0
$\begingroup$

Suppose we have a noise process $F(t)$. Suppose $F$ is stationary and we know its autocorrelation function $$\rho_F(\tau) \equiv \langle F(0)F(\tau) \rangle$$ where the average is over realizations of the process. We then construct a new noise process $G(t) \equiv \cos(\Omega t)F(t)$. Obviously, $G$ is not stationary. The autocorrelation function of $G$ is \begin{align} \langle G(t_1) G(t_2)\rangle &= \langle F(t_1) \cos(\Omega t_1)F(t_2) \cos(\Omega t_2) \rangle \\ &= \langle F(t_1) F(t_2) \rangle \cos(\Omega t_1) \cos(\Omega t_2) \\ &= \rho_F(t_2 - t_1) \cos(\Omega t_1) \cos(\Omega t_2) \, .\\ \end{align}

The spectral density of $F$ can be usefully defined as $$S_F(\omega) = \int \frac{d\omega}{2\pi} \rho_F(\tau) e^{-i \omega \tau} \, . $$ However, since $\langle G(t_1) G(t_2) \rangle$ depends on $t_1$ and $t_2$ and not just their difference, we cannot easily define $S_G(\omega)$. This is distressing because for deterministic signals we expect modulation by a sinusoid to just shift the spectral density, so we expect there should be a simple formula for $S_G(\omega)$.

What, if any, is a meaningful way to construct the spectral density of $G$?

$\endgroup$
  • $\begingroup$ Add a random phase $\Theta \sim [0,2\pi)$ to the argument of the cosine. $\endgroup$ – Dilip Sarwate Jun 15 '15 at 2:14
  • $\begingroup$ @DilipSarwate why? $\endgroup$ – DanielSank Jun 15 '15 at 2:15
  • $\begingroup$ Because it works, and makes the process wide-sense-stationary. $\endgroup$ – Dilip Sarwate Jun 15 '15 at 2:15
  • $\begingroup$ @DilipSarwate Because what works? Arbitrarily changing the problem to get a nice property isn't useful unless there's a good reason for the modification. If someone asks how to build a house you can't give them a doll house and claim victory. $\endgroup$ – DanielSank Jun 15 '15 at 2:18
  • 2
    $\begingroup$ I don't really understand the two downvotes. This is a clearly formulated question showing some effort on the OP's side. Unfortunately, I've seen many much worse questions on this site recently, so for me this one improves the average standard. $\endgroup$ – Matt L. Jun 15 '15 at 13:13
3
$\begingroup$

This is a standard problem in (among others) the theory of analog or digital communication systems. If $F(t)$ is real-valued, then the standard gimmick is indeed - as already pointed out in a comment by Dilip Sarwate - to add a random phase to the carrier phase. This is not only a trick but it actually reflects reality in the sense that we often do not know (and do not care to know) the exact phase of the carrier with respect to the signal $F(t)$. So what we actually want is NOT the power spectrum of $F(t)\cos(\omega_0 t)$ (which does not exist), but the power spectrum of the process $F(t)\cos(\omega_0 t+\theta)$ with some unknown phase $\theta$, the obvious model of which is a random variable with a uniform distribution in $[0,\pi)$. This random phase is of course constant over time, but its value is chosen at random reflecting our uncertainty about the origin of the time axis. With this random phase the resulting process is wide-sense stationary (WSS) (if $F(t)$ is), and the power spectrum looks exactly as one would expect.

Now comes the interesting part. If $F(t)$ is a WSS complex-valued process and if we're interested in the power spectrum of the process

$$G(t)=\text{Re}\{F(t)e^{j\omega_0 t}\}\tag{1}$$

then we could also use the random phase trick, but under some reasonable assumptions we don't even need it. The process $G(t)$ in (1) occurs e.g. in systems using quadrature modulation. It is straightforward to show that the autocorrelation of $G(t)$ is

$$\begin{align}R_G(t,\tau)&=E\{G(t+\tau)G^*(t)\}=\\&=\frac12\text{Re}\{R_F(\tau)e^{j\omega_0\tau}\}+\frac12\text{Re}\{R_{FF^*}(\tau)e^{j\omega_0(\tau+2t)}\}\tag{2}\end{align}$$

which becomes independent of $t$ if $R_{FF^*}(\tau)=0$, i.e. if the cross-correlation of $F(t)$ and $F^*(t)$, vanishes. If $F(t)=A(t)+jB(t)$, it can be shown that this is the case if

$$R_A(\tau)=R_B(\tau)\tag{3}$$

and

$$R_{AB}(\tau)=-R_{AB}(-\tau)\tag{4}$$

So if conditions (3) and (4) are satisfied, we don't even need a random carrier phase in order to define a power spectrum of $G(t)$ as defined in (1). Conditions (3) and (4) mean that the real and imaginary parts of $F(t)$ have the same autocorrelation, i.e. the same power spectrum, and that their cross-correlation is an odd function, i.e. their cross-power spectrum is purely imaginary. This last condition also implies $R_{AB}(0)=0$, i.e the real and imaginary part are uncorrelated when sampled at the same time. It turns out that for many practical problems these conditions are at least approximately satisfied.

Given that conditions (3) and (4) are satisfied, the autocorrelation function of $G(t)$ is given by

$$R_G(\tau)=\frac12\text{Re}\{R_F(\tau)e^{j\omega_0\tau}\}\tag{5}$$

and its power spectral density is given by the Fourier transform of (5):

$$S_G(\omega)=\frac14\left[S_F(\omega-\omega_0)+S^*_F(-\omega-\omega_0)\right]\tag{6}$$

where $S_F(\omega)$ is the power spectral density of $F(t)$.

$\endgroup$
  • $\begingroup$ You mentioned (not coincidentally) quadrature communication systems. In fact this question was motivated by trying to understand how analog noise affects the result of an IQ demodulation system. I've been trying to ask focused questions directed at each step of the process, but if you could spare a few minutes in a chat room I have a feeling you could really help me sort this out in one go. $\endgroup$ – DanielSank Jun 15 '15 at 22:28
  • $\begingroup$ I can naively compute $\langle |\tilde{I}(\omega)|^2 \rangle$. Using the convolution theorem etc. I find $\langle |\tilde{I}(\omega)|^2 \rangle = \langle | \tilde{V}(\omega+\Omega)|^2 \rangle + \langle |\tilde{V}(\omega-\Omega)|^2 \rangle$. In doing this I dropped two cross terms which I reasoned are zero because $\tilde{V}$ at various frequencies should be uncorrelated. Is there something bogus about this reasoning? Again, it would be great to discuss this in the chat room linked in my prior comment. $\endgroup$ – DanielSank Jun 15 '15 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.