If we send a square waveform through an antenna, will we get square electromagnetic waves with electric and magnetic fields looking like squares? Also, since there is an abrupt/almost jump in amplitude, will we get very high frequency sine waves like predicted by the Fourier transform?
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asked Apr 17, 2019 at 7:22
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12\$\begingroup\$ A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth. \$\endgroup\$– Peter SmithApr 17, 2019 at 7:50
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2\$\begingroup\$ antennas have finite bandwidth \$\endgroup\$– analogsystemsrfApr 17, 2019 at 13:00
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2\$\begingroup\$ Infinite bandwidth and zero impedance \$\endgroup\$– user16222Apr 17, 2019 at 20:35
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\$\begingroup\$ If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes? \$\endgroup\$– user253751Apr 17, 2019 at 22:25
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3 Answers
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As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll reach frequencies your antenna can't transmit, for various reasons. If you look at a chart of the electromagnetic spectrum, you will find that radio waves above a certain frequency are called "light", and your antenna probably can't reach those frequencies no matter how good it is.
(But, indeed, if you have an antenna that is capable of transmitting over a wide bandwidth -- that is, from very low to very high frequencies -- and you send some approximation of a square wave over it, you will see very high frequencies appear, just as predicted by the Fourier transform.)
There's also another problem: You can't quite actually approach a true square wave shape from any finite sum of sine waves, no matter how many. This problem is much more theoretical, and unlikely to actually come up in practice, but it's called the Gibbs phenomenon. It turns out that no matter how high in frequency you go, your approximation of a square wave will always overshoot at the big jumps from low to high and high to low. The overshoot will get shorter and shorter in time, the better your approximation (the higher in frequency you go.) But it will never go down in magnitude; it converges to about 9% of the size of the jump.
answered Apr 17, 2019 at 8:00
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16\$\begingroup\$ You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument. \$\endgroup\$– DerManuApr 17, 2019 at 12:50
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5\$\begingroup\$ The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.) \$\endgroup\$ Apr 17, 2019 at 13:53
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4\$\begingroup\$ The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform. \$\endgroup\$ Apr 17, 2019 at 13:53
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1\$\begingroup\$ @copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases. \$\endgroup\$– supercatApr 17, 2019 at 18:36
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4\$\begingroup\$ Technically, any antenna will emit light if you get it hot enough \$\endgroup\$– Nate S.Apr 17, 2019 at 22:48
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No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.
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3\$\begingroup\$ Continuous functions don't exist in the real world either, because of quantum effects. \$\endgroup\$– supercatApr 17, 2019 at 18:37
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4\$\begingroup\$ Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree. \$\endgroup\$ Apr 17, 2019 at 21:57
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In a more general case compared to the answers above, nothing can be stoped or started in zero time ie instantly. To do so would imply an infinitely high frequency component which would translate to infinite energy. The constraining factors are speed of light limitation of Special Relativity and Quantum Mechanics Uncertainty Principle.
The sharper the transition you want, the more energy you have to pump into the system
answered Apr 17, 2019 at 10:05