I have a general question about sampling theory. Most of this comes
from college coursework, and our friend Google.
If my highest frequency of interest, F(in) was 500 Hz, and my
frequency of sample, F(s), was 8 kHz then I meet the Nyquist
criterion. However, I still need to have an anti-aliasing filter (low-
pass, reconstruction filter, etc.) to properly attenuate any frequency
components/ images above F(s)/2 assuming baseband sampling (Nyquist
Zone 1).
If my frequency of sample is 8 kHz, then my images should be centered
around 8 kHz: F(s) +/- F(in). So my images would be 7.5 kHz and 8.5
kHz well outside F(s)/2. Images will also appear around every
multiple of F(s) as well. But the images from every Nyquist Zone will
still fold back into my baseband, correct? So even though I am 'over
sampling' I still need an anti-aliasing filter.
Many delta-sigma converters have high base sampling rates; e.g., 192
kHz, and can oversample 128*F(s), 256*F(s), 384*F(s), etc. Even here,
the concept is the same. What I have always understood is
oversampling spreads out the quantanization noise (Noise shaping?),
and relaxes the anti-aliasing filter requirements. Regardless, a anti-
aliasing filter is needed because images will fold back without one.
Do I have the basics down?

"Kingcosmos"


> I have a general question about sampling theory. Most of this comes
> from college coursework, and our friend Google.
> If my highest frequency of interest, F(in) was 500 Hz, and my
> frequency of sample, F(s), was 8 kHz then I meet the Nyquist
> criterion. However, I still need to have an anti-aliasing filter (low-
> pass, reconstruction filter, etc.) to properly attenuate any frequency
> components/ images above F(s)/2 assuming baseband sampling (Nyquist
> Zone 1).


** OK so far .....


> If my frequency of sample is 8 kHz, then my images should be centered
> around 8 kHz: F(s) +/- F(in).


** Nonsense.

Images only occur when the signal being sampled has frequency components at
MORE than 1/2 the sampling rate.

Very sharp anti-aliasing filters are normally set to "cut off" just below
that frequency - a signal that makes it through such a filter is AOK.



>So my images would be 7.5 kHz and 8.5
> kHz well outside F(s)/2. Images will also appear around every
> multiple of F(s) as well.


** Oh dear ......

> But the images from every Nyquist Zone will
> still fold back into my baseband, correct? So even though I am 'over
> sampling' I still need an anti-aliasing filter.


** No.

If you are sure that no signal components ( including noise) approach 1/2
the sampling rate - forget it.


> Many delta-sigma converters have high base sampling rates; e.g., 192
> kHz, and can oversample 128*F(s), 256*F(s), 384*F(s), etc. Even here,
> the concept is the same. What I have always understood is
> oversampling spreads out the quantanization noise (Noise shaping?),
> and relaxes the anti-aliasing filter requirements. Regardless, a anti-
> aliasing filter is needed because images will fold back without one.
> Do I have the basics down?


** Not at all.



........ Phil

On Aug 10, 3:21 am, "Anonymous" <m...@privacy.net> wrote:
> "Kingcosmos" <waynelit...@sbcglobal.net> wrote in message
>
> news:1186723512.533750.217830@i38g2000prf.googlegroups.com...
>
> >I have a general question about sampling theory. Most of this comes
> > from college coursework, and our friend Google.
> > If my highest frequency of interest, F(in) was 500 Hz, and my
> > frequency of sample, F(s), was 8 kHz then I meet the Nyquist
> > criterion. However, I still need to have an anti-aliasing filter (low-
> > pass, reconstruction filter, etc.) to properly attenuate any frequency
> > components/ images above F(s)/2 assuming baseband sampling (Nyquist
> > Zone 1).
> > If my frequency of sample is 8 kHz, then my images should be centered
> > around 8 kHz: F(s) +/- F(in). So my images would be 7.5 kHz and 8.5
> > kHz well outside F(s)/2. Images will also appear around every
> > multiple of F(s) as well.
>
> An image in mixing schemes refers to unwanted frequencies being
> mixed (sampled) into your wanted frequencies and seeming to be part
> of your wanted frequencies even if they weren't before.. In your reference
> to
> 7.5 and 8.5, you seem to be referring to the wanted sidebands and
> not the image.
>
> You're coming in with a signal bandwidth of 500 Hz. Your intention
> is that there will be no energy above that frequency, but you cannot
> guarantee it because there maybe interference and noise present. If
> you do not take steps to remove that interference and noise, then it
> will also be sampled. Now you will have a noisy sampled signal.
>
> But But But......a noisy signal is bad enough, but there will only be images
> present if the noise extends above 4kHZ for which you will need your
> anti-aliasing filter.

Thanks for the response.

Well, from what I read in ADI's Data Conversion Handbook (chap. 2)
what you call 'wanted sidebands' are what they are refering to images
around every multiple of F(s). Let me go read it again for clarity,
maybe I missed something in their explanation.

I would ask Phil to clarify, but he does not have patience for 'stupid
questions' and frankly I do not have patience for assholes who forget
to take their medication. I wish Google Groups had an option to kill-
file.

Kingcosmos wrote:
> I have a general question about sampling theory. Most of this comes
> from college coursework, and our friend Google.
> If my highest frequency of interest, F(in) was 500 Hz, and my
> frequency of sample, F(s), was 8 kHz then I meet the Nyquist
> criterion. However, I still need to have an anti-aliasing filter (low-
> pass, reconstruction filter, etc.) to properly attenuate any frequency
> components/ images above F(s)/2 assuming baseband sampling (Nyquist
> Zone 1).

Unless you know ahead of time that the signal has no content up there --
anti-alias filtering is only necessary if there's something that will
alias down to baseband.

> If my frequency of sample is 8 kHz, then my images should be centered
> around 8 kHz: F(s) +/- F(in). So my images would be 7.5 kHz and 8.5
> kHz well outside F(s)/2.

Well, kinda, and kinda not. Depending on how you model sampling there
is either no frequencies above Fs/2 after sampling, or the spectrum
repeats itself every Fs. In the first case those "images" aren't there;
in the second case they just replicate baseband.

> Images will also appear around every
> multiple of F(s) as well. But the images from every Nyquist Zone will
> still fold back into my baseband, correct? So even though I am 'over
> sampling' I still need an anti-aliasing filter.

See my first comment.

> Many delta-sigma converters have high base sampling rates; e.g., 192
> kHz, and can oversample 128*F(s), 256*F(s), 384*F(s), etc. Even here,
> the concept is the same. What I have always understood is
> oversampling spreads out the quantanization noise (Noise shaping?),
> and relaxes the anti-aliasing filter requirements. Regardless, a anti-
> aliasing filter is needed because images will fold back without one.
> Do I have the basics down?
>
Oversampling by itself doesn't spread out quantization noise.
Oversampling plus a sigma-delta modulator does. See
http://www.wescottdesign.com/articles/sigmadelta.html for a simplified
explanation.

Oversampling _does_ relax the requirements on the anti-alias filter,
because the difference between the frequencies you must block out and
the frequencies you want to keep gets bigger. If you are going to
oversample and decimate digitally, however, you still need anti-aliasing
filters in the digital domain -- these are usually called "decimation
filters", but the function is the same.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Fri, 10 Aug 2007 19:48:29 +0100, "Anonymous" <me@privacy.net>
wrote:

>Go back to the old medium wave stations running AM (Amplitude Modulation)
>
>You get the carrier which is the main frequency. Now, even though AM
>suggests
>that the carrier will change in amplitude, it does not. If the DJ were to
>whistle
>into the mike at 1kHz, you'll get two sidebands, at frequencies of +/- 1kHz
>in addition to the main carrier. The reason for this is that AM is really a
>multiplication of the form (1+sin (modulation)) * sin (carrier).

I'm not sure what you mean about the carrier not changing in
amplitude. The energy in the spectrum at the carrier frequency
falls as the energy in the sidebands rises. (Otherwise, we'd have
a neat method of power generation!) This is a real change in
amplitude: To duplicate this exact waveform as a summation of
3 sinusoids, the carrier component would need to have its
amplitude reduced compared to the unmodulated case.

Those who want to get a good hands-on feel for AM (or FM,
PWM, etc) can download my Daqarta software and use the
free built-in signal generator. You can set up various
carrier and modulator frequencies and depths, and view
and measure the resultant waveforms and spectra (and
hear the results on your sound card).

There is no need to purchase Daqarta for this... the signal
generator (and everything else that doesn't use sound card input
signals) will keep working after the trial period expires, and you are
welcome to use it as long as you like. (PS: On some systems,
users have reported no trial period at all. If that happens to
you, let me know via the Contact page on the site and I will
create a special trial key for you. The upcoming version, in
a few weeks, should correct this problem.)

Best regards,




Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
http://www.daqarta.com
Scope, Spectrum, Spectrogram, Signal Generator
Science with your sound card!

"Bob Masta"

> I'm not sure what you mean about the carrier not changing in
> amplitude. The energy in the spectrum at the carrier frequency
> falls as the energy in the sidebands rises. (Otherwise, we'd have
> a neat method of power generation!)


** Not at all.

An AM signal gains power as the modulation percentage is increases.

With square wave modulation of 100 % depth, the average power level
doubles - as the carrier is boosted to double voltage level for 50 % of
the time.

AM radio stations typically allow upwards modulation beyond 100 % making
the increase even more.



........ Phil

On Sat, 11 Aug 2007 20:36:35 +0100, "Anonymous" <me@privacy.net>
wrote:

>
>"Bob Masta" <NoSpam@daqarta.com> wrote in message
>news:46bda393.920733@news.sysmatrix.net...
>>>You get the carrier which is the main frequency. Now, even though AM
>>>suggests
>>>that the carrier will change in amplitude, it does not. If the DJ were to
>>>whistle
>>>into the mike at 1kHz, you'll get two sidebands, at frequencies of +/-
>>>1kHz
>>>in addition to the main carrier. The reason for this is that AM is really
>>>a
>>>multiplication of the form (1+sin (modulation)) * sin (carrier).
>>
>> I'm not sure what you mean about the carrier not changing in
>> amplitude. The energy in the spectrum at the carrier frequency
>> falls as the energy in the sidebands rises.
>
>No.
>
>Not so.
>
>If you expand out the expression above, you will find a
>term sin(carrier) of unchanging amplitude.
>
>

Whoops! My apologies to all! I had forgotten that I needed
to use a slightly modified form of the AM equation to prevent
clipping of the output. (Digital systems are not forgiving in
this respect.) I have been using the modified forn for so long
now that I'd forgotten about this difference with the standard
form. The two different equations are discussed at
http://www.daqarta.com/dw_aa0d.htm

One nice feature of the modified version is that as you
increase modulation depth above 100%, the output
smoothly goes to pure multiplication at 200% depth.

Again, my apologies for misleading anyone.

Best regards,


Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
http://www.daqarta.com
Scope, Spectrum, Spectrogram, Signal Generator
Science with your sound card!