# Hilbert Transform

## 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.

## Answers and Replies

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