# Hilbert transform

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

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 Sun, Aug 17, 2014 at 11:04 AM, jason sam [email protected] wrote:

The Hilbert transforms a real signal into an analytic signal. Think
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