CS144 7-3 Clocks and Clock Recovery
Introduction to Computer Networking, Stanford University
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Unit 7: Lower Layers 중 Clocks and Clock Recovery 파트 수업 필기 정리
Clocks
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data is transmitted using a 'clock'
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the receiver needs to know when to sample the arriving data
when the clocks are running at different speed- how different are these clocks?
- units of parts per million
- ± 100 ppm = ±0.01%but usually the clocks are not drifting quickly, so offsets are relatively fixed
- how different are these clocks?
Asynchronous Communication
- make sure the packet not too long (control p)
- if we know the tolerants between the clocks to reliably decode the packet
Synchronous Communication
used in more practical situations
- Encoding the clock with the data
- Recovering the clock
- Getting the data into the receiver's clock "domain"
- Sender sends 10mega bits per second (using 10MHz clock) through network link
- in Transmit Clock Domain, Clock Recovery Unit recovers the 'sender's clock'
- put the data into Elasticity Buffer
- Receiver(RX) processes it with its own clock
How to send clock info encoded?
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make sure there's sufficient transitions in it,
otherwise the clock recovery unit could think it as half or doubled signal
Manchester (En)Coding
- 0 at downwards transition
- 1 at upwards transition
- d.c. balance? → set a threshold voltage level between 0 and 1 transition, so the recovery unit distinguishes transitions better
- But in this way, we need to insert more transitions than we need (doubling the bandwidths in the worst case)
4B5B Encoding
4-bit data = 4 bits of original data
5-bit data = 5 bits of encoded data
- this encoding method guarantees that there are some certain transitions
- you can always be sure that outgoing data has enough transition in it
- Fewer transition than Manchester coding, but people have developed the circuits so now it's most commonly used
Problem Set
Question 1
An infrared remote control uses asynchronous communications to send 64 bits of data using a 1MHz clock. The clocks at the two ends differ by at most 10kHz.
(10kHz is 1% of 1MHz. 1% = 10,000 ppm)
True Statement
If the receiver’s clock is 10kHz faster than the transmitter’s clock, then we will count a bit twice.
Question 2
True statements about Manchester coding?
Answers
- Used to introduce transitions between 0 and 1 to aid clock recovery
- Guarantees a transition during every bit time
Question 3
You friend discovers another form of block coding: 4B/6B, and she asks you to compare it Manchester coding and 4B/5B. Which of the following might you discover in your analysis?
Answers
- The overhead of 4B/6B will be less than Manchester coding but more than 4B/5B.
- 4B/6B will allow for more control codes than 4B/5B and Manchester coding.