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Hilbert Transform

  • #1
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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? :confused:

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
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7,499
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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