Fiber capacity has received increasing attention as the amount of transmitted data has increased dramatically. At the end of the last century, DWDM technology began to be used in trunk communications and rapidly spread. Although the capacity of the DWDM system at that time was only 40 × 2.5G, the number of channels supported by DWDM in the laboratory even exceeded 1,000 waves, and the single-wave channel rate soared to an astonishing 160G (the super 1,000 waves and the single-wave 160G are two independent events). It is widely believed that with WDM technology, the capacity of optical fiber may be unlimited.
Today, the capacity of the DWDM system has also been gradually upgraded to 80 × 200G, and 80 × 400G systems are gradually being commercialized. However, at this time, researchers have found that the capacity of optical fiber is more and more difficult to break through, whether it is the number of channels supported by the DWDM system, or the rate of a single wave is limited by certain factors.
Calculation Of Fiber Capacity
The capacity of an optical fiber can usually be calculated in two ways.
1. Dimension of signal multiplexing
From the dimension of signal multiplexing, the capacity C of an optical fiber is equal to the product of the number of channels N and the single channel rate. For coherent communication systems, the single-channel rate is related to the number of polarizations of the optical signal (in both x and y directions), the baud rate Rs (symbol rate), and the number of bits per symbol log2 (M). The capacity C of an optical fiber can be calculated by equation 1.
C = 2-N-Rs-log2 (M)
M in Eq. 1 is the number of modulation orders in multi-order modulation. However, Eq. 1 ignores some constraints, for example, the number of channels N and baud rate Rs are limited by the channel bandwidth B, and the number of modulation orders M is limited by the signal-to-noise ratio SNR. Therefore, the result calculated according to Eq. 1 may be different from the actual situation.
2. Shannon’s limit
According to Shannon’s second theorem, the capacity of coherent optical communication system with polarization multiplexing can be calculated by Eq. 2. In Eq. 2, B is the channel bandwidth and SNR is the signal-to-noise ratio.
C = 2-B-log2 (1+SNR) Equation 2
In Eq. 2, C/B = 2-log2 (1+SNR), which represents the spectral efficiency in bits/s/Hz.
Strictly speaking, the capacity of an optical fiber can only be calculated by Eq. 2. However, Eq. 2 is not easy to understand, therefore, Eq. 2 needs to be combined with the multiplexing dimension of the signal in Eq. 1 to make it easier to understand.
What Is Shannon’s Theorem?
1. Shannon’s first theorem
Shannon’s first theorem, also known as the distortion-free source coding theorem or the variable-length code source coding theorem, is to transform the original source symbols into new code symbols so that the code symbols obey the equal probability distribution as much as possible so that the amount of information each code symbol carries is maximized, and thus the source information can be transmitted with as few code symbols as possible.
Let the discrete memoryless source X contain N symbols {x1,x2,…,xi,… ,xN}, and the source emits K heavy symbol sequences, then this source can emit N^k different symbol sequence messages, where the probability of occurrence of the jth symbol sequence message is PKj, and the length of the binary code group resulting from the encoding of its source is Bj, and the average length of the code group B is
B=PK1B1+PK2B2+…+PN^kBN^k
When K tends to infinity, the relationship between B and H(X) is B/K=H(X) (K tends to infinity)
2. Shannon’s second theorem
Shannon’s second theorem is also known as the noisy channel coding theorem. When the information transmission rate of a channel does not exceed the channel capacity, an arbitrarily high transmission reliability can be achieved by using a suitable channel coding method, but if the information transmission rate exceeds the channel capacity, reliable transmission is not possible.
Let a certain channel have r input symbols, s output symbols, and a channel capacity of C. When the information transmission rate of the channel R
3. Shannon’s third theorem
Shannon’s third theorem is known as the distortion-preserving source coding theorem under the distortion-preserving criterion or the lossy source coding theorem. Provided that the code length is sufficiently long, it is always possible to find a source coding such that the coded information transmission rate is slightly greater than the rate distortion function and the average distortion of the code is not greater than a given permissible distortion, i.e., D'<=D. Let R(D) be the information rate distortion function of a discrete memoryless source, and choose a finite number of distortion functions such that for any permissible average distortion D>=0, and any small a>0, and any sufficiently long code length N, there must exist a source code W with M<=EXP{N[R(D)+a]}, and an average distortion of the coded code D'(W)<=D+a.
Impact Of The Shannon Limit On Fiber Capacity
1. Effect of channel bandwidth B
The product of the number of channels N and the baud rate Rs in Eq. 1 has a relationship with the channel bandwidth B.
1.1 Number of channels N
The maximum value of the incoming optical power is limited by the nonlinear effect of the fiber, so the number of channels N carried in the same fiber is also limited.
1.2 Baud rate Rs
The enhancement of baud rate Rs will be constrained by 2 aspects.
(1) Limitations of electrical bottlenecks. The improvement of baud rate will lead to a series of intractable problems such as electrical signal loss, power dissipation, electromagnetic interference, and so on. At this stage, the highest baud rate that can be realized by commercial chips is about 128Gbaud, which can take into account the advantages of capacity, power consumption and cost.
(2) The limitation of channel spacing S. The channel bandwidth B divided by the number of channels in the channel N is the channel spacing S. According to Nyquist’s first criterion, the baud rate Rs is always less than or equal to the channel spacing S. Some of the baud rates and channel spacings of the ultra-100G DWDM system are shown in Table 1.
When the single channel rate is certain, increasing the baud rate can correspondingly reduce the modulation order of signal coding, thus increasing the transmission distance of the system.
1.3 Channel bandwidth B
According to the definition of the channel spacing S, S = B/N, then N = B/S, substituting into Eq. 1, we have:
C = 2-B-Rs/S-log2 (M) Equation 3
If η = Rs/S is defined, then:
C = 2-η-B-log2 (M) Equation 4
i.e., N-Rs = η-B (η ≤ 1), indicating that the variation of the number of channels N and the baud rate Rs is always limited by the channel bandwidth B.
Ideally, η = 1, then Eq. 4 can be written as:
C = 2-B-log2 (M) Eq. 5
Eq. 5 and Eq. 2 are formally identical, with 2-log2 (M) denoting the spectral efficiency. From Eq. 5, it can be seen that the capacity of the fiber is proportional to the channel bandwidth B.
2. Effect of spectral efficiency
Spectral efficiency is realized by using different signal codes. The signal coding of ultra 100G coherent optical communication system usually includes: QPSK, 8QAM, 16QAM …… The number in front of QAM is the modulation order, and the modulation order of QPSK is 4. Higher order modulation requires higher channel signal-to-noise ratio, as shown in Fig. 1, and accordingly, the maximum transmission The higher the modulation order, the higher the signal-to-noise ratio of the channel, as shown in Figure 1.
As can be seen from the figure, the spectral efficiency curve becomes flatter and flatter as the SNR increases. So, although the SNR will continue to increase as the technology advances, the impact on fiber capacity is becoming more and more limited.
Put At The End
The impact of the Shannon limit on the capacity of ordinary simplex singlemode fiber (G.652.D or G.654.E) was analyzed above, and there is limited room for capacity enhancement of ordinary simplex singlemode fiber, both in terms of channel bandwidth and spectral efficiency, but the growth rate of the communication service is still strong, and the capacity of the ordinary simplex single-mode fiber is being challenged by rapid growth of the communication service.
The current optical communication system mainly realizes information transmission through amplitude modulation/demodulation of optical signals because the technology for realizing such modulation/demodulation is mature and the cost is relatively low.
To achieve higher-capacity information transmission through optical signals, theoretically, higher-order modulation/demodulation techniques can be used, such as 64th-order, 256th-order, and 512th-order Quadrature Amplitude Modulation (QAM) of the optical wave, but that would require: i. A spectrally very pure optical carrier (what kind of limitations are encountered by such optical carriers when they are transmitted in optical fibers has not been documented in a very adequate research) ii. Extremely weak optical signal modulation / demodulation techniques and devices, these two conditions are currently no relatively low-cost viable solutions.
In addition, under the current conditions, with very large resources to achieve in the single optical wave conditions to enhance the capacity, as opposed to increasing the number of channels transmitted in the optical fiber, increase the number of optical fiber cores and other measures, cost and reliability are not appropriate.
As the service life of fiber optic cables is up to 20 years or more, the capacity of next-generation optical fiber should be at least 10 times higher than that of the existing ordinary simplex singlemode fiber in order to meet the development of communication services. What kind of fiber can be considered next-generation fiber? Multi-core fiber counts? We will see.