# Hilbert transform from analytic signal

Show that the Hilbert transform of $h(t) = m(t) \cos(2 \pi \nu_c t)$ is

$$\hat{h} (t) = m(t) \sin(2 \pi \nu_c t),$$

where $m(t)$ is a real valued, band-limited function (i.e. we have Fourier transform $M(\nu) = 0$ for $|\nu| > \nu_m$) and $\nu_c > \nu_m.$

Attempt

I did some research and I found out that this result follows directly from 'Bedrosian's theorem'. But I am required to compute this by first finding its analytic signal $h_a$ whose imaginary part would then be the Hilbert transform. Here is my expression for $h_a (t):$

$$h_a (t) = 2 \int^\infty_0 \Big[ H(\nu) \Big] e^{j 2 \pi \nu t} \ d\nu = 2 \int^\infty_0 \Big[ \frac{1}{2} (M(\nu + \nu_c) + M(\nu - \nu_c)) \Big] e^{j 2 \pi \nu t} \ d\nu.$$

I have used the 'modulation property' of the Fourier transform to get to the RHS. So how can I proceed with the integration when we do not have an explicit expression for $m(t)$?

Note that the condition on the relation between the bandwidth of $m(t)$ and the modulation frequency ensures that $h(t)$ is a band pass signal. It can be written as

$$h(t)=\frac12 m(t)e^{j2\pi \nu_c t}+\frac12 m(t)e^{-j2\pi \nu_c t}\tag{1}$$

From our assumption $\nu_m<\nu_c$ it follows that the two expressions on the right-hand side of $(1)$ do not overlap in the frequency domain. Consequently, the analytic signal is simply given by the first term on the right-hand side of $(1)$ (scaled by a factor of $2$), which is the part of the spectrum of $h(t)$ occurring at positive frequencies:

$$h_a(t)=m(t)e^{j2\pi\nu_c t}\tag{2}$$

The Hilbert transform of $h(t)$ is given by the imaginary part of $(2)$:

$$\hat{h}(t)=\text{Im}\{h_a(t)\}=m(t)\sin(2\pi\nu_c t)\tag{3}$$

EDIT: The equation for $h_a(t)$ given in your question is also correct. If you want to continue that way, it's important to see that due to $\nu_c>\nu_m$, $M(\nu+\nu_c)$ equals zero for $\nu>0$. So the integral can be written as

$$h_a(t)=\int_0^{\infty}M(\nu-\nu_c)e^{j2\pi\nu t}d\nu\tag{4}$$

Likewise, $M(\nu-\nu_c)$ is zero for $\nu<0$, so we can extend the lower integration limit in $(4)$ to $-\infty$:

$$h_a(t)=\int_{-\infty}^{\infty}M(\nu-\nu_c)e^{j2\pi\nu t}d\nu=\mathcal{F}^{-1}\{M(\nu-\nu_c)\}=m(t)e^{j2\pi\nu_c t}\tag{5}$$

which is of course identical to $(2)$.

• Thank you so much for your help, it makes perfect sense now. May 4, 2016 at 9:11