How to extract IF (instantaneous frequency) from Discrete Hilbert Transform? The continuous time Hilbert transform is
$$\hat x(t) := x(t) + j\left( p.v. \left\{\frac{1}{t\pi} \ast x \right\} (t)\right)$$

where
$$ \theta(t) = \tan^{-1}\left( \frac{ p.v.  \left\{\frac{1}{t\pi} \ast x \right\} (t)}{x(t)} \right)$$
and
$$
         \omega(t) = \theta'(t) 
$$

and since $\theta(t)$ is smooth, you can actually calculate the instantaneous frequency. But in the discrete time domain, the angle function is discrete, so given that in the dt domain, your hilbert transform filter is 

$$ h_{HT}[n] = \left\{
      \begin{array}{ll}
       \frac{1}{n\pi}  &amp; \mbox{if}\ \  n \neq 0 \\
        0               &amp; \mbox{otherwise} \\
      \end{array}
     \right.$$

How do you get the approximate frequency? or are you really just getting $\Delta\theta[n]$? Sorry if this is me just getting hung up on something trivial, but most references don't seem to clearly address this.

$$\theta[n] = \tan^{-1}\left(\frac{(h_{HT} \ast x)[n]}{x[n]}\right)$$
$$\omega[n] = \left\{
       \begin{array}{ll}
        \theta[n] \ast \frac{1}{T_s}(\delta[n] - \delta[n-1])  &amp; ? \\
        \theta[n] \ast \frac{1}{2T_s}(\delta[n+1] - \delta[n-1])  &amp; ? \\
        ???
       \end{array}
      \right.
$$

I'm looking for what is typically done. The "Standard" approach.
I have a nice solution for you that I'm always using whenever I need to compute the instantaneous frequency of a discrete analytical signal and which works much better then the common approach using $tan(.)$ where you need to introduce thresholds and fiddle around with adding/subtracting $2\pi$. In addition to that it is even faster.

Given a signal $x(t)$ and the corresponding analytical signal $$\xi(t) = x(t) + i\mathcal{H}x(t)$$ with $\mathcal{H}$ being the hilbert transform operator the instantaneous frequency $\omega(t)$ can be computed as followed. We have $\omega(t)=\dot{\varphi}(t)$ with $\varphi(t)=\Im log\xi(t)$ being the phase of the analytical signal $\xi$ ($\Im$ denotes the imaginary part). Combining these equations we get
$$
\omega(t)=\Im\frac{d}{dt}log\xi(t)=\Im\frac{\dot{\xi}(t)}{\xi(t)}
$$
This can also be seen be as follows:
$$
  \Im\frac{\dot{\xi}}{\xi} = \Im\frac{ \dot{A}e^{i\varphi} + iAe^{i\varphi}\dot{\varphi} }{Ae^{i\varphi}}
                       = \Im\Big[ \frac{\dot{A}}{A} + i\dot{\varphi} \Big]
                       = \dot{\varphi}
$$
Discretizing this equation you can now compute the discrete instantaneous frequency from the discrete analytical signal:
$$ \omega[n]\approx\Im\frac{\xi[n+1]-\xi[n]}{\xi[n]} $$
or second order:
$$ \omega[n]\approx\Im\frac{\xi[n+1]-\xi[n-1]}{2\xi[n]} $$

Of course this is not the exact solution as it is still requires a discrete differentiation of the signal. But this could be done using standard techniques.