# Spectral density of signal written as quadratures

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.

• The convolution theorem and linearity of the Fourier transform will guide you, if you're bold and apply them to your equation. May 20, 2015 at 19:05
• 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 $I(t)$. May 20, 2015 at 19:10
• @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? May 20, 2015 at 20:41
• @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. Jun 16, 2015 at 23:08
• If whoever down-voted this would care to offer some constructive criticism it would be helpful. Jun 17, 2015 at 17:36

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.

• 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. Jun 16, 2015 at 22:59
• +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. Jun 16, 2015 at 23:10
• 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 $R_A(t,\tau) \equiv \langle A(t)A(t+\tau)\rangle$. Jun 16, 2015 at 23:50
• Submitted an edit to make everything consistent. Jun 17, 2015 at 1:01
• @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). Jun 18, 2015 at 7:57