AW: Phase and Delay (was: Phase vs. Delay)

[Haible Juergen]

	>It's certainly true that a phase shift circuit makes a
	>fine vibrato.  But I'll claim that that this vibrato effect happens
	>because the phases of the component sines are changing and not
because
	>of any Doppler-like pitch shift due to storing the signal in a
medium
	>and pulling it out at a different rate.

I'd say it's not an either / or ; it's just a different view of the same
thing.
There is *something* stored in the capacitors (and inductors) of
a filter circuit. It's just that the waveform of the original is not
preserved, or in other words "the medium shows dispersion".

	>   At least *something* will show up in the output immediately. But
	>   the shape of your input will be altered, as different spectral
	>   components of your input signal will get different delay. That's
	>   how I think it works, at least.
	>
	>Well, let's see...  Imagine a single phase shift stage in front of
	>you, the classic opamp circuit.
	>
	>At low frequencies you'll have near 0 degrees phase shift, there's
no
	>delay because the signal goes straight through.
	>
	>At high frequencies there's 180 degrees phase shift.  Well not
exactly
	>phase shift; it's a straightforward polarity inversion.  No delay
	>there either.

What exactly does it mean when we speak of applying
one frequency f, and looking at phase and / or delay ?
It means that a sine wave of this frequency was switched
on in the infinite past. Otherwise we can't look at one single
frequency. Now we look at the input and output signal at
times >= 0. Asume we have a positive zero crossing at the input
at t = 0. The "same" zero crossing will appear at the output
after a certain time t1 > 0. Now we can refer to this time
directly (i.e. as delay), or in relation to the period of the sine
wave (i.e. as phase). Both descriptions are equivalent.

The problem is that you cannot know *which* zero crossing
is which, from just looking at input and output signal. The
zero crossing at the output at t1 might also be the result of
a zero crossing at the input an t = 0 - N * 1/f. So the delay
of the filter might be any N * 1/f + t1. You can measure 
phase (with its inherent ambiguity), but you cannot measure
the delay in this configuration. Not being able to measure it
doesn't mean it's not there, of course. And if you know a little
about the filter under test (number of poles etc.) you can 
decide which N is the right one, and thus decide on the 
right delay, too.

But accepted: That's not a typical experiment you would do
to measure a delay. And it's obvious why the *concept* of
frequency domain is generally described with amplitude
and phase rather than amplitude and delay.

But now let's look at time domain again. 

	>At low frequencies you'll have near 0 degrees phase shift, there's
no
      >delay because the signal goes straight through.

Let's make things easier and go to very low frequency, to
0Hz / DC.
We're switching on a DC source at the input at t=0.
(the same concept works for *any* frequency, but it's
easier to look at it for DC).
You're right: There is an (almost) immediate response on the 
output. But this is not a result of DC (or any other constant
low frequency applied) at all.
Switching the DC source on actually means a *step* that
contains all frequencies (and not just a single one). What
you see at the output immediately are the highest components
of the step's spectrum, and in fact this first part of the step
response is *inverted* (in the topology that has no inversion
for DC). After some time, the output signal crosses zero
and then approximates the DC input voltage.

Unfortunately, I cannot directly look at the step response
and see the group delay. (The step response is infinite,
but the delay on very low freq signals is finite ...) But one can
see a certain trend at least: There is an immediate
response with sharp corners <=> group delay approaches
zero for high frequencies. Output approaches input after
after some time <=> gain at low freq = 1, but there is some
delay ... things like that. I bet the system theory specialists
can read out much more, but not me ...

JH.