Hilbert transform of $\cos(\phi(t))$. I am attempting to derive the Hilbert transform of $\cos{\phi(t)}$. I understand that the transform is given by
\begin{align*}
H[\cos(\phi(t)] = \frac{1}{\pi} \ p.v. \ \int_{-\infty}^{\infty} \frac{\cos(\phi(t))}{t - \tau} \ d\tau
\end{align*}
however, I am not sure exactly how to approach the problem; i.e what complex contour to take or possible simplifications. I believe the solution should be $\sin(\phi(t))$, can anyone tell me whether this is correct?
 A: No, it is not correct. $\mathcal{H}\{\cos \phi(t)\} = \sin \phi(t)$ does not always hold for arbitrary phase function $\phi(t)$.
Indeed, $\cos \phi(t)$ and $\sin \phi(t)$ would be a Hilbert transform pair,  if and only if $\exp(i\phi(t)) = \cos \phi(t) + i \sin \phi(t)$ is analytic,  i.e. the Fourier spectrum of $\exp(i \phi(t))$ is one-sided:
$$
\mathcal{F}_{t\rightarrow f}\left\{ \exp(i\phi(t)) \right\} (f) = 0,  ~\forall f < 0. 
$$
Here is an example. Let $\phi(t)$ be the phase function of a sinusoidal frequency-modulated signal, say
$$
\phi(t) = 2\pi 5t + \epsilon \sin(2\pi 10t), 
$$
(where $\epsilon$ is a positive constant). Using Bessel functions of the first kind (see, e.g. (Benson, 2006) pp.50~53),  we can expand $\cos \phi(t)$ and $\sin \phi(t)$ as Fourier series:
$$
\begin{aligned}
\cos \phi(t)  &= \cos(2\pi 5t+\epsilon \sin(2\pi 10t)) \\
&= \sum_{n=-\infty}^{+\infty} J_n(\epsilon) \cos(2\pi(5+10n)t), \\
&= \sum_{n=0}^{+\infty} (J_n(\epsilon) + J_{-n-1}(\epsilon)) \cos(2\pi(5+10n)t), \\
\sin \phi(t)  &= \sin(2\pi 5t+\epsilon \sin(2\pi 10t)) \\
&= \sum_{n=-\infty}^{+\infty} J_n(\epsilon) \sin(2\pi(5+10n)t) \\
&=  \sum_{n=0}^{+\infty} (J_n(\epsilon) - J_{-n-1}(\epsilon)) \sin(2\pi(5+10n)t). 
\end{aligned}
$$
However, noting that $\mathcal{H}\{\cos (2\pi f_0 t) \} = \sin (2\pi f_0 t)~(\forall f_0>0)$,  the Hilbert transform of $\cos \phi(t)$ is given by
$$
\begin{aligned}
\mathcal{H}\{ \cos \phi(t) \} &= 
\sum_{n=0}^{+\infty} (J_n(\epsilon) + J_{-n-1}(\epsilon)) \mathcal{H}\{\cos(2\pi(5+10n)t) \} \\
&= \sum_{n=0}^{+\infty} (J_n(\epsilon) + J_{-n-1}(\epsilon)) \sin(2\pi(5+10n)t),
\end{aligned}
$$
which clearly shows that $\mathcal{H}\{\cos \phi(t)\} \ne \sin \phi(t)$. Indeed, $\exp(i \phi(t)) = \sum_{n=-\infty}^{+\infty} J_n(\epsilon) \exp(i 2\pi(5+10n)t)$ consists of harmonic components with negative frequencies and thus it is not analytic.

In terms of the difference between $\mathcal{H}\{\cos \phi(t)\}$ and $\sin \phi(t)$, we have a more general result given by Nuttal theorem (Nuttal et al., 1966).

Nuttal theorem. Denote the original signal as $x(t) = a(t) \cos[2\pi f_0 t + \phi(t)]$, and the expected quadrature signal as $x_q(t) = a(t) \sin[2\pi f_0 t + \phi(t)]$. Then the $L_2$ difference between $x_q(t)$ and $\mathcal{H}\{x(t)\}$ can be computed by
$$ \int_{-\infty}^{+\infty} | x_q(t) - \mathcal{H}\{x\}(t) |^2 dt = 2\int_{-\infty}^{-f_0} |F(f)|^2 df, $$
where $F(f)$ is the Fourier spectrum of the complex signal $a(t) \exp(i\phi(t))$.

Then the question here is just a special case $a(t) \equiv 1, f_0 = 0$. Therefore, the less energy contained in the negative half of the spectrum of $\exp(i\phi(t))$, the better $\mathcal{H}\{\cos \phi(t)\}$ can approximate $\sin \phi(t)$.

References

*

*Benson, D. (2006). Music: A Mathematical Offering.

*Nuttall, A., & Bedrosian, E. (1966). On the quadrature approximation to the Hilbert transform of modulated signals.

