# Question about Hilbert transform of a cosine signal

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: 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.

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:

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