Could data transmission power consumption be lessened by transmitting more zeroes?

Question

According to the IEA 250 TWh of power was consumed in 2019 due to data transmission globally, about 1% of worldwide electrical power use.

I would like to pose a very simple question regarding power consumption in data transmission. Does it use more power to transmit a zero or a one? - Is there a difference?

If there is a difference, then could power consumption be lessened by inverting Ethernet frames to optimise for more "ones" or "zeros", and the addition of a bit to indicate if the frame has been inverted or not.

Assuming there could be a power saving to be made, is there a way this could be implemented in a backwards compatible way?

Answer 1

TL;DR: no, there's no such scheme that we wouldn't already be using. There's reasons, below.

Information theory tells us that we have to transmit the least bits (using less energy than transmitting more bits) if we use source coding to compress the input data – making 0 and 1 equally likely.

The job of channel coding is to then take these equally likely bits and find a transmission scheme that is optimal for the end-to-end system – typically optimal as in least bit error rate for a given transmit power, or least power needed for a fixed bit error rate. There can be many other parameters to take into consideration, but these are the main things we usually look at when we optimize channel coding for long-haul high-rate communications, which use the most power.

So, what you propose is "already done", and there's 80 years of extensive theory and practice in communications engineering going into it.

For example, we know that schemes who are off to signal one bit value and transmit something for the other are in almost all cases power-wise inefficient, really. The medium of transmission is an electromagnetic wave – be it the radio interface of your phone, be it the field between the wires in a twisted pair, or be it the optical fiber for >= 100 Gbit/s links. And these have a phase, which allows us to transmit, say, amplitudes of -0.5/+0.5 instead of 0.0/1.0, and get the same "distance" between noisy received symbols at the receiver. However, the average power used by the first scheme is \$0.5^2=\frac14\$, whereas in the second case it's \$\frac12\left(0^2+1^2\right)=\frac12\$. This BPSK (binary phase-shift keying) vs OOK (on-off keying) example serves to illustrate that there's beauty in making things symmetrical – and then, you lose the "bit that has lower energy" argument altogether.

Now, there's not only symbol sets that have a constant power; on the contrary, in high-rate communications, we do use sets that have very high ranges of different powers. However, if you start "shaping" the probability distribution of these symbols, you run into a problem:

Say, you had a constellation with 1024 different possible transmit symbols (1024-QAM, for example). If you simply take 10 input bits and pick the symbol with that number, your single symbol transports 10 bits of information! Easy. That also means every symbol is equally likely, as every 10 bit sequence of bits is equally likely.

Now, you come along and say, you want to optimize for power, so the higher-amplitude symbols should be occurring less often than the lower-amplitude ones. Turns out that under that condition, each symbols does no longer carry 10 bits; 10 bits per symbol is the maximum you can get across with 2¹⁰=1024 symbols, and that happens when you choose the probabilities of all the symbols identically. So, to transmit the same, say, 1 million bits, where in the equidistributed scheme you needed 100 thousand symbols, you now need more. How much more depends on how exactly you shape the probability¹.

Now, so to be more power-efficient per symbol you transmit, you need to transmit more symbols!

It gets worse: at the receiver, a decision which symbol you've sent has to be made. This gets significantly more involved when the symbols are not equally distributed. Receiver signal processing and channel decoding contribute significantly to communication power demand. With significant, I mean, easily up to half of the overall system consumption is spent in the receiver, not the transmitter, which has to bring the symbols physically onto the transmission channel!

So, this is a path that usually leads nowhere.

It does lead somewhere if your channel is not nice and linear, and higher signal powers lead to more distortion. This is what we see in highest-rate (think 400 Gbit/s upwards) fiber links, where you'll find probabilistic shaping used to maximize the mutual information between transmitter and receiver. It really doesn't apply to simpler use cases today, and the community has been pretty good at mathematically proving that the situations where it does yield a gain are really not these use cases with lower data rates.

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¹ We actually have formulas to describe that: the maximum you could get out of a source \$X\$ with such shaped symbol set probability \$(P(x_i))_{i=1,\ldots,1024}\$ is the source's entropy:

$$H(X) = -\sum_{i=1}^{1024} P(x_i) \log_2(P(x_i))$$

With a bit of analysis you'll find that has a global maximum for \$P(x_1)=P(x_2)=\ldots=\frac1{1024}\$, as probabilities have to always add up to 1. The value of the entropy at that is \$H(X) = -1024\cdot \frac1{1024}\log_2\left(\frac1{1024}\right) = -(-10)=10\$ (bit).

Answer 2

It depends completely on what interface and encoding is used for data, if there is any difference of transmitting ones or zeroes, so there is no general answer.

For your extremely complex case of Ethernet, it depends on which Ethernet you mean.

For example, 10Mbps Ethernet uses Manchester encoding, so there is no difference if you send a frame full of ones or zeroes, the signal looks identical except for the phase of the square wave.

100Mbps Ethernet will always transmit IDLE symbol when link is up and there are no frames transmitted. And moreover, the data is scrambled with a pseudorandom noise generator to make the actual transmitted data to not affect much how the data looks on the wires.

Gigabit Ethernet is rather similar. And that's only the common types of Ethernet over copper PHYs, as you have also Ethernet over optical interfaces.

For your TV remote, a protocol like Sony SIRCS does send longer light pulses for logic 1 bits, so yes, that is an example that does use more power for transmitting ones than zeroes. But you can't make changes that are backwards compatible with it.

Answer 3

I only know one case. Here transmitting more ones saves power.

GDDR4 and DDR4 memories have a feature called Data Bus Inversion. The data lines are pulled up with resistors (terminated to high) and driven low with MOSFET switches. In this case driving low takes more power. So if the byte to be transmitted has many zeros, the driver will invert all bits in the bytes along with a marker in order to drive more ones and save power.

The RAM bus is special in that it is high-speed, parallel, not differential, and multi-drop. Slow buses don't need pullup resistors that terminate the lines and thus they take little power to transmit any bit pattern. Many interfaces that aren't so slow are terminated at the driver and open at the receiver. They only take energy to toggle the bits. Faster interfaces usually have differential current-steering drivers. They take the same power to transmit any bit pattern. Thus I am not aware of any other use case of Data Bus Inversion.

Answer 4

could power consumption be lessened by inverting Ethernet frames to optimise for more "ones" or "zeros"

Pretty much the majority of long-haul data transmissions are synchronous. These consume the majority of the total power. Synchronous means that clock and data are embedded. In turn, this means that on average, the high bit-count equals the low-bit count hence, no saving.

Apart from that, the majority of long-haul data is transmitted differentially over copper to obtain an adequate level of protection against noise and surges. When data is transmitted differentially, one line will be high while the other is low and, this state inverts each time data changes hence, there is no net difference in power consumed.

Answer 5

Other people have pointed out that in most cases the energy used to transmit a zero is the same as the energy used to transmit a one. However, if that is not the case, then you are correct that it is more energy-efficient to transmit more of one or the other.

If the probability of a 1 symbol is p and the probability of a 0 symbol is 1 − p, then the average amount of information (entropy) communicated per bit is −_p_ log p − (1 − p) log(1 − p). This function is zero at p = 0 and p = 1, and has a maximum of log 2 (one bit of information) at p = 1/2.

Let E₀ and E₁ be the amount of energy needed to transmit a 0 and 1, respectively; then let r = E₀ / (E₀ + E₁). For example, if the energy for the symbols is equal then r = 1/2, and if a 1 requires twice as much energy as a 0 then r = 1/3.

The average amount of energy needed to transmit a symbol is proportional to r(1 − p) + (1 − r)p, and the average amount of energy needed to transmit one bit of information (on average) is [r(1 − p) + (1 − r)p] / [−_p_ log p − (1 − p) log(1 − p)]. This expression has a minimum when r / (1 − r) = log(p − 1) / log p. The graph of r versus p looks like:

So, if 0s take less energy (r → 0), then you should transmit fewer 1s (p → 0) and vice versa.

For example, in the scenario where a 1 costs twice as much as a 0 (r = 1/3), then p = (3 − sqrt(5)) / 2 ≈ 0.3819…

Answer 6

It's only true if your consider OOK modulation. In OOK, a signal emission corresponds to a 1 and no signal to a 0. However it's not widely used because of hardware complications.

With BPSK modulation, a 0 is send as a -1. It makes hardware design easier as you always have the same energy level to manage. => For the emission, you can make your amplifier work near compression point. => For the reception, you just have to monitor the rms power ( -1×-1 = 1x1 = 1) to adjust your gain.

With higher order modulation, rms power will again vary but in a better way than OOK.

It's as always a balance to find.