I was following this interesting post by a new user Rubem Pacelli and got stuck at Proakis' referenced definition (see Section 4-1-4 starting on page 159 here). The math, all repeated further below, makes sense to me but my intuition doesn't and was hoping to better connect the two. It is likely a deeper detail I am missing with Stochastic Processes and my related assumptions. I provide my intuition first below, which completely makes sense to me, and then below that Proakis' derivation which then turns my world upside down. My question is what is the disconnect in my thought process? (I highly doubt Proakis is wrong since this is the 4th edition, and also agree with all the math he has given).

My intuition (with some math):

The real and imaginary components of a complex baseband signal are preserved in the equivalent real passband signal centered about a carrier frequency according to:

$$ x(t) = x_i(t) \cos(\omega_c t) - x_q(t) \sin(\omega_c t) \label{1}\tag{1}$$


$$ x_c(t) = x_i(t) + j x_q(t)\label{2}\tag{2}$$


$x(t)$: (real) passband signal

$x_c(t)$: (complex) baseband signal

$\omega_c$: (real) the carrier frequency as $2\pi f_c$ where $f_c$ is in Hz.

$x_i(t)$: (real) the real component of the complex baseband signal

$x_q(t)$: (real) the imaginary component of the complex baseband signal

At this point I believe the above result would apply regardless of $x_i(t)$ and $x_q(t)$ being deterministic signals or stochastic signals. This may be the root of my problem or lack of understanding since Proakis shows otherwise. Before I continue to Proakis' reference, let me bottom line the above to say that if I applied a real stochastic signal to the quadrature component only of the baseband signal, I would expect to see that same stochastic signal (relative to the carrier) in the passband signal, just as we would with a deterministic signal. Proakis derivation (or my misunderstanding of it) seems to suggest that it gets completely cancelled.

Bottom line of Proakis's derivation:

In Proakis' derivation he gets the following result matching formula 4-1-45 on page 160 for Digital Communications 4th edition.

$$\phi_{nn}(\tau) = \phi_{xx}(\tau)\cos(2\pi f_c \tau) - \phi_{yx}(\tau)\sin(2\pi f_c \tau)$$


$\phi_{nn}(\tau)$: Auto-correlation of the passband signal

$\phi_{xx}(\tau)$: Auto-correlation of the real component ($x_i(t)$) of the complex baseband signal

$\phi_{yx}(\tau)$: Cross-correlation of the real and imaginary components of the complex baseband signal

So right away, this is in complete contrast with the intuition. The passband signal is the autocorrelation of the real component only of the complex baseband signal (!?). The cross correlation is zero when the real and imaginary parts are independent, so in that case where they are we would completely lose the quadrature component of the baseband signal. I don't get this.

Specifically why is this not as I give below, matching the form and intuition from $\ref{1}$

$$\phi_{nn}(\tau) = \phi_{xx}(\tau)\cos(2\pi f_c \tau) - \phi_{yy}(\tau)\sin(2\pi f_c \tau)$$

Assuming a stationary process, if we take the Fourier Transform of the result we get the power spectral density. So it appears that Proakis' result is suggesting that the noise process on the quadrature component of the complex baseband signal does NOT affect the spectrum. This part is in complete conflict with my intuition.

I can almost see an equivalence when the random process for the in phase and quadrature components are independent and identically distributed- since the power spectral density for each would match. He does state this indirectly with equation 4-1-40 that $\phi_{xx} = \phi_{yy}$ which would make that so. However that is not my issue as is seems the derivation as presented from equations 4-1-42 to 4-1-44 would similarly proceed without that restriction. To reiterate my question with this context: if you had a random process for the real component that is still independent but not identically distributed as the imaginary component, and specifically where $\phi_{xx} \neq \phi_{yy}$ (for an extreme example what if we nearly zeroed out the real component?), how would the derivation as given from 4-1-42 to 4-1-44 capture the effect of the different distributions? Or more simply this suggests the quadrature noise is suppressed and if so, why is this different from the case for deterministic signals where we can control the real and imaginary components independently without loss??

Further Details For Intuition on \ref{1} and \ref{2} for the interested:

It's easy to see how we get the results given for \ref{1} and \ref{2} if we consider the process of converting a complex baseband signal to a real passband signal as multiplying by $e^{j\omega t}$, which shifts the spectrum to the right, and then taking the real part (which will provide all the information content in the complex baseband signal to a single antenna or wire, and in the process discard to redundant imaginary component).

Complex Baseband and Passband Spectrums

Note the Complex Passband shown as $x_c(t)e^{j\omega_c t}$ when expanded into real and imaginary components would be:

$$x_c(t)e^{j\omega_c t} = (x_i(t)+jx_q(t))(\cos(\omega_c t) + j\sin(\omega_c t))$$

And the real portion of the above would be:

$$Re\{x_c(t)e^{j\omega_c t}\} = x_i(t)\cos(\omega_c t) - x_q(t)\sin(\omega_c t)$$

If you expand out $\cos(\omega_c t)$ and $\sin(\omega_c t)$ using Euler's formula to separate the negative frequency components (terms of $e^{-j\omega_c t}$) and positive frequency components (terms of $e^{j\omega_c t}$), you get the expression as I show it in the third spectrum in the graphic above.

  • 1
    $\begingroup$ Dan, does this short answer clear up things a bit? $\endgroup$
    – Matt L.
    Commented Nov 4, 2022 at 7:21
  • 1
    $\begingroup$ And probably this one too. $\endgroup$
    – Matt L.
    Commented Nov 4, 2022 at 7:27
  • $\begingroup$ You should call $x_c(t)$ the "analytic signal" of $x_i(t)$. Or is $x_q(t)$ not necessarily the Hilbert Transform of $x_i(t)$? $\endgroup$ Commented Nov 5, 2022 at 4:58
  • $\begingroup$ @robertbristow-johnson: $x_c(t)$ is just the complex baseband signal, which is generally not analytic. It would be analytic for SSB. $\endgroup$
    – Matt L.
    Commented Nov 5, 2022 at 10:29
  • $\begingroup$ When it is frequency translated it is clearly the analytic signal (the complex passband in the middle) of the real signal with the spectrum at the bottom of the plot. So the very top plot is the spectrum for the “frequency translated analytic signal” of the time domain signal with spectrum shown in the bottom of the plot; which as Matt points out isn’t an analytic signal itself (has negative freq components). Your links are indeed very helpful Matt, and relevant. $\endgroup$ Commented Nov 5, 2022 at 14:04

4 Answers 4


I've been a bit hesitant about adding another answer to the existing ones, but since no answer has been accepted yet, and since from Dan's own answer it seems to me that there might still be a few misunderstandings, I finally decided to write up a few things, some of which may be redundant.

First of all, it's important to be clear about the different kinds of signals, and the respective conditions for their wide-sense stationarity. Let $X(t)$ denote the complex baseband signal, with its real-valued in-phase and quadrature components $I(t)$ and $Q(t)$, respectively:


Assuming that $I(t)$ and $Q(t)$ are jointly wide-sense stationary (WSS), it's straightforward to show that $X(t)$ is also WSS:


where $m_X$, $m_I$, and $m_Q$ denote the mean values of the respective processes. The autocorrelation of $X(t)$ is given by


So if $I(t)$ and $Q(t)$ are jointly WSS, then the mean of $X(t)$ is a constant, and its autocorrelation does not depend on $t$, which makes $X(t)$ WSS.

The next process we'll consider is the complex modulated signal


Its mean is


which can only be independent of $t$ if $m_X=0$, i.e., if $m_I=m_Q=0$. The autocorrelation function is


Consequently, the modulated complex process $Y(t)$ is WSS under the additional condition that $I(t)$ and $Q(t)$ are zero mean.

Finally, we're interested in the real-valued bandpass process given by


From $m_X=0$, it follows that $m_Y=0$, and, consequently, $m_Z=0$. The autocorrelation of $Z(t)$ is


The autocorrelation of the real-valued bandpass process $Z(t)$ can only become independent of $t$ if we can make $R_{YY^*}(t,\tau)$ independent of $t$. The cross-correlation $R_{YY^*}(t,\tau)$ is given by


which only becomes independent of $t$ if $R_{XX^*}(\tau)=0$. That cross-correlation can be expressed as


Consequently, the condition $R_{XX^*}(\tau)=0$ is equivalent to the conditions $R_I(\tau)=R_Q(\tau)$ and $R_{IQ}(\tau)=-R_{IQ}(-\tau)$. With these additional conditions, the real-valued bandpass process $Z(t)$ is WSS. In that case, its autocorrelation is given by


where I've used $(8)$, $(6)$, and $(3)$, and the conditions $R_I(\tau)=R_Q(\tau)$ and $R_{IQ}(\tau)=-R_{IQ}(-\tau)$.


  • the complex-valued baseband process $X(t)$ is WSS if the in-phase and quadrature components $I(t)$ and $Q(t)$ are jointly WSS.

  • the complex-valued modulated process $Y(t)=X(t)e^{j\omega_ct}$ is WSS if in addition to above conditions, the in-phase and quadrature components are zero mean processes.

  • the real-valued bandpass process $Z(t)=\textrm{Re}\{Y(t)\}$ is WSS if the additional conditions $R_I(\tau)=R_Q(\tau)$ and $R_{IQ}(\tau)=-R_{IQ}(-\tau)$ are satisfied.

  • $\begingroup$ Thanks Matt- Digesting this now. Did I misunderstand something in my own answer, or are you referring to my statements of remaining questions/ confusions? (Just checking is something I wrote was incorrect as I understood it) $\endgroup$ Commented Nov 9, 2022 at 12:15
  • $\begingroup$ equation 5, isn't the mean of $e^{j\omega_c t}=0$, so $m_Y(t)=0$? Or is it the point that it would only be 0 when $t$ is at a full cycle or infinity, which isn't a restriction here, so cyclostationary? So leading to other thoughts that it would be independent of $t$ when sampled at $\omega_c$ (knowing we're not introducing sampling here but confirming my thoughts/understanding) $\endgroup$ Commented Nov 9, 2022 at 12:21
  • $\begingroup$ Really good, really clean. Thank you! I especially liked the process used in equation 8. And the remaining "intuition" I was seeking is now clear to me if I consider the "Heyser corksrew" view of $e^{j\omega t}$ (Fig 6 at andrewduncan.net/air) Basically if the statistics of I and Q in the baseband are not equal, the statistics of the passband will keep modulating between the statistics of I and that of Q, and therefore can only be cyclostationary. When the are equal it modulates between the same and therefore stationary. $\endgroup$ Commented Nov 9, 2022 at 12:39
  • 1
    $\begingroup$ @DanBoschen: Concerning Eq. (5), $e^{j\omega_ct}$ is deterministic, so we pull it out of the expectation operator, and it remains as a time-dependent quantity. $\endgroup$
    – Matt L.
    Commented Nov 9, 2022 at 12:47
  • 1
    $\begingroup$ @DanBoschen: As for your answer, one thing I thought might still be unclear was your second bullet point "If the complex signal is stationary, then the real and imaginary components will have an identical auto-correlation ...". This is not the case. The complex signal (baseband as well as passband) can be stationary without that condition. You do need that condition for the real-valued bandpass signal to be WSS. $\endgroup$
    – Matt L.
    Commented Nov 9, 2022 at 12:50

Work in progress Please hold all comments, complaints, questions, brickbats and downvotes unti I am done and have removed this request.


Suppose $\{X(t)\colon -\infty < t < \infty\}$ and $\{Y(t)\colon -\infty < t < \infty\}$ are two continuous-time real-valued zero-mean wide-sense stationary (WSS) low-pass random processes with autocorrelation functions $R_{X}(\tau)$ and $R_{Y}(\tau)$ respectively, and power spectral densities $S_{X}(f)= \mathcal F\big[R_{{X}}(\tau)\big]$ and $S_{Y}(f)= \mathcal F\big[R_{{Y}}(\tau)\big]$ respectively. The low-pass assumption will be taken to mean that $S_{X}(f) = S_{Y}(f) = 0$ for all $|f| > \frac W2$. We also assume that the processes are jointly wide-sense stationary, that is, $E[X(t)Y(t+\tau)]$ is also a function of $\tau$ alone, and so can be denoted as $R_{X,Y}(\tau)$ instead of the more prolix $R_{X,Y}(t,t+\tau)$.

The OP constructs a bandpass process $\{N(t)\colon -\infty < t < \infty\}$ defined by $$N(t) = X(t)\cos(2\pi f_c t) - Y(t)\sin(2\pi f_c t),~ -\infty < t < \infty,\tag{1}$$ where $f_c \gg W$ and asks about its properties. Well, firstly let's determine whether $\{N(t)\}$ is a WSS process. Since $E[X(t)]=E[Y(t)]=0$, it is clear that $E[N(t)]=0$ also. On the other hand, \begin{align} E[N(t)N(t+\tau)] &= E\bigr[ \big(X(t)\cos(2\pi f_c t) - Y(t)\sin(2\pi f_c t)\big) \\ & \hspace{.3in}\cdot \big(X(t+\tau)\cos(2\pi f_c (t+\tau)) - Y(t+\tau))\sin(2\pi f_c (t+\tau))\big)\bigr]\\ &= R_{X}(\tau)\cos(2\pi f_c t)\cos(2\pi f_c (t+\tau)) + R_{Y}(\tau)\sin(2\pi f_c t)\sin(2\pi f_c (t+\tau))\\ & -R_{X,Y}(\tau)\cos(2\pi f_c t)\sin(2\pi f_c (t+\tau))-R_{X,Y}(-\tau)\sin(2\pi f_c t)\cos(2\pi f_c (t+\tau)) \end{align} which in general is a function of both $t$ and $\tau$.

In general, $\{N(t)\}$ is not a WSS process.

However, if $\{X(t)\}$ and $\{Y(t)\}$ have identical autocorrelation functions (that is, $R_{X}(\tau)=R_{Y}(\tau)$), then the first line in the second equality above simplifies to $R_{X}(\tau)\cos(2\pi f_c\tau)$ (or to $R_{Y}(\tau)\cos(2\pi f_c\tau)$ for those who march to the beat of a different drummer). Similarly, if $R_{X,Y}(\tau)$ is an odd function of $\tau$ (that is, $R_{X,Y}(-\tau) = -R_{X,Y}(\tau)$), then the second line above simplifies $-R_{X,Y}(\tau)\sin(2\pi f_c \tau)$ (which fortunately is also an even function of $\tau$) leading to $$E[N(t)N(t+\tau)] = R_{X}(\tau)\cos(2\pi f_c\tau) -R_{X,Y}(\tau)\sin(2\pi f_c \tau) = R_{N}(\tau), \tag{2}$$ that is, $\{N(t)\}$ as defined in $(1)$ is a zero-mean WSS process with autocorrelation function as given in $(2)$. The most commonly considered special case is when $\{X(t)\}$ and $\{Y(t)\}$ are independent processes (and so the crosscorrelation function is $0$) leading to

\begin{align}R_{N}(\tau)&= R_{X}(\tau)\cos(2\pi f_c\tau)\tag{3}\\ S_{N}(f) &= \frac 12 S_{X}(f-f_c) + \frac 12 S_{X}(f+f_c)\tag{4} \end{align} where we have assumed that $\{X(t)\}$ and $\{Y(t)\}$ are continuous-time real-valued independent zero-mean jointly wide-sense stationary processes with identical autocorrelation functions and low-pass power spectral densities with support $\big[-\frac W2, \frac W2\big]$

At this point, it is conventional for a host of nitpickers to point out that it suffices to have $\{X(t)\}$ and $\{Y(t)\}$ be uncorrelated processes -- independence is an unnecessarily strong requirement -- but I insist on independence which allows for consideration of possible nonlinear operations on $\{N(t)\}$ without messing up things from the get go, but ymmv.

Bandwidth and Power: Note that $(4)$ is saying that the common lowpass spectrum of the $\{X(t)\}$ and $\{Y(t)\}$ processes has been translated half to center at $f_c$ and half to center $-f_c$ to occupy spectral bands $[f_c-\frac W2,f_c+\frac W2]$ and $[-f_c-\frac W2,-f_c+\frac W2]$. Thus, the bandwidth of $\{N(t)\}$ is $W$, twice the bandwidth of the lowpass processes $\{X(t)\}$ and $\{Y(t)\}$. On the other hand, the power of the $\{N(t)\}$ process (which is just the area under $S_{N}(f)$ (equivalently the value of $R_{N})0)$)) is just $R_{X}(0)=R_Y(0)$. That is, the two input processes to the "modulator" have total power $2R_{X}(0$ while the output process $\{N(t)\}$ has only half the power. This very "inefficient modulation process" is occurring because $\cos(2\pi f_ct)$ and $\sin(2\pi f_ct)$ are not unit-power signals; they each have power is $\frac 12$ watts or volts$^2$. In summary, $(1)$ results in a bandpass process with twice the bandwidth and half the total power of the low-pass processes used in the creation of the bandpass process.

Finally, suppose that $\{X(t)\}$ and $\{Y(t)\}$ are white noise processes and thus $$S_X(f) = S_Y(f) = \begin{cases}\frac{N_0}{2}, & |f| \leq \frac W2,\\0, & |f| > \frac W2,\end{cases}$$ and thus have power $N_0\dfrac W2$ each.


I think that you are implicitly seeing the quadrature components of the baseband process as two independent degrees of freedom.

The whole derivation by Proakis goes exactly in the opposite direction. It starts from a condition on the passband stochastic process (i.e. stationarity) and derives the constraints that the components of the baseband process must obey to ensure that. The fact that $\phi_{xx}(\tau) = \phi_{yy}(\tau)$ is a necessary consequence of the stationarity condition.

EDIT: This constraint is somehow eating up one degree of freedom from the baseband process.

If you drop the stationarity condition, 4-1-40 and 4-1-41 must not hold any more.

  • $\begingroup$ Thanks for this response. I get that, my question is specific to this and regardless of that point -- I can without that restriction follow the math from his equations 4-1-42 to 4-1-44 and reach a similar result as long as the real and imaginary components are independent. (There is likely a very simple math error here on my part, but I don't yet see it...If I get more time I will add those equations specifically to my post with the derivation proceeding as such so it's even clearer) $\endgroup$ Commented Nov 4, 2022 at 1:52
  • $\begingroup$ Also I don't see how that is a consequence of stationarity alone: Can't we have a stationary process on the real component, and a different but stationary random process on the imaginary component with both autocorrelations not equal (as a simple example completely different power levels)-- the resulting complex process will also be stationary, right? $\endgroup$ Commented Nov 4, 2022 at 1:56
  • $\begingroup$ No, we can't, because that would violate 4-1-44. I agree that it may be not intuitive, but the math is clear. $\endgroup$
    – Vito
    Commented Nov 4, 2022 at 7:22
  • $\begingroup$ @DanBoschen: That's the point, the bandpass process will not be stationary unless the autocorrelations of the real and imaginary components are identical. So in that case you couldn't even define the autocorrelation of the bandpass process. $\endgroup$
    – Matt L.
    Commented Nov 4, 2022 at 7:31
  • $\begingroup$ Why not? You can, but it would be a time-dependent autocorrelation $\phi_{nn}(t,\tau)$. $\endgroup$
    – Vito
    Commented Nov 4, 2022 at 7:47

This post and this post by MattL were helpful and related.

Predominantly my confusion stemmed from the following details I didn't completely understand (or forgot) from stochastic complex signals and the details of computing an autocorrelation:

  • The complex conjugate product in the autocorrelation for a complex signal will result in like terms (real x real, or imag x imag) mapping to real components and cross terms (real x imag or imag x real) mapping to imaginary components.

  • (correcting this statement based on Matt's answer and clarification in a comment under his answer) If the complex baseband signal is stationary, then the real and imaginary components of the complex baseband signal must have an identical auto-correlation, and opposite sign cross-correlation in order for the equivalent real passband to be stationary (Rubem tried early to convince me of this and finally see the intuition for this that I seeked that I have added to the bottom of this answer). This is referred to as being a circularly complex signal.

Detailing the above:

Consider the auto-correlation of the complex WSS baseband signal $x_c(t)$ repeated below as:

$$x_c(t) = x_i(t) + j x_q(t)$$


$$\phi_{x_c x_c}(\tau) = E\{ x_c^*(t) x_c(t+\tau)\}$$

It may be clearer to see the result by determining the auto-correlation of this baseband signal and THEN for that result determine the equivalent passband signal. In the product in the autocorrelation of $x_c(t)$ the autocorrelations of the real components and imaginary components will both map to a real result, while the cross-correlations of the real and imaginary components will map to a complex result. In simplified form this is just the result of a complex conjugate multiplication where we see the similar result, such as:

$$z_1^*z_2 =(I_1-jQ_1)(I_2+jQ_2) = [I_1 I_2 + Q_1 Q_2] + j [I_1Q_2 -I_2Q_1]$$

So similar to the simple complex product of two arbitrary signals as given above, we arrive at the expected value for the following product:

$$E\{x_c^*(t) x_c(t+\tau)\}$$

$$= E\{ (x_i(t) - j x_q(t)) ( x_i(t+\tau) + j x_q(t+\tau))\}$$

$$ =[E\{(x_i(t)x_i(t+\tau)\} E\{(x_q(t)x_q(t+\tau)\}] + j [E\{(x_q(t)x_i(t+\tau)\} - E\{(x_i(t)x_q(t+\tau)\}]$$

$$ = [\phi_{x_ix_i}(\tau) + \phi_{x_q x_q}(\tau)] + j [\phi_{x_qx_i}(\tau) - \phi_{x_ix_q}(\tau)]$$

From which the equivalent passband signal consistent with equation (1) in my OP would then be:

$$ [\phi_{x_ix_i}(\tau) + \phi_{x_q x_q}(\tau)] \cos(\omega_c t) - [\phi_{x_qx_i}(\tau) - \phi_{x_ix_q}(\tau)] \sin(\omega_c t)$$

Thus both real and imaginary terms of the baseband noise do map to the passband signal, but because of the second property I mentioned in the introduction that applies to circular complex stochastic processes (that I still seek and intuition for), $\phi_{x_ix_i} = \phi_{x_qx_q}$ and $\phi_{x_qx_i} = \phi_{x_ix_q}$, then the above, in this case, reduces to:

$$ 2\phi_{x_ix_i}(\tau) \cos(\omega_c t) - 2\phi_{x_qx_i}(\tau) \sin(\omega_c t)$$

Also notice the factor of two in the result. This is intuitive in that the autocorrelation is a power quantity, and half the power of the complex baseband signal is in the real component and half is in the imaginary component. So if we are to describe the autocorrelation in terms of the real component only, we need to double that to be consistent with the total power. (The power spectral density of the real or imaginary component of the baseband signal is typically notated as $N_o/2$, and thus doubling that would provide the total power density of the signal as $N_o$.). So this leads to the question why the factor of 2 does not appear in Proakis' derivation that approaches this from the passband signal. (I will add that detail here once I resolve that if someone else doesn’t address this in their own answer; it should be clear when working back from the power spectral density and making the total power the same for either complex baseband or passband representations.)

Intuition (at least for me) of the mathematical result that the the real and imaginary components of the baseband complex signal must have an equal autocorrelation and opposite signed cross-correlation in order for the equivalent real passband to be stationary:

In Andrew Duncan's paper The Analytic Impulse, he depicts what he called the "Heyser Corkscrew" in Figure 6 and copied below.

Heyser Corkscrew

To be clear, for the intuition, this is representing the carrier specifically, the $e^{j\omega_c t}$ with its real and imaginary components (and in the construction to a real passband, we ultimately select the real component). To connect to the topic here, consider another stochastic baseband complex signal that is multiplied with this, and we see how in that process the real and imaginary components will modulate (twirl in this graphic) onto the real passband. Thus the statistics of the real passband will modulate between the statistics of I (the real component) and Q (the imaginary component) of the complex baseband stochastic process. If the statistics of I and Q in the baseband signal are not the same, the result will be cyclo-stationary (not stationary). When the autocorrelations match and cross correlations are of opposite sign, then there will be no change in the statistics due to this modulation, and we also see how the term "circularly complex" applies.


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