|Subject:||Re: [Discuss-gnuradio] Hilbert transform|
|Date:||Sun, 17 Aug 2014 17:53:52 +0200|
|User-agent:||Mozilla/5.0 (X11; Linux x86_64; rv:24.0) Gecko/20100101 Thunderbird/24.5.0|
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.
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
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