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Harmonic composition of a square wave

Discussion in 'Electronic Basics' started by davidd31415, Jun 24, 2005.

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  1. davidd31415

    davidd31415 Guest

    I know you can make a square wave from the sum of sinusoidals, but does
    this mean that if you look at a sine-wave that wasn't made by using
    sinusoids (perhaps using a switch or an oscillating crystal to turn the
    signal on and off) on a spectrum analyzer that you would see all of the
    harmonics required to make up the square wave?
     
  2. Not true, actually. See http://mathworld.wolfram.com/GibbsPhenomenon.html
    Depending on the analyzer and its settings, you will see
    harmonics as they would be computed in the usual way
    for obtaining the Fourier series. Whether the "sine-wave"
    (or square wave, for that matter) was made by composing
    sinusoids or not, its harmonics depend only on its shape.
     
  3. I assume you mean that if you look at a *square* wave that wasn't made
    by using sinusoids...

    The answer is yes. How the waveform is created has nothing to do with
    its harmonic content. Only its shape determines its harmonic content.
    The faster the rise and fall edges and the squarer the corners, the
    higher the frequencies contained in the package.
     
  4. davidd31415

    davidd31415 Guest

    Ahh yes, glad the context gave that away, *square* indeed.

    So how would the sampling rate of an oscilloscope be related to what a
    square wave ends up looking like on the scope? Is there a rule of
    thumb for the sampling rate of an oscilloscope when sampling square
    waves?
     
  5. At the very least, the scope cannot show a rise time (or any thing
    else) less than the time between samples. It can connect two samples
    with a straight line. At two samples per square wave, it displays a
    triangle.
     
  6. The rule I use is to have at least a factor of 5. I get a relatively
    "true" picture of the square wave with a scope (and probing setup, of
    course) that passes at about 3db down at 5X the square wave frequency.
    That gives you the 1X, and 3X fairly good and enough of the 5X to get
    decent presentation.

    ....

    If you want to see about what the impact of such a decision would be
    on a square wave, I believe you can use Excel (or some other program)
    to compute the impact on a particular square wave.

    SUM [ (1/n)*SIN(n*w)/SQRT(1+(n/5)^2) ], n=1,3,5,7, ...

    with w=2*PI*f

    This equation, if I've got it right, assumes that the voltage is
    (1/SQRT(2)) at the scope's 3db down frequency. You can see this part
    in the above equation, where you see /SQRT(1+(n/5)^2). At n=5 (5X the
    frequency of the square wave, which is by definition 1/5th of the
    scope's bandwidth, you get /SQRT(2). I think this properly scales the
    values as the frequency is increased or decreased. The leading (1/n)
    part of the equation is the Fourier scaling for the components of the
    square wave. Combined together and summed for some moderately sized
    'n', I think this should approximate what the scope will show you,
    including its roll-off behavior. Of course, I'm open to being wrong.

    But regardless, my rule is 5X for the analog scopes. Works for me.

    Jon
     
  7. Bob Masta

    Bob Masta Guest

    You might want to have a look at DaqGen, my freeware
    sound card signal generator for Windows. You can play
    around with waveforms and see the effects via the built-in
    spectrum analyzer.

    Best regards.


    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
    Home of DaqGen, the FREEWARE signal generator
     
  8. Charles Jean

    Charles Jean Guest

    ___
    This post got me to thinking about a related subject. I'm a hobbyist,
    and my math is limited to one year of calculus, but I would like to
    see if I have a correct conception of what's going on here. I can see
    that any periodic function can be put through the Fourier transform to
    obtain an infinite series of sin and/or cos terms to completely
    describe the original function. This applies to electronic circuits,
    musical instruments, vibrational analysis of bridge decks, etc.
    So, when one sees a "perfect square wave" on the oscope, it is
    actually always a mixture of sine waves of f(fundamental-the frequency
    of the square wave as seen on the oscope),3f,5f,7f...... frequencies.
    A more complex wave like that produced by a violin string would look
    different than either a sine wave or a square wave, because the
    mixture of waves producing it are not at the amplitude/frequency
    required by the Fourier transform to produce a square wave. If I were
    to see what looks like a very low distortion sine wave on the oscope,
    I can infer that this is a "true sine" wave, with very little
    contribution from any higher harmonics, and not some weird lucky mix
    of higher sin/cos frequencies that are significant compared to the
    fundamental? Or would the use of a spectrum analyzer be required to be
    sure?
    For circuit elements like capacitors and inductors, whose reactance
    varies with frequency, what happens when dealing with a square wave?
    what frequency does one use in the reactance formulas, knowing that
    you're dealing with a mixture of them? I would instinctively just put
    in the fundamental frequency, but is this right? TIA for clearing any
    of this up for me.
     
  9. colin

    colin Guest

    Dont forget any practicaly generated squarewave isnt going to need an
    infinite series of sine waves to fully define it.

    a squarewave isnt necessarily made up of sinewaves its just very useful
    indeed to be able to consider it as a series of sinewaves.

    a pure sinewave cant be split into other sinewaves. you can probably tell on
    the scope if its say a 90% pure sinewave or more.

    if you feed a squarewave into a circuit with a non flat frequeucny response
    you can think of it by seperating it into the harmonics, then seeing how big
    each harmonic is afterwards but also what phase, and then try and
    reconstruct the waveform.

    it all depends what you want to do with the reactive circuit, sometimes you
    want to filter out the fundemental so you would chose that as the frequency,
    but you might also want to use one of the other harmonics instead and filter
    out one of those and you have a frequency multiplier. unless of course its
    an inductor for a power supply in wich case you would want to filter out as
    much as posible.

    Colin =^.^=
     
  10. Charles Jean wrote:
    8<
    Yes, you can infer that. Another way to this conclusion is if you
    pass any other periodic waveform through a good low pass filter that
    passes the periodic fundamental but greatly attenuates the second and
    higher harmonics, you always get very nearly the same sine wave out of
    the filter.
    For analysis of what happens, either you:

    Assume a linear circuit, calculate the effect of each component
    frequency, individually, and using the assumption of linearity, add
    the various component frequency effects together to get the overall
    effect or just deal with what happens to each harmonic, separately.

    Use the instantaneous (differential) descriptions of all the
    components and integrate the result (usually using numerical
    approximations).

    Most Spice programs allow both the amplitude and phase versus
    frequencies that you specify (the first method) or the time response
    to an arbitrarily stimulus (the second method) that simulates what you
    would see on a scope.
    As many of them as you are interested in. Your interest may fade
    because the amplitudes become insignificant, or because you have some
    reason to suspect that frequencies above some point are so attenuated
    or ignored by other parts of the system that they ate moot.
    If the waveform is a pretty clean sinusoid, that will give you a
    pretty good approximation of what is going on. If the waveform is a
    pulse with an on time 1% of the period, you will learn almost nothing
    useful.
     
  11. Bob Myers

    Bob Myers Guest

    True, but that's just another way of saying that there are no truly
    "perfect" square waves in practice, since finite limits on bandwidth
    always mean that you can never get to zero rise/fall time.

    No, it really, really is. Any periodic signal IS composed of sinusoidal
    components; the frequency domain (i.e., what you see on the screen of
    a spectrum analyzer) is just as valid as the time domain (which is what you
    see on the screen of an oscilloscope).


    Bob M.
     
  12. colin

    colin Guest

    yes indeed you dont get those nasty little discontinuities wich cuase the
    ringing as someone mentioned. :) its also the squarenss of the corners wich
    are never perfectly square.
    ah, I meant made up as in 'constructed from'. (ie adding together
    individualy generated sinewaves). wich is what i thought the OP was getting
    confused about, if you look at an oscillator (such as LC or crystal rather
    than relaxation type) generaly it will start off producing a fairly pure
    sinewave that builds up in amplitude, when the amplitude exceeds the supply
    rails it will be clipped, and hence the top and bottom 'round part' will be
    removed. so the harmonics arise from the bits that are missing from what
    would otherwise be a very large sinewave.

    You have to be carefull, if you look at the spectrum of a squarewave thats
    been through an all pass filter wich adds 90' phase shifts above a certain
    frequency it will be different on an oscilloscope although it has the same
    ampliitude of harmonics, so it will look the same on a spectrum analyser
    unless its a digital one wich displays phase too, but phase information on a
    FFT display can sometimes be eratic.

    Basicaly I was just trying to say its valid to work in only time domain
    completly if you need too, wich might be the case if your looking at a
    digital waveform, however even with digital waveforms you will often need to
    consider the frequency domain too, especialy if its the loop of a PLL for
    instance, or the effects of the highest harmonics on crosstalk, power supply
    decoupling reactance, emi etc.

    If your working with RF then you would normaly only think in terms of the
    frequency domain.

    Its good to have the fexibilty to think in either one or the other or both
    at the same time.

    Colin =^.^=
     
  13. Bob Masta

    Bob Masta Guest

    On Sat, 25 Jun 2005 01:35:44 GMT, Charles Jean


    The trick part of your question is "looks like a very low distortion
    sine wave on the scope". I have found that it is very difficult to
    judge distortion from a scope trace, below a few percent. And
    certain distortion combinations can be even harder to detect
    visually, maybe up to 10% or so, if all they do is fatten the
    sine wave a bit. So you definitely need a spectrum analyzer to
    know about low distortion levels.

    But in truth there is nothing in the spectrum that is not
    in the waveform. In theory, you could have a *really*
    large-screen high-resolution scope face and an overlay
    of a perfect sine wave to match up with, and you could
    detect distortion down to as low a level as you want.
    Just not very practical!

    Best regards,



    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
    Home of DaqGen, the FREEWARE signal generator
     
  14. Charles Jean

    Charles Jean Guest

    ___
    Thanks for all the great insights you folks provided! They cleared up
    some of the fog. I thought John's idea of putting the signal through a
    good low-pass filter was a quick, semi-quantitative test for the
    presence of significant harmonics was especially neat. Any good
    references on how to design/build a decent one? Is it possible to
    make one that has a variable cut-off frequency? Or is there a circuit
    out there somewhere for a "poor-boy" RF spectrum analyzer?(0.5-30
    MHz). Please remember that any thing above elementary calculus leaves
    me with puzzled look on my face-this includes differential equations!
    Thanks again for the great responses.

    Charlie
     
  15. I don't know about .5 to 30 MHz, but if you stay down in the audio
    spectrum, you can make some very narrow band pass filters with opamps
    that are easily tunable over a wide range while holding fairly fixed
    Q. The type that is easiest to adjust is probably a state variable or
    bi quad configuration. You can search Google for lots of design info.
    Most versions also have a low pass output and some have a high pass
    and notch output, so you can do lots of experiments with them. For
    instance, with the notch, you can remove the fundamental and see what
    other harmonics are left from the wave.

    Any of these can be made with a good quad opamp.

    http://pdfserv.maxim-ic.com/en/an/AN1762.pdf
     
  16. Shannons theorem says that in order to see a signal with frequency n,
    the sampling rate needs to be at least 2*n. Practical application:
    Humans hear frequencies of up to 20 kHz, so the sampling frequency of a
    CD player needs to be at least 40 kHz. Frequencies higher than 20 KHz
    are filtered out before sampling by a low pass, in order to prevent
    artefacts (beating of sampling and signal frequency).

    This also means that digital equipment can only truthfully represent
    sine waves. Any other signal form contains high order harmonics, which
    need to be cut of at some point. The closer a square wave gets to the
    sampling rate, the more harmonics are removed, and the "rounder" the
    signal will appear.

    Thus how much higher than the signal frequency the sampling rate needs
    to be depends on how good you want the signal form represented, but 10
    times is good enough for most purposes.
     
  17. Fred Abse

    Fred Abse Guest

    That's Nyquist's sampling theorem, surely.

    Shannon's law deals with channel capacity and signal to noise ratio:

    C = B log(base2)(1+S/N)
     
  18. redbelly

    redbelly Guest

    Yep. And if the 2*f sampling is 90 degrees out of phase with the
    signal (namely, sampling at the zero-crossings), then the signal will
    not appear even then.

    Mark
     
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