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It may be a really simple question, but I'm not sure about this one:

Given a complex white Gaussian noise process with iid real and imaginary parts and a double sided power spectral density of $N_0/2$. What is the PSD of this random process?

Is it just: $$S_n(f) = \frac{N_0}2 + j \frac{N_0} 2\qquad ?$$ Or am I wrong?

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  • $\begingroup$ t or f, frequency? $\endgroup$ – Stanley Pawlukiewicz Dec 19 '18 at 14:57
  • $\begingroup$ Thank you, it is Sn(f). But how do I compute that? $\endgroup$ – Phinie Dec 19 '18 at 15:10
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    $\begingroup$ The PSD is simply $N_o$. Power is not a complex quantity. The PSD of the real component is $N_o/2$, the PSD of the imaginary component is equally $No/2$ and the two components sum in power. It is white so constant $N_o$ for all frequencies. $\endgroup$ – Dan Boschen Dec 19 '18 at 15:28
  • $\begingroup$ I am clarifying in my answer below, as there is also a difference between the single sided and double sided quantities to pay attention to! $\endgroup$ – Dan Boschen Dec 19 '18 at 15:42
  • $\begingroup$ So I could also say it like that: $$R_n(u) =<n^*(t)n(t+u)>=s_n^2 \delta (u)$$ and $$s_n^2 = N_0$$ $\endgroup$ – Phinie Dec 19 '18 at 16:23
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@Dan Boschen gives a precise answer in the comment. I would like to add to that a simple derivation using the formal definition as

$$S_{xx}(f) = E \vert x(t) \vert^2$$

Where $E$ stands for expectation and the process $x(t)$ is the one you describe in OP. Since it is complex, we can decompose it as $$x(t) = r(t) + j.i(t)$$ Where $r(t),i(t)$ are the real and imaginary parts of the process, which are themselves Gaussian. Now applying the PSD definition on top, we get

$$S_{xx}(f) = E \vert x(t) \vert^2= E \vert r(t) \vert^2 + E \vert i(t) \vert^2 $$ Given that $r(t),i(t)$ are both zero mean, then each of their corresponding terms above are nothing other than their variances, i.e. $\frac{N_0}{2}$ hence we get

$$S_{xx}(f) = \frac{N_0}{2} + \frac{N_0}{2}=N_0 $$

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  • $\begingroup$ Thank you a lot! It really helped me understand it a bit better... $\endgroup$ – Phinie Dec 19 '18 at 16:14
  • $\begingroup$ So I can say that the PSD is equal to the Variance of the random process? $\endgroup$ – Phinie Dec 19 '18 at 16:20
  • $\begingroup$ Yes and not only that. The PSD is uniform over all frequencies, which is observed using the last equation. $\endgroup$ – Ahmad Bazzi Dec 19 '18 at 16:21
  • $\begingroup$ To be clear, Sxx(f) would be the Single Sided PSD, not the Double Sided PSD. Not sure if the OP was looking for single or double but worth clarifying as it also causes an No/2 confusion in addition to real vs imaginary. $\endgroup$ – Dan Boschen Dec 19 '18 at 16:40
  • $\begingroup$ But why can I say that the variance of r(t) $\sigma^2=N_0/2$? Shouldn't the variance be infinite since $R_n(\tau)=\sigma^2 \delta(\tau)$ and $R(0)=E(|n(t)|^2)$? $\endgroup$ – Phinie Dec 19 '18 at 18:03
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With reference to $N_o$ this usually is the symbol for the power spectral density of thermal noise, where $N_o = kT$, where k is Boltzmann's Constant and T is the temperature in Kelvin. The thermal noise signal is a complex, white Gaussian distributed noise, with half of the power in the real component and half the power in the imaginary component (but power in itself is not a complex quantity). Therefore if the total power density is $N_o$, the power density of the real component would be $No/2$ and the power density of the imaginary component would also be $N_o/2$.

Time domain view

Considering the complex signal we can represent the spectral density as "Single-Sided" or "Double-Sided". With a Single-Sided power spectral density, the frequency axis starts at 0 and extends to infinity, making use that the the negative frequency components have the same density as the positive. We do not need to show this redundant information, but we do need to double the quantity that is shown to account for the power that extends into the negative frequencies. If we include the negative frequency axis, as done in a Double-Sided power spectral density plot, then the density is given as $N_o/2$.

Frequency Domain View

In the end in either case the power density is $N_o$ Watts/Hz. In the Single Sided case the power in 1 Hz of bandwidth is completely shown on the positive axis, while in the Double Side case, half the power is in the positive axis and half is in the negative axis.

Thus in this we have made two considerations in how the power may be distributed; either in the positive and negative frequency axis if we present Single or Double-sided spectrums, and in the real and imaginary axis if we consider how the power in the full complex signal ($N_o$) is distributed between real and imaginary components. For example, if we were only to consider the real component, the power spectral density would be $N_o/2$ and then if we viewed that as a Double Sided PSD, the PSD would be $N_o/4$.

Single and Double Sided power spectral densities are frequency encountered when working with phase noise, which may provide a further intuitive way of seeing this: The double sided PSD for phase noise is given as $\scr{L}_\phi(f)$ and represents what we would read directly off of a spectrum analyzer, where we can observe both the upper and lower sidebands of the spectrum (although we typically only report one side on the plot which causes confusion in if it should be called single or double sided; one side is shown but it is a double sided power spectral density quantity--- we just don't need to show the other side since it is identical). The single sided power spectral density $S_\phi(f)$ represents the power from both sidebands as given on a plot with a positive frequency axis only (since the negative sideband is equivalent to the positive, we can put it all on the positive frequency axis simply by doubling in power what we read off of the spectrum analyzer). So $S_\phi(f)$ represents all the power and is used in related calculations where we integrate over a band of frequency offsets, while $\scr{L}_\phi(f)$ is convenient in that we can read it directly off of test equipment (spectrum analyzers).

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