"Here's another way of looking at it, putting a sound wave through a hole
in the wall can't produce Doppler shift, no matter how many tones are in the
waveform, and a speaker is effectively an artificial hole in the wall, in
that the effective sound source isn't the speaker any more than it is the
hole in the wall. Does anyone here think that if you stretched a thin
diaphragm over a hole in a soundproof wall and had a band playing behind it,
the diaphragm would cause Doppler Distortion? The speaker, provided it isn't
exceeding its linear limits, is effectively exactly the same thing for all
practical purposes. Instead of being driven by the sound source in the other
room, it's driven by the electrical equivalent of the sound source in the
other room. Can any of you provide an explanation of how an acoustic wave
driving a diaphragm and passing the soundwave
through it is in any way different than the diaphragm being driven by a
motor being supplied with the exact electrical analog of that acoustic wave?
And by that I mean that the difference will be such that the electrically
driven one will produce Doppler distortion while the acoustically driven one
doesn't."
It seems to me that it boils down to the above situation, if the diaphragm
covering the hole in the wall driven directly by the acoustic wave in the
other room doesn't produce Doppler shift, then the diaphragm being driven by
the exactly equivalent electrical wave won't either. Since the diaphragm
over the hole is moving exactly like the acoustic wave coming through the
hole, I don't see how any form of Doppler shift could be introduced, as
obviously, the hole itself won't introduce Doppler shift. Note, I'm
referring only to Doppler shift, not the lowpass effect of the hole, the
inertia of the diaphragms, etc, which may affect the sound, but will have
absolutely nothing to do with Doppler shift.


This is a very interesting thought experiment! But I believe I have
the answer.

A diaphragm's movement caused by a sound wave on the other side of the
wall is *not* the same movement as a speaker's produced by the same
sound in electrical wave form.

Let's set up this situation:
-- Room with sound sources - Diaphragm over hole - air on other
side
-VS-
-- electrical wave exactly replicating the waveform from those sound
sources - speaker cone - air in front of speaker.

First, what happens in the room:

Let's start out with an incredibly low frequency, loud source in the
room. We'll assume that this causes a waveform in the air with huge
pressure gradients that causes the diaphram to move in-out-in-out
relatively slowly with a very large amplitude.

While this is happening, we then start up a second source in the room,
which is high frequency, and lower amplitude. Assuming the air is 100%
linear, the two waveforms will add together perfectly.

However, when the higher frequency waveform (series of pressure
changes in the air) reaches the diaphragm, the diaphram is in *motion*
due to the low frequency source. This motion is relative to the high
frequency source, and therefore, instead of adding a perfect
representation of the high frequency waveform on top of the
diaphragm's movements, a *DOPPLER SHIFTED VERSION* of that high
frequency waveform appears on the diaphragm!

In other words, not only do speakers exhibit doppler distortion, but
microphone diaphragms would, as well! (if one does, then the other
should, too. This only makes sense). We generally don't care about it
with microphones, because the diaphragms move so slightly that it's
basically not there at all.

But in this thought experiment, it's there!

Now, on the other side of the wall, the diaphragm is vibrating a
doppler-distorted version of the original sounds. On this side, it
effectively acts like a speaker, and the waveform it's vibrating at
(the doppler distorted version of the original) is *again* doppler
distorted when going into the next room.

Now, I haven't done the math out on this, but I would suspect that any
doppler effects caused by the diaphragm "picking up" the waveform
would be *EXACTLY CANCELED OUT* by the doppler effects of the
diaphragm vibrating the air in the next room.

Net result: No doppler distortion through a diaphragm over a hole in
the wall!


Now, the speaker, on the other hand, exhibits the doppler distortion
going from speaker-air, with no "compensating" distortion on the
recording end, and therefore still exhibits doppler distortion. Well,
unless the microphone used the exact same transducer as the speaker to
record, and it had the same excursion -- then the doppler effect of
the microphone would cancel out the doppler effect of the speaker, and
the result would be an un-doppler-distorted recreation of the original
acoustic sound. HOWEVER, the resultant sound is *still* a doppler
distorted version of the *electrical* waveform, technically.

Ken