Since this question is book-oriented, I will kindly ask you to accompany it with a book that is considered by many researchers the bible of Digital Communication: Proakis - Digital Communications, one from the 4th ed, and another from the 5th ed.

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

$$S_x(f) = \left\{ \begin{array}{cl} \frac{N_0}{2}, & \vert f \pm f_0 \vert < W/2 \\ 0, & \text{Otherwise} \\ \end{array}\right., \tag{0}$$ where $f_0$ is the carrier frequency and $W$ is the bandwidth of $x(t)$. What is the PSD of the lowpass equivalent of $x(t)$? If you are a Communication Systems Engineer, I am almost sure that you answered it without hesitating: it is $$S_{x_l}(f) = \left\{ \begin{array}{cl} N_0, & \vert f \vert < W/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.,$$ where $x_l(t)$ is the lowpass equivalent of $x(t)$. I suppose you will answer it as I see it everywhere, in any book or any paper. It turns out that it is not necessarily $N_0$.

Until the 4th ed., Proakis had considered a normalization in the autocorrelation function definition

$$R_x(\tau) = \frac{1}{2}E\{x^*(t)x(t+\tau)\}.$$

In the author's own words (on pg 76 of the 4th ed): "The factor of $\frac{1}{2}$ in the definition of the autocorrelation function of a complex-valued stochastic process is an arbitrary but mathematically convenient normalization factor". Indeed, such normalization makes the lowpass equivalent of the autocorrelation function be equal to the autocorrelation function of $x_l(t)$ (I know, that is tortuous, but check both books and you will notice it), which is reasonable.

On the 5th ed (where the book was heavily refactored), Proakis has defined

$$R_x(\tau) = E\{x^*(t)x(t+\tau)\},\tag{1}$$

it changes many mathematical definitions I (and you too, probably) had learned about random signal analysis in communication systems. One of theses is that (see EXAMPLE 2.9–1. of the 5th ed.) $$S_{x_l}(f) = \left\{ \begin{array}{cl} 2N_0, & \vert f \vert < W/2 \\ 0, & \text{Otherwise} \\ \end{array}\right. \tag{2}. $$

I often use different statistical signal processing techniques from related areas (adaptive filters, machine learning, numeric optimization, etc...) in my systems, and they always define the autocorrelation function in the canonical way (equation (1)). Therefore, the mathematical definitions of the 5th ed. is awesome for me. However, the canonical form leads to the equation (2), which I've never seen someone defining $S_{x_l}(f)$ like that, thus making me insecure about how to define it.

The question is: Is it reasonable to define this PSD as $2N_0$?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Peter K.
    Nov 6, 2022 at 20:49
  • $\begingroup$ I've locked this question until some civility returns to the discussion. Please continue the discussion -- civilly, without personal abuse like @RubemPacelli directs -- in chat and come to a resolution. $\endgroup$
    – Peter K.
    Nov 6, 2022 at 20:51

3 Answers 3


With regards to White Gaussian Noise, $N_o$ is used to represent the total power spectral density within a band of interest. There is no arbitrary scaling with this (otherwise there would be no consistency with all the $C/N_o$ curves that we use for establishing the bit error rate given SNR). Scaling that is applied depends specifically on its use with the motivation of making the result consistent with total power in every case. The math is such that 1/2 + 1/2 = 1. It really need not be made much more complicated than that.

$N_o$ is a power spectral density of white noise, meaning power over a unit bandwidth, and typically units of Watts/Hz. To simplify the explanation with no loss of clarity, let's consider a waveform that occupies just 1 Hz of bandwidth ($W=1$) as I proceed with examples below so that I don't have to keep carrying $W$ along. In this case the total power in that same bandwidth due to the White Gaussian Noise will be $N_o$ Watts.

Now we can consider if we want to represent the signal as a real passband signal (as we would do when it is modulating an RF carrier), or as the equivalent complex baseband signal (as we would do when creating modern waveforms in DSPs prior to up-conversion). In both cases we will represent the frequency domain using positive and negative frequencies, and the waveforms themselves are equivalent in total power and spectral occupancy: What this means is as a passband signal we occupy 1 Hz both in the positive frequency axis and 1 Hz in the negative frequency axis, and as a complex baseband signal the occupancy would span from $f=-1/2$ to $f=+1/2$.

For the case of the complex baseband signal: The noise $N_o$ will have real and imaginary components (we refer to the real components as I for "in-phase" and the imaginary components as Q for "quadrature"). The noise in this case for a typical complex baseband Gaussian White Noise process will be split equally across I and Q, so within our 1 Hz band of interest (that would span from $f=-1/2$ to $f=+1/2$ in the complex baseband signal), we will have $N_o/2$ for the total noise power in $I$ and $N_o/2$ for the total noise power in $Q$. The total noise (surprise, surprise!) will be $N_o$.

For when we are representing our narrowband signal as a real passband signal, then this signal occupies a 1 Hz at both the positive and negative frequencies. The signal and the noise are split equally in power between the positive and negative frequencies. A very simple and concise example of this with regards to a signal is to consider a simple cosine wave and Euler's formula:

$$\cos(\omega_c t) = \tfrac{1}{2}e^{j\omega_c t} + \tfrac{1}{2}e^{-j\omega_c t}$$

We clearly see from this the positive and negative frequency representation ($e^{j\omega_c t}$ being a single positive frequency tone) and how the power is split equally between the two for this case of a signal. The result for our total noise $N_o$ would be the same! So half of the noise power as $N_o/2$ is occupying the 1 Hz band in the negative frequency axis, and the other half of the noise power as $N_o/2$ is occupying the 1 Hz band in the positive frequency axis. The total noise (surprise, surprise!) will be $N_o$.

There are also cases where we are only interested in the real portion of a complex baseband signal (for example, demodulating BPSK coherently). In this case, we can discard the quadrature component and all of its noise, and the resulting noise power spectral density will be $N_o/2$.

I don't yet see any good reason, other than to cause complete confusion, in representing the total noise power spectral density as $2N_o$ given the convention of having $N_o$ represent the total noise power spectral density in a given band of interest.

  • $\begingroup$ @RubemPacelli Welcome to DSP.SE. This is a friendly forum and we generally help each other out and learn in the process. I did read your references and specifically the paper before commenting. Also I and am not arguing that you can arrive at $2N_o$ if you use arbitrary scaling. I was making a point here on the importance of not arbitrarily scaling $N_o$ in certain applications which I detailed. My final concluding paragraph is of practical significance. (There could be a good reason for the total noise power in a band of interest to be $2N_o$, I just don't yet see any good reason to do so) $\endgroup$ Nov 3, 2022 at 13:53
  • $\begingroup$ Actually it is on the other way around: If you take the arbitrary scaling, you get $N_0$, if you don't, you get $2N_0$. That is the problem. It conflicts with the theory of areas related to signal processing, as the scaling of $\frac{1}{2}$ is arbitrary and does not agree with the autocorrelation function from other books. $\endgroup$ Nov 3, 2022 at 14:01
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    $\begingroup$ //The math is such that 1/2 + 1/2 = 1. It really need not be made much more complicated than that.// ------ yupper. Add that energy in the positive frequencies to the identical amount of energy in the negative frequencies, and you have the total energy. $\endgroup$ Nov 4, 2022 at 16:44

is there an error in this definition on book? If you look carefully this section, you are going to see that Proakis denotes $x(t)$ as the real bandpass signal. If $S_x(f)$ covered only the positive frequencies, $x(t)$ would be complex, which contradicts my previous statement.

I don't know about the book. My undergrad communications was Carlson, my graduate communications text was Wosencraft and Jacobs and I had a statistical communications text that was Van Trees or somebuddy. I also had a really formal and fundamental signals and systems course using a rare book by Papoulis. Never had Proakis.

If $x(t)$ is real, then the spectrum $X(f)$ is even-symmetric in the real part (and in the magnitude) and odd-symmetric in the imaginary part (and in the phase response). That means if you have some thing up there at $f_0$, then there is also a mirror image of the thing at $-f_0$.

Now if half of your energy is up at $f_0$, we call that half of the energy "$\frac{N_0}{2}$" and it adds to the other half of the energy at $-f_0$ and you have a total energy of $N_0$.

Also if

$$ x(t) = x_l(t) \cos(2 \pi f_0 t) $$

Then you'll see the amplitude (I didn't say "energy") of $X_l(f)$ cut in half and the two half spectrums repeated at $-f_0$ and $+f_0$.

  • $\begingroup$ Regardless of which revered textbook you used in your studies, I woukld suggest that the white noise process that the OP has defined has bandwidth 2W extending from $f_0-W$ to $f_0+W$ (cf. $$S_x(f) = \left\{\begin{array}{cl}\frac{N_0}{2}, & \vert f - f_0 \vert < W \\ 0, & \text{Otherwise} \\\end{array}\right.,$$ which is an exact quote (copy-and-paste) from the OP's question. $\endgroup$ Nov 2, 2022 at 3:50
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    $\begingroup$ All bandlimited signals are real; there are no complex-valued bandpass signals. We can represent a real-valued bandpass (RF) signal as I and Q baseband components modulated onto cosine and sine carriers. Ditto for noise. There is no complex-valued bandpass noise, white or otherwisem but we can represent bandpass noise with I and Q low-pass equivalents. $\endgroup$ Nov 2, 2022 at 4:09
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    $\begingroup$ @robertbristow-johnson As Proakis has said: "A bandpass signal is a real signal whose frequency content, or spectrum, is located around some frequency $\pm f_0$" which is far from zero. So Dilip is right, indeed, there is no complex-valued bandpass signals. $\endgroup$ Nov 2, 2022 at 9:33
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    $\begingroup$ I'm sorry Ruben, the mathematics speak far more loudly than either Dilip or Proakis. ANY spectrum that is not Hermitian symmetric (that is real-even, imag-odd) must have a non-zero imaginary part. Given $$S_x(f) = \left\{ \begin{array}{cl} \frac{N_0}{2}, & \vert f - f_0 \vert < W \\ 0, & \text{Otherwise} \\ \end{array}\right.,$$ the only way for $x(t)$ to be purely real is for $f_0$ to be zero. $\endgroup$ Nov 2, 2022 at 17:14
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    $\begingroup$ Is Dilip's comment about physical realization? I don't see what's nonexistent about $\mathcal{F}^{-1}\{u(\omega) - u(\omega - 1)\}$. I also find the -1 unwarranted. $\endgroup$ Nov 3, 2022 at 5:41


Yes, it is sensible to state that $$S_{x_l}(f) = \left\{ \begin{array}{cl} 2N_0, & \vert f \vert < W/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.,$$ as long as you define the equations coherently.

Long question

I've finally found the solution! This problem seems to be an old issue that annoyed many researchers out there. There is even a very good paper [1] that address the normalization problem and how it affects all the definitions. The mathematical analysis you gonna find here comes from Proakis 5th ed., but mainly from the paper [1]. I will choose to not make the normalization since it agrees with the autocorrelation function definition from related areas (adaptive filters, machine learning, numeric optimization, etc...).

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 $$S_x(f) = \left\{ \begin{array}{cl} \frac{N_0}{2}, & \vert f \pm f_0 \vert < W/2 \\ 0, & \text{Otherwise} \\ \end{array}\right.. \tag{5}$$

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 $R_{x_i}(\tau)$)

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

Similarly, recording that $x_l(t) = x_i(t) + j x_q(t)$ and using the equations (1) and (2), we have that (without normalizing it by $\frac{1}{2}$!!) [1]:

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

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

$$ \boxed{R_{x_l}(\tau) = 2 R_{x_i}(\tau)} \tag{7} $$

Which gives

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

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.

As you could notice, This deduction is far from obvious, and the issue over the normalization worsen everything.

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    $\begingroup$ Nicely done! :-) $\endgroup$
    – Peter K.
    Nov 2, 2022 at 20:11
  • $\begingroup$ I would have thought equation 3 should be: $R_{x}(\tau) = R_{x_i}(\tau) \cos{2\pi f_0 \tau} - R_{x_q}(\tau) \sin{2\pi f_0 \tau}$ Do you have a page number / formula number for your reference where that is expanded / derived by Proakis? As written it suggests the quadrature component of the complex baseband noise has no impact on the equivalent real passband signal - but we know that isn't true, so something doesn't add up there or I am misunderstanding the statement. And what is the purpose of your equation 2, aren't they both equal to 0 given I and Q are IID? $\endgroup$ Nov 3, 2022 at 3:21
  • $\begingroup$ @DanBoschen "Do you have a page number / formula number for your reference where that is expanded / derived by Proakis?" here $\endgroup$ Nov 3, 2022 at 10:20
  • $\begingroup$ "As written it suggests the quadrature component of the complex baseband noise has no impact on the equivalent real passband signal - but we know that isn't true." that is true, please read my references before making conclusions. $\endgroup$ Nov 3, 2022 at 10:26
  • $\begingroup$ @DanBoschen thank you for reading it!! Yes, I did the writing by hand. At the beginning totally incredulous, believing that there is an error as I always had in my mind that the PSD is $N_0$, and not $2N_0$. For my susprise, Proakis 5th is right. Then I find this paper, I redo again and, well, that is right too!! Just then I came here and opened this post. $\endgroup$ Nov 3, 2022 at 14:13

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