3
$\begingroup$

I'm walking through the published solutions of my homework and I'm struggling interpreting them. In them I was given a random Gaussian process $\{X_t\}$ and a random variable

$$\{Y_t\} = X_t\cos(2\pi f_0t+\Theta), \quad \Theta \sim \mathcal{U}_{[0,2\pi]}, $$ i.e., $\Theta$ is a random phase.

It is also given that $S_x(f)$ is defined as $0$ for $|f| \geq F_0/2$, and that $f_0 = 100F_0$

  • The solution assumes $\{X_t\}$ and $\{Y_t\}$ are Joint Wide Sense Stationary, why is that the case?
  • afterwards the question defines $\{Z_t\} = Y_t\cos(2\pi f_0t+\Theta)$ and asks to find the optimal linear filter of $\{X_t\}$ out of $\{Z_t\}$, and finding the error.

I've found: $$S_z(f) =\frac{1}{4} S_x(f) + \frac{1}{16}S_x(f+2f_0) + S_x(f-2f_0)$$ and $$S_{zx}(f) = \frac{1}{2} S_x(f)$$ The solution based on these two results states $H(f) = 2$, and that the error would be 0 for $|f| \leq F$ - I'm struggling to see why.

$\endgroup$
7
  • 1
    $\begingroup$ Does jointly WSS imply that the processes are individually WSS too? If so, is $\{Y_t\}$ a WSS process? It seems that $\{X_t\}$ is given to be a WSS process since its PSD is specified. $\endgroup$ Jul 25, 2023 at 13:27
  • $\begingroup$ @DilipSarwate I always understood "jointly {stationarity constraint}" to be a requirement on the process $Z_t := X_t Y_t$ AND the joint CDF $F_{X_{t_1}, X_{t_2}, \ldots, Y_{u_1}, Y_{u_2}, \ldots}(x_1, x_2, \ldots, y_1, y_2,\ldots)$. So, for WSS, the boundedness constraints on the first two moments apply to $Z$, the correlation properties to the joint PDF. That way, you can have something like a dependent process $Y$ "compensating" the "annoying" properties of e.g. a Cauchy-distributed process $X$; that is a relevant tool when modelling things like "clipping". $\endgroup$ Jul 25, 2023 at 14:00
  • 1
    $\begingroup$ @MarcusMüller Some people don't agree with your assertion that jointly WSS npricesses don't need to be individually WSS. See, for example, probabilitycourse.com/chapter10/10_1_4_stationary_processes.php $\endgroup$ Jul 26, 2023 at 13:53
  • 1
    $\begingroup$ my textbook defines WSS as a process for which $\mu_x$ isn't dependent on time, and the autocorrelation is dependent only on time differences and not time itself. two processes are called JWSS if they fulfill requirement for WSS individually, and $E[Y_{t+\tau}X_{t}]$ is only dependent on $\tau$. apologies for the delayed response $\endgroup$ Jul 26, 2023 at 15:50
  • 1
    $\begingroup$ @Piratemetaldrinkingcrew no harm done! well, your book agrees with Dilip's definition, not with mine :) $\endgroup$ Jul 26, 2023 at 17:49

1 Answer 1

4
$\begingroup$

Now that the OP has responded with his textbook's definition of jointly wide-sense-stationary (WSS) processes as those that are individually WSS and whose cross-correlation function $R_{X,Y}(t_1,t_2)$ depends only on the difference $t_1-t_2$ of the arguments, let's consider his questions.

  1. The solution assumes $\{X_t\}$ and $\{Y_t\}$ are Joint Wide Sense Stationary, why is that the case?

Well, $\{X_t\}$ is assumed to be WSS since its PSD is specified (as being band-limited to $\left(-\frac{F_0}{2},+\frac{F_0}{2}\right)$. Now, \begin{align} E[Y_t]&=E[X_t\cos(2\pi f_0t + \Theta)]\\ &= E[X_t]E[\cos(2\pi f_0t + \Theta)] & \Theta, X_t~ \text{independent}\\ &= E[X_t]\cdot\int_0^{2\pi}\cos(2\pi f_0t + \theta)\cdot \frac 1{2\pi} \mathrm d\theta & \Theta \sim \mathcal U[0,2\pi]$\\ &= 0 \end{align} while \begin{align} E[Y_sY_t] &= E[X_s\cos(2\pi f_0s + \Theta)X_t\cos(2\pi f_0t + \Theta)]\\ &= E[X_sX_t]E[\cos(2\pi f_0s + \Theta)\cos(2\pi f_0t + \Theta)]\\ &= R_X(s-t)\cdot\int_0^{2\pi}\cos(2\pi f_0s + \theta)\cos(2\pi f_0t + \theta)\cdot \frac 1{2\pi} \mathrm d\theta\\ &= \frac 12 R_X(s-t)\cos(s-t) \end{align} and so $\{Y_t\}$ is a WSS process too. Turning to the cross-correlation function, we have \begin{align} E[X_sY_t] &= E[X_sX_t\cos(2\pi f_0t + \Theta)]\\ &= E[X_sX_t]E[\cos(2\pi f_0t + \Theta)]\\ &= R_X(s-t)\cdot\int_0^{2\pi}\cos(2\pi f_0t + \theta)\cdot \frac 1{2\pi} \mathrm d\theta\\ &= 0 \end{align} and so $\{X_t\}$ and $\{Y_t\}$ are uncorrelated WSS processes, and thus are (trivially) jointly WSS processes. Note that this is just modulation of the (low-pass) $\{X_t\}$ process onto a high-frequency carrier ($f_0 = 100F_0$) to produce the band-pass process $\{Y_t\}$ whose PSD is nonzero only in the frequency bands of width $F_0$ centered around $\pm f_0$.

Turning to the second question, $Z_t$ is the output of the mixer that demodulates $\{Y_t\}$ back down to baseband, and the optimal filtering of this mixer output to complete the demodulation process is just discarding the double-frequency terms. An ideal LPF with cut-off $\frac{F_0}{2}$ minimizes the noise in the demodulator output.

I have no clue what $H(f)=2$ means in the OP's book solution. I suspect that what is meant is an ideal LPF of bandwidth $\frac{F_0}2$ and passband gain $2$. Passing the $\{Z_t\}$ process (whose PSD is $S_Z(f) =\frac{1}{4} S_X(f) + \frac{1}{16}S_X(f+2f_0) + S_X(f-2f_0)$ through this filter produces a process with PSD $S_X(f)$, which is (mean-square) equivalent to $\{X_t\}$.

$\endgroup$
3
  • $\begingroup$ thank you! why does a band limited PSD mean that a process is WSS? as for $H(f)$ It's the transmission function of the wiener filter. I think I understand what you mean by discarding higher frequency terms - thank you for your detailed and clear reply! $\endgroup$ Jul 29, 2023 at 14:22
  • 1
    $\begingroup$ Non-WSS processes don't have PSDs in the usual sense that one interprets PSD as the Fourier transform of the (univariate) ACF. For a Non-WSS process, the ACF is a function of two time variables, and the notion of PSD is more complicated. So, given the PSD of $\{X_t\}$ without any other bells and whistles indicating that the process is not a WSS process and must be handled differently, it is reasonable to assume that $\{X_t\}$ is indeed a WSS process. If you don't agree with this assumption, my whole answer becomes irrelevant. (Obviously the creator of your HW solutions agrees with me!) $\endgroup$ Jul 29, 2023 at 14:39
  • $\begingroup$ That's a really good answer - thank you! I never thought of it that way $\endgroup$ Jul 29, 2023 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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