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Background:

When a wide-sense stationary (WSS) random process $x(t)$ is passed through a linear, time-invariant (LTI) filter with impulse response $h(t)$ to produce $y(t)$, the following relationship holds: $$ S_y(f) = |H(f)|^2 S_x(f) = H^*(f) H(f) S_x(f), $$ where $S_x, S_y$ are the power spectral densitities (PSD) of $x$, and $y$, respectively, and $H(f)$ is the Fourier transform of $h(t)$ (ordinary frequency, unitary). In the time-domain, this relationship can be shown to be: $$ R_y(\tau) = h^*(-\tau) * h(\tau) * R_x(\tau), $$ where $R_x,R_y$ are the autocorrelations of $x$, and $y$, respectively.

Question:

If $v(t)$ is a wide-sense cyclostationary process with autocorrelation $R_v(t, \tau)$, is cyclic with period $T_0$, and is passed through the same linear filter to produce $w(t)$: $$ w(t) = h(t) * v(t), $$ will $w(t)$ also be cyclostationary with cyclic period $T_0$? (Assuming $h(t)$ is BIBO stable.) If so, is there a "simple" relationship that expresses $R_w(t, \tau)$ in terms of $h$ and $R_v$?

EDIT:

For a wide-sense cyclostationary process $v(t)$, the cyclic autocorrelation $R_v^{n/T_0} (\tau)$ is defined as: $$ R_v^{n/T_0} (\tau) = \int_{-T_0/2}^{T_0/2} R_v(t, \tau) e^{-j n 2\pi t/T_0} dt $$

A relationship between $R_v^{n/T_0} (\tau)$ and $R_w^{n/T_0} $ would also be acceptable for this question.

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Your first two equations generalize to the spectral correlation and cyclic autocorrelation. The most general results are (12) and (13) on my blog post on input-output relations for $n$th-order cyclic cumulants and $n$th-order cyclic polyspectra: signal processing operations and cyclostationary signals.

You can find the generalization of your first equation in several places, including Professor Gardner's book "Statistical Spectral Analysis", Eq (90) in Chapter 11:

$$S_y^\alpha(f) = H(f+\alpha/2) H^*(f-\alpha/2) S_x^\alpha(f)$$

And yes, the output of the filter is also cyclostationary with the same period provided that the transfer function $H(f)$ is such that $H(f+\alpha/2)H^*(f-\alpha/2)$ is not zero for at least some $f$ for which $S_x^\alpha(f)$ is also not zero. In other words, the filter could remove so much of the energy of the signal that the remainder is no longer cyclostationary.

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  • $\begingroup$ Not only an answer, but one from a legend in the field, too! Thanks Chad! $\endgroup$ – Robert L. Sep 14 '18 at 19:55

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