at 13:30
  • I understand that mathematically, but I can not imagine that. A sinusoid always has power even if finite in duration, and of course every frequency is represented as a sinusoid with certain duration in the signal. – Noha Dec 08 '20 at 14:08
  • If you make the duration finite, you can no longer exactly be sure whether it was at a frequency $f_0$. In fact, by limiting your sinusoids "existence", you just windowed it with a rectangle – meaning the spectrum is now sinc-shaped, and no longer a dirac. – Marcus Müller Dec 08 '20 at 19:05
  • I know that the spectrum of a limited sinusoid is a sinc-shaped, but this doesn't mean that it has power. Of course it has power. I think any signal is composed of different sinusoids with different durations and amplitudes. – Noha Dec 08 '20 at 21:21
  • "composed of" is a dangerous term: composed of uncountably many weighted sinusoids? Sure, that's a sensible interpretation of what the Fourier transform is. Composed of a countable set of discrete sinusoids: No! Only infinitely long periodic signals are. – Marcus Müller Dec 08 '20 at 21:30
  • The magnitude spectrum of Fourier transform tells us how much a certain frequency is found in a signal relative to the other frequencies found in that signal. If the frequency is not found, the magnitude spectrum at that spectrum is zero. If it is found (in the form of sinusoid) with small amplitude and limited duration, the magnitude spectrum at that frequency will be small. – Noha Dec 09 '20 at 09:19
  • @Noha well said, but we need to be very careful here: You're mixing the terms of an energy spectrum of an observed signal with a finite length with a power spectral density, which is a stochastic property of an infinitely long weak-sense stationary process. If we apply the same methodology to both – saying "my signal exists only for my observation length, but I integrate over infinity and find the limit of integral divided by integration length", then all signals have 0 power everywhere. Don't mix these terms! – Marcus Müller Dec 09 '20 at 09:24
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White noise is a conceptual signal more than real world signal.
In the context of estimation it is the signal which can't be estimated based on its past.
In the context of Frequency domain it is the one with constant value (On average) for any of its bins.

Now, for continuous signals, it implies it has infinite energy, hence it is only a mathematical concept. As no such thing in real life.

Royi
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  • Would you say that we can’t have discrete signals in “real life”? – Dan Boschen Mar 06 '23 at 15:39
  • @DanBoschen, I am not sure what you mean. Noise is a phenomenon of data which can't be predicted (Wald Decomposition). That's all I am saying. – Royi Mar 06 '23 at 17:21
  • But we can generate discrete samples of white noise, so in a discrete process white noise samples can exist “in real life”. If you agree I think it is an interesting point to make. (“Exist” become philosophical I’m afraid but interesting point that discrete time signals can indeed be white in frequency in terms of having a flat PSD over all frequencies, unlike continuous time signals to your energy limitation point) – Dan Boschen Mar 06 '23 at 17:34
  • We can create noise artificially (Let's not get into how good), but usually it is just a signal which obeys the way we model noise: A signal which can not be predicted (In the White noise case). The context of real world signal above I meant how we model not artificial signals. For instance, in communication, what we're after and can infer is signal, all other htings are noise which make our task harder. – Royi Mar 07 '23 at 09:40
  • Yes thanks, thank makes sense. – Dan Boschen Mar 07 '23 at 10:28
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  1. Yes. Indeed.
    Remark The way I interpret your question is: "What's the the value of the PSD (Which you refer as power) frequency". The answer to that is that many people dealing with White Noise try to understand if they can intuitively think it is built by infinite sum of Harmonic Signals. Which actually the definition of White Noise: It requires all basis functions in order to build it. Each with the same power (On average).
  2. In case you'd see such signal it will indeed have infinite power. Yet you can only encounter Band Limited White Noise which is white within the frequencies it was sampled. See How to Simulate AWGN (Additive White Gaussian Noise) in Communication Systems for Specific Bandwidth.
Mark
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  • Nope, indeed not: at any frequency $f_0$ of white noise, the power is 0, since $$\int\limits_{f_0}^{f_0} G(f),\mathrm df=0$$ for all bounded functions $G(f)$, so especially for a constant value PSD $G(f)=N_0/2$.
    • – Marcus Müller Dec 05 '20 at 10:22
  • I agree if we try to think in the formal way of Measure Theory. I think in the case above the OP meant just if for each bin he would chose the value would be $ {N}_{0} / 2 $. – Mark Dec 05 '20 at 11:00
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    I'll allow myself to disagree there: they specifically say "(not in a range of frequencies)"! – Marcus Müller Dec 05 '20 at 11:29
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    @Mark: "... can we say that the power of each frequency in the signal is exactly $_0/2$?" The only answer to this is really "no", because that question is based on a misunderstanding on what power spectral density means. You can't define power for a single frequency component, unless there is a sinusoid at that frequency, which corresponds to a Dirac delta impulse in the power spectrum. But that is not the case for white noise. – Matt L. Dec 05 '20 at 11:35
  • Again, The way I interpret the question of the OP is the value of a single point of the function of the PSD. So he used the term Power. He didn't ask what is the only signals which have a Direc Delta in each of their bins. If we have the PSD as a function which its values are called power at f then the OP answer is correct. At each frequency the power, as a value of the function, is $ {N}_{0} / 2 $. – Mark Dec 05 '20 at 13:22
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    @Mark really, no! The value of the PSD is not a power, and the difference between it being a power density or a power is really important, especially considering how the question was phrased with, referring to power on a single exact frequency. Any single exact frequency of white noise has exactly 0 power! – Marcus Müller Dec 05 '20 at 21:10
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    @Mark: The value of the PSD is $N_0/2$ at each frequency, correct. However, this does absolutely not mean that the power at each frequency equals $N_0/2$. No matter how many times you repeat that, it remains completely wrong. – Matt L. Dec 06 '20 at 12:41
  • I am not saying it is the power. I am saying I interpreted the meaning of the Op saying power as the value of the PSD. Guys, I have MSc in Math in the field of Measure Theory. I have done this integrals in more formulate properties then you show here (The above isn't even well defined integral, it is should be defined by a limit, but anyhow). So let's formality aside. I interpreted the OP the way I did. He can read and clarify by himself. By the way if you want to be "Formal". What's the energy of a single realization of White Noise at given frequency? What will be the average of that? :-). – Mark Dec 06 '20 at 16:21
  • The above questions are not questions of mine. They are just food for thought for those want to give some formalities a try. The answer will be very interesting for you. – Mark Dec 06 '20 at 16:22
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    @Mark: Please impress us with good answers instead of letting us know your degree. – Matt L. Dec 06 '20 at 19:16
  • Please impress me by something more than the integral over a countable set is zero. Come on, you're just waving there as I wrote something unknowingly instead of thinking for a second why is that? – Mark Dec 06 '20 at 20:05
  • @Mark now we're talking! "What's the energy of a single realization of White Noise at given frequency?" 0! "What will be the average of that?" 0! You said it yourself: a zero measure set's expected integral is still 0, even under stochastic consideration. Really, there's zero energy for zero bandwidth - and that's both mathematically sound, as you notice yourself, as well as physically logical. – Marcus Müller Dec 08 '20 at 13:33
  • @MarcusMüller, That's wrong. Single realization of White Noise means it is infinitely long signal. Its energy at certain frequency is given by the inner product of it with the base (Exponent of that frequency). It will give the value at the frequency of its Fourier Transform. This is not a countable set. The average of those values (Their absolute value) is $ {N}_{0} / 2 $. – Mark Dec 08 '20 at 14:54
  • Mark, if you're arguing through infinitely long signals, you need to divide your energy by an infinitely long duration to arrive at power. You'll notice that this won't work out to prove any specific power at any specific frequency. You really need no measure theory for any of this – pure physical observation proves that: the energy is the (magnitude of the) inner product of the noise realisation with an infinitely long complex sinusoid. Since we're assuming the noise to be white, the expected point-wise product is zero (otherwise the noise wasn't uncorrelated), and hence, the inner product=0. – Marcus Müller Dec 08 '20 at 19:07