Harmonic composition of a square wave

Not true, actually. See

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.

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?

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.

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.

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

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

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.

On 23 Jun 2005 18:30:07 -0700, "davidd31415" wrote:

___ 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.

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 =^.^=

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.

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.

you

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 =^.^=

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

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.

___ 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

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.

Fred Abse wrote:

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

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)