# Understanding homework solution - why are $\{X_t\}$ and $\{Y_t\}$ joint WSS, and finding Wiener filter + error

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.

• 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. Jul 25, 2023 at 13:27
• @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". Jul 25, 2023 at 14:00
• @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 Jul 26, 2023 at 13:53
• 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 Jul 26, 2023 at 15:50
• @Piratemetaldrinkingcrew no harm done! well, your book agrees with Dilip's definition, not with mine :) Jul 26, 2023 at 17:49

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 $$\{X_t\}$$.

• 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! Jul 29, 2023 at 14:22
• 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!) Jul 29, 2023 at 14:39
• That's a really good answer - thank you! I never thought of it that way Jul 29, 2023 at 21:21