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Suppose a sequence $\{a_k\}_{k=0}^{\infty}$ has power spectral density

$$ S_{\mathbf{a}}(f) := \lim_{n\to \infty}\mathbb{E}\left[\frac{1}{f_0 n}\left| \sum_{k=0}^{n-1} 2\pi f_0 a_k \exp(-2\pi ikf/f_0) \right|^2 \right] $$

where $f_0$ is the frequency of an "ideal" clock so that the rising edge of the clock occurs at $t_k = k/f_0 + a_k$.

Suppose we can only draw $m$ samples from the infinite sequence. How fast does the quantity

$$ S_{\mathbf{a}}^{(m)}(f) := \mathbb{E}\left[\frac{1}{f_0 m}\left| \sum_{k=0}^{m-1} 2\pi f_0 a_k \exp(-2\pi ikf/f_0) \right|^2 \right] $$

converge to $S_{\mathbf{a}}(f)$? Can general statements be made? (If general statements cannot be made, I am content with results restricted to sequences that are roughly flicker.)

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  • $\begingroup$ What kind of phase you speak of? As long as I understand, PSD does not preserve phase... maybe I missed something $\endgroup$ Oct 5, 2023 at 15:12
  • $\begingroup$ I have changed to the terminology "power spectral density" rather than phase noise. $\endgroup$
    – user14717
    Oct 5, 2023 at 18:22
  • $\begingroup$ I would prefer if you might adopt a notational convention that is more common within the electrical engineering discipline. What is the "fundamental period", $T_0$? Is it the sampling period? Or the period of some periodic function that you haven't told us about? $\endgroup$ Oct 5, 2023 at 19:15
  • $\begingroup$ @robertbristow-johnson: Hopefully clarified. $\endgroup$
    – user14717
    Oct 5, 2023 at 20:29
  • $\begingroup$ $t_k = k/f_0 + a_k$ doesn't make any sense at all. $\endgroup$ Oct 5, 2023 at 21:45

1 Answer 1

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I am restating the question using standard EE DSP notation as we might find in O&S (and correcting errors regarding spurious appearance of $2 \pi$):


Suppose a discrete-time, finite power signal

$$x[n] \triangleq x(n/f_\mathrm{s}) \qquad \qquad \text{(where } x(t) \text{ is the continuous-time signal being sampled)}$$

has power spectral density

$$ S_{x}(f) = \lim_{N\to \infty}\mathbb{E}\left[\frac{1}{N}\left| \sum\limits_{n=-N/2}^{N/2-1} x[n] e^{-j 2 \pi n f/f_\mathrm{s}} \right|^2 \right] $$

where $f_\mathrm{s}$ is the frequency of an "ideal" clock (the sample rate) so that the rising edge of the clock occurs at $t_n = n/f_\mathrm{s} + x[n]$ (...this expression makes no sense).

Suppose we can only draw $M$ samples from the infinite sequence. How fast does the quantity

$$ S_{x}^{(M)}(f) = \mathbb{E}\left[\frac{1}{M}\left| \sum\limits_{n=-M/2}^{M/2-1} x[n] e^{-j 2 \pi n f/f_\mathrm{s}} \right|^2 \right] $$

converge to $S_{x}(f)$? Can general statements be made? (If general statements cannot be made, I am content with results restricted to sequences that are roughly flicker.)


The next thing that the OP should realize is the standard definition of the Discrete-Time Fourier Transform (DTFT):

$$ X\big(e^{j\omega}\big) \triangleq \sum\limits_{n=-\infty}^{\infty} x[n] e^{-j \omega n} $$

Which defines the normalized angular frequency $\omega$ in terms of continuous-time frequency $f$ as

$$ \omega \triangleq 2 \pi \frac{f}{f_\mathrm{s}} $$

THAT is "standard notation" in the DSP subdiscipline within the electrical engineering discipline.

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    $\begingroup$ +1 for formalizing notation. $\endgroup$ Oct 6, 2023 at 20:37
  • $\begingroup$ x[n] is not a sampling of a continuous time function. It is truly a sequence in the mathematical sense. $\endgroup$
    – user14717
    Oct 7, 2023 at 4:34
  • $\begingroup$ $x[n]$ is another notation for $x_n$. It's a sequence of numbers. Now if your sequence of numbers represent something at a precise time $t_n=\frac{n}{f_\mathrm{s}}$ that's sampled continuous-time signal. Then your normalized angular frequency is related to frequency in the continuous -time domain (Hz) as $$ \omega = 2 \pi \frac{f}{f_\mathrm{s}}$$ You need no other reason to have the sample rate $f_\mathrm{s}$ in the equations. $\endgroup$ Oct 7, 2023 at 4:53

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