Is it possible to transmit square waves EM signal through air?

Question

Can we transmit any signal other than sinusoidal through air, if yes how and if no why ?

You generally don't want to transmit anything but a sinusoid. Anything else is composed multiple sinusoids. A square wave has the primary frequency, and infinite harmonics in specific phase and amplitude to that primary. All of the harmonics will interfere with transmitters and receivers on higher frequencies. — JRE, Aug 28 '17 at 15:43
@JRE while this is of course not wrong, it doesn't really make much sense by itself. Just because any signal can be Fourier decomposed doesn't mean it is composed of sinusoids. The trig functions happen to constitute a mightily useful basis of the $L^2(\mathbb{R})$ Hilbert space, that's all. In fact you could also decompose such $L^2$ signals into e.g. Haar wavelets, in that view it would actually be the other way around: a sine is composed of many rectangular-like wiggles! — leftaroundabout, Aug 28 '17 at 18:47
Incidentally, I see this is right now being discussed on Physics.SE. — leftaroundabout, Aug 28 '17 at 21:54
The reason we use sine waves to describe EM signals is because they are the simplest solutions to Maxwell's equations for wave propagation, with just an amplitude and frequency to describe them. — Mark Lakata, Aug 28 '17 at 22:28
As an even more fascinating line of thinking, consider that with known technologies, we cannot even create a square wave on a wire. Every square wave we create is distorted in some way. We simply use such waves in ways which are not affected by said distortion, so we can pretend they were perfect square waves when we model them. — Cort Ammon, Aug 28 '17 at 23:28
You certainly don't want to do take the naïve approach of trying to transmit a signal the literally "looks like a square wave" at any significant power over a long distance. Unless you use a some modulation technique to limit the bandwidth of the transmitted, you will be creating a huge amount of interference to other users of radio transmission - which is illegal in most countries. Of course a lab experiment using very lower power that only has a range of a few millimeters is unlikely to cause any problems - unless two people in the lab try to do the same experiment at the same time! — alephzero, Aug 29 '17 at 01:24

Curd · Answer 1 · 2017-08-28T19:16:41.467

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A square wave can be seen as a superposition of sine waves (of odd multiple frequencies of the lowest frequency). I.e. a 1MHz square wave is composed of 1MHz, 3MHz, 5MHz, 7MHz etc. sine waves.

So as long as everything behaves linearly and there is no dispersion (i.e. no frequency dependency of \$n\$; both conditions are satisfied in case of vacuum) theoretically it is possible.
Be aware, however, that it requires a (theoretically inifinite) wide bandwidth.

edited Aug 28 '17 at 19:16

answered Aug 28 '17 at 15:43

Curd

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Linear is not sufficient to actually preserve the square shape, unless you also restrict it to first-order differential operators (in which case the only linear equation is the advection equation). The typical 2nd order wave equations may have nontrivial dispersion relations, and that can completely smear the flanks of the square wave. It so happens that the Maxwell equations (in air and for RF) have $\omega^2 = c^2\cdot k^2$, so that's not an issue here, but this looks quite different in e.g. fibre optics. – leftaroundabout Aug 28 '17 at 18:53
yes. I'll add that. – Curd Aug 28 '17 at 19:15

score 7 · Answer 2 · edited Aug 28 '17 at 19:38

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Seems like a simple question but it's really not.

First.. there is no such thing as a square wave. A perfect square wave requires and infinite number of harmonics and bandwidth, so even transmitting a square wave unmolested down a bit of wire is actually impossible. Doing it in the RF domain requires multiple or very high bandwidth antennas and quickly becomes impractical.

Second, in reality we never just send a sine wave. If all we send was a pure carrier.. we are not in fact sending any signal. In order to add a signal to the carrier, the carrier needs to be modulated. The act of modulation means it is no longer a pure sine wave.

So in other words, other than perhaps a few simple radio beacons, we almost never transmit a pure sine-wave. What we do send are sine-waves that are modulated a little within a reasonable range of the carrier frequency to be efficiently transmitted and receivable at the other end.

edited Aug 28 '17 at 19:38

Bruce Abbott

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answered Aug 28 '17 at 15:54

Trevor_G

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Thanks : BTW, one more query :) .. What about the signals that are sent by the satellite, are they not digital or are they later converted to digital when it reaches our antenna using ADC ? – Dhiraj Dhakal Aug 28 '17 at 16:08
@DhirajDhakal satelite signal is a digitally modulated carrier signal. – Trevor_G Aug 28 '17 at 16:11
OK but digitally modulated carrier signals are still analog waveforms aren't they? – Dhiraj Dhakal Aug 28 '17 at 16:18
@DhirajDhakal as I said.. yes and no. They look like simple sine-waves on the scope but the modulation can take many shapes. – Trevor_G Aug 28 '17 at 16:24
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Everything in the real world is an analog waveform. An analog waveform can encode and represent an abstract digital data stream. – Russell Borogove Aug 28 '17 at 16:58
Turning a switch off and on would produce a square wave. – Jasmine Aug 28 '17 at 20:07
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@Jasmine sorry, only if said switch is connected to something with absolutely zero capacitance and inductance. i.e. never. – Trevor_G Aug 28 '17 at 20:09
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@DhirajDhakal, Waveforms are neither analog nor digital. They are just waveforms. "Analog" and "digital" are the names of different ways of processing a signal. If you use a phase-locked-loop to demodulate an FM signal, that's an analog process. If you first convert the waveform into a sequence of numbers, and then use a computer algorithm to extract the information, that's a digital process. – Solomon Slow Aug 28 '17 at 20:24

score 7 · Answer 3 · answered Aug 28 '17 at 16:10

To send a signal through the air as an RF signal you need to modulate the carrier: -

In the example above we have amplitude modulation (AM) and frequency modulation (FM). As you might be able to see in this simple example, if the baseband signal (in black) is too high in frequency compared to the carrier you get a basic problem in that the carrier can't change its amplitude fast enough (or its frequency quick enough) to suit the baseband signal.

A square wave that is transmitted via modulation will need to be bandwidth limited. Here is an example that demonstrates how far as you can take AM fed with a square wave baseband signal: -

After using an envelope detector the recovered square wave will look pretty sloppy as shown on the bottom picture above. So basically there is no need to transmit a full bandwidth square wave because many types of detectors will fail to recover the original signal properly.

I guess the background idea of OP could have been that there wouldn't be any need for demodulation (i.e. no need for envelope detector in this case) if the signal is transmitted directly as rectangular signal. — Curd, Aug 28 '17 at 19:29

Michael Seifert · Answer 4 · 2017-08-28T21:08:14.177

Any periodic signal of any shape can be thought of as a sum of an infinite number of sine waves. For example, a perfect square wave can be thought of as the sum of a sine wave of frequency 1 MHz, another sine wave at 3 MHz with 1/3 the amplitude of the first wave, another sine wave at 5 MHz with 1/5 the amplitude of the first wave, another sine wave at 7 MHz with 1/7 the amplitude of the first wave, and so on up to infinity. (All of of these sine waves have to be in phase with each other as well.)

However, a signal that leaves the antenna as a "perfect square wave" will not generally be received as a "perfect square wave". This is because each of the sine waves that makes up your square wave will be attenuated as it travels, and (here's the important bit) the amount of attenuation that each sine wave depends on its frequency. In other words, if your receiver is a distance from your transmitter such that the sine wave at 1 MHz is attenuated by 25%, then the one at 3 MHz will be attenuated by 28%, the one at 5 MHz will be attenuated by 29%, and so forth. (The numbers here are made up; the important part is that they will not all be the same, since attenuation varies with frequency.) Moreover, the speed of a sine wave's propagation depends ever so slightly its frequency as well; so the sine waves that were nice & in phase when they left the transmitter would no longer be in phase when they arrived.

All of these effects taken together mean that even if your antenna could emit a perfect square wave, the received signal would not be a perfect square wave; the constituent sine waves that make it up would get out of phase during their travel, and their amplitudes would be in the wrong proportions. Effectively, the atmosphere acts like a (very weak) low-pass filter on your signal during its travel. A sine wave at a single frequency avoids this problem, as it only has one amplitude and one phase to speak of. It's only when you talk about multiple superposed sine waves whose relative amplitudes and phases are important that the signal becomes distorted.

For a square wave at radio frequencies, the received signals would still be pretty darn close to a square wave; the effects I've mentioned above really only become meaningful above frequencies of 1 GHz. The limitations of the receiving electronics are probably more important than this for any physically reasonable setup. But so long as there's matter in the way between your transmitter and your receiver, it's impossible to get 100% fidelity.

@MichaelKjörling: By the time-honored technique of making them up; I'll edit the answer to make it clearer that I've done so. The important part is that the attenuations wouldn't all be exactly the same. — Michael Seifert, Aug 28 '17 at 21:07
Making something up to show a point isn't necessarily wrong (though I'd expect to see less of it on Electrical Engineering SE than on some other sites in the network), but please do try to avoid wordings like "will be" if you do, as those can easily be taken to imply that the statement is one of fact. Thanks for clarifying. — user, Aug 28 '17 at 21:09
The percentages actually come from the Fourier analysis of a square wave. Adding up all of the odd harmonics - the fundamental + the 3rd harmonic at 1/3 amplitude + the 5th harmonic at 1/5 amplitude, etc. approximates a square wave. Years ago, while working in the exhibits department of a science museum, I built a physical device to demonstrate this. Also, have an EE degree. See here for more: http://mathworld.wolfram.com/FourierSeriesSquareWave.html — GreyBeardedGeek, Aug 28 '17 at 22:43
@GreyBeardedGeek: I'm pretty sure Michael Kjörling was referring to my hypothetical attenuation percentages, not the actual amplitudes of the sine waves in the signal. — Michael Seifert, Aug 28 '17 at 22:46

score 0 · Answer 5 · answered Aug 28 '17 at 23:19

Well, a simple harmonic oscillator (ideal pendulum, Helmholtz resonator, LC circuit, etc.) puts out a continuous sine wave. Simple harmonic motion equals sine curves in the time domain. Real-world oscillators will contain distortion, noise, clipping, which adds some percentage of harmonics or broadband white-noise. (Even turning the oscillator on will contribute step-function harmonics, while an ideal single-frequency oscillator must operate for infinite time.)

As with perfect zero-ohm inductors and infinite-ohm insulators, the ideal sinewave generator doesn't actually exist.

Going further, "perfect sine-wave" is associated with a "single pure frequency" found on the mechanical, acoustic, or electromagnetic spectra. The ideal sine-wave is a radio concept: a single infinitely-narrow CW channel on your radio dial. Or, with an optical spectrograph, a single emission-line with zero linewidth is the same as an electromagnetic sinewave.

No, lasers don't emit perfect sinewaves. That would require that the laser have infinite coherence-length (coherence time.)

score 0 · Answer 6 · answered Aug 29 '17 at 06:12

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You can send what you want. The mathematical derivative is what you receive. So, in this case, you will receive a few spikes. ;)

answered Aug 29 '17 at 06:12

Mick

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Is it possible to transmit square waves EM signal through air?

6 Answers6