Re: How does an inverter affect phase?

Martin Eisenberg wrote:

Jani Huhtanen wrote:

It seems that applying ideal HT twice inverts the signal, but
only if the signal is not periodic. How it does that, I don't
know, because impulse response of a HT does not have a DC
component. It seems that for a sparse spectrum (i.e., perioidic
signal) HT cannot invert a signal because pure DC component
(such as in f(t)=1+sin(t)) does not get recovered from its
hilbert transform while HT _does_ invert a non-periodic
waveform (such as sinc(x)).

I should add that I experimentally tested the ideal HT by
calculating the analytical HTs of functions with Maple.

So what did you actually get, and in what version did you use? Here,
Maple 7 leaves 'inttrans[hilbert](1,t,u)' unevaluated when
transforming 1+sin(t), and doesn't do 'int(1/(t-u), t=-infinity
..infinity, CauchyPrincipalValue) assuming u::realcons' either.


Martin


I don't remember the version (probably 9). I'll check it when I get back to
work.

However, here are some of the results from memory:

# with(inttrans);
# hilbert(1+sin(t),t,x);
=> cos(x)

# hilbert(sin(t),t,x);
=> cos(x)

# invhilbert(cos(x),x,t) [1]
=> sin(t)

# hilbert(sin(t)/t,t,x);
=> (cos(x)-1)/x

Last one is particularly interesting as it is the impulse response of a
bandlimited HT.

[1] I don't remember if the function was invhilbert or inversehilbert or
something similar.

I now have a partial understanding why HT behaves like this. This is because
there is only infinitesimally small energy in DC component for, say,
sinc(t). While, for example, 1+sin(t) has a finite energy at DC (i.e.,
spectrum has a delta at DC). Thus "taking away" a finite amount of energy
results in "visible" changes in the resulting function, while taking
infinitesimally small amount of energy is not noticeable (roughly
speaking). Still I don't quite grasp how even that infinitesimally small
energy emerges in inversehilbert transform.

ERRATA:
In a previous post I told that I substracted the DC component out of sinc.
This is obviously impossible as the substraction would have to be
infinitesimally small.

--
Jani Huhtanen
Tampere University of Technology, Pori
.

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