# Bit rate in function of radio frequency

I would like to transmit data over the ionosphere with a specific frequency (that will enable the radio waves to bounce on the ionosphere), this frequency might be about 30 MHz. I am wondering how much data will I be able to transfer with this frequency ?

• Your practical limits will likely be regulatory (limits on power, bandwidth, modulation) rather than technical. May 1, 2014 at 16:25

To calculate this you can use the Shannon–Hartley theorem.

$$C = B \log_2 \left( 1+\frac{S}{N} \right)$$

where

C is the channel capacity in bits per second;
B is the bandwidth of the channel in hertz (passband bandwidth in case of a modulated signal);
S is the average received signal power over the bandwidth (in case of a modulated signal, often denoted C, i.e. modulated carrier), measured in watts (or volts squared);
N is the average noise or interference power over the bandwidth, measured in watts (or volts squared);
and S/N is the signal-to-noise ratio (SNR) or the carrier-to-noise ratio (CNR) of the communication signal to the Gaussian noise interference expressed as a linear power ratio (not as logarithmic decibels).

In a simple case, your maximum bandwidth will be the frequency you use. More likely it will be whatever frequency you modulate your signal at.

If your system were perfect, that would also be your channel capacity. But, it won't be perfect, ever. There will be noise and signal loss, more than you want.

As an engineer, I like to draw straight lines and pretend they're close enough to the real world. For a situation like this, where a whole mess of small (and a few large) losses come into play, I would make a Fermi estimate.
I understand that's not easy. One of the biggest problems for aspiring engineers is they're not willing to just guess. This prevents them from moving forward, likely failing, and returning to make a better guess. Just get some paper and sit down in a place where you can make some noise and say "eehhuuhhhmm, fifty-six... ish?". Write that down. Now calculate it. Does it make any sense? Use whatever information you have; if your calculations show that you can set up your own radio station with some AA batteries then get a new piece of paper.
In the end the system you're trying to model is really complex; clearly, it depends on undulating ionospheric plasma and the time of day. You'll likely be able to guess your values within an order of magnitude, which might be enough information to get you to start building something.

• Thank you ! I have hard time to find what is S/N equal to, because there are many things to take in account such the distance (I would like to transmit data over at least 8000 km), and the power of my antenna (I read somewhere that only a few watts are sufficient to transmit data over thousands of kilometers. What could I do to find this ? May 1, 2014 at 7:05
• @Samuel The ionosphere is the dominate feature of the channel which causes a lot of loss, fading and multipath. In addition noise from lightning can completely destroy a channel path as in the upper atmosphere it can be nearly continuously discharged. Classical SNR, bandwidth/noise floors are rarely the limiting factors. May 4, 2014 at 19:55
• @user6972 You could be right depending on your definition of 'classical SNR'. However, as far as SNR sans 'classical', fading, multipath, and lightning can be included. There is nothing wrong with describing a noise term that is a function of some probability distribution, like a Poisson distribution for lightning. May 4, 2014 at 20:12
• @Samuel many of the ionosphere effects are non-linear and can't be used in a standard noise model. Especially when they are so highly dynamic and change constantly. It is a very different animal than normal channels. May 4, 2014 at 21:28
• @user6972 We seem to be having a miscommunication. These 'classical SNR', 'standard noise model', and 'normal channels' ideas are, as you say, inadequate. However, they appear to be straw men that you have brought with you. Even in this very complex system, the concept of SNR still exists and can be modeled. Perhaps you should propose an alternate name for it if you think the term is reserved for classical, linear, and standard noise (a very limited scope). I ask that you so so in your own answer though. May 4, 2014 at 21:42

I can't tell you how much the ionosphere will reflect back your signal so you'll need to do some digging to find that out. What I can help you with is the generally accepted formula for free-space transmissions namely link loss - this is how many dBs of attenuation you could expect theoretically if two antennas were communicating in free-space. The formula is: -

Link Loss (dB) = 32.5 + 20$log_{10}$(F) + 20$log_{10}$(d)

where F is MHz and d is distance between the two antennas (kilo metres).

Your transmit distance is 8000 km (at 30 MHz) and these numbers give a link loss of 32.5dB + 29.5dB + 78dB = 140dB. Put another way, if your transmitter power is 1 watt (30 dBm), you could expect a power of -110 dBm at your receiving antenna.

How much power does your receiver require? There is another generally accepted formula that states: -

Power required in dBm is -154dBm + 10$log_{10}$(data rate) dBm

So, if your data rate is 1k bits/second, you should be able to adequately receive (with reasonably low error rates) a power of -124 dBm.

So far, at 8000 km you should be able to receive -110 dBm from a 1 watt transmitter and for a 1kbps transmission your receiver needs a minimum of -124 dBm. This implies you have 14dB in-hand but, anyone working in radio will tell you that's not enough for continued all day long reception and that something like 30dB margin is more acceptable.

However, you are targeting ionospheric conditions being right and are probably not 100% bothered about really decent continuous data at low bit error rates so maybe it's enough.

Improvements can be made by using directional antennas but at 30 MHz you are probably going to use dipoles at each end and these will give you a slight increase in margin of about 3dB. Higher power transmissions are also one area that could improve things.

Good luck.

• Of thank you ! If I choose a 10W antenna instead of a 1W antenna, will I receive the signal 10 times better ? May 2, 2014 at 14:11
• 10watts is 10dB more than 1watt. This means it is stronger. May 2, 2014 at 16:22
• @Andy bouncing signals off the ionosphere has to be included in the path loss and this is very high and changes constantly. Your estimations using the link loss formula won't work well at all. May 3, 2014 at 3:10
• @user6972, dude, please read the first line of my answer and note what I said. Also if you have down voted the answer because you didn't see my pening line then pls consider reinstating that vote. May 3, 2014 at 10:16
• @user6972 You wrote, in your answer, 30 kHz instead of 30 MHz and used a small "h" for "Hz" but do you see me griping about this or that; now get a life. If you want to do a full explanation of how much the ionosphere attenuates stuff go right ahead in YOUR answer and stop trying to moan at me because I chose to put a disclaimer on a part of MY answer I have no knowledge or experience of. May 4, 2014 at 20:14

30 Khz frequency is a bit too high for a 8000km ionosphere skip unless you get just the right conditions. Take a look at the HF propagation models and prediction programs.

Estimating range is based on the link budget which is based on the sensitivity of your receiver. The S/N or C/N ratio is based on your choice of bandwidth, modulation type and noise figure of the receiver's front end.

To answer your question directly, I can only offer a comparison. HF Digital packet radios can sometimes achieve data rates around 1400 bits/s during ideal conditions using forward error correction and complex fading algorithms at about 150W of power they sometimes reach 2500km. By reducing the data rates and narrowing the bandwidth higher distances can be achieved due to the increase of the receiver's sensitivity.

EDIT:

Your main obstacle is going to be overcoming the ionosphere losses and fading, not C/N ratios or data bandwidths. If you want to experiment with this I suggest you build a receiver that can receive the NCDXF/IARU beacons in morse code, which is a simple form of digital modulation. There are many software programs that can demodulate the audio, you just need to build a simple receiver/antenna that can convert the RF to audio (or buy a cheap receiver).

These beacons transmit at 100W to ID themselves, then 100W symbol, 10W symbols, 1W symbols, 0.1W symbols. Based on what you can receive you can estimate the channel strength. While you can transmit data at 1W or less at very low rates (narrow bandwidths) like morse code does, digital packet radios usually need about 100W or more and good ionosphere conditions to get a reliable digital link (which is related more to the fading, bandwidth and C/N issues)

They are located around the world so you can see how the frequency is propagating around the world 24 hours a day at different frequencies.

You can learn more about the system with these two papers (1 & 2). There are also sites that have on-line monitoring systems that you can see how the propagation is between points. In addition some of this data is feed back into the propagation forecasting tools that are out there.

I should also add that the non-linear plasma in the ionosphere causes a lot of fading and multipath issues which really hamper higher speed digital transmissions. These distortions are not included in the link budgets presented here, but if you have a modulation format with complex modulation (like with higher speeds of packet radio) these effects will also degrade your signal significantly.

One other important caveat is that this frequency band 30Mhz and lower is EXTREMELY noisy. Buildings, computers, city noise, engines (alternators/generators) will completely wipe out most of your hard work. The band is mostly usable if you are in a remote location with very little industrial activity.

• Do you mean that I need 150W power ? Oo May 2, 2014 at 14:12
• @TrevörAnneDenise look at the propagation models that I linked to and you'll find you'll need a lot of power or a very large directional antenna to get any kind of digital modulation to go 8000km. May 3, 2014 at 3:13