Question about Hilbert transform of a cosine signal

Question

In the paper entitled "On instantaneous frequency", it is claimed that the Hilbert transform of a signal in the form of $A(t)cos(\phi(t))$ would result in $A(t)sin(\phi(t))$, where $A(t)$ is instantaneous amplitude and $\phi(t)$ is the phase function. As an example, consider the following: $$t = [0, 20],\;\textrm{time interval in seconds, sampled each 0.01 sec}$$ $$A(t) = e^{-0.1(t - 10)^2},\;\textrm{instantaneous amplitude}$$ $$\omega = [0.1,\ 0.5,\ 1.0,\ 1.5],\;\textrm{angular frequency in rad/sec}$$ $$f(t) = A(t)\cos(\omega t)$$ $$g(t) = A(t)\sin(\omega t)$$ $$h(t) = \frac{1}{\pi}P\int_{-\infty}^{+\infty}\frac{f(\tau)}{t - \tau}dt,\;\textrm{Hilbert transform of } f(t)\textrm{, where } P \textrm{ is the Cauchy principal value}$$ It is expected that $g(t)$ and $h(t)$ be the same according to the fact claimed in mentioned paper. I used function "hilbert" of MATLAB to compute Hilbert transform of $f(t)$ for aforementioned $\omega$ values. Results show that for $\omega$ close to zero, $g(t)$ and $h(t)$ are different and as $\omega$ gets higher, the difference decreases. I was wondering if I am making a mistake or something is wrong with function "hilbert" of MATLAB? The following figure shows the results:

Answer 1

Make sure that you understand the conditions under which

$$\mathcal{H}\big\{A(t)\cos(\omega_0 t)\big\}=A(t)\mathcal{H}\big\{\cos(\omega_0 t)\big\}=A(t)\sin(\omega_0t)\tag{1}$$

holds. Eq. $(1)$ holds if $A(t)$ is a low-pass signal with a cut-off frequency smaller than $\omega_0$. This implies that $A(t)\cos(\omega_0t)$ is a band-pass signal with no energy at DC. Eq. $(1)$ is a special case of Bedrosian's theorem.

In your example, $A(t)$ is not band-limited, and, consequently, $(1)$ doesn't hold. However, if you increase the modulation frequency $\omega_0$, the overlap of the spectra of $A(t)$ and of the carrier becomes smaller, because $A(t)$ does have a low pass character. So for large $\omega_0$, Eq. $(1)$ is approximately satisfied, which is exactly what you see in your plots.

Also take a look at this related answer.

Answer 2

The claim is generally false. This is studied in details in a 1997 paper by B. Picinbono: On instantaneous amplitude and phase of signals

Let $m(t)$ be a positive function corresponding to the information o be transmitted. By multiplying the carrier frequency signal $cos(\omega_0 t)$ by $m(t)$, we obtain the signal $x(t) = m(t) > cos(\omega_0 t)$ and it is commonly admitted that is $m(t)$ the instantaneous amplitude of the signal $x(t)$. This appears in many textbooks

This leads to the conclusion that the definitions given previously, even if they are widely used, are incoherent because they do not associate with a given real signal a well-defined pair of functions that are the instantaneous amplitude and phase of $x(t)$.

The basis of the the claim resides in the Bedrosian theorem, that asserts that, in some conditions, the Hilbert transform $\mathcal{H}$ can apply separately on a product

$$\mathcal{H}[x_1(t)x_2(t)] = x_1(t)\mathcal{H}[x_2(t)]$$

The classical condition is of "distinct frequency supports": that $x_1(t)$ is strictly band-limited above $B$ (ie zero-spectrum for $\nu> B$, and $x_2(t)$ is strictly band-limited below $B$ (ie zero-spectrum for $\nu< B$). This condition does not hold with the Gaussian function, since it is not band-limited. However, when you increase $\omega$, the overlap between the Gaussian and the carrier sine has less energy, and the claim becomes "less false" in simulations.

Further, if $a(t)$ and $cos[φ(t)]$ have distinct supports (as above), one has: $$\mathcal{H}({a(t) \cos[\phi(t)]}) = a(t)\mathcal{H}({\cos[\phi(t)]})$$ but one cannot say that:

$$\mathcal{H}({\cos[\phi(t)]}) = \sin[\phi(t)]\;\textrm{ (false in general)}$$