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Sampling Theory question

Discussion in 'Electronic Basics' started by Kingcosmos, Aug 10, 2007.

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  1. Kingcosmos

    Kingcosmos Guest

    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?
     
  2. Phil Allison

    Phil Allison Guest

    "Kingcosmos"


    ** OK so far .....


    ** 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.



    ** Oh dear ......

    ** No.

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


    ** Not at all.



    ........ Phil
     
  3. Anonymous

    Anonymous Guest

    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.
     
  4. Kingcosmos

    Kingcosmos Guest

    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.
     
  5. Tim Wescott

    Tim Wescott Guest

    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.
    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.
    See my first comment.
    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
     
  6. Anonymous

    Anonymous Guest

    It helps at this point if you have a background in telecoms; even a Ham
    Radio licence would help you. This is one of the difficulties of DSP; you
    really need a grounding in quite a number of areas of electronics.

    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).

    If the DJ speaks into the mike....or.....applies your 500 Hz bandlimited
    signal, you'll get one sideband which is a copy of your original spectrum
    but shifted up from 0 Hz to the carrier frequency. The other sideband is
    an inverted copy of the original spectrum but decreasing downwards
    from the carrier.

    So, AM is the multiplication of your signal by the carrier frequency.

    -----ooooo-----

    Come now to sampling. Instead of a carrier frequency of just one sine wave,
    you have a series of very short pulses. But, this comes down to (by Fourier
    Analysis; another bit of electronics you need under your belt) to a lot of
    carriers,
    each a multiple of the frequency of the basic sampling rate. Each one of
    these
    carriers becomes Amplitude Modulated by your original signal. (This time,
    sampling is seen as multiplication in the first instance and not as
    modulation)
    and you get the picture that you described of multiple carriers each with
    sidebands
    above and below each carrier.

    This, I suspect is the picture that you describe as "images". Now, although
    this is
    a correct use of English, in that each sideband is an image of your original
    baseband
    signal, it is not what is meant by "imaging" or "image frequencies" when
    modulating
    one signal by another.

    OK, an example off the shelf (and be prepared for some silly errors as I'm
    thinking as I'm typing.

    Let us suppose that your sampling at 8kHz, which is what you suggested,
    and that you've your baseband signal of 500 Hz Bandwidth, but that you've
    also got a signal at 7.8kHz, which is crucially above the Fs/2 that everyone
    talks about. When we multiply them all by the sampling signal, what do we
    get?

    (I've deliberately chosen 7.8kHz to guarantee to be able to illustrate
    imaging into your wanted 500 Hz bandwidth)

    Well first of all, the sampling operations is sin(modulation) * sin(carrier)
    and
    not (1+sin(modulation)) * sin (carrier) as per the medium wave station, so
    we
    only get our copies of the sidebands around each multiple of the sampling
    frequency. WE DO NOT GET THE UNCHANGING CARRIERS.

    Right from the example, (0.5kHz +7.8kHz) * 8kHz will give us 200 Hz
    (which is the lower sideband from 7.8kHz), 7.5kHz (which is the lower
    sideband
    from 0.5kHz), 8.5kHz (which is the upper sideband from 0.5kHz) and 15.8kHz
    (which is the upper sideband from 7.8kHz)

    Now, it seems at first sight as though there isn't a problem but so far
    we've only
    considered the first harmonic at 8kHz. We have to consider _ALL_ the
    harmonics
    including the "zeroth" harmonic of DC, because they're all there in the
    sampled
    signal.

    Lets look at the DC or baseband. Originally we just wanted your 0.5kHz
    signal, but now as the result of the modulation at 8kHz, we've also got
    a signal at 200Hz resulting from the sampling of the 7.8kHz signal.

    This 200Hz signal IS WHAT IS KNOWN AS AN IMAGE.You can't
    filter it out once you've got it because it's right in your bandwidth.

    OK, now look at what happens at the next harmonic of the sampling
    frequency at 16 kHz......you can work out the wanted numbers
    from 16+/- 0.5 and 16+/- 7.8.

    But.....the 8kHz modulation has created another image at 15.8 kHz,
    which appears as an image as a 200 Hz from (16 - 15.8) in what
    should be the lower sideband of the 16 kHz carrier.

    And so it goes on for every subsequent harmonic of the original sampling
    frequency.

    Right, I'm getting bored with this, it's very tedious! There's one final
    point
    to make and it is to introduce where the Nyquist criterion comes from.

    Consider two of the carrier, say the 8kHz and 16kHz already discussed.

    Provided no upper sideband of the 8kHz carrier is above 12 kHz, and
    provided no lower sideband of the 16kHz carrier is below 12 kHz, then
    there won't be any aliasing. ie, the bandwidth of the sidebands has to
    be less than the distance between two carriers, or, put another way,
    the maximum frequency in the baseband must not exceed half
    the sampling frequency.....as I said, you need to have a bit
    of a background in radio.

    (I am a Radio Ham with the callsign G....4....S....D....W)
     
  7. Bob Masta

    Bob Masta Guest

    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
    www.daqarta.com
    Scope, Spectrum, Spectrogram, Signal Generator
    Science with your sound card!
     
  8. Anonymous

    Anonymous Guest

    No.

    Not so.

    If you expand out the expression above, you will find a
    term sin(carrier) of unchanging amplitude.
     
  9. Phil Allison

    Phil Allison Guest

    "Bob Masta"

    ** 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
     
  10. Anonymous

    Anonymous Guest

    I wonder if Bob was talking about a balanced modulator which produces
    DSB suppressed carrier, and not about DSB with full carrier which is the
    stuff
    of AM on the medium wave?

    An easy mistake to make these days.
     
  11. Anonymous

    Anonymous Guest

    No.

    Now I'm talking a load of GeorgeWBush.

    In the absence of modulation, there's no carrier from
    a balanced modulator, and none appears in response
    to modulation.

    Oops!
     
  12. Bob Masta

    Bob Masta Guest

    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
    www.daqarta.com
    Scope, Spectrum, Spectrogram, Signal Generator
    Science with your sound card!
     
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