1. CHAPTER 2
HARDWARE BUILDING BLOCK ENCODING FRAMING ERROR DETECTION

2. HARDWARE BUILDING BLOCK
Node: Assume general-purpose (programmable) computers; e.g., workstations. Sometimes replaced with special-purpose hardware
Finite memory
Connects to network via a network adaptor
Fast processor, slow memory

3. HARDWARE BUILDING BLOCK
Link : Analog and Digital

4. Link Digital

5. Communication Modes
Simplex : only 1 way transmission, sender and receiver can not transmit data at the same time
Half duplex : 2 way transmission, but sender and receiver can not transmit data at the same time. If sender wants to transmit data then only sender’s path is active
Full duplex : 2 way transmission, both sender and receiver can transmit data at the same time

6. Link …cont’d Sometimes you install your own: Media Bandwidth Distance
Cat 5 UTP
10 – 100 Mbps
100 m
50-ohm coax (ThinNet)
200 m
75-ohm coax (ThickNet)
500 m
Multimode fiber
100 Mbps
2 km
Single-mode fiber
100 – 2400 Mbps
40 km

7. Link …cont’d Sometimes leased from the phone company: Services
Bandwidth
ISDN
64 Kbps T1
1,544 Mbps T3
44,763 Mbps
STS-1
51,840 Mbps
STS-3
155,251 Mbps
STS-12
622,080 Mbps
STS-24
1, Gbps
STS-48
2, Gbps

8. Encoding A. Overview Signals propagate over a physical medium.
Digital signals
Analog signals
Data can be either digital or analog; we're interested in digital data.
Problem: Encode the binary data that the source node wants to send to the destination node into the signal that propagates over to the destination node

9. Encoding
bits

10. Encoding NRZ (Non-Return to Zero) Problem: Consecutive 1s or 0s
Low signal (0) may be interpretted as no signal
High signal (1) leads to baseline wander
Unable to recover clock

11. Encoding Make a transition from the current signal to encode a one
NRZI (Non-Return to Zero Inverted)
Make a transition from the current signal to encode a one
stay at the current signal to encode a zero
solves the problem of consecutive ones.
But doesn’t solve the problem of consecutive zeros

12. Encoding
Manchester
Merging the clock with the signal by transmitting the exclusive-OR of the NRZ-encoded data and the clock
bit 1 encoded as high-to-low transition
bit 0 encoded as low-to-high transition
bit rate is half of baud rate, so the encoding considered only 50 % efficient

13. Encoding (cont’d)

14. Encoding
4B/5B
Idea: Every 4 bits of data is encoded in a 5-bit code, with the 5-bit codes selected to have no more than one leading 0 and no more than two trailing 0 (i.e., never get more than three consecutive 0s).
Resulting 5-bit codes are then transmitted using the NRZI encoding.
Achieves 80% efficiency.

16. Exercise 1.
Show the 4B/5B encoding, and the resulting NRZI signal for the following bit sequence :

17. Framing A. Overview Problem: Breaking sequence of bits into a frame
Must determine first and last bit of the frame
Typically implemented by network adaptor
Adaptor fetches (deposits) frames out of (into) host memory

18. Framing B. Byte-Oriented Protocol
Sentinel approach : a) BISYNC b)IMP-IMP

19. Byte-Oriented Protocol
Problem: ETX character might appear in the data portion of the frame.
Solution:
escape the ETX character with a DLE character in BISYNC
escape the DLE character with a DLE character in IMP-IMP

20. Byte-Oriented Protocol
Byte Counting Approach : DDCMP
Problem: Count field is corrupted (framing error).
Solution: Catch when CRC fails

21. Framing C. Bit-Oriented Protocol
HDLC: High-Level Data Link Control (also SDLC and PPP)
Delineate frame with a special bit-sequence:

22. Bit-Oriented Protocol
Bit Stuffing
Sender: any time five consecutive 1s have been transmitted from the body of the message, insert a 0.
Receiver: should five consecutive 1s arrive, look at next bit(s):
if next bit is a 0: remove it
if next bits are 10: end-of-frame marker
if next bits are 11: error

23. Framing D. Clock-based Framing SONET: Synchronous Optical Network
ITU standard for transmission over fiber
STS-1 (51.84 Mbps)
Each frame is 125ms long.

24. Exercise Suppose the following sequence of bits arrives over a link:
Show the resulting frame after any stuffed bits have been removed. Indicate any errors that might have been introduced into the frame

25. Error Control
Error detection
Error correction

26. Error Detection Original message : 10011010 C(x) = x3+ x2 + 1
message to be sent?

27. Cyclic Redundancy Check
Sender:
multiply M(x) by xk; example: x10 + x7 + x6 + x4 ( );
divide result by C(x) (1101);
Send = , since this must be exactly divisible by C(x);
Want to ensure that C(x) does not divide evenly into polynomial E(x).

28. Cyclic Redundancy Check

29. Cyclic Redundancy Check
Message to be sent is :
Receiver :
Divide by 1101

30. Cyclic Redundancy Check
Detection :
All single-bit errors, as long as the xk and x0 terms have non-zero coefficients.
All double-bit errors, as long as C(x) has a factor with at least three terms.
Any odd number of errors, as long as C(x) contains the factor (x + 1).
Any “burst” error (i.e sequence of consecutive errored bits) for which the length of the burst is less than k bits.
Most burst errors of larger than k bits can also be detected.

31. Cyclic Redundancy Check
Common polynomials for C(x):

32. Exercise
Suppose we want to transmit the message and protect it from errors using the CRC polynomial x3 + 1.
Use the polynomial long division to determine the message that should be transmitted
Suppose the leftmost bit of the message is inverted due to noise on the transmission link. What is the result of the receiver’s CRC calculation? How does the receiver know that an error has occurred?

33. Error Detection B. Two-Dimensional Parity
Idea : adds bit 1 to data for balancing the total sum of bit 1
Odd Parity : total sum of bit 1 of (data + parity) is odd
Even Parity : total sum of bit 1 of (data + parity) is even

34. Error Detection C. Internet Checksum
Idea : sum (ones complement) all the words which transmitted and send the summation result to destination node
The summation result called checksum
The destination node has to sum the received data, compares the result with checksum value sent by source node
If the result doesn’t match, it means that the data has error (include checksum)
Disadvantage : can’t detect miss position bits