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Suppose I have a real-valued random signal $f(t)$ written as $$f(t) = I(t)\cos(\Omega t) - Q(t)\sin(\Omega t)\, .$$ What is the relationship between the spectral densities of $I$ and $Q$ and the spectral density of $f$?

One approach is to use the fact that $$\int_0^\infty \frac{d\Omega}{2\pi} S_f(\Omega) \cos(\Omega \tau)= \langle f(t)f(t+\tau)\rangle_t \, .$$ But so far I haven't understood how to make progress.

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    $\begingroup$ The convolution theorem and linearity of the Fourier transform will guide you, if you're bold and apply them to your equation. $\endgroup$ – Jazzmaniac May 20 '15 at 19:05
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    $\begingroup$ Instead of convolution, some might prefer to use the modulation theorem to deduce the spectrum of $I(t)\cos(\Omega t)$ from the spectrum of $I(t)$. $\endgroup$ – Dilip Sarwate May 20 '15 at 19:10
  • $\begingroup$ @DilipSarwate I'm totally familiar with convolution etc and I can work through it no problem. However, I'm intrigued by the "modulation theorem". I'm guessing this is a nice version of the fact that multiplication by a sinusoid shifts all the frequency content. Is there some nice statement of this to which you are referring? $\endgroup$ – DanielSank May 20 '15 at 20:41
  • $\begingroup$ @DilipSarwate Your comments on this other question indicate that $I(t) \cos(\Omega t)$ doesn't have a spectral density unless certain extra assumptions are made. $\endgroup$ – DanielSank Jun 16 '15 at 23:08
  • $\begingroup$ If whoever down-voted this would care to offer some constructive criticism it would be helpful. $\endgroup$ – DanielSank Jun 17 '15 at 17:36
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This answer is related to my answer to this question of yours. The short answer is that in general the random process $f(t)$ has no power spectral density because it is not wide sense stationary (WSS), even if $I(t)$ and $Q(t)$ are.


I will first introduce some notation. I use the following definition of the autocorrelation function of a (possibly complex) random process $X(t)$:

$$R_X(t,\tau)=E\{X(t+\tau)X^*(t)\}$$

where $E\{\cdot\}$ denotes the expectation operator, and $^*$ denotes complex conjugation. Note that this definition with the complex conjugate is consistent with the usual definition of the inner product of complex functions.

The cross-correlation of two (possible complex) random processes $X(t)$ and $Y(t)$ is defined by

$$R_{XY}(t,\tau)=E\{X(t+\tau)Y^*(t)\}$$

Consequently, the cross-correlation of $X(t)$ and its complex conjugate $X^*(t)$ is given by

$$R_{XX^*}=E\{X(t+\tau)X(t)\}$$

If $X(t)$ is wide-sense stationary (WSS), then its autocorrelation function only depends on the time difference $\tau$ and is written as $R_X(\tau)$. Similarly, if two processes $X(t)$ and $Y(t)$ are jointly WSS, their cross-correlation function only depends on the time difference $\tau$ and is written as $R_{XY}(\tau)$.

For WSS processes, the power spectrum is defined as the Fourier transform of their autocorrelation function:

$$S_X(\omega)=\int_{-\infty}^{\infty}R_X(\tau)e^{-j\omega\tau}d\tau$$

Similarly, for two jointly WSS processes, their cross-spectral density is given by the Fourier transform of their cross-correlation function:

$$S_{XY}(\omega)=\int_{-\infty}^{\infty}R_{XY}(\tau)e^{-j\omega\tau}d\tau$$


Now for the answer of the question. To see that $f(t)$ is generally not WSS, define a complex-valued WSS process $X(t)$ by

$$X(t)=I(t)+jQ(t)\tag{1}$$

with real-valued and jointly WSS processes $I(t)$ and $Q(t)$. The process $f(t)$ is then given by

$$f(t)=\Re\{X(t)e^{j\Omega t}\}=\frac12(X(t)e^{j\Omega t}+X^*(t)e^{-j\Omega t})\tag{2}$$

The auto-correlation function of the real-valued process $f(t)$ is

$$R_f(t,\tau)=E\{f(t+\tau)f(t)\}\tag{3}$$

Plugging (2) into (3) gives (after some straightforward manipulations)

$$R_f(t,\tau)=\frac12\Re\{R_X(\tau)e^{j\Omega\tau}\}+ \frac12\Re\{R_{XX^*}(\tau)e^{j\Omega(\tau+2t)}\}\tag{4}$$

From (4) it is clear that the auto-correlation function of $f(t)$ depends on $t$, and, consequently, the power spectral density of $f(t)$ is not defined. However, if $R_{XX^*}(\tau)=0$ and if $E\{X(t)\}=0$ then $f(t)$ is WSS and has a power spectral density.

The cross-correlation of $X(t)$ and $X^*(t)$ is given by

$$\begin{align}R_{XX^*}(\tau)&=E\{X(t+\tau)X(t)\}\\&=E\{(I(t+\tau)+jQ(t+\tau))(I(t)+jQ(t))\}\\&=\ldots\\&=R_I(\tau)-R_Q(\tau)+j(R_{IQ}(\tau)+R_{IQ}(-\tau))\end{align}\tag{5}$$

where $R_I(\tau)$ and $R_Q(\tau)$ are the auto-correlation functions of $I(t)$ and $Q(t)$, respectively, and $R_{IQ}(\tau)$ is the cross-correlation of $I(t)$ and $Q(t)$. From (5) it is obvious that the condition $R_{XX^*}(\tau)=0$ is satisfied if

$$\begin{align}R_I(\tau)&=R_Q(\tau)\\ R_{IQ}(\tau)&=-R_{IQ}(-\tau)\end{align}\tag{6}$$

is satisfied, i.e. if $I(t)$ and $Q(t)$ have the same auto-correlation function (and thus the same power spectrum), and if $R_{IQ}(\tau)$ is an odd function, the latter implying $R_{IQ}(0)=0$, i.e. $I(t)$ and $Q(t)$ are uncorrelated when sampled at the same instant.

The auto-correlation function $R_X(\tau)$ can be expressed in terms of $R_I(\tau)$, $R_Q(\tau)$, and $R_{IQ}(\tau)$:

$$\begin{align}R_X(\tau)&=E\{X(t+\tau)X^*(t)\}\\&=\ldots\\&=R_I(\tau)+R_Q(\tau)-j(R_{IQ}(\tau)-R_{IQ}(-\tau))\end{align}\tag{7}$$

If the conditions (6) are satisfied, this becomes

$$R_X(\tau)=2(R_I(\tau)-jR_{IQ}(\tau))\tag{8}$$

Pluggin (8) into (4) gives for the auto-correlation function of $f(t)$

$$R_f(\tau)=R_I(\tau)\cos(\Omega\tau)+R_{IQ}\sin(\Omega\tau)\tag{9}$$

(assuming the conditions (6) are satisfied). From (9) the power spectral density of $f(t)$ can be expressed by the power spectral densities of $I(t)$ and $Q(t)$, and by the cross-power spectral density of $I(t)$ and $Q(t)$:

$$S_f(\omega)=\frac12[S_I(\omega-\Omega)+S_I(\omega+\Omega)]+\frac{1}{2j}[S_{IQ}(\omega-\Omega)-S_{IQ}(\omega+\Omega)]\tag{10}$$

which is the final result. Note that $S_f(\omega)$ is of course real-valued because due to the second condition in (6), the cross-power spectral density $S_{IQ}(\omega)$ is purely imaginary and, consequently, the second term on the right-hand side of (10) is real-valued.

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  • $\begingroup$ In Eq. (4) I think $R_X$ and $R_{XX^*}$ should be switched. Is this correct? The star on the second $X$ in Eq. (5) also seems to have disappeared. $\endgroup$ – DanielSank Jun 16 '15 at 22:59
  • $\begingroup$ +1 but I think the notation might be confusing or there are errors. Some times the correlation functions are written with a single subscript as in $R_X$, but other times with two $R_{XX^*}$. It would be helpful to be consistent and always use two, particularly since Eq. (7) seems to define the version with a single subscript as the expectation of a variable multiplied by its complex conjugate, which is unexpected. I'm not sure yet how much of this is confusing notation and how much might be typos. $\endgroup$ – DanielSank Jun 16 '15 at 23:10
  • $\begingroup$ Ah, maybe when you write two subscripts it means e.g. $R_{AB}(t, \tau) \equiv \langle A(t) B^*(t+\tau)\rangle$ whereas with one subscript the function is assumed real and we define it as $R_A(t,\tau) \equiv \langle A(t)A(t+\tau)\rangle$. $\endgroup$ – DanielSank Jun 16 '15 at 23:50
  • $\begingroup$ Submitted an edit to make everything consistent. $\endgroup$ – DanielSank Jun 17 '15 at 1:01
  • $\begingroup$ @DanielSank: I think that everything is correct in my answer, but now I understand your misunderstanding. I'll add to my answer some explanation concerning my notation (which is quite common in the DSP literature). $\endgroup$ – Matt L. Jun 18 '15 at 7:57

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