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Hilbert Transform Homework: Show Envelope is |m(t)|
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[QUOTE="roam, post: 5460690, member: 120460"] [h2]Homework Statement [/h2] 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)|.## [h2]Homework Equations[/h2] 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)## [h2]The Attempt at a Solution[/h2] 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 [B](i)[/B] 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. [/QUOTE]
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Hilbert Transform Homework: Show Envelope is |m(t)|
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