# What is the effect on Hilbert transform of signal after multiplication by sinusoid?

Question :

Find Hilbert transform of $$[u(t-a)-u(t-b)]\cos2\pi f_{0}t\\\\$$ such that $$\\0

my attempt:

we know Hilbert transform of

$$[u(t-a)-u(t-b)]\xrightarrow{\mathcal H} \dfrac{1}{\pi}\ln\left|\dfrac{t-a}{t-b}\right|$$

but after multiplying left side by sinusoid $$\cos 2\pi f_{0}t$$ how the right hand side will alter??. Can we expect good results ( like shifting in spectrum by $$\pm f_{0}$$ which takes place in Fourier transform)

Let $$x(t)=\big[u(t-a)-u(t-b)\big]$$, and $$s(t)=x(t)\cos(2\pi f_0t)$$. If we can assume that $$f_0\gg 1/(b-a)$$, then we can approximate the analytic signal of $$s(t)$$ (i.e., the signal with all negative frequency components removed) by

$$s_a(t)\approx x(t)e^{j2\pi f_0t}\tag{1}$$

The Hilbert transform of $$s(t)$$ is the imaginary part of the analytic signal, i.e., we have the following approximation:

\begin{align}\mathcal{H}\{s(t)\}&=\textrm{Im}\{s_a(t)\}\\&\approx x(t)\sin(2\pi f_0t)\\&=\big[u(t-a)-u(t-b)\big]\sin(2\pi f_0t)\tag{2}\end{align}

The same result is easily obtained by applying Bedrosian's theorem, which assumes that $$x(t)$$ and $$\cos(2\pi f_0t)$$ have no spectral overlap. This is approximately satisfied if the same condition as above is assumed to be true: $$f_0\gg 1/(b-a)$$.

This condition basically says that the signal $$s(t)$$ is a band pass signal, which implies that $$x(t)$$ is a low pass signal. The Hilbert transform of such amplitude modulated signals always equals the original low pass signal $$x(t)$$ multiplied by the Hilbert transform of the carrier:

\begin{align}\mathcal{H}\big\{x(t)\cos(\omega_0t+\phi)\big\}&=x(t)\mathcal{H}\big\{\cos(\omega_0t+\phi)\big\}\\&=x(t)\sin(\omega_0t+\phi)\tag{3}\end{align}

The exact Hilbert transform of the given signal can only be computed by evaluating the integral

\begin{align}\mathcal{H}\big\{s(t)\big\}&=p.v.\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{s(\tau)}{t-\tau}d\tau\\&=p.v.\frac{1}{\pi}\int_{a}^{b}\frac{\cos(2\pi f_0\tau)}{t-\tau}d\tau\tag{4}\end{align}

but I don't think that that was the idea of the exercise.