Radio antennas that are much shorter than the wavelength

From my limited experience with ham radio when I was a kid, I expect transmitting and receiving antennas to have lengths that are on the same order of magnitude as the wavelength, and in fact I recall having to mess around to compensate for the fact that a given antenna wouldn't be properly resonant over an entire frequency band. This also seems to match up with what we see with musical instruments, where, e.g., a saxophone's tube is half a wavelength and a clarinet's is a quarter.

For commercial FM radio with a frequency of 100 MHz, the wavelength is about 3 m, so I can believe that some of the receiving antennas I've seen are a half-wave or quarter wave. But for AM radio at 1000 kHz, the wavelength is 300 m, which is obviously not a practical length for a receiving antenna.

Can anyone explain this in physics terms, hopefully without making me break out my copy of Jackson and wade through pages of spherical harmonics? Does AM reception suffer from the length mismatch, e.g., by being less efficient? Does it benefit from it because it's so far off resonance that the frequency response is even across the whole band? Is there a dipole approximation that's valid for AM only? For both AM and FM? If the sensitivity is suppressed for the too-short antenna, is there some simple way to estimate the suppression factor, e.g., by assuming a Breit-Wigner shape for a resonance?

This question touched on this issue, but only tangentially, and the answers actually seem inconsistent with the observed facts about AM. Also related but not identical: Radio communication and antennas

• IIRC AM antennas need to be a loop whereas FM can be open. Perhaps this has something to do with it? – Brandon Enright May 18 '13 at 21:53
• – Mike Dunlavey May 19 '13 at 1:50
• Some possibly relevant material here: en.wikipedia.org/wiki/Antenna_%28radio%29#Resonant_antennas . In general, dipole radiation goes like $d^2\omega^4$, when the dipole is small compared to a wavelength. Out of the factor of $\omega^4$, an $\omega^2$ part is due to the mismatch between the size of the dipole and the wavelength, which makes the phase almost the same at both ends of the dipole. By reciprocity, I think something similar would hold for a receiving antenna. If $\omega$ is 100 times smaller than it "should" be for AM, it seems like you might pay a price of $100^2=10^4$. – Ben Crowell May 19 '13 at 15:27
• Also check out en.wikipedia.org/wiki/Loop_antenna#Small_loops . It seems they're very inefficient but the loss is acceptable because the noise sources are quite high so the signal-to-noise ratio isn't affected much. – Brandon Enright May 19 '13 at 17:00
• @BenCrowell no I meant high. The Wikipedia article suggests that for AM the noise floor can be 55db above thermal noise so even if the antenna has a 50db loss it doesn't have a significant effect on the signal-to-noise ratio. – Brandon Enright May 19 '13 at 19:58

http://www.antenna-theory.com/antennas/shortdipole.php is a website with useful info., including formulas.
To oversimplify, it seems to say that once the antenna is a tenth or less of the wavelength, the exact ratios don't matter so much. The antenna is inefficient, but it works for both sending and receiving. If you can detect the signal, of course you can amplify it as much as you want.

• I think putting this together with @BrandonEnright's comment, it's starting to make sense to me. When you receive, the antenna's lack of sensitivity cuts the signal strength, but it also cuts the noise, so S/N is OK. An inefficient low-power (receiving) system is not as bad as an equally inefficient high-power (transmitting) system. – Ben Crowell May 19 '13 at 19:43
• @BenCrowell that's part of the story. Other part is weird: convert the antenna to a resonator, which will emit intense EM waves. Place a variable coil across your dipole, or easier, varicap across your small loop. Tune it to the distant transmitter, and depending on copper loss, a giant local EM field will appear. (Ideally use liquid He! Or at least thick copper, and massive ferrite.) The receiver's emissions punch a hole in the incoming EM, leaving a downstream shadow. The missing energy ends up in the antenna. N.b. atoms employ this trick (strongly absorbing 500nM waves by 0.1nM atoms.) – wbeaty Jun 16 '18 at 15:45

The key to the efficiency of an antenna (whether for transmitting or receiving - the two processes are essentially reciprocal) is resonance, and impedance matching with the source / receiver. The size also matters in terms of the relationship between power and current.

A nice analysis of the impact of size of an antenna on the power/current relationship is given at this site. Summarizing:

The current in a dipole antenna goes linearly from a maximum at the center to zero at the end. Because the amplitude of the generated E-field from a given point is proportional to the current at that point, the average power dissipated is (equation 3A2 from the above link):

$$\left<P\right>=\frac{\pi^2}{3c}\left(\frac{I_0 \ell}{\lambda}\right)^2$$

(note - this is in cgs units... more about that later). For the same current, as you double the length of your (much shorter than $\lambda/4$) antenna, you quadruple the power.

Directly related to this concept of power is the concept of radiation resistance: if you think of your antenna as a resistor into which you are dissipating power, then you know that

$$\left<P\right> = \frac12 I^2 R$$

and combining that with the above equation for power, we see that we can get an expression for the radiation resistance

$$R = \frac{2\pi^2}{3c}\left(\frac{\ell}{\lambda}\right)^2$$

This is still in cgs, which will drive most electrical engineers nuts. Converting to SI units (so we get resistance in Ohms) we just need a scale factor of $10^{9}/c^2$ (with $c$ in cgs units...); thus we get a simple approximation for radiation resistance in SI units (I now go from $c=2.98\times10^{10}~\rm{cm/s}$ to $c=2.97\times10^8~\rm{m/s}$):

$$R = \frac{2\pi^2 c}{3\times 10^{-7}}\left(\frac{\ell}{\lambda}\right)^2$$

which agrees nicely with the expression given at this calculator for an electrically short dipole (note - their expression is for $\ell_{eff}$ which is $\ell/2$ for a short dipole; and they use slightly rounded numbers which is OK since there are some approximations going on anyway).

But if we are driving with a 50 Ohm cable, and our antenna represents a much smaller "resistance", then most of the power would be reflected and we don't get a good coupling of power into the antenna (remember - because of reciprocity, everything I say about transmitting is true for receiving... but intuitively the transmission case is so much easier to grasp). So to get good efficiency, we need to make sure there is an impedance match between our antenna and the transmitter / receiver.

If you know what frequency you are working at, impedance matching can be done with a simple LC circuit: the series LC represent a low impedance to the antenna, but a high impedance to the receiver. In the process, they convert the large current in the antenna into a large voltage for the receiver (source of image and detailed explanation) This is an example of resonant matching: it works well at a specific frequency. One can use signal transformers to achieve the same thing over a wider range of frequencies - but this loses you some of the advantages of resonance (all frequencies are amplified equally).

It remains to be shown what the real effect is of reducing antenna size on the received signal. For this, the most extensive reference I could find was this MIT open course lecture. Starting on page 121, this shows that the effective length of a dipole determines how much of the incoming energy can be "harvested", and it again shows that the power is proportional to the square of the size. So an antenna that is twice as short will collect four times less power. But that means it will also collect four times less noise. As long as most of the noise in the system comes from "outside", the ratio (SNR) will be the same, and you don't suffer from the smaller antenna.

This changes once the antenna becomes so small that other sources of noise become significant. It is reasonable to think that this will happen when the conductive (lossy) resistance of the antenna becomes comparable to the reactive (radiation resistance). But since the former scales with the length of the antenna, and the latter with the square of the length, it is obvious there will be a size at which the non-ideal effects will dominate.

The better the conductors, and the better the amplifiers, the smaller the antenna can be.

Summary

So yes, the power transmitted drops with the square of the length, making a short antenna less efficient as a transmitter (and therefore, as a receiver). Much of the time, though, you care about signal to noise ratio - is there more signal than noise coming from your antenna?

For this, we need to look at the Q of the antenna (bandwidth). The higher the Q, the more gain you have at just the frequency of interest (because of resonance); while "noise" is a wide-band phenomenon, "signal" is a narrow-band one, so a high Q amplifies the signal without amplifying (all the) noise. If we can make an antenna with a high Q, then it doesn't matter so much that it is short.

• @uhoh in general, Reciprocity demands that antennas behave roughly the same for xmt and rcv. (To absorb waves, antennas must emit waves, otherwise they wouldn't couple to EM at all.) Absorption is emission: diffraction patterns with the missing energy of the EM node-shadows ending up inside the receiver. Thus a good emitter automatically becomes a good absorber. Electrically-small, high-Q receiving antennas work well, but they must be tuned. The tuning capacitor in old AM radios is 2-section: one for the oscillator in the superhetrodyne circuit, and one across the coil on the ferrite rod. – wbeaty Jun 16 '18 at 15:55
• @wbeaty A realistic radio transmitter and receiver are not the same circuit just run in opposite directions. Radio transceivers are not black bodies, so "absorption is emission" does not get you to "transmitters are receivers'. Reciprocity is a handy concept, but to get this question right, I think it's better to go beyond the spherical cow stage. That said, I'll take another look in a day or two to read through this whole thing again. – uhoh Jun 16 '18 at 21:53
• @uhoh antennas, atoms, motors, resistors, they all absorb energy, not volts. AM radios don't respond to voltage. Instead antennas act resistive, absorbing mWatts. If needed, we can step it up to large volts, or step it down to large mA, via coupling #turns. But also I agree: detailed analysis of the receiving/absorbing process tells us much, while all the cool ideas get concealed if we just use Reciprocity. The AM resonant coil fires out a "shadow beam!" Weird, but it's just a diffraction pattern, which shows that an absorber can be microscopic, if it can emit high power while absorbing. – wbeaty Jun 17 '18 at 18:23
• @wbeaty what frustrates me is that I am beginning to pick up on your enthusiasm for the spherical cow/radio approach. Yes the AM radio resonator is an absorbing obstacle, it removes power from the incident wave, and you can model this in the far field by thinking of it broadcasting a shadow beam. Very nice! – uhoh Jun 17 '18 at 22:00
• @uhoh Yes, most antennas are actually spherical cows; simple. If you need complexity, analyze longer antennas (not quarter or halfwave dipoles.) Or of course Yagi-Uda arrays, amateur beam antennas, log-periodic. The "director" elements of Yagi antennas are the Classical Physics version of stimulated emission! Like a row of pumped, resonant chromium ions contributing to an existing collimated light beam. Hence the weirdness, and difficulty of explaining Yagi in prose rather than equations. – wbeaty Jun 19 '18 at 14:35

The requirements for transmitting antennae are much higher than for receiving antennas. Transmitting antennas must optimally radiate, so that the signal is not obscured by other stations with better antennas. If a receiver antenna is too short and far away from resonance, all received stations are uniformly weaker. What matters is that the desired signal is not less than the limit of sensitivity of the receiver. This is true for AM and FM.

The main reason is that in Ham radio you care about transmitting, in that case you need to make sure the antenna is in the right length so you get a standing wave inside the antenna. You can read about standing wave ratio here.

If you are just receiving then you could use any wire, loop antennas are practical for low frequency transmissions in AM, where you care about the change of amplitude.

• Hmmm...but don't exactly the same concerns apply to reception as to transmission, due to reciprocity? en.wikipedia.org/wiki/Antenna_%28radio%29#Reciprocity – Ben Crowell May 19 '13 at 15:16
• Actually, small AM receivers use a high-Q resonant antenna coil, which behaves like a much larger antenna, but requires a variable capacitor. If a tiny coil builds up an RF e-field extending for ten of meters, then that antenna has an effective size (EA) of 10M. Such receivers were in use before the invention of amplifiers (vac tubes.) The tiny antenna-coil resonator intercepts relatively enormous energy, which greatly increases S/N ratio. Modern receivers might just employ low-noise FET amplifiers, with no need for a variable capacitor knob to adjust the LC resonator antenna frequency. – wbeaty Jun 16 '18 at 15:34

Many ham antennas include coils that help the antenna appear to be the right length for the frequency in use, there are also trapped antennas there the coil will block frequencies above a specific point and the frequency drops the coil will allow the energy to pass to the antenna element on the other side so at high freq you have a shorter antenna and as you go down the antenna gets longer and longer.

RF and antennas are a science unto themselves so I'll dumb this down, (no offense, I still struggle with it).

Without going into a lot of the physics, the shorter the antenna is with regard to the wavelength for that frequency, the lower the efficiency. Efficiency is lost in many ways; resistance of the antenna itself, dialectic losses, (non-conductors used in the construction of the antenna still have a minute amount of conductivity resulting in loss due to heat) and impedance mismatch loss.

The most common wavelengths for antenna's for transmitting are 1/4, 1/2 and 5/8. The reasons these lengths work is due to resonance. Here's a very simple explanation of resonance. Finally, as antennas get shorter, their efficiency goes down. This is key.

Let's say you have a perfectly round pond with a perfect bottom and perfect symmetry and perfect, gradual shoreline. You drop a steel ball directly into the center of it. The wave makes it's way to the shoreline. When it gets there, it bounces (reflects) back toward the center. (Yes, you could actually demonstrate this in a sink or something similar). The outgoing waves may or may not be in sync with the returning waves. If they are out of sync, as you can imagine, they kind of cancel each other out and you no longer have a nice, clean sine wave radiating from the center out to the edge.

If the pond were the perfect size, say a perfect multiple of the length of the wave, then the reflected wave would complement the outgoing wave when they met and would actually enhance each other as they cross and make them bigger, (but eventually they would "peter out" unless you dropped another steel ball in the center).

Radio waves work the same way. When the antenna is trimmed to the right length, the wave sits on the antenna wire/element, in sync with reflected waves. (RF doesn't simply go to the end of the antenna and radiate; the waves actually reflect back and forth, until, eventually all of the RF is radiated, just like in the "tuned" pond, the waves would complement each other, staying in sync, until they run out of energy).

If you can grasp that, then you kind of have an idea of how an antenna is tuned to resonance. (Not all antenna's are resonant and they don't have to be, but that's another discussion).

So, we're down to AM broadcast radio. Now we are dropping a HUGE steel ball (transmit wavelength) into a really small pond, (our antenna). As you can imagine, the distance from the center to the shore is going to be a very small fraction of the wave length. So, for sake of argument, it will only take 1/50th of a wavelength to reach the shore. And needless to say, the reflected wave is going to be a train wreck and there's no chance of them being, "in sync". Recall earlier I mentioned efficiency. In our first example, we had a very efficient "antenna" (the pond size) and were able to maximize and utilize the full potential of the wave because it reinforced itself. But, in this case, we have next to no efficiency, however the wave still reached the shoreline, but, it wouldn't go very far. The way around this is simply to use a ton of power and "muscle" the signal out there. The efficiency on the receiving end, (the edge of the pond) doesn't matter; we don't need the waves to reinforce each other because we're using a ton of power, (a really large, heavy steel ball).

To be sure, AM broadcast station antenna's ARE efficient as the tower itself is the radiator! High-powered stations (clear channel) will frequently have quarter-wave towers. They'll also use capacitance "hats" on top to simulate extra length. So, to answer your question, the reason those short antennas work is by brute force; they have no choice. However, whereas now, you may lose a class C AM signal within 50 miles, if you had a receiving antenna with the same efficiency as the transmitting antenna, you'd be able to receive that same signal at perhaps 20 times that distance.

Again, forgive the analogies but they're not too far off the mark on a fundamental level.

The antenna can be modified by connecting inductance or capacitance and also resistance devices to change its resonant frequency.

The time constant $\tau=RC$ or $\tau=R/L$ is aprox matched to the wavelength of the resonant signal.

When an analogue radio is tuned it is these components that are adjusted to alter the resonance.

Its still best for the length of the antenna to be such that the induced charge motion that creates a maximum effect in either voltage or current at the "detector". A standing wave antinode (https://en.wikipedia.org/wiki/Standing_wave) does not move, so that can used. One antenna cannot be optimal in this respect for all frequencies though.

It's only for a negligible conductor tuned to its optimal frequency that the radio wave will push the charges up and down the entire length. (This is the most efficient solution, though.)

Both AM and FM radio suffer efficiency losses from this effect; but attachment of resonant circuit devices to an antenna can compensate. The energy loss is the work done in these devices, for which there are standard equations, for example for a capacitor work is $$W=CV^2/2$$ The total work expression needs to be integrated over time according to the sinusoidal voltage displacement/signal.