复旦大学：《计算机网络 Computer Networking》课程电子教案（PPT课件讲稿）06 Physical Layer（Cont）& Data Link Layer

15-441 Lecture 6 Physical Layer( Cont)& Data link layer Based on slides from previous 441 lectures ecture 6 15-441e2008

Lecture 6 15-441 © 2008 15-441 Lecture 6 Physical Layer (Cont) & Data Link Layer Based on slides from previous 441 lectures 1

Last time Physical Layer Modulation Application Bandwidth Presentation Session Nyquist/Shannon Transport Multiple×ing Network M edla Datalink Physical

Last Time • Physical Layer – Modulation – Bandwidth – Nyquist/Shannon – Multiplexing – Media Application Presentation Session Transport Network Datalink Physical

Today(& Tomorrow) 1. Physical layer Application (Encoding Presentation 2. Datalink layer Session introduction Transport framing error Network coding, MAC, Datalink switched networks Physical 3. Broadcast-networks home networkin

Today (& Tomorrow) 1. Physical layer. (Encoding) 2. Datalink layer introduction, framing, error coding, MAC, switched networks. 3. Broadcast-networks, home networking. Application Presentation Session Transport Network Datalink Physical

From Signals to Packets Packet Transmission Sender Receiver Packets 010001010101110010101010101110111000000111101010111010101010110101101 Header/Body Header/Body Header/Body Bit stream 00101110 "Digital" Signal Analog Signal ecture 6 15441@2008

From Signals to Packets Analog Signal “Digital” Signal Bit Stream 0 0 1 0 1 1 1 0 0 0 1 Packets 0100010101011100101010101011101110000001111010101110101010101101011010111001 Header/Body Header/Body Header/Body Sender Receiver Packet Transmission Lecture 6 15-441 © 2008 4

How Encode? Seems obvious, why take time with this? 0 0 00 85 0 -85 ecture 6 15441@2008

How Encode? •Seems obvious, why take time with this? Lecture 6 15-441 © 2008 V 0 .85 -.85 0 1 0 0 0 1 1 0 1 5

Why y Encode? 01 o 1 How many more ones? ecture 6 15441@2008

Why Encode? Lecture 6 15-441 © 2008 0 1 0 1 How many more ones? 6

Non-Return to zero(nrz 0 0 00 85 0 -85 1->high signal; 0-> low signal Long sequences of 1's or os can cause problems Sensitive to clock skew, i.e. hard to recover clock Difficult to interpret o' s and 1s ecture 6 15-441e2008 7

Non-Return to Zero (NRZ) •1 -> high signal; 0 -> low signal •Long sequences of 1’s or 0’s can cause problems: – Sensitive to clock skew, i.e. hard to recover clock – Difficult to interpret 0’s and 1’s V 0 .85 -.85 0 1 0 0 0 1 1 0 1 Lecture 6 15-441 © 2008 7

Why do we need encoding? Keep receiver synchronized with sender Create control symbols besides reqular data symbols E.g. start or end of frame, escape Error detection or error corrections Some codes are illegal so receiver can detect certain classes of errors Minor errors can be corrected by having multiple adjacent signals mapped to the same data symbol . Encoding can be done one bit at a time or in multi-bif blocks, e. g, 4 or 8 bits Encoding can be very complex, e.g. wireless ecture 6 15441@2008 8

Why Do We Need Encoding? •Keep receiver synchronized with sender. •Create control symbols, besides regular data symbols. – E.g. start or end of frame, escape, ... •Error detection or error corrections. – Some codes are illegal so receiver can detect certain classes of errors – Minor errors can be corrected by having multiple adjacent signals mapped to the same data symbol •Encoding can be done one bit at a time or in multi-bit blocks, e.g., 4 or 8 bits. •Encoding can be very complex, e.g. wireless. Lecture 6 15-441 © 2008 8

Non-Return to zero Inverted (NRZI) 01000 0 85 0 -85 1→ make transition:o→ signal stays the same Solves the problem for long sequences of 1's but not for o’s ecture 6 15441@2008 9

Non-Return to Zero Inverted (NRZI) •1 → make transition; 0 → signal stays the same •Solves the problem for long sequences of 1’s, but not for 0’s. V 0 .85 -.85 0 1 0 0 0 1 1 0 1 Lecture 6 15-441 © 2008 9

Ethernet Manchester Encoding 0 85 0 -85 1us Positive transition for o, negative for 1 Transition eve ycle communicates clock ecture 6 15441@2008 0

Ethernet Manchester Encoding •Positive transition for 0, negative for 1 •Transition every cycle communicates clock V 0 .85 -.85 0 1 1 0 .1s Lecture 6 15-441 © 2008 10