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From: jbuck@janus.Berkeley.EDU (Joe Buck)
Newsgroups: comp.dsp
Subject: Re: 180 deg phase shift
Message-ID: <42081@ucbvax.BERKELEY.EDU>
Date: 8 May 91 18:11:16 GMT
References: <1644@fs1.ee.ubc.ca> <1991May7.152409.3933@njitgw.njit.edu> <1991May8.022953.781@appmag.com>
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In article <1991May8.022953.781@appmag.com> todd@appmag.com (Todd Day) writes:
>First of all, 180 deg phase shift is not the same as simply inverting
>the signal.  A phase shift implies a time delay of some sort.

Todd, you're too smart for your own good, and so are the people talking
about Hilbert transforms, even though they are, in a sense, right.

>Second of all, he *may* have meant a 180 deg phase shift for each of the
>frequencies that add up to his composite signal.

OK, this means that |H(j omega)| should be 1 and arg(H (j omega)) should
be pi for all omega (H is the transfer function).  That means that
H(j omega) should be -1 for all omega, so simply inverting the signal
really does shift by 180 degrees at all frequencies.

Now, for the Hilbert transform people.  The Hilber transform is a filter
whose response is -j for positive frequencies and j for negative frequencies,
and it causes a 90 degree phase shift.  Now let's cascade two of them.
We square the transfer function to obtain the total transfer function.
Note that j^2 = (-j)^2 = -1, so again we get a simple inversion.

>  The output signal will
>definitely NOT look much like the input signal.  You can prove this to your
>self by drawing a time axis and delaying a 100 Hz and 50 Hz signal by 180
>degrees.

You screwed up your plot, Todd.  Otherwise you'd find that the right
answer is an inversion.  Since each individual component is negated by
a 180 degree shift, clearly the sum is.

The original poster clearly had a homework problem with a trivial answer,
and all these "experts" screwed it up.  Embarrassing...

--
Joe Buck
jbuck@janus.berkeley.edu	 {uunet,ucbvax}!janus.berkeley.edu!jbuck