As shown, 100 base-t Ethernet link adopted after 3 coding of MLT - why is only about 31.25 Mhz bandwidth?

The yellow line in figure part said: MLT - 3 cycles can send 4 bit data, combining three waveform in the figure, where reflected a 4 bit data to a cycle?

CodePudding user response:

Although not didn't read a book, but the text already wrote have been very clear, have a beating, the data without data still remains the same, the following code is obviously jumped 4 times, that is, there are four packets,
For half of the original also is easy to understand, in the middle of the data is 1, if the next packet without data recovery is 0, then it is sent 4 bit beat eight times, and the back, next package if no data is just keep still, a few packets to beat a few times, cut by half

CodePudding user response:

Cycle of 5-10100 is a cycle of 11100 and for a cycle, to professional bosses help solve

CodePudding user response:

For the last post, actually slowly understand the process of evolution, the first tube he did ya of data all shake up to me, didn't have data to shake the data also shake, if for 5 bits for one cycle, he want to shake a 5,
The following improvements, have data to shake, not data recovery 0, continue to send or no data has been unchanged, 5 bit sent two packets, for example on the graph as long as the vibrating 4, data or for a long time to send to their advantage, shaking the magnitude of the little,
Backmost more lazy, as long as there is data I have been shaking, as long as there is data I shake it, 5 bit, for example, until meet two without packet, 5 bit2 packets I as long as the two vibrating with respect to ok, or, the better the longer data packet, niche humble opinion when didn't see the

CodePudding user response:

A cycle refers to the above A B C D E F G which paragraph?

CodePudding user response:

reference 1/f, ice cream, jelly and response:

although not didn't read a book, but the text already wrote have been very clear, have a beating, the data without data still remains the same, the following code is obviously jumped four times, that is, there are four packets,
For half of the original also is easy to understand, in the middle of the data is 1, if the next packet without data recovery is 0, then it is sent 4 bit beat eight times, and the back, next package if no data is just keep still, a few packets to beat a few times, shorten the half

A cycle refers to the above A B C D... Which period?

CodePudding user response:

Don't confuse the clock frequency and the frequency of data
Ethernet and protocol is detection rate
The two sides confirmed the physics and the synchronous clock frequency
Figure in the clock frequency is the same this is physical agreed by both sides is not in the actual transmission

31.25 M is the cycle of data

CodePudding user response:

refer to 6th floor doufuxadf response:

don't confuse the clock frequency and the frequency of data
Ethernet and protocol is detection rate
The two sides confirmed the physics and the synchronous clock frequency
Figure in the clock frequency is the same this is physical agreed by both sides is not in the actual transmission

31.25 M is the cycle of data

In figure 2 a waveform (NRZ) is 125 MB/s, 62.5 M sending the data of frequency, understand it?

The third wave (MLT - 3) the frequency of a 31.25 Mhz data sent this how to understand or how to see out?

CodePudding user response:

refer to 7th floor myomnipotence response:

Quote: refer to the sixth floor doufuxadf response:

don't confuse the clock frequency and the frequency of data
Ethernet and protocol is detection rate
The two sides confirmed the physics and the synchronous clock frequency
Figure in the clock frequency is the same this is physical agreed by both sides is not in the actual transmission

31.25 M is the cycle of data

In figure 2 a waveform (NRZ) is 125 MB/s, 62.5 M sending the data of frequency, understand it?

The third wave (MLT - 3) the frequency of a 31.25 Mhz data sent this how to understand or how to see out?

Don't be misled by figure

Clock is a stopwatch
A switch stopwatches 1
B switches stopwatches 2
What is the speed of the AB company good stopwatch timing starts when
The rest of the AB stopwatch counting each see
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Bandwidth
The car (1) clock pickup (2 parking 2 clock) heavy card (4 clock)

The speed is 100
Is the car traffic 100 bandwidth 100
Are small goods traffic bandwidth 50 fifty
25 is heavy traffic bandwidth 25
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CodePudding user response:

This is what the book ah, the feeling is the electrical engineering and, although the underlying stuff in less than, still want to buy look at learning how to learn

CodePudding user response:

references 9/f, ice cream, jelly and response:

this is what the book ah, feeling is the electrical engineering,, although the underlying stuff in less than, still want to buy look at learning how to learn

The name of the book illustration network hardware, three rounds of xian a book