I remember long range radio stations which still worked 30yrs ago. I could listen to the whole Europe. I also think I saw movies where radio enthusiasts could talk all over the world..

I never looked up how it works. But I had impression that it's quite simple to send a radio signal around the globe (you just need very tall antenna).

However I now am trying to figure out how to do it exactly and there is nothing much..

My questions are to envisage a technical solution for 5,000 km range with short on-off signal:

  • what kind of wavelength can travel that far
  • what kind of transmitter power is required

Another question:

  • I understand that such long propagation range is only possible by reflection from ionised atmosphere layers - how do you estimate actual path length then ?
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    $\begingroup$ Hello and welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Commented Jun 2, 2020 at 17:33
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    $\begingroup$ I changed the tag on this question, as “long path” has a specific meaning in amateur radio $\endgroup$
    – Scott Earle
    Commented Jun 2, 2020 at 23:37
  • $\begingroup$ How short is "short on-off signal"? Is "short" 30 seconds? 1 microsecond? It makes a huge difference in the effective range of the signal. $\endgroup$ Commented Jun 4, 2020 at 12:56
  • $\begingroup$ That's what I am actually trying to figure out - minimum signal length at max feasible power consumption and max frequency to create enough power at RX side to be able to detect it (ie avoiding disaster what Marcus was talking about) [note: max and max - not typo] $\endgroup$ Commented Jun 4, 2020 at 18:26
  • $\begingroup$ @BoppityBop but that's a drastically different question than what you're asking here. Please, really make true your promise to ask a new question, where you describe exactly what the acceptable boundary conditions for your problem are. We can't optimize a max/max problem - there's no bounded solution to it. You really need to ask a question where you describe what you need to achieve, in particular, and what the things that limit you are. $\endgroup$ Commented Jun 5, 2020 at 13:52

4 Answers 4


This question is clearly asking about commercial shortwave broadcast radio, but since this is the amateur radio forums, here's an answer with an amateur radio slant instead of shortwave broadcast. But this still applies to shortwave broadcast.

Generally, anything below 30MHz (wavelength roughly longer than 10m) will bounce (or more accurately, refract) off the ionosphere. However, the location and nature of the ionosphere changes periodically and randomly based on time of day and solar activity, which shifts exactly what frequencies bounce best. Generally, 14MHz works best during day, and 4MHz works best in deep night, but this may vary based on actual conditions. 30MHz works best during high solar activity, which occurs roughly 5 years out of every 11. These are broad generalizations, and more detailed accurate answers can be found in extensive literature on propagation and on live propagation measurement websites.

Presuming amateur radio use, the antenna doesn't actually have to be that large or high. It does help to be close to 1/4 wavelength, and larger antennas tend to work better, but there are many shortened antennas that still work acceptably well. Radiation angle is actually much more important (for transmission) than height -- you want to aim at the horizon to get maximum distance, but not so low that it's absorbed by trees and buildings first. It turns out that low antennas typically have a lower radiation angle, which is also why broadcast shortwave antennas start at ground level. (Radiation angle matters for reception too, but it's not as critical.)

Amateur radio can use power as low as 5w, but 100w is more typical for long range communications. Slower and narrow bandwidth modulations can use lower power, so, for instance, CW, PSK31, and FT8 typically use 30w, but can go much lower and still propagate to the other side of the world. Again, this is very sensitive to "band conditions", so if the ionosphere is not right for the frequency, even very high power will not propagate well, and if it is right, 5w might do as well as 100w. Or, if there's a lightning storm, the noise from the lightning might drown out your transmissions.

Contrast amateur radio power (1w - 1500w in the US) with commercial broadcast station power (up to 2Mw) -- amateur radio receivers are typically more sensitive (and expensive) than commercial shortwave broadcast receivers, and amateur radio operators typically use much larger antennas. Inexpensive commercial shortwave receivers frequently use a very small internal loopstick antenna (about the size of your hand), and if a larger external antenna is used, can pick up much weaker signals. These two differences necessitate the higher power for broadcast stations.

There are multiple bands appropriate for shortwave broadcast, between 2MHz and 26MHz. Each of these bands has different propagation characteristics that make them usable at different ranges and times of day, same as for amateur radio.

  • $\begingroup$ Commercial assumes a way to make money. Its not my point. Re: frequency - are you saying 100w transmitter can send a MHz signal across the Atlantic and it will be heard? What kind of signal/noise on receiving side we are talking about in this case? $\endgroup$ Commented Jun 4, 2020 at 12:22
  • $\begingroup$ Commercial is a licensing class, and to some extent, a minimum amount of funding required to run the station, and doesn't really indicate profit. See also the legal definition of broadcast vs. transmit. Amateur radio regularly uses 100w to talk around the world. From the east coast of US I can easily reach Japan, Australia, New Zealand using FT8 on 30w. Power helps, but good antennas and lucky "weather" conditions help more. (See: spaceweather.com) Signal/noise ratio varies wildly, anything from crystal clear could be next door to mostly static. $\endgroup$
    – user10489
    Commented Jun 5, 2020 at 4:44
  • $\begingroup$ Amateur radio can use power as low as 5w, but 100w is more typical for long range communications. That is a statement of the past - the vast majority (in terms of numbers) of long-reach contacts is done using FT-8 with definitely less than 100 W of power, exactly as you say! $\endgroup$ Commented Jun 5, 2020 at 13:48
  • $\begingroup$ 100w is for wide band communctions like SSB (phone/voice). FT-8 and the other modes I listed are all narrow band digital modes. Also, FT-8 is a full duty cycle mode, so technically 30w FT-8 has similar energy to 100w SSB. $\endgroup$
    – user10489
    Commented Jun 8, 2020 at 0:53

Kind of in a hurry, so only a quick answer:

That was longwave (LW) radio (150 kHz to ca 380 kHz), or medium wave (MW) (530 kHz to 1600 kHz).

Wavelength $\lambda$ is related to frequency as $\lambda = \frac{c_0}{f}$, with $c_0=3\cdot10^8\,\frac{\text m}{\text s}$ the speed of light, and $f$ the frequency.

LW still exists, not very popular for broadcast these days since higher frequency installations are practically full-coverage, unless you're at sea. In most of central Europe, MW has been shut off, because there was no sweet spot between LW and the VHF bands (30 to 300 MHz, which includes the "usual" 88 to 108 MHz FM broadcasting band with music and news stations).

Newer longwave installations don't use the (noise-sensitive) AM (amplitude modulation) use DRM (Digital Radio Mondiale).

Existing longwave transmitters use lots of power. Typical MW AM stations have 50 kW to 2 MW (!) in power.

India has relatively recently build a transmission system for DRM, to cover most of the subcontinent, and it uses about 1 MW – much less than a AM system achieving the same would have to use.

you just need very tall antenna

The antenna needs to be typically a quarter wavelength in size (roughly), so often complex arrays of tall antennas are necessary to achieve that. They extend, usually, down to the ground, so they are not "Mounted at height", they are simply "high".

what kind of wavelength can travel that far

Generally, in free space, all wavelengths travel infinitely far, but you're right, only certain ones fall in a region where you can exploit the bending properties of the ionosphere, for example.

Also, the power density per area (i.e. how much RF power hits 1 m²) goes down with the square of the distance and with the square of the wavelength. (however, that is at least partially compensated by the fact that an antenna of a given size can be made to have a higher effective area for a higher frequency. The increase in "picking up power" area is also square with the inverse of wavelength).

So, the fact that LW reached and reaches so far is:

  1. atmospheric and ground-conducted wave phenomena, as you've mentioned
  2. the relative ease at which a highest-power amplifier for these frequencies can be built (1940's engineers simply couldn't build a 100 MHz amplifier that does 500 kW of output power; the technology didn't exist.)
  3. The low expectations for transmission quality on the existing bands.

So, it's a mixture of physical, technical and historical reasons.

how do you estimate actual path length then ?

Using a propagation model. There's quite a few of these, some are very complex and include sun activity, weather and angles of antennas, others are more rules of thumbs.

Also, notice that there's not necessarily a single path. In fact, that's usually not the case, or only the case if you restrict yourself to a very narrow bandwidth. As an example:

The aforementioned DRM system (deployed with the intent of reaching > 1 billion people!) is designed with a mode that can tolerate 7.5 ms of delay spread, i.e. a signal travelling the longest path reaches the receiver 7.5 ms after it already reached it via the shortest path. The tolerable path length difference, thus, has to be up to $7.5 \cdot10^{-3}\,\text{s}\,\cdot\,c_0 =2.25\cdot10^6\text{m}=2250\,\text{km}$. So, it's really impossible to talk about "path length", you need to account for the multiple different paths.

The DRM receiver actually calculates the way the multipath environment looks like – and compensates the effect of that;
which is the main reason why DRM works better at the same power:

Imagine you listen to someone talking. However, that's overlaid with an echo coming in 7.5 ms later; that simply means that when his voice hits 266.67 Hz – that's a middle C in musical notation – the echo happens to be the same tone, but half a tone period later, so that it cancels out!
We call that frequency-selectivity of the multipath channel.

So, DRM has built-in RF "echo cancellation", and hence works better, and you thus can get better audio (remember Longwave broadcast, that was always very narrowband, and sounded muffled, and there was no chance to get music). Or, you can use the same advantage to get the same audio quality, but with less power; or with the same power, but further away.

I saw movies where radio enthusiasts could talk all over the world.

Right place to mention this – radio amateurs tend to do such things (and others).

Now, when you want to do this, you don't need a 1 MW amateur band AM transmitter. You can actually "work the world" with far less power, assuming you don't try to get something as "bandwidth-hungry" as music across. In fact, with maybe 0.1 W, you can get very far on a continent, if you use a very robust transmission scheme, a "mode":

Just as AM and DRM are two methods of transporting broadcast audio, there's methods for just transporting textual methods that need far less power at the receiver, and hence far less power at the transmitter. "WSPR" and "FT-8" are such modes.

For example, to transmit WSPR, you technically need nothing but something that can switch a voltage on and off at the desired rate; and people built such things and made talks about such things, using nothing but a raspberry Pi (a 35€ computer).


...solution for 5,000 km range with unmodulated signal:

  • what kind of wavelength can travel that far

  • what kind of transmitter power is required

  • ionized atmosphere layers how do you estimate actual path length

It is possible to communicate 5000-km over a wide range of frequencies. The effective radiated power required (transmitter power x antenna gain) depends on propagation of a given wavelength over a given path at a given moment. To first order, propagation depends on:

  • Wavelength: frequencies that are not absorbed or passed through the ionosphere
  • Path: over land or sea, conductive or lossy terrain, smooth (reflective) or mountainous (dispersive), etc.
  • Time of day: the sun ionizes the atmospheric layers that reflect or absorb radio waves
  • Time of year: day length and degree of ionization vary
  • Status of the ~11-year long sunspot cycle

Free tools are available to estimate the probability of being able to communicate over a given path. My personal preference is the PropView component of the DX Lab Suite, but there are others, and web sites like VOACAP Online allow you to identify "best" frequency between geographical coordinates.


5,000 km range with unmodulated signal

Either anything can do that, or nothing, depending on your perspective.

Generally, slower modulations can propagate a farther distance and still be successfully demodulated. This is because transmitting a particular symbol for a longer time accumulates more energy in the symbol, and thus it is easier to detect.

An unmodulated signal is just one symbol for an infinite time. So the energy in it is infinite, and so it will always be detectable, provided one is willing to wait long enough. Of course some propagation modes will work better than others, and so the unmodulated signal will be detected sooner rather than later. But even propagation modes that are very poor will still work, eventually.

On the other hand, an unmodulated signal contains no information, so what does it even mean to receive it? Also, for the signal to be unmodulated it must have no start and no end, so how is that even possible to transmit?

  • $\begingroup$ Can't you transmit a sine wave for 1 wavelength period? Would it not be detected at the receiver? $\endgroup$ Commented Jun 3, 2020 at 4:20
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    $\begingroup$ @BoppityBop as I tried to explain: If the detector detected that, then there's a very high likelihood would also detect noise as signal. Comparison: imagine you build a detector for the sound of a needle falling and hitting the floor. That's technically not hard. Making that work at a rock concert, however, is impossible, because everything else is so much louder. $\endgroup$ Commented Jun 3, 2020 at 7:11
  • $\begingroup$ Yes I understood your argument. I was arguing some of the Phil's points. To answer yours - yes but.. If you know time frame (say 4hrs a day) and signal power estimate at the receiver and wave shape and signal precise duration - I believe - you can make a practical solution which somewhat better than you think. Obviously only test can tell if it works as I expect or it fails as you expect. But nonetheless - I am exploring the idea. $\endgroup$ Commented Jun 3, 2020 at 10:14
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    $\begingroup$ @BoppityBop If you aren't transmitting, and then you start, and then you stop, that's not an unmodulated signal. That modulation is called on-off keying, and the effective range is a function of (among other things) how long you leave the transmitter on. $\endgroup$ Commented Jun 3, 2020 at 19:04
  • $\begingroup$ My knowledge of modulation didn't extend beyond frequency modulation. I edited the question. Thank you $\endgroup$ Commented Jun 4, 2020 at 12:16

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