Keywords: network engineering; Twisted pair; Fourier formula;
Suppose 1:
① Five types of unshielded twisted pair, length100m;
(2) signals with transmission speed of 100 MB/s on twisted pair;
③ the constant amplitude attenuation bandwidth of twisted pair is BHz (as shown in figure 3): b = c-a. ..
Deduction 1:
In Ethernet with transmission speed of 100 MB/s, the coding mode is 4 b/5 b (that is, in order to solve the synchronization problem in transmission, the actual input data of 4 b is actually coded by 5B [3] code group), so the number of state changes of the square wave signal actually transmitted on the communication line per second is 125M, that is,/kloc-.
If the signal changes 125M times per second, it takes 8/ 125m s for the state to change eight times.
According to the Fourier series formula, the period t of the first harmonic is: 8/125ms; The frequency f of the first harmonic is 125m/8 Hz.
After a square wave signal (the number of state changes per second is 125M) passes through five twisted-pair wires with the length of 100 m, the number of harmonics that can be attenuated by constant amplitude N 1 (bars) is as follows:
Conclusion 1:
Because it is assumed that 1 conforms to the wiring standard of five kinds of twisted pair, when the harmonic number passed by constant amplitude attenuation is N 1, the receiving equipment can recognize it.
Hypothesis 2:
① Five types of unshielded twisted pair with a length greater than 100 m;
(2) signals with transmission speed of 100 MB/s on twisted pair;
③ The constant amplitude attenuation bandwidth of twisted pair is narrowed to B/2 Hz.
Deduction 2:
With the increase of twisted pair length and the narrowing of constant amplitude attenuation bandwidth b, the number of constant amplitude attenuation harmonics that can pass through will also decrease. As shown in fig. 2, when N2 is less than 4, the output signal will be difficult to be recognized by the receiving device.
However, under assumption 2, it is difficult to ensure that constant amplitude attenuation passes through a sufficient number of harmonics.
According to the formula n = b/f, when b is reduced, only by reducing f (that is, reducing the transmission speed of signals on twisted pair) can N=B/f remain unchanged.
Hypothesis 3:
① Five types of unshielded twisted pair with a length greater than 100 m;
(2) transmitting the signal with the speed reduced to 10 MB/s on the twisted pair;
③ The constant amplitude attenuation bandwidth of twisted pair is narrowed to B/2 Hz.
Deduction 3:
10 MB/s and 100 MB/s are the most common data transmission speeds in Ethernet protocols, so we can consider reducing the data transmission speed to10 MB/s.
In 10 MB/s Ethernet, Manchester coding is adopted (that is, in order to solve the synchronization problem in transmission, the actual input data of 1 b is actually encoded with the code group of 2 b), so the number of state changes of the square wave signal actually transmitted on the communication line is 20M, that is, 20 MB/s.
If the signal changes 20M times per second, it takes 8/20 ms for the state to change eight times.
According to the Fourier series formula, the period t of the first harmonic is: 8/20 ms;
The frequency f of the first harmonic is 20M/8 Hz.
After the square wave signal (the number of state changes is 20M times per second) passes through this twisted pair, the number of harmonics N3 that can pass through with equal amplitude attenuation is: the number of harmonics N2 that can pass through with equal amplitude attenuation is:
Even if the constant amplitude attenuation bandwidth is reduced by half due to the increase of cable length, if the data transmission speed is reduced to 10 MB/s, the number of harmonics N3 that can pass under assumption 3 will be 3. 1 (under normal conditions).
In this case, the receiving device can completely recognize the signal.
Hypothesis 4:
① Five types of unshielded twisted pair with a length greater than 100 m;
(2) signals with transmission speed of 10 MB/s on twisted pair;
③ The constant amplitude attenuation bandwidth of twisted pair is narrowed to b/kHz;
④ Harmonic number N4 = N 1 passed by constant amplitude attenuation.
Deduction 4:
Conclusion 4:
Under hypothesis 4, even if the constant amplitude attenuation bandwidth of the cable is reduced to 16% (1/6.25), as long as the speed of data communication is reduced, the number of harmonics received under normal conditions (assuming 1) can be guaranteed to reach N 1, that is, the receiver can correctly identify it.
3 Conclusion
According to the above analysis, the author thinks that twisted pair can be used for non-standard wiring when designing lines with a distance less than 200 m, but the following methods should be adopted:
(1) In order to ensure the speed of the whole network, an adaptive switch with the interface of10100 MB/s can be used, but the switch port of this twisted pair connection is forced to be set to 10 MB/s in full duplex mode.
(2) Use five or more pairs of twisted-pair wiring, improve the construction technology, and try to avoid strong interference sources to reduce the decline of cable performance parameters.
In practical engineering, a computer connected by a twisted pair with a length of about 200 meters was tested:
① Internet access using IE browser is normal;
② Using ping program, the packet loss rate of 10 min is 0%. The results show that the above analysis based on Fourier is reasonable.
In addition, this method can also be applied to the early class 3 twisted pair or the situation that it cannot be used in the 10/ 100 m adaptive network because of the low cable quality and poor wiring technology, which can reduce the cost and the trouble of rewiring. Of course, if the length of twisted pair is further increased, the attenuation of total power by parameters such as DC impedance may cause the receiving equipment to fail to detect the signal, so this means of realizing communication by slowing down can only be used in a certain range.
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