# Transmission speed of modern digi-modes

What are the bit per second ratings and signal-to-noise-ratio ratings of various modern, popular, good performance and well supported digi-modes for HF low power (0.1 watt) long distance (500 to 10000km) communication?

My brief understanding is that BPS is a trade off for SNR. PSK31 is 31 BPS. JT9 is slower, 13 characters per 50 seconds, but with better SNR rating.

I can't answer your "bits-per-second in actual usage" question, because I've never worked at HF myself.

My brief understanding is that BPS is a trade off for SNR.

SNR (Signal-to-Noise Ratio) is what you effectively see on the receiver, not something that you can directly influence by your choice of bps – but they are, indeed strongly linked:

Generally, for a higher bit rate, you need higher bandwidth. Higher bandwidth means that your receive filter needs to be wider. Higher filter bandwidth means higher equivalent noise bandwidth. And noise power in your receive signal is proportional to that bandwidth. So, if you've got to signals, both with a signal power of x Watt, and one is 10 kHz wide, and the other is 10 MHz wide, then the filter bandwidth for the 10 MHz one will necessarily "let through" thousand times the noise, and SNR for the 10 kHz signal (same signal power!) will be 30dB better (=factor of thousand, due to noise being only one thousandst).

I said "Generally, for a higher bit rate, you need higher bandwidth".

How does that happen?

Well, mathematically proven is that you cannot shove arbitrarily much data over any link. That's called the Channel Capacity, and it's actually measured in bits per second!

Let us bask in the formulas for a second. However, I need to stress this: This is a theoretical boundary for what is possible with a given link. It does not say that any transceiver does actually achieve that rate. However, there's no transceiver that could be faster.

\begin{align} C&= B\cdot \log_{2}\left(1+\frac SN\right)\\ \text{with:}\\ C:& \text{ the channel capacity}\\ B:& \text{ bandwidth}\\ S:&\text{ signal power}\\ N:&\text{ noise power} \end{align}

So, as you can see, the amount of bits you can push through a channel per second is proportional to its bandwidth $B$. And also, it goes with the logarithm of the SNR – which itself only increases linearly with the signal power $S$.

That's the first takeaway here: If you want to double the possible bitrate of a link, you can double the bandwidth and double the transmit power to keep SNR constant,or you can just use much much more power. That's the reason why non-ham high rate systems have high bandwidth, but rarely much power.

You asked about SNR for a specific HF scenario. So let's roll!

• TX power: $P_{TX}=0.1\text{ W}$
• operational frequency: Let's say we're using one of the subbands of the 15m band $f=c_0\cdot\lambda=300\cdot 10^6\frac{\text m}{\text s}\cdot 14.184\text{ m}=21.15\text{ MHz}$, $c_0$ is the speed of light, $\lambda$ our wavelength
• Bandwidth: highest legal bandwidth (to my knowledge) $B=2.8\text{ kHz}$
• Distance: Let's do this for $d_0= 500\text{ km}$, or expressed in wavelengths $\lambda$, $\frac{d_0}{\lambda} =35250$.

We don't know our antennas (yet?), so let's just roll with antenna gains $G_{TX}$ and $G_{RX}$ for the transmit and receive antennas, respectively.

So, due to free space loss, the power $P_{RX}$ reaching the receiver is

\begin{align} P_{RX}&= P_{TX}\left(\frac{\lambda}{4\pi d_0}\right)^2\,G_{TX}G_{RX}\\ &\text{which, in our special case, is, entering numbers}\\ &=0.1\text{ W} \left(\frac{1}{4\pi 35250}\right)^2\, G_{TX}G_{RX}\\ &\approx 5.1 \cdot 10^{-13}\text{ W}\,G_{TX}G_{RX}\\ &\text{this is not handy. Let's put it as dB:}\\ &\approx -123 \text{ dBW}+ G_{TX}\,[\text{dB}] + G_{RX}\,[\text{dB}]\\ &=-93 \text{ dBm} + G_{TX}\,[\text{dB}] + G_{RX}\,[\text{dB}]\\ \end{align}

At the receiver, we get a really good filter and filter exactly to $B=2.8\text{ kHz}$. We know that thermal noise power in any receiver is given by

\begin{align} P_N &= kBT\\ \text{with}\\ P_N:&\text{ noise power}\\ k:&\text{ Boltzmann's constant}\\ T:&\text{ absolute system temperature} \end{align}

Now, I've learned by heart that noise power in 1 Hz is always -174 dBm at room temperature, or

$$P_N = -174\frac{\text{dBm}}{\text{Hz}} + B\,[\text{dBHz}]$$.

Our $28\text{ kHz}\approx 34.5\text{ dBHz}$, so $P_N=(-174+34.5)\text{ dBm}= -139.5\text{ dBm}$. That allows us to calculate SNR:

\begin{align} \frac SN &= \frac{P_{RX}}{P_N}\\ &= -93 \text{ dBm} + G_{TX}\,[\text{dB}] + G_{RX}\,[\text{dB}] &-& -139.5\text{ dBm} \\ &= 46.5\text{ dB} + G_{TX}\,[\text{dB}] + G_{RX}\,[\text{dB}]\text{.} \end{align}

So assuming you have no antenna gain whatsoever, i.e. your antenna is an isotropic radiator, you still get an SNR of $\text{SNR}_{500\text{ km}}=46.5\text{ dB}$, which is pretty awesome! (By the way, we're neglecting interference here.)

While we're at it, let's derive the Channel Capacity for this case:

\begin{align} C_{500\text{ km}} &= 2.8\text{ kHz} \log_2 \left( 1+ 46.5\text{ dB}\right)\\ &= 2800 \frac1{\text s}\log_2 \left( 1+ 10^{4.65}\right)\\ &\approx 43.25\frac{\text{kb}}{\text s} \end{align}

However, we must assume that the digital mode you want to use also has to work with the 10,000 km link. From the formula for the free space loss above we see that receive power reduces with the square of distance. Since 10,000 km is 20 times the distance of 500 km, the receive power will be

\begin{align} P_\text{RX, 10,000 km}&= \frac1{20^2}P_\text{RX, 500 km}\\ &=\frac1{400}\cdot-93\text{ dBm}\\ &=-26\text{ dB}-93\text{ dBm}\\ &=-119\text{ dBm} \end{align}

So we suffer from a 26 dB worse SNR, i.e.

$$\text{SNR}_{10,000\text{ km}}=46.5\text{ dB}-26\text{ dB}= 20.5\text{ dB}$$

With a 20.5 dB SNR, our Channel capacity reduces to:

\begin{align} C_{10,000\text{ km}} &= 2.8\text{ kHz} \log_2 \left( 1+ 20.5\text{ dB}\right)\\ &= 2800 \frac1{\text s}\log_2 \left( 1+ 10^{2.05}\right)\\ &\approx 19\frac{\text{kb}}{\text s} \end{align}

Remember, this is a maximum achievable rate, and it assumes you do a lot of stuff right: filtering, modulation/demodulation, and channel coding, and even then, it doesn't say how to build a transceiver that achieves that rate. It's really just a theoretical limite.

Ham systems are –intentionally, and by tradition– not as complex and optimized as commercial communication systems and hence tend to have a much lower spectral efficiency – you'll be happy if you can get more than one kilobit across.

• Nice work, but your HF propagation is incorrect. You've assumed a receiver temperature of 300K. At HF it could be 1000x more than this. Rather use something like VOACAP to find receive SNR for given Tx power and locations. Jun 12 '16 at 2:12
• Good summary of HF noise here: ab4oj.com/icom/nf.html Jun 12 '16 at 2:19
• @tomnexus right! I come from higher-frequency operations, where noise is practically exclusively generated in the receiver, as Johnson-Nyquist noise in components. You're right, for waves that easily penetrate the atmosphere, cosmic background radiation plays an important role! Jun 12 '16 at 8:15

Signal to noise ratio rating shows how protocol can cope in bad signal conditions, meaning if this ratio is lower protocol can handle worse signals. Some protocols can handle signals lower than noise.

Bits per second rating shows speed of data transfer protocol can provide. In practice, lower signal to noise ratio is achieved by slowing down data transfer.