Thanx Marcus and Tom fr ur explanations. I will read further and ask any questions if i have.
Tom Rondeau <t...@trondeau.com> wrote: >On Sun, Aug 17, 2014 at 11:04 AM, jason sam <user0...@gmail.com> wrote: > >Hi, >I have made a simple flowgraph as attached.I have on query that when i >observe the signal coming out of the 'Hilbert transform' block using a >time sink then its imaginary part is shown to be zero.According to the >theory the hilbert transform of a signal x(t) is: >x(t)+jx~(t) >where x~(t) is the quadrature phase component of x(t).Then why is the >signal from the hilbert block has zero imaginary part?? >Regards, >Ali > > > >The Hilbert transforms a real signal into an analytic signal. Think about your >case this way: you start with a real sine wave, so in the frequency domain, >you have a delta function at +f and -f. But if you have that same sine way as >a complex number, then you'll only have a delta at +f. A sine wave travels >along the unit circle, but in which direction? A complex (analytic) signal >gives you the value and the direction, like a vector instead of a scalar. So >we've reduce the ambiguity of the solution by providing the direction: >clockwise or counter clockwise. > > >The Hilbert transforms the signal from real to complex by removing the values >in the negative frequency. In fact, most HIlbert transforms (like the one here >in GR) are just high-pass filters with the passband starting at 0 Hz that >provide this conversion process. > > >I wrote a post showing the Hilbert transform effects without actually >explaining it. Still, it might be helpful to understand it: > > >http://www.trondeau.com/blog/2013/9/26/hilbert-transform-and-windowing.html > > >Tom > > > >
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