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?

  • $\begingroup$ t or f, frequency? $\endgroup$
    – user28715
    Commented Dec 19, 2018 at 14:57
  • $\begingroup$ Thank you, it is Sn(f). But how do I compute that? $\endgroup$
    – Phobos
    Commented Dec 19, 2018 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. $\endgroup$ Commented Dec 19, 2018 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! $\endgroup$ Commented Dec 19, 2018 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 $$s_n^2 = N_0$$ $\endgroup$
    – Phobos
    Commented Dec 19, 2018 at 16:23

3 Answers 3


@Dan Boschen gives a precise answer in the comment. I would like to add to that a simple derivation using the formal definition as

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

Where $B$ is the bandwidth, and $E$ stands for expectation and the process $x(t)$ is the white noise process described in OP. (Since the noise density is flat in frequency as white, we can simply multiply by $B$ to get the total power as given by $ E \vert x(t) \vert^2$. For pure white noise $B$ extends to infinity, but that doesn't prohibit this simple derivation from proceeding.

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

$$S_{xx}(f)B = E \vert x(t) \vert^2= E \vert r(t) \vert^2 + E \vert i(t) \vert^2 $$ 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. $\frac{BN_0}{2}$ hence we get

$$S_{xx}(f)B = \frac{BN_0}{2} + \frac{BN_0}{2}=B N_0 $$

  • $\begingroup$ Thank you a lot! It really helped me understand it a bit better... $\endgroup$
    – Phobos
    Commented Dec 19, 2018 at 16:14
  • $\begingroup$ So I can say that the PSD is equal to the Variance of the random process? $\endgroup$
    – Phobos
    Commented Dec 19, 2018 at 16:20
  • $\begingroup$ Yes and not only that. The PSD is uniform over all frequencies, which is observed using the last equation. $\endgroup$ Commented Dec 19, 2018 at 16:21
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    $\begingroup$ 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 $R(0)=E(|n(t)|^2)$? $\endgroup$
    – Phobos
    Commented Dec 19, 2018 at 18:03
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    $\begingroup$ $S_{xx}(f)$ is not the same as $E|x(t)|^2$. $E|x(t)|^2$ is the total_power_ of the process which equals the area under the PSD curve, that is, $$E|x(t)|^2 = \int_{-\infty}^\infty S_{xx}(f) \,\mathrm df.$$ $\endgroup$ Commented Jan 11, 2022 at 21:35

With reference to $N_o$ this usually is the symbol for the power spectral density (PSD) of thermal noise, where $N_o = kT$, where k is Boltzmann's Constant and T is the temperature in Kelvin. With regards to a complex baseband signal, 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 PSD from the real component alone would be $No/2$ and the PSD from the imaginary component alone would also be $N_o/2$.

Time domain view

Considering a real signal we can represent the PSD as "One-Sided" or "Two-Sided", but a complex signal would always have a "Two-Sided" PSD. With a One-Sided PSD, 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 Two-Sided PSD plot, then the density is given as $N_o/2$. For the case of a complex time domain waveform, the positive and negative frequencies in the spectrum are independent of each other, and therefore we must show a plot as Two-Sided and with the frequency axis extending to $\pm \infty$ (or at least out to the occupied bandwidth in both positive and negative frequency directions) to show all information about the signal's PSD.

Frequency Domain View

In the end in either case the PSD is $N_o$ Watts/Hz. In the One-Sided case the power in 1 Hz of bandwidth is completely shown on the positive axis, while in the Two-Sided case, half the power is in the positive axis and half is in the negative axis. Regardless of this due to the redundancy from complex conjugate symmetry for the spectrums of real signals, we still only need to show the positive frequencies only, such that a Two-Sided PSD would be 3 dB lower (1/2 the power) of the One-Sided PSD for the same signal even though both plots show the positive frequency axis only.

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 One or Two-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 Two-Sided PSD, the PSD would be $N_o/4$.

One and Two-Sided PSD's are frequency encountered when working with phase noise, which may provide a further intuitive way of seeing this: The Two-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 One or Two-sided; one side is shown but it is a double sided PSD quantity--- we just don't need to show the other side since it is identical). The One-Sided PSD $S_\phi(f)$ 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).

Not understanding this or accounting for this properly can lead to a 3 dB error when trying to determine SNR or total noise within a bandwidth of interest.

  • $\begingroup$ I have been thinking. How is $N_0/2$ always $k_BT$. Doesn't the PSD also depend on the time duration of the thermal noise as well. I would have thought that a finite duration excerpt of white noise has a smaller PSD but spreads out in bandwidth. $\endgroup$ Commented Nov 18, 2021 at 17:40
  • $\begingroup$ @LewisKelsey It is a noise density which normalizes time (W/Hz or dBm/Hz etc). So the total noise is absolutely dependent on the bandwidth as you suspected. $\endgroup$ Commented Nov 18, 2021 at 17:42
  • $\begingroup$ From what I've worked out PSD is the energy at the frequency, so still depends on time, so when is it $k_BT$? at what time duration? $\endgroup$ Commented Nov 18, 2021 at 18:07
  • $\begingroup$ “density” by definition is an every per unit bandwidth; typically per Hz (or per root-Hz when given in magnitude instead of power quantities. White Noise has no energy at one exact frequency, it is a distribution so we must define the bandwidth for the reasons you are finding $\endgroup$ Commented Nov 18, 2021 at 18:25
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    $\begingroup$ For instance the DFT itself is a good example of this: the result at one bin is not due to the energy at that one frequency but the distributed energy resulting in the “resolution bandwidth” or “equivalent noise bandwidth” of that one bin. We should avoid an ongoing chat/discussion here in these comments since that is discouraged by the moderators but if you think one of us or both of us are still confused I’d be happy to open a chat area where we can discuss (heading out now but could do this later) $\endgroup$ Commented Nov 18, 2021 at 18:32


it is

$$S_{x_l}(f) = \left\{ \begin{array}{cl} 2N_0, & \vert f \vert < B/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.,$$ or $$S_{x_l}(f) = \left\{ \begin{array}{cl} N_0, & \vert f \vert < B/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.,$$ depends whether you do a normalization, vide my original post

Long question

The answer

Let $$ x(t) = \text{Re}[x_l(t) e^{j2\pi f_0 t}]$$ be a filtered white Gaussian noise, where $x_l(t) \in \mathbb{C}$ is the complex envelope (also called the lowpass equivalent) of $x(t)$. The Power Spectral Density (PSD) of $x(t)$ is given by

$$S_x(f) = \left\{ \begin{array}{cl} \frac{N_0}{2}, & \vert f \pm f_0 \vert < B/2 \\ 0, & \text{Otherwise} \\ \end{array}\right., \tag{0}$$ where $f_0$ is the carrier frequency and $B$ is the bandwidth of $x(t)$.

Additionally, let $x_i(t)$ and $x_q(t)$ be the phase and quadrature components of $x_l(t)$, i.e., $x_l(t) = x_i(t) + j x_q(t)$. Assuming the stationarity of $x(t)$ leads to the following properties [2, 3]: $$ R_{x_i}(\tau) = R_{x_q}(\tau) \tag{1} $$ and $$ R_{x_i,x_q}(\tau) = -R_{x_q, x_i}(\tau) \tag{2} $$

Recording that $x(t) = x_i(t) \cos{2\pi f_0 t} - x_q(t) \sin{2\pi f_0 t}$ and using the equations (1) and (2), we have that [2]:

$$ R_{x}(\tau) = R_{x_i}(\tau) \cos{2\pi f_0 \tau} - R_{x_q,x_i}(\tau) \sin{2\pi f_0 \tau} \tag{3} $$

Since $x_i(t)$ and $x_q(t)$ are independent processes, $R_{x_q,x_i}(\tau) = 0$ and the equation (3) reduces to

$$ \boxed{R_{x}(\tau) = R_{x_i}(\tau) \cos{2\pi f_0 \tau}} \tag{4} $$

That is the autocorrelation function of the bandpass signal, $x(t)$. But remember that $$S_x(f) = \left\{ \begin{array}{cl} \frac{N_0}{2}, & \vert f \pm f_0 \vert < B/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.. \tag{5}$$

If you are good at signals and systems, you should already have noticed that (remember that $S_{x_i}(f)$ is the Fourier transform of $R_{x_i}(\tau)$)

$$S_{x_i}(f) = \left\{\begin{array}{cc} N_0 & \vert f \vert < B/2 \\ 0 & \text{Otherwise}. \end{array}\right. \tag{6}$$

Similarly, recording that $x_l(t) = x_i(t) + j x_q(t)$ and using the equations (1) and (2), we have that (we are normalizing it by $\frac{1}{2}$!!) [1]:

$$ R_{x_l}(\tau) = \frac{1}{2}E[x_l^*(t)x_l(t + \tau)] = R_{x_i}(\tau) + j R_{x_q,x_i}(\tau) \tag{7} $$

Again, since $x_i(t)$ and $x_q(t)$ are independent, it follows that $R_{x_q,x_i}(\tau) = 0$.

$$ \boxed{R_{x_l}(\tau) = R_{x_i}(\tau)} \tag{7} $$

Which gives

$$S_{x_l}(f) = \left\{ \begin{array}{cl} N_0, & \vert f \vert < B/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.. \tag{8}$$

Notice that, although $x_l(t)$ is complex, $R_{x_l}(\tau)$ and $S_{x_l}(f)$ are reals [1]!!! More specifically, they are reals and evens: $S_{x_l}(f)$ is square function, and $R_{x_l}(\tau)$ is a sinc.

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    $\begingroup$ I already predicted the amount of downvotes, but I suspect it's due to who is answering, and not to errors in the answer. If I am wrong, please, let the reason of downvotes in the comments so that I can increase the quality of my answer :) $\endgroup$ Commented Dec 21, 2022 at 13:03

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