Hilbert Transform Homework: Show Envelope is |m(t)|

[roam]

Homework Statement

For a real, band-limited function ##m(t)## and ##\nu_v > \nu_m,## show that the Hilbert transform of

$$h(t) = m(t) cos(2\pi \nu_c t)$$

is

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

and therefore the envelope of ##h(t)## is ##|m(t)|.##

Homework Equations

Analytic signal for a a real function ##f(t)## with Fourier transform ##F(\nu)## is given by

##f_a (t) = 2 \int^\infty_0 F(\nu) \exp(j 2 \pi \nu t).##

Modulation property of the Fourier transform

##f(t) \cos(2\pi \nu_0 t) \iff \frac{1}{2} \left( F(\nu + \nu_0) + F(\nu - \nu_0) \right)##

The Attempt at a Solution

The problem asks to compute the analytic signal from the Fourier transform of ##h(t)##, so I think I need to use the equation above for finding the analytic signal:

$$h_a (t) = 2 \int^\infty_0 \Big[ H(\nu) \Big] e^{-j 2 \pi \nu t} \ d\nu = 2 \int^\infty_0 \Big[ \int^\infty_{-\infty} m(t) \cos(2 \pi \nu_c t) e^{-j2 \pi \nu t} dt \Big] e^{-j 2 \pi \nu t} \ d\nu \ (i)$$

Furthermore, we are told that ##m(t)## is band-limited, which means ##M(\nu) = 0## for ##|\nu| > \nu_m.## So perhaps we can use the above property to write (i) as

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

Is this correct so far?

If this is correct, how do I simplify this further to find the imaginary part \hat{h}(t) (Hilbert transform)? Do I simply need to write the exponential as sine and cosine?

Any help would be greatly appreciated.

 

[mighty2000]

Thank you!Yes, you are on the right track. To simplify further, you can use the fact that the Fourier transform is symmetric, i.e. ##M(-\nu) = M(\nu)##. This means that the second term in the brackets will cancel out with the first term, leaving you with just ##M(\nu + \nu_c)## in the integral. Then, you can use the modulation property of the Fourier transform to write this as ##\frac{1}{2}M(\nu)##, giving you the final result of ##h_a(t) = m(t)\sin(2\pi\nu_c t)##.