Re: How does an inverter affect phase?

I wrote:

...
Some senseless rambling ahead:

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 would really like to heae someone explain the behaviour of a
infinite-precision (ala DSP riddle) Hilbert transformer on a DC component
when applied twice. Does the DC emerge from its infinitesimally close
neighborhood (which is absent in case of periodic signals)? Isn't this
what you were referring to Oli?

Because, if I substract the DC out of sinc (i.e., resulting in "punctured"
frequency response) and hilbert transform it twice, I get a inverse sinc
_with_ DC. Is this "paradox" result of the integral transforms defined
through Riemann integral? Does the indifference of Riemann integral, for
point changes in function, also result as indefference for point changes
in spectrum?


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

--
Jani Huhtanen
Tampere University of Technology, Pori
.

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