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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