Luiz Carlos wrote:

>>How does precision enter in? Quantize sin(t)+sin(Pi*t) to one bit, and
>>it still never repeats. On the other hand, the steps in any quantized
>>signal imply infinite bandwidth, and since no filter can be perfect,
>>you'll always get some X-ray exposure. Watch out for that MP3 player!
> 
> 
> Jerry,
> 
> I din't say to quantize it after. I said to generate it with limited
> precision.

Yes, yes. When the difference between the calculated signal and a 
periodic one becomes too small to be represented, the real-world signal 
becomes cyclic. Thanks for the insight. A good one!

> And I was thinking about the recursive implementation (pole over the
> unit circle):y[n]=2*cos(w)*y[n-1]-y[n-2].
> 
> I think the amount of X-rays generated by a MP3 player will not hurt
> me, but the sound ... :)

I'm with you there!

   ...

>>A real example, or a mathematical one?  They aren't the same, you know. 
>>That's what this thread has come to be about.
>>
>>Jerry
> 
> 
> Yes, I know. I'm just getting used to idea that real signals are not
> "mathematically" bandlimited.  It changes my perception of the world!

Everything I learn expands my world view somehow. Welcome to the club!

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<EqydnQa_B8Yj74zdRVn-sw@centurytel.net>...

> A lot of things... :)

Fred, I need some time to digest all you said.
I'll come back later.

Luiz Carlos

> How does precision enter in? Quantize sin(t)+sin(Pi*t) to one bit, and
> it still never repeats. On the other hand, the steps in any quantized
> signal imply infinite bandwidth, and since no filter can be perfect,
> you'll always get some X-ray exposure. Watch out for that MP3 player!

Jerry,

I din't say to quantize it after. I said to generate it with limited
precision.
And I was thinking about the recursive implementation (pole over the
unit circle):y[n]=2*cos(w)*y[n-1]-y[n-2].

I think the amount of X-rays generated by a MP3 player will not hurt
me, but the sound ... :)

> Ingrate! I hand you an ice-cream cone, and you throw it on the ground
> because it has no sprinkles! :-)

Sorry about the ice-cream cone, but it was just a cone, no ice-cream
at all! :)
(Just to write something!)

> A real example, or a mathematical one?  They aren't the same, you know. 
> That's what this thread has come to be about.
> 
> Jerry

Yes, I know. I'm just getting used to idea that real signals are not
"mathematically" bandlimited.  It changes my perception of the world!

Luiz Carlos.

Fred Marshall wrote:

> Yes, that's a good example.  I don't know what "pseudo-periodic"
> means .... 
> 
> Fred

I don't know if it's an "official" definition, but one way to think 
of pseudo-periodicity in the case of sin(t) + sin(w*t) with 
irrational w is: The rising zero crossings of both components come 
arbitrarily close to each other; but besides at t=0 they never 
exactly coincide, unlike with commensurate frequencies.

The second part of that statement is trivial to show. I just tried to 
prove the first assertion, but my ansatz* was sloppy and I'm not 
motivated to rectify it, though it seems it would work out.

* I hope you understand that. I mean that which one writes down first 
to derive things from. I remember reading the term, but couldn't find 
it or an English translation in any dictionary at my disposal.


Martin

"Luiz Carlos" <oen_no_spam@yahoo.com.br> wrote in message
news:3fd8f66b.0401230509.38272b12@posting.google.com...
> Martin,
>
> Somebody here said: sin(x)/x. (Now obvious!)
> So, I'll ask for something a little bit different:
> I want an example for a causal signal that has bandlimited spectrum.

Luiz Carlos,

If you compare the relationship between the Laplace Transform and the
Fourier Transform you should get some insight.  Or, the z-transform and the
Finite Discrete Fourer Transform.

Laplace is for causal things - like real systems whose outputs don't precede
their inputs or predict the future.
Fourier deals with things that don't have to be causal - which is what we
often do in DSP while admitting errors in the process.

The theoretical answer is there's no such thing as a causal signal (e.g.
impulse response) that is also bandlimited.  The practical answer is in the
eyes of the beholder.  You get to define what it means.

So, you can have a sine burst which is causal in the sense that it's zero
everywhere else.
And, you can compute a Finite (not Discrete) Fourier Transform of it - with
a sampled spectrum resulting .. I think.
And, depending on the length of the time record, the resolution will vary
but all the frequencies will be there because the Fourier Transform is of
infinite extent.
(If the sine burst happens to have an integral number of periods in it then
you get a different sort of result that only has zero samples at the higher
frequencies - so let's not go there).
However, if you sample it first, then you have caused spectral aliasing that
can't be removed.
(The missed information in the sampling is in those nasty little corners or
edges where it starts and ends).
Thus admitting errors in the process.
And, obviously, some temporal sample length is usually "good enough" for our
purposes and we use the FFT -> Finite Discrete Fourier Transform.

Let's see, I hope I haven't used the wrong terms - I haven't done this in a
while:
Fourier Transform uses an integral from -infinity to +infinity.
Finite Fourier Transform uses an integral over a limited range like -T to
+T.
Discrete Fourier Transform is a sum from -infinity to +infinity of sampled
data or a vector.
Discrete Finite Fourier Transform is a sum over a limited range like -T to
+T of sampled data or a vector.
The FFT algorithm implements a Discrete Finite Fourier transform.
AND it goes a step further by convention:
It only computes this Discrete Finite Fourier Transform on a discrete set of
frequency points.
This generates a set of points on which one may perform an
Inverse Discrete Finite Fourier Transform which we call the IFFT.
(It appears that physicists refer to a Finite Fourier Transform when we
might call it a Discrete Fourier Transform or a Discrete Finite Fourier
Transform).

Obviously, one can compute any of these.
If a function itself is limited to nonzero values in the range -T to +T,
then its Fourier Transform and its Finite Fourier Transform are the same.
If a function is defined as an infinite series of samples or an infinite
vector, if you will, then its Fourier Transform and Discrete Fourier
Transform are the same.
If a function is defined as a finite series of samples, then its Fourier
Transform and its Discrete Finite Fourier Transform are the same.
If a Discrete Finite Fourier Transform is only computed at discrete
frequencies then I don't know what to call it!!  Discrete Out, Discrete In,
Finite Fourier Transform?
Anyway, the latter can get you into trouble:

1) Often, functions aren't defined or exist that are of finite extent (i.e.
they are of infinite extent or are "long") but we take a time chunk and
Fourier Transform them anyway.  This causes spectral aliasing - (whether we
transform the data to see the result or not).  Why does is cause spectral
aliasing?  Because to multiply by a rectangular window in time is to
convolve with a sinc or Dirichlet in frequency.  This convolution mixes
frequency components together - which results in new frequencies being
generated, thus aliasing.
Perhaps folks would be more comfortable saving the term "aliasing" for
sampling effects that cause "folding" and call this "spectral spreading".
Yes, maybe that's better.

Similar to (1):
2) Functions that aren't defined or exist that are of finite extent in
frequency (i.e. they are of infinite extent or are "wideband") but we take a
spectral chunk of them anyway.  This causes temporal aliasing or, perhaps
better, temporal spreading.
(Actually, in practice, this probably means we are going to sample the time
data as if the spectral limit were there.  The assumed spectral limits would
cause temporal aliasing but we didn't actually limit the spectrum.  Rather
we sampled as if the spectral limits exist and that temporal sampling causes
spectral aliasing - and I do mean aliasing here).

3) Often, functions are defined as a series of samples in time and let's
assume that this is an analytical definition that applies for all
time -infinity to +infinity or there is a data source that is continually
spitting out samples.  It may be that these samples already embody spectral
aliasing because the sample rate and the underlying data source / analytical
function have a bandwidth =>fs/2=1/2T.

Similar to (3):
4) Often, functions are defined as a series of spectral samples and let's
assume that this is an analytical definition that applies for all
frequencies -infinity to +infinity.  It may be that these samples already
embody temporal aliasing because the sample interval Omega and the
underlying analytical function have a temporal extent => 1/2*Omega.
An example of this is when we do convolution using FFT/multiply/IFFT.
- we start with temporal extent of N*T for two arrays where T is the sample
interval.
- we compute FFTs of length N with spectral resolution of 1/N*T.
- multiply the two FFTs together: an array of N samples with resolution of
1/N*T results.
- IFFT the result: an array of N samples with resolution of T results.
Usually we say this causes temporal aliasing because we didn't zero-pad the
arrays before the FFT to a length of N+N-1.
Another way to look at it is this:
The result of the convolution in time has to be N+N-1 samples in length -
which corresponds to a temporal length of (N+N-1)*T seconds.  With this
temporal extent, the frequency samples have to be closer together than
1/(N+N-1)*T Hz.
However, our multiplication in frequency was done with a resolution of 1/N*T
Hz.
The spectrum is undersampled causing temporal aliasing.
Just another way to look at it using the sampling theorem instead of
constructions of circular convolutions in time.
It suggests that we could do the multiply in frequency and then interpolate
before the IFFT.  That makes sense because zero-padding in time is
equivalent to a type of interpolation in frequency.
One could ask:
In this case, is it fundamentally different to interpolate in frequency
after the FFT as compared to zero-padding in time before the FFT?  I think
the answer is no because one could choose to interpolate using the same
transform pair of functions.  Zero padding in time can be viewed as
windowing followed by convolution with a picket fence.  The dual of this in
frequency is convolution with a sinc or Dirichlet and multiplication by a
picket fence which increases the spectral sample rate / decreases the
spectral sample interval.
[not addressing the number of operations required, just the nature of the
results].

Maybe I've gone too far afield but these ideas are important to the
understanding of where we create "errors" and, therefore, where to look for
problems.

Fred

Luiz Carlos wrote:

>>For instance, sin(t) + sin(Pi*t) is only pseudo-periodic because the 
>>frequency ratio is irrational.
>>
>>
>>Martin
> 
> 
> Martin,
> 
> I din't like this one!
> It is periodic, it has an infinite period!
> If you say infinite period is not allowed, I'll say you can't generate
> sin(t)+sin(Pi*t), because you need infinite precision to generate it.
> If not, you'll have a finite period.

How does precision enter in? Quantize sin(t)+sin(Pi*t) to one bit, and
it still never repeats. On the other hand, the steps in any quantized
signal imply infinite bandwidth, and since no filter can be perfect,
you'll always get some X-ray exposure. Watch out for that MP3 player!

>>You needn't look far: excite a filter with noise.
>>
>>Jerry
> 
> 
> Jerry,
> 
> I want the mathematical formula for this noise (or pseudo-noise). Not
> statistical parameters!

Ingrate! I hand you an ice-cream cone, and you throw it on the ground
because it has no sprinkles! :-)

> Ok, I know this is not very much practical, but I think it
> interestesting!

> Martin,
> 
> Somebody here said: sin(x)/x. (Now obvious!)
> So, I'll ask for something a little bit different:
> I want an example for a causal signal that has bandlimited spectrum.

A real example, or a mathematical one?  They aren't the same, you know. 
That's what this thread has come to be about.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Ronald H. Nicholson Jr. wrote:

> ... things below the noise floor can sometimes be measured,
> given enough time.

Huff! That could open a whole new thread: "Where is the floor?" In one
machine I built, the signal was a ramp amounting to 3/4 count/second,
riding on 120 counts P-P of noise. That's a 44 dB N/S ratio. I know it
wasn't just whistling in the dark because I closed a loop around it and
the system worked like a charm.

(That's the job that aroused my interest in DSP. I figured that It would 
be easier to read about it than to make it up as I went along, but I had 
no time then.)

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

In article <r58110dsd1gqv9avbb9qun77nksduqstt9@4ax.com>,
Allan Herriman  <allan.herriman.hates.spam@ctam.com.au.invalid> wrote:
>Mathematical zero - the thing is exactly zero.
>Engineering zero - the thing is less than the noise floor.

Except that things below the noise floor can sometimes be measured,
given enough time.

However there is also:
quantum zero - the thing is smaller than Planck's constant / 4 * pi
where the mathematical model is no longer correct.


IMHO. YMMV.
-- 
Ron Nicholson   rhn AT nicholson DOT com   http://www.nicholson.com/rhn/ 
#include <canonical.disclaimer>        // only my own opinions, etc.

Allan Herriman wrote:
> 
> On Thu, 22 Jan 2004 09:53:18 -0600, jim <"N0sp"@m.sjedging@mwt.net>
> wrote:
> 
> >
> >
> >Allan Herriman wrote:
> >>
> >>
> >> How do you know you *don't* get radio interference, illumination and
> >> X-rays when you strike a bell?  Maybe they're just at a really low
> >> level, too low for you to detect.
> >
> >What if you just pushed the bell with your finger? What if the bell were
> >made of marshmellows? Would you still expect the frquency response to go
> >to infinity?
> >        Under what conditions does striking a bell become mathematically
> >perfect?
> 
> I think you missed my point.  (Jerry got it though.)
> 

Yes, I got the point. But, by your own admition, your assertions are
based only on faith since there is no measurable proof. Anyway, an ideal
impulse should have a flat response for all frequencies.
	

> Mathematical zero - the thing is exactly zero.
> Engineering zero - the thing is less than the noise floor.
> 
> No analog filter has (or can have) infinite attenuation in its
> stopband over a finite range of frequencies (except if the attenuation
> is infinite for all frequencies), and no time-limited analog signal
> can have zero power over any finite range of frequencies (except if
> the signal is zero for all time).
> 
> So, push the marshmallow bell with your finger and get a non-zero
> amount of X-rays.
> 
> ... but that's using the mathematical definition of zero.
> 
> In the engineering sense (and common sense!), you don't get anything
> at all (except a sticky finger).
> 
> Regards,
> Allan.


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Martin,

Somebody here said: sin(x)/x. (Now obvious!)
So, I'll ask for something a little bit different:
I want an example for a causal signal that has bandlimited spectrum.

Luiz Carlos.