"Scott Gardner" wrote in message

On Sun, 23 Nov 2003 04:04:38 -0500, "Arny Krueger"
wrote:

"Scott Gardner" wrote in message

On Sat, 22 Nov 2003 21:07:24 -0500, "Arny Krueger"
wrote:

"ScottW" wrote in message
news:AmUvb.4608$ML6.2599@fed1read01
"Erik Squires" wrote in
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So, here's my question. If I digitize a 15 kHz signal, using a
44.1 kHz sampling rate, I'm going to get about 3 samples / cycle.

Are the normal digital filters good enough to reproduce a 15 kHz
singal with varying amplitude?

Deconstruct this signal to frequency domain and it won't be a
pure 15 kHz.

Wrong.

There has to be another component imposed on the 15 kHz that
"varies the amplitude".

Wrong.

Three samples is sufficient to define a sine wave that has unique
frequency, phase and amplitude. In fact, just slightly more than
two samples is sufficient.


Arny,
I am probably looking at this the wrong way, using an
oversimplified model, but I can't see how a sine wave can be
completely defined by three points.

It's a theorem that has never been disproved that says that it takes
slightly more than 2 points to adequately define a unique sine wave.

I'm picturing a sine wave plotted with time along the x-axis,
and amplitude along the y-axis. If I tell you that the amplitudes
at zero seconds, 1 second, and 2 seconds are all zero, I've given
you three different points along the wave.

Right, but the frequency of that wave has to be outside the range
for which the theorem applies.

From this, the period can be
measured and the frequency derived from that, but I don't see how
I've given you enough information to calculate the amplitude.
Let me know what I'm missing.

You are missing the fact that the frequency of a signal with three
points that are zero is too high for the Nyquist theorem to apply.
In fact, the frequency of the signal has to be exactly half the
sample rate.

Do the three points have to
have non-zero amplitude for them to be used to define the waveform?

At least one of the points has to be non-zero, and this will be true
if the frequency of the signal is even just slightly below half the
sample rate. At exactly half the sample rate, the signal can have
any amplitude and have three zero samples. It's a well-known
boundary condition.

Okay, that makes sense to me now. Since the three points I
chose were at zero, one and two seconds, my sampling rate was
therefore 1 Hz. Since the amplitudes at all three points were zero,
they therefore represented the start, crossover, and end of a waveform
with a period of two seconds, or a frequency of 1/2 Hz. Since the
frequency of my waveform (1/2 Hz) was greater than or equal to half of
my sampling rate (1 Hz), there wasn't enough information there to
determine the waveform characteristics.

Bingo!

Thanks for the clarification. I still remembered how to do
polynomial curve fitting, based on the number of local minima and
maxima in the plot of the polynomial, but if I ever learned
trigonometric curve fitting, I had obviously forgotten it.

Thanks for correctly perceiving my explanation.

There are at least three forms of curve-fitting that I've seen used pretty
frequently:

(1) Polynomial
(2) Exponential
(3) Trigonometric

I seem to recall that all that is required is that the curve-fitting
methodology be based on orthogonal functions, and that's only a requirement
if you want a general, unique solution.