# Can we break the Shannon capacity?

I have a friend working in wireless communications research. He told me that we can transmit more than one symbol in a given slot using one frequency (of course we can decode them at the receiver).

The technique as he said uses a new modulation scheme. Therefore if one transmitting node transmits to one receiving node over a wireless channel and using one antenna at each node, the technique can transmit two symbols at one slot over one frequency.

• I am not asking about this technique and I do not know whether it is correct or not but I want to know if one can do this or not? Is this even possible? Can the Shannon limit be broken? Can we prove the impossibility of such technique mathematically?

• Other thing I want to know, if this technique is correct what are the consequences? For example what would such technique imply for the famous open problem of the interference channel?

Any suggestions please? Any reference is appreciated.

• It sounds like quite a leap from "transmitting two symbols instead of one in some unit time" to "breaking the Shannon limit". Did your researcher friend say anything about disproving Shannon? – Nick T Mar 24 '14 at 21:24
• Signal constellations definitely are needed to improve throughput (of information). You're not getting anywhere near the Shannon limit without them. They're nothing new... and Shannon completely considered them, forward error correction, and a host of other factors when deriving his limit. – Ben Voigt Mar 25 '14 at 13:41

Most certainly not. While there has been some claims to break Shannon here and there, it usually turned out that the Shannon theorem was just applied in the wrong way. I've yet to see any such claim to actually prove true.

There are some methods known that allow for transmission of multiple data streams at the same time on the same frequency. The MIMO principle employs spatial diversity to achieve that. Comparing a MIMO transmission in a scenario that offers high diversity with the Shannon limit for a SISO transmission in an otherwise similar scenario might actually imply that the MIMO transmission breaks Shannon. Yet, when you write down the Shannon limit correctly for the MIMO transmission, you again see that it still holds.

Another technique to transmit on the same frequency at the same time in the same area would be CDMA (Code Division Multiple Access). Here, the individual signals are multiplied with a set of orthogonal codes so that they can be (perfectly in the ideal case) separated again at the receiver. But multiplying the signal with the orthogonal code will also spread its bandwidth. In the end, each signal employs much more bandwidth than it needs and I've never seen an example where the sum of the rates was higher than Shannon for the whole bandwidth.

While you can never be sure that breaking Shannon is actually impossible, it is a very fundamental law that stood the test of time for a long time. Anyone claiming to break Shannon has most likely made a mistake. There needs to be overwhelming proof for such a claim to be accepted.

On the other hand, transmitting two signals on the same frequency at the same time in the same area is easily possible using the correct method. This is by no means an implication that Shannon is broken.

• Interestingly, when I saw the MIMO technique, I thought the same, that it looked like a way to break Shannon capacity, but I suspected that Shannon limit is not so easily broken. Could you explain further, or provide a link, how Shannon limit applies in MIMO? I'd love to read about it. Thanks. – siritinga Mar 26 '14 at 20:19
• On breaking Shannon, it's also possible that they made certain assumptions. For example, compressed sensing says that if the signal is sparse in some basis, then the signal can be reconstructed perfectly after sampling at less than the Nyquist frequency. en.wikipedia.org/wiki/Compressed_sensing#Overview – Scott Mar 31 '14 at 1:18
• I don't know about telecom modulation, but in data compression I think it can be beaten. Shannon's limit basically states that to encode X values, you need at least X bits, but I think if you were to move the window around, you could reduce it from X. – MarcusJ Jun 24 '15 at 16:30
• You can't really beat it without cheating (e.g. one-to-one codes as used in W. Szpankowski and S. Verdú, "Minimum Expected Length of Fixed-to-Variable Lossless Compression Without Prefix Constraints," IEEE Trans. on Information Theory , vol. 57, no. 7, pp. 4017-4025, July 2011). – Batman Jul 31 '15 at 19:35

The capacity of a channel should be viewed as analogous to the speed limit on a highway. It is possible to travel at a speed greater than the posted limit on a highway but it is not possible to achieve good gas mileage while doing so. Similarly, it is possible to transmit data at rates higher than the capacity of the channel (in fact, unlike highways, there are no cops who will try to stop you from doing so) but it not possible to transmit at such high rates with very small error probability. If we don't care about BER, it is possible to send "data" over the channel at arbitrarily high rate. Of course, most of what the receiver will get is sheer garbage, but we have agreed that BER is not important. For example, in a pulse amplitude modulation (PAM) system with fixed maximum transmitter power, we can use binary modulation to transmit (maximum power) pulses of amplitude $\pm A$ in each signaling interval of duration $T$ and achieve a data rate of $T^{-1}$ bps, or we could use quaternary modulation and transmit pulses of amplitude $\pm A$ or $\pm A/3$ to get a data rate of $2T^{-1}$ bps or octonary modulation with pulses of amplitude $\pm A$, $\pm \frac{5}{7}A$, $\pm \frac{3}{7}A$, $\pm \frac{1}{7}A$ to get a data rate of $3T^{-1}$ bps, etc. The BER gets progressively worse as the number of levels increase and they are spaced closer and closer together, but hey, we agreed that BER is not a concern; data rate is.

What information theory tells us is that if we restrict ourselves to communication schemes that have data rates smaller than the channel capacity, then we can achieve any given BER no matter how small. The schemes will be very complex, exorbitantly expensive to implement, and have long delays (latency) if the desired BER is very small, but they exist and can be found (though the search might require immense effort). But the capacity of a channel is not like the velocity of light in physics: a fundamental limit that cannot be exceeded. It is possible to transmit at rates higher than the capacity, just not reliably.

• I get what you are saying, but I think it would be more accurate to say that the information cannot go above the Shannon limit. Sure, the data goes up if you accept errors but the information either stays the same or, much more likely, goes down. – Jim Clay Mar 24 '14 at 20:28
• The strong converse for the channel coding theorem for DMC's tells us that the error probability goes to 0 exponentially at rates below capacity and to 1 exponentially at rates above capacity in $n$, so the capacity result is sharp in a nice sense. I guess it should also be stated that zero error capacity (combinatorial problem) and Shannon capacity (probabilistic problem) aren't the same thing. – Batman Mar 25 '14 at 3:55
• This should have been the accepted answer. – QMC Jun 7 '17 at 1:52
• Hi: Does anyone know a "classic" paper or book that explains this material in a reasonable way for someone with zero background in it ? ( background is statistics ). thanks. – mark leeds Jul 11 '18 at 14:35

I know of 3 ways to exceed Shannon -

1) MIMO exceeds Shannon. Technically each MIMO channel is limited by Shannon, but the sum of the channels exceeds the limit. The practical limit is the ability to distinguish each MIMO channel.

2) Dr. Solyman Ashrafi (CTO at MetroPCS) owns a patent for a technique using naturally orthogonal wavelets (or Hermite functions), and has assigned it to his company called QuantumXtel. Each wavelet is bound by Shannon, but you can stack wavelets. There are some issues to be worked out, but UTD made a prototype some years ago. I'm not sure what's going on with that now.

3) Dr. Jerrold Prothero owns a patent for a technique using non-periodic symbols, and has started to company called Astrapi to develop them into a practical solution. He claims that Shannon's Law is incomplete because it only considers periodic functions, and has created a new theorem (that incidentally reduces back down to Shannon in the case of periodic only functions). The paper is available for peer review. The new function is based on slew rate and sampling rate, and may allow far more data to be passed than currently.

Who knows? Maybe one of these will actually work. At least no-one here is a kook.

The Shannon Capacity is derived by applying the well known Nyquist signaling. In the case of a frequency selective channel, it is known that OFDM is a capacity achieving strategy. The OFDM applies the conventional Nyquist signaling.

In early 1970's Faster than Nyquist (FTN) signaling is motivated by Mazo to allow sending more than 1 symbol per symbol period (i.e., implicitly to get a capacity higher than the Shannon limit). And it is stated that approximately 2X Capacity can be achieved with FTN.

Recently one work which is an orthogonal FTN (OFTN) is suggested which aims to get a capacity higher than the conventional Shannon's Capacity. However, this work is still valid for the following cases

1. Frequency selective channel with i.i.d multipath taps (L) and moderate to high SNR. For fixed SNR, the gap between OFDM and OFTN is higher for higher L. The complexities of OFTN and OFDM are somehow comparable.
2. The receiver must have at least L antennas.

I dont think we beat the Shannon Limit; but spectral efficiency can certainly be improved using coding techniques - as proved by higher data rates in 4G and 5G