What is the Hilbert transform of $e^{-jwt}$ ? I know how to do it via convolution , but i am looking for a less formal, more intuitive explanation such as the phase shift it imparts
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Helpful? – Cosmas Zachos May 09 '20 at 22:19
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The Hilbert transform is the convolution with $pv(\frac1{\pi t})$ equivalently it is $F^{-1}(i\ sign(v) F(h))$. Here $F(h) = 2\pi \delta(v+\omega)$ so you are looking at $F^{-1}(2i\pi sign(v) \delta(v+\omega))=F^{-1}(-2i\pi \delta(v+\omega))=-i e^{-i\omega t}$ – reuns May 11 '20 at 02:13