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Re: [Discuss-gnuradio] Hilbert transform


From: Marcus Müller
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

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


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