Re: [Discuss-gnuradio] Hilbert transform

Hi Ali,

I see an imaginary part whenever the real part drops from 1 to 0; which is to be expected, since the Hilbert filter is a high pass one.

Background: Remember, this is the digital world. There is no Hilbert transform here -- there could only be a discrete Hilbert transform. And even that is a convolution with an infinite series and cannot be done. So the Hilbert transform is a FIR approximation.

Greetings,
Marcus

On 17.08.2014 17:04, jason sam 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  

_______________________________________________  Discuss-gnuradio mailing list  Discuss-gnuradio@gnu.org  https://lists.gnu.org/mailman/listinfo/discuss-gnuradio