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Why digital signals have "infinite bandwidth"? Frequency is the number of cycles per second. I have seen before the formula that bandwidth=Maximum Frequency - Minimum Frequency. I remember I was told something like because of the vertical straight line for the digital signal, the bandwidth is infinite. But I still don't understand how is the bandwidth infinite this way?

For periodic digital signals, there could still be a number of digital cycles per second, wouldn't it? Then how is there a maximum and minimum frequency when the number of cycles per second is always constant throughout? Bandwidth would then be BW=Max_Freq - Min_Freq = 0.

And for non-periodic digital signals, there may not be a fixed number of cycles per second. So there is a maximum and minimum frequency. But still, the bandwidth is still not infinite, isn't it?

In both cases, how is the bandwidth infinite?

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  • \$\begingroup\$ A true square wave is an infinite sum of sine waves (see Fourier Series). These cannot actually be created in practice. \$\endgroup\$ – NickHalden Aug 28 '11 at 17:40
  • \$\begingroup\$ My answer on your previous question may answer this question. \$\endgroup\$ – Kellenjb Aug 28 '11 at 17:44
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An ideal digital signal has infinitely steep edges. We can compose this signal from sines, one fundamental and a number of harmonics.

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Neither of those separate sines has infinite steepness. The only way to get our steep edge is by adding an infinite number of harmonics.

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  • \$\begingroup\$ Can I then say that the bandwidth is the number of "layers" of analogue harmonic frequencies required to map the digital signal? Then the formula Bandwidth=Frequency_Highest - Frequency_Lowest does not look like it would tell the number of "layers" of harmonic frequencies required. \$\endgroup\$ – xenon Aug 28 '11 at 18:33
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    \$\begingroup\$ @xEnOn, you can say that the bandwidth is primarily determined by your digital signals rise time. The time it takes to slope from one state to another. Almost all of the high frequency spectral content will be determined by the rising edge shape and speed. \$\endgroup\$ – Kortuk Aug 28 '11 at 19:49
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A true square wave has a vertical line for it's face.

That is, an increase from Vmin to Vmax in zero seconds.

Anything (except zero) divided by zero is infinity.

In reality it takes time for the voltage to rise from Vmin to Vmax, and so there is a finite limit to the bandwidth.

What you're confusing is the signal frequency, or period. The speed of high and low pulses is not the bandwidth of the signal, but the speed of the data communication.

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    \$\begingroup\$ it could be 1. I think -2000 is just as likely : ) \$\endgroup\$ – Kortuk Aug 28 '11 at 21:50
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    \$\begingroup\$ "The expression \$\frac{0}{0}\$, which may be obtained in an attempt to determine the limit of an expression of the form \$\frac{f(x)}{g(x)}\$ as a result of applying the lim operator independently to both operands of the fraction, is a so-called "indeterminate form". That does not simply mean that the limit sought is necessarily undefined; rather, it means that the limit of \$\frac{f(x)}{g(x)}\$, if it exists, must be found by another method, such as l'Hôpital's rule." - Wikipedia \$\endgroup\$ – Majenko Aug 28 '11 at 21:53
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    \$\begingroup\$ "Division: \$\frac{0}{x} = 0\$, for nonzero x. But \$\frac{x}{0}\$ is undefined, because 0 has no multiplicative inverse (no real number multiplied by 0 produces 1), a consequence of the previous rule; see division by zero." - Wikipedia \$\endgroup\$ – Majenko Aug 28 '11 at 21:54
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    \$\begingroup\$ A square wave does not have vertical lines, because then it would not be a function. It is piecewise continuous: it stops at one value and continues on another. \$\endgroup\$ – Kaz Nov 6 '12 at 23:57
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    \$\begingroup\$ @YasirAhmed Rather, a square wave is something that (under the strictest definition) doesn't actually occur in a real circuit. \$\endgroup\$ – Kaz Feb 22 '18 at 22:06
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Constructing a discontinuous voltage waveform as a sum of sine waves would require an infinite number of sine waves. A "perfect" digital signal will switch instantaneously between VSS and VDD (or vice versa); such instantaneous switching events would represent discontinuities in the waveform.

In practice, chips neither produce perfect digital signals on their outputs nor require them on their inputs. Examination of an output signal on a good enough scope will typically reveal that it is a little bit "smoothed off", and most chips will tolerate having input signals which are substantially "smoothed off", provided that they don't linger or bounce around in the region between 1/4 and 3/4 VDD. Indeed, some chips are designed to deliberately smooth off their output waveforms (sometimes by a programmable amount) and/or accept inputs which are smoothed off even to the point of being a bit "mushy".

It's worth noting that while something like a 1Hz perfect square wave might be expressed as the sum of continuous sine waves ranging from 1Hz to 1MHz and beyond, it is very unlikely that any apparatus which is designed for receiving a 1MHz signal would perceive a 1Hz square wave as having a continuous 1MHz component. The 1Hz square wave would contain, among other things, a 999,999Hz component whose strength was 1/999,999 of the fundamental, and a 1,000,001Hz component whose strength was 1/1,000,001 of the fundamental. The apparatus that was trying to receive a "1Mhz" signal would detect those components, and many others, to varying degrees; during each one-second interval, there would be times when they would all be in phase, and times when about half would be in phase and half out. The apparatus would thus perceive a variable amount of "1MHz" signal--most likely detecting a substantial amount near the moments when the input was switching (because all of the detected waves would be in phase), and a much smaller amount at other times (because the detected waves would have a mixture of phases). A really sharp 1Hz square wave, driving a strong antenna, would thus not cause continuous interference on a 1MHz transmission, but rather would yield a 2Hz "tick tick tick".

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  • \$\begingroup\$ A square wave function is not discontinuous. It is piecewise continuous. Continuity is in fact one of the necessary criteria needed for the existence of a Fourier transform, IIRC. \$\endgroup\$ – Kaz Nov 6 '12 at 23:59
  • \$\begingroup\$ @Kaz: While functions which are not continuous everywhere could be subdivided into those which are piecewise-continuous and those which aren't, I've usually heard both subcategories described as "discontinuous". Indeed, while I could contrive some functions whose value is defined everywhere, but which aren't piecewise continuous (e.g. let f(x) be 1 for rational values of x and 0 for irrational values), I can't think of any such functions I'd regard as "useful". There could be mathematical functions which had no value for various ranges of x, but... \$\endgroup\$ – supercat Nov 12 '12 at 16:04
  • \$\begingroup\$ ...I'm not sure such things could be called "waveforms". \$\endgroup\$ – supercat Nov 12 '12 at 16:07

protected by Dave Tweed Mar 14 '15 at 1:06

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