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.


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 $R_v$?


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.


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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