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I am 17, and I am new to electronics, and I've learned everything online and expect to continue to do so with all the resources. I have dug around and can't find concise answers on this question ...

How exactly are radio waves propagated, and how can I build a simple circuit pair from which one can send the radio waves and the other can intercept them?

I have read different things in different sources, and I'll link them all here:

1.http://www.nrao.edu/index.php/learn/radioastronomy/radiowaves

The aforementioned site claims that radio waves are essentially EM (knew that), but mentions photons. Photons are the essence of all EM, but in a simple circuit there is just current flows by the battery. How would I produce photons from a one-way current?

2.http://www.qrg.northwestern.edu/projects/vss/docs%20/Communications/3-how-do-you-make-a-radio-wave.html

That site above claims that you can "make a radio wave" simply by having an electric field, which is an electric circuit. So, by that logic, any electric circuit is producing radio waves as is? In that case, a homopolar motor would technically produce radio waves as well(it is a complete circuit, yes)? So then the radio waves will propagate in a pattern depending on how many times the circuit goes on and off, so I could encode data by patterns just by removing and placing the battery back to the circuit? I don't get it. Can anyone clarify that article more?

What I wanna do is make two simple circuits out of copper, and produce a radio wave that the other circuit will intercept and use an AND-gate to turn on an LED wirelessly.

However, I do not understand exactly how radio waves are propagated!

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    \$\begingroup\$ You don't need to understand how radio waves propagate to build a circuit - I've got by for years doing that! \$\endgroup\$ – Andy aka Jan 6 '14 at 20:57
  • \$\begingroup\$ @Andyaka I know that! I just want to understand how to work with radio waves because I get frustrated when I can't figure it out. To better put it, I would like to do a project that sends with radio waves so I can get the feel of how this transmission works. \$\endgroup\$ – Gloria Jan 6 '14 at 21:06
  • \$\begingroup\$ How about static magnetic and electric fields? What about alternating magnetic and electric fields? So far so good? If not you have to do the basics first to be able to grasp alternating electromagnetic waves and how they get from the near field to the far field. I have no-idea what your knowledge base is so I can't tell what level to approach this question at or even if I'm qualified to answer and I'll have a go at most things! \$\endgroup\$ – Andy aka Jan 6 '14 at 22:07
  • \$\begingroup\$ You want "to turn on an LED wirelessly", creative-science.org.uk/mobile_LED.html As to "How exactly are radio waves propagated", when current flows, a magnetic field is created perpendicular to it. The magnetic field expands quickly, then when the current decreases, stops, or reverses, that magnetic field starts to collapse, but not all of the magnetic field returns. That ever expanding lost magnetic field is a radio signal. \$\endgroup\$ – Optionparty Jan 6 '14 at 22:48
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    \$\begingroup\$ To make a radio wave, you need to have radio frequencies. At zero frequency (DC, battery), it takes an infinite transmitter to transmit a zero amount of power, and an infinite receiver to receive the zero power. Yes, you start by switching the power/battery at radio frequency. \$\endgroup\$ – david Jan 7 '14 at 2:42
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Don't worry about photons unless you want to venture into quantum physics. A photon is the quantum of electromagnetic radiation, which is also a wave. I've yet to find an application in RF engineering where quantum effects are relevant.

In all electronic circuits, there are two fields: an electric and a magnetic. The electric field is associated with voltages, and the magnetic with currents.

We have components that make strong electric fields: capacitors.

We also have components that make strong magnetic fields: inductors.

In each of these components, we think of one kind of field as dominant. But consider what happens if we rapidly change the magnetic field through an inductor, say by passing a strong permanent magnet through it: a voltage will exist between the terminals of the inductor. This voltage is an electric field. We call this Faraday's law of induction.

A similar thing can happen to a capacitor. To change the electric field, there must be a current. Or if you manage to change the electric field, you will find a current somewhere. Manipulating the electric field inside a capacitor is rather more difficult than dropping a magnet through a coil, but if you can construct an appropriate experimental apparatus, you will find this is true.

Thus, a changing electric field can create a magnetic field. A changing magnetic field can create an electric field.

Electromagnetic radiation is these two fields creating each other in free space. The electric field changes, creating a change in the magnetic field just in front of it, creating a change in the electric field just in front...

3D EM radiation graph

To get these fields to radiate away in free space like this, you must create both, in phase, perpendicular to each other. This is why a capacitor is not a good antenna: it creates a strong electric field, but the magnetic field is relatively small. It radiates a little bit, but mostly the energy is stuck in the electric field, unable to radiate away because it has no magnetic field to carry it away from the capacitor. Same is true of an inductor, with current and voltage, magnetic and electric exchanged. See Why is an inductor not a good antenna?

Antennas are just leaky inductors or capacitors. Many antennas are equally both at the same time, such that their impedance is purely resistive at the design frequency, rather than inductive or capacitive. Through clever geometry, they create magnetic and electric fields perpendicular and in-phase, which then radiate away.

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    \$\begingroup\$ @WEB_DOT_COMFOUR: That's too much of an edit, I think. You are putting words into Phil's mouth but they are words that he didn't say. In fact, he started his post with "Don't worry about photons ...". The edit function is for fixing errors. You should undo the edit and create your own answer and reference Phil's. \$\endgroup\$ – Transistor Jan 25 '18 at 18:17
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Radio waves are produced when the electric field rapidly changes: there has to be an alternating current.

An electric field spreads out into space. When you change an electric field, the distant parts of it do not change instantly. The change is limited by the speed of light. If you fluctuate the electric field, you therefore create a wave.

You can think of it as space being permeated everywhere by an electric field; your circuit just creates a disturbance in it, like disturbing the surface of water. The disturbance travels away at the speed of light, like ripples in a pond. If your circuit just has steady DC flowing through it, the disturbance occurs just when you switch it on and when you switch it off.

(Indeed, electric equipment causes interference when it turns on and off: relays, switches, the commutation of electric motor brushes, or anything that generates sparks: all radiate and can interfere with radio communication, or with sensitive equipment.)

Radio-transmitting circuits are optimized for radiating; they deliberately do things that designers try avoid in circuits that must minimize their radiation (which is most circuits). Transmitters amplify some high frequency AC, and energize an antenna.

There are many kinds of antennas and how they all work is a big topic. One example of an antenna is simply a dipole of half a wavelength: two long conductors pointing in opposite directions, each a quarter wavelength long.

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  • \$\begingroup\$ An answer without mention of the magnetic field is not wrong, it's just incomplete. \$\endgroup\$ – Andy aka Jan 6 '14 at 22:43
  • \$\begingroup\$ @Andyaka I had that in there, but I erased it before posting to keep it short. It doesn't help the intuition about how the radiation propagates, since the magnetic and electric oscillations are in phase. \$\endgroup\$ – Kaz Jan 6 '14 at 23:52
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Radio waves were not explained until James Clerk Maxwell described electricity and magnetism with what are now called Maxwell's Equations. They use a form of vector calculus and are far from simple. For your question, it boils down to acceleration. A flowing current does not produce radio. The electrons have to accelerate, like going back and forth. Electrons move through wires very slowly but you can shake them back and forth very quickly over very short distances with an alternating electric field, by applying AC to the wire. The electrons are reversing direction and will radiate. A changing electric field produces a magnetic field and a changing magnetic field produces an electric field. Somewhat as if the electric and magnetic fields are pinched off from the wire and fly away at the speed of light.

You can also get acceleration by going in a circle (changing direction in general) and there are transmitters that work that way. Not with a wire in circle, with electrons in a vacuum going very fast in a circle from a strong magnetic field. There are nice magnets that do this job in older microwave oven circuits. Search "magnetron".

The simple way to demonstrate radio is to duplicate the original experiments with a spark gap transmitter and loop of wire with a small gap to see a spark from the received power. Do a search on spark gaps and radio waves. If you make one, beware that people will pick up your experiments on AM radios in all directions.

A surprising fact of nature is revealed by Maxwell's equations and it is what makes radio useful for long distance communication. We would expect anything that radiates in all directions to have power (intensity) that drops with the square of the distance - as in 1/(r^2). If radio detection was based on this it would be next to useless. But, as the power does drop with the square, the amplitude is proportional to the square of the power and drops as 1/r. And it is the amplitude of the field that we detect in radio (or the motion induced in electrons in a wire antenna). If you are 1km from a transmitter and go to a point 100km away, the signal amplitude is only 1/100 as strong - a value amplifiers can easily handle. If radio were based on power, the value would be 1/10000. You can imagine the problem sending signals 5000km (1/25,000,000) or to the Moon if we depended on amplitude.

I would ignore photons. Unlike radio, a photon has energy determined by frequency and you don't need quantum mechanics for radio.

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    \$\begingroup\$ What you have said about radio propagation is incorrect. The power of a radio wave emanating from a point source drops off as the square of the distance. This is due to the radio wave traveling out in all directions so that spherical spreading applies. If you vary your distance from 1 km to 100 km , the received power will be reduced by a factor of 10000 (the square of 100). Fortunately, receiving circuits can be made to be very sensitive so that useful information can be extracted from transmitters located at large distances from the receiver (including from the earth to the moon). \$\endgroup\$ – Barry Jan 6 '14 at 22:22
  • \$\begingroup\$ Once a radio wave has begun the energy does not alternate between magnetic and electric fields. You are wrong in this respect. Magnetic and electric fields are perfectly aligned in time. This is what gives free space a constant 377 ohms impedance - E has to be in phase with H at all time or it aint radio. \$\endgroup\$ – Andy aka Jan 6 '14 at 22:42
  • \$\begingroup\$ @Barry: or made sensitive enough to receive a signal from a 23 Watt transmitter that is 7,000,000,000 miles away travelling at 37000 MPH! \$\endgroup\$ – RedGrittyBrick Jan 6 '14 at 23:22
  • \$\begingroup\$ @Barry You are correct. I reversed power and amplitude and will edit. 1/r is correct for amplitude and for example, amplitude is what a diode detector needs. \$\endgroup\$ – C. Towne Springer Jan 7 '14 at 0:46
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    \$\begingroup\$ @Andyaka From Wikipedia - E and B are perpendicular to each other and the direction of wave propagation and in phase. The changing magnetic field creates a changing electric field through Faraday's law. In turn, that electric field creates a changing magnetic field through Maxwell's correction to Ampère's law. This perpetual cycle allows these waves to move through space at velocity c. - Not crazy about this explanation though and need to edit my answer. \$\endgroup\$ – C. Towne Springer Jan 7 '14 at 3:29
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Signal power does drop as a square function for E fields because the area covered by the emitted signal increases as a square of the distance, radius.

The point about photons, I think... The key is that photons are quanta at a frequency classified at light, where radio waves are quanta at a frequency below light. But I really don't know. Where is Richard Feynman when you need him...

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  • \$\begingroup\$ Whether it's "light" or "radio", electromagnetic radiation can be studied using either the classical theory (waves) or the quantum theory (photons). Like in many areas of physics, the classical theory is good enough for most things, but the quantum theory is needed to understand some observations that aren't explained by the classical theory. In the radio bands, there are very very few cases where the quantum theory is needed, but that doesn't mean that rf energy is not quantized --- only that the quanta are very small (very low energy per photon). \$\endgroup\$ – The Photon Jan 7 '14 at 1:52
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    \$\begingroup\$ Your answer perpetuates a myth. In fact, the electric and magnetic fields in EM radiation are in phase. They do not collapse and regenerate out of phase, feeding each other: it is not like LC resonance at all. \$\endgroup\$ – Kaz Jan 7 '14 at 17:36
  • \$\begingroup\$ See here: en.wikipedia.org/wiki/Electromagnetic_radiation#Wave_model "A common misconception is that the E and B fields in electromagnetic radiation are out of phase because a change in one produces the other ..." \$\endgroup\$ – Kaz Jan 7 '14 at 17:38
  • \$\begingroup\$ OK I edited my answer Kaz. I was taught the other way. Would not want to perpetuate a myth. \$\endgroup\$ – dfowler7437 Jan 7 '14 at 22:33

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