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digital modulation

Discussion in 'Mathematics and Physics' started by PG1995, Sep 4, 2013.

  1. steveB

    steveB Well-Known Member Most Helpful Member

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    " The complexity is related to the number of abrupt changes in the signal"

    Well, a square wave has 4 abrupt changes per period, while a triangle wave as 3 abrupt changes per period. So, this is your definition of "complex" it seems? Having a definition makes the discussion less vague.

    It also seems relevant to distinguish between different types of abrupt changes. The square wave has discontinuity in the signal itself, while the triangle wave has discontinuity in the derivative of the signal.

    "a triangular signal requires less number of harmonics for faithful representation compared to a square wave."

    Well, both the square wave and the triangle wave have the same exact number of harmonics. They are both infinite in number and at exactly the same frequencies. The only difference between them is the amplitudes of the harmonics. As far as "faithful representation" you again need a definition of that term to avoid being vague. I assume you are saying that if you are limited to a finite number of harmonics, the approximated triangle wave will look more like a true triangle wave, than the approximated square wave will look like the true square wave. I think this is true from an intuitive point of view, but without a definition, it's not very scientific.

    "comparatively square wave is a complex signal"

    Ok, since you defined the "complex signal" to be one with greater number of abrupt changes, I can't argue. But, what about the type of "abrupt changes".

    So what if you have a signal with abruptness type like a triangle wave (derivative discontinuity), but with greater number of abrupt changes per cycle? Which will be better represented with a finite number of harmonics.

    Anyway, my point is that, yes you can think intuitively about harmonic content by considering these things, but ultimately a mathematical calculation tells you everything with no ambiguity. Very often our intuition works well and other times it will fail miserably.
     
    Last edited: Mar 13, 2015
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  2. PG1995

    PG1995 Active Member

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    Thank you, Steve.

    You have made very important points.

    1: Well, a square wave has 4 abrupt changes per period, while a triangle wave as 3 abrupt changes per period.
    2: It also seems relevant to distinguish between different types of abrupt changes. The square wave has discontinuity in the signal itself, while the triangle wave has discontinuity in the derivative of the signal.
    3: So what if you have a signal with abruptness type like a triangle wave (derivative discontinuity), but with greater number of abrupt changes per cycle? Which will be better represented with a finite number of harmonics.

    In my opinion (again committing felony of being vague!), an abrupt change in the signal itself, like a discontinuity in the case of square wave, would be considered 'more intense' than the abrupt change or discontinuity in derivative of a signal as for a triangle wave.

    Let's say that a triangle wave has 4 abrupt changes per period compared to 3 abrupt changes in a square wave. Which will be better represented with a finite number of harmonics? The abrupt changes of a square wave are more intense but less intense abrupt changes of triangle wave are greater in number. This is where intuition fails and you need mathematics to help you.

    From the discussion above, I believe that we can conclude one important point. If two signals have similar type of 'abrupt changes' then number of those abrupt changes per period could help us determine the frequency spectrum.

    In the given case of ASK, PSK, and FSK modulation, we can roughly define the period as time during which waveform leaving a modulator remains the same. In case of 4-ASK modulation, the waveform remains the same for twice the duration of B-ASK modulation.

    JimB: I'm not trying to prove anything and I totally agree with what you have said. But I found this text (source) and thought that should share it with you. Thanks.

    Regards
    PG
     

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  3. JimB

    JimB Super Moderator Most Helpful Member

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    I see what the text is saying, and it makes sense if you are approaching the problem from the point of view:
    "We have a signal, how can we send it by radio"
    But if you approach the problem from the point of view:
    "We have a radio system, how can we send signals with it?"
    It becomes a strange thing to say.

    However, to stretch those statements into this:
    Is just taking things way too far!

    Does anyone else out there in ETO land agree with me, or am I just a sill old sod with a bee in his bonnet?

    JimB
     
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  4. dave

    Dave New Member

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  5. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    JimB, you may be an old sod (as am I), but I agree...

    PG1995, the only significant factor controlling antenna size (and I'm assuming that to mean either length or dimensions), is the carrier frequency. With AM, carrier frequency is, of course, utterly unaffected by that modulation. And with FM, what little frequency shifting that is created by that modulation is so insignificant (kHz vs. MHz) that antenna dimension change would be ridiculous to attempt or consider.

    In short, neither modulation techniques nor wave-shapes have anything to do with antenna dimensions.
     
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  6. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    Modulation Bandwidth has a tradeoff for SNR in any method AM,FM,PM, QAM.QPSK, n-QAM etc...

    The spectral density in Bits/Hz and SNR effects are well documented.

    As we know from Communication Theory there is always a tradeoff between bit error rate (BER) and SNR in the channel which is a tradeoff with Bits/Hz BW. 56k/64k Modems use 3400 Hz BW for ~ 18 bits/Hz

    There is also a tradeoff with modulation type and fading.

    ADSL and cable modems do the same as dialup modems except use many staggered modulated channels with DSP to decode the spectrum. <<< the reason for many staggered channels rather than 1 sub-carrier is the group delay must be compensated for each channel is different by training on a test signal, then the phase amplitude is normalized and optimal for lowest BER. This is why cable and ADSL modems take many seconds to synchronize but then they have excess signal strength above noise threshold to make use of the high bandwidth compression in Bits/baud or bits/Hz. This would not be effective for example with cell phones where sensitivity loss is excessive.

    You can draw arbitrary waveforms and see the spectral effect by changing n, number of harmonics.
    http://www.falstad.com/fourier/ or choose any std. waveform.
     
    Last edited: Mar 15, 2015
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