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The Hilbert transform

where $\hat{q}[n] = \mathscr{H}\big\{ q[n] \big\}$ and the analytic signal is

$$ \omega_0[n] + \Im m \{\epsilon[n]\} - \Im m \{\epsilon[n-1]\} = \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \right\} $$

But, because this error is really added to the phase, averaging really helps. Suppose we have a moving average filter working on the estimate of $\omega_0[n]$:

$$\begin{align} \overline{\Omega_M}[n] &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \arg \left\{ \frac{x_\mathrm{a}[n-m]}{x_\mathrm{a}[n-m-1]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \times \frac{x_\mathrm{a}[n-1]}{x_\mathrm{a}[n-2]} \times ... \frac{x_\mathrm{a}[n-(M-1)]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \omega_0[n-m] + \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \sum\limits_{m=0}^{M-1} \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \Big( \Im m \big\{\epsilon[n]\big\} - \Im m \big\{ \epsilon[n-M] \big\} \Big) \\ \\ &= \overline{\omega_0[n]} + \xi[n] \\ \end{align}$$

The Hilbert transform is:

where $\hat{q}[n] = \mathscr{H}\big\{ q[n] \big\}$.

The analytic signal is

$$ \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \right\} = \omega_0 + \Im m \{\epsilon[n]\} - \Im m \{\epsilon[n-1]\} $$

But, because this error is really added to the phase (and not to the frequency), averaging really helps. Suppose we have a moving average filter working on the estimate of $\omega_0[n]$:

$$\begin{align} \overline{\Omega_M}[n] &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \arg \left\{ \frac{x_\mathrm{a}[n-m]}{x_\mathrm{a}[n-m-1]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \times \frac{x_\mathrm{a}[n-1]}{x_\mathrm{a}[n-2]} \times ... \frac{x_\mathrm{a}[n-(M-1)]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \omega_0[n-m] + \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \omega_0[n-m] + \frac{1}{M} \sum\limits_{m=0}^{M-1} \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \sum\limits_{m=0}^{M-1} \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \Big( \Im m \big\{\epsilon[n]\big\} - \Im m \big\{ \epsilon[n-M] \big\} \Big) \\ \\ &= \overline{\omega_0[n]} + \xi[n] \\ \end{align}$$

$$\begin{align} \overline{\Omega_M}[n] &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \arg \left\{ \frac{x_\mathrm{a}[n-m]}{x_\mathrm{a}[n-m-1]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \times \frac{x_\mathrm{a}[n-1]}{x_\mathrm{a}[n-2]} \times ... \frac{x_\mathrm{a}[n-(M-1)]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \omega_0[n-m] + \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \sum\limits_{m=0}^{M-1} \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \Big( \Im m \big\{\epsilon[n]\big\} - \Im m \big\{ \epsilon[n-M] \big\} \Big) \\ \end{align}$$

$$ \overline{\epsilon^2} = \frac{2 \, \overline{q^2[n]}}{M^2 A^2} $$

$$ \sqrt{\overline{\epsilon^2}} = \frac{\sqrt{2 \, \overline{q^2[n]}}}{M \, A} $$

$$ 10\log_{10}\left(\,\overline{\epsilon^2}\,\right) = 10\log_{10}\left( \frac{\overline{q^2[n]}}{A^2/2} \right) - 20\log_{10}(M) $$

The first term right of $=$ is S/N ratio. If $M=$100 (which is in the examples of the other answers), then the last term is -40 dB. It's a curiosity to me how they (I dunno, Royi, Overlord, or Cedron) got the CRLB to be nearly twice the number of dB (looks like -65 dB when SNR=0). Me suspects they might have used "$20\log_{10}(\cdot)$" when they should have used $10\log_{10}(\cdot)$.

$$\begin{align} \overline{\Omega_M}[n] &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \arg \left\{ \frac{x_\mathrm{a}[n-m]}{x_\mathrm{a}[n-m-1]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \times \frac{x_\mathrm{a}[n-1]}{x_\mathrm{a}[n-2]} \times ... \frac{x_\mathrm{a}[n-(M-1)]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-M]} \right\} \\ \\ &= \frac{1}{M} \sum\limits_{m=0}^{M-1} \omega_0[n-m] + \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \sum\limits_{m=0}^{M-1} \Im m \big\{\epsilon[n-m]\big\} - \Im m \big\{ \epsilon[n-m-1] \big\} \\ \\ &= \overline{\omega_0[n]} + \frac{1}{M} \Big( \Im m \big\{\epsilon[n]\big\} - \Im m \big\{ \epsilon[n-M] \big\} \Big) \\ \\ &= \overline{\omega_0[n]} + \xi[n] \\ \end{align}$$

$$ \overline{\xi^2[n]} = \frac{2 \, \overline{q^2[n]}}{M^2 A^2} $$

$$ \sqrt{\overline{\xi^2[n]}} = \frac{\sqrt{2 \, \overline{q^2[n]}}}{M \, A} $$

$$ 10\log_{10}\left(\,\overline{\xi^2[n]}\,\right) = 10\log_{10}\left( \frac{\overline{q^2[n]}}{A^2/2} \right) - 20\log_{10}(M) $$

The first term right of $=$ is S/N ratio. If $M=$100 (which is in the examples of the other answers), then the last term is -40 dB. It's a curiosity to me how they (I dunno, Royi, Overlord, or Cedron) got the CRLB to be nearly twice the number of dB (looks like -65 dB when SNR=0). At first I thought they might have used "$20\log_{10}(\cdot)$" when they should have used $10\log_{10}(\cdot)$, but that doesn't explain why we agree on the slope of the line. I really do not understand where this 25 dB of extra performance comes from with the number of samples in the average $M=$100. I'm hoping someone can explain that.

So the estimate of the sliding mean of instantaneous frequency is equal to the actual sliding mean of instantaneous frequency plus and error term with variance of

$$ \sigma_\epsilon^2 = \frac{2 \, \overline{q^2[n]}}{M^2 A^2} $$

$$ \sigma_\epsilon = \frac{\sqrt{2 \, \overline{q^2[n]}}}{M \, A} $$

I'm gonna leave it here and let it take some incoming critique and maybe modify it later. The method remains unchanged, it's just that this gives us an idea about how the added noise causes an added error to the frequency measurement with $M+1$ samples (and $2L$ additional samples, to compute the Hilbert transform).

So the bigger the amplitude of your signusoid, $A$, the better. The smaller of the amplitude of your AWGN, $\sigma_q = \sqrt{\overline{q^2[n]}} $, the better. And the more values of instantaneous frequency in the average, $M$, the better. This averaging is really low-pass filtering, and I believe that other LPF methods, like a 1st-order recursive (or "leaky integrator"), will work just as well.

So the estimate of the sliding mean of instantaneous frequency is equal to the actual sliding mean of instantaneous frequency plus an error term with mean square that is

$$ \overline{\epsilon^2} = \frac{2 \, \overline{q^2[n]}}{M^2 A^2} $$

$$ \sqrt{\overline{\epsilon^2}} = \frac{\sqrt{2 \, \overline{q^2[n]}}}{M \, A} $$

Expressed in dB, this looks like:

$$ 10\log_{10}\left(\,\overline{\epsilon^2}\,\right) = 10\log_{10}\left( \frac{\overline{q^2[n]}}{A^2/2} \right) - 20\log_{10}(M) $$

The first term right of $=$ is S/N ratio. If $M=$100 (which is in the examples of the other answers), then the last term is -40 dB. It's a curiosity to me how they (I dunno, Royi, Overlord, or Cedron) got the CRLB to be nearly twice the number of dB (looks like -65 dB when SNR=0). Me suspects they might have used "$20\log_{10}(\cdot)$" when they should have used $10\log_{10}(\cdot)$.

I'm gonna leave it here and let it take some incoming critique and maybe modify it later. The method remains unchanged, it's just that this gives us an idea about how the added noise causes an added error to the frequency measurement with $M+1$ samples (and $2L$ additional samples, to compute the Hilbert transform).

So the bigger the amplitude of your sinusoid, $A$, the better. The smaller of the amplitude of your AWGN, $\sqrt{\overline{q^2[n]}}$, the better. And the more values of instantaneous frequency in the average, $M$, the better. This averaging is really low-pass filtering, and I believe that other LPF methods, like a 1st-order recursive (or "leaky integrator"), will work just as well.