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Fall 2000 Datacom 1 1
OutlineHardware Building BlocksEncodingFramingError DetectionSliding Window Algorithm
Point-to-Point Links
Fall 2000 Datacom 1 2
Network Connections
Memory latency is more important than CPU speed
Fall 2000 Datacom 1 3
Encoding
Bits
NRZ
0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0
Fall 2000 Datacom 1 4
The Electromagnetic Spectrum
Fall 2000 Datacom 1 5
Cables and Fiber Link Bandwidths
Cable Typical Bandwidth Length
Category 5 twisted pair
10-100 Mbps 100 m
Thin-net Coax 10-100 Mbps 200 m
Thick-net Coax 10-100 Mbps 500 m
Multimode fiber 100 Mbps 2 km
Single mode fiber 100-2400 Mbps 40 km
Fall 2000 Datacom 1 6
Leased Line Service BandwidthService Bandwidth
DS 1 1.544 Mbps=24*64 Kbps
DS 3 44.736 Mbps=30 * DS1
STS-1 (Synchronous Transport Signal also Optical Carrier (OC))
51.840 Mbps
STS-3 155.250 Mbps =3* STS-1
STS-12 622.080 Mbps
STS-24 1.244160 Mbps
STS-48 2.488320 Mbps
Fall 2000 Datacom 1 7
Services at the home
Service Bandwidth
POTS 28.8-56 Kbps
ISDN 64-128 Kbps
xDSL 16 Kbps-55.2 Mbps
CATV 20-40 Mbps
Fall 2000 Datacom 1 8
Asymmetric Digital Subscriber Line (ADSL)
Fall 2000 Datacom 1 9
Very high data rate DSL(symmetric)
Fall 2000 Datacom 1 10
Wireless Links
Project Orbit (km) Number of Satellites
Uplink freqs.
Downlink
Freq.
ICO 10,355 10 2170-2200 MHz
1980-2010 MHz
Globalstar 1,410 48 L-band S-band
Iridium 780 66 L-band L-band
Teledesic 1350 288 Ka-band Ka-band
Fall 2000 Datacom 1 11
Wireless LANs
• HIPERLAN (High Performance European Radio LAN--5.2 GHz and 17 GHz
• IEEE 802.11---2.4 GHz• PICONET
– Bluetooth (Ericsson, Nokia, IBM, Toshiba, Intel)
– 2.45 GHz, 1 Mbps, 10 M
Fall 2000 Datacom 1 12
Problem: Consecutive 1s or 0s
• Low signal (0) may be interpreted as no signal• High signal (1) leads to baseline wander• Unable to recover clock
Fall 2000 Datacom 1 13
Alternative Encodings
• Non-return to Zero Inverted (NRZI)– make a transition from current signal to encode a one;
stay at current signal to encode a zero– solves the problem of consecutive ones
• Manchester– transmit XOR of the NRZ encoded data and the clock– only 50% efficient.
Fall 2000 Datacom 1 14
Coding Examples
Fall 2000 Datacom 1 15
Encodings (cont)
• 4B/5B– every 4 bits of data encoded in a 5-bit code– 5-bit codes selected to have no more than one leading 0
and no more than two trailing 0s – thus, never get more than three consecutive 0s– resulting 5-bit codes are transmitted using NRZI – achieves 80% efficiency
Fall 2000 Datacom 1 16
Framing
• Break sequence of bits into a frame• Typically implemented by network adaptor
Frames
BitsAdaptor Adaptor Node BNode A
Fall 2000 Datacom 1 17
Approaches• Bit-Oriented Approach-High level Data Link
Control (HDLC)
– delineate frame with special pattern: 01111110
– problem: special pattern appears in the payload
– solution: bit stuffing
• sender: insert 0 after five consecutive 1s
• receiver: delete 0 that follows five consecutive 1s
Header Body
8 16 16 8
CRCBeginningsequence
Endingsequence
Fall 2000 Datacom 1 18
Bit stuffing example
Fall 2000 Datacom 1 19
Approaches (cont)
• Byte oriented Protocols• Byte Counting- based
– include payload length in header– e.g., Digital Data Communication Message Protocol (DDCMP)
– problem: count field corrupted– solution: catch when CRC fails
SY
N
Header Body
8 8 4214 168
SY
N
Cla
ss CRCCount
Fall 2000 Datacom 1 20
Byte stuffing example
Fall 2000 Datacom 1 21
Approaches (cont)
• Clock-based– each frame is 125us long– e.g., SONET: Synchronous Optical Network– STS-n (STS-1 = 51.84 Mbps)
Overhead Payload
90 columns
9 rows
STS-1Hdr STS-1Hdr STS-1Hdr
STS-3cHdr
Fall 2000 Datacom 1 22
Error Detection and Correction
• Two dimensional Parity• Internet checksum• Cyclic Redundency Check
Fall 2000 Datacom 1 23
Two Dimensional Parity
Fall 2000 Datacom 1 24
Cyclic Redundancy Check
• Add k bits of redundant data to an n-bit message– want k << n
– e.g., k = 32 and n = 12,000 (1500 bytes)
• Represent n-bit message as n-1 degree polynomial– e.g., MSG=10011010 as M(x) = x7 + x4 + x3 + x1
• Let k be the degree of some divisor polynomial– e.g., C(x) = x3 + x2 + 1
Fall 2000 Datacom 1 25
CRC (cont)
• Transmit polynomial P(x) that is evenly divisible by C(x) – shift left k bits, i.e., M(x)xk
– subtract remainder of M(x)xk / C(x) from M(x)xk
• Receiver polynomial P(x) + E(x)– E(x) = 0 implies no errors
• Divide (P(x) + E(x)) by C(x); remainder zero if:– E(x) was zero (no error), or
– E(x) is exactly divisible by C(x)
Fall 2000 Datacom 1 26
Selecting C(x)
• 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) contains 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 error 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
• See Table 2.6 on page 102 for common C(x)
Fall 2000 Datacom 1 27
CRC Example
Fall 2000 Datacom 1 28
CRC Example
Fall 2000 Datacom 1 29
CRC Example
Fall 2000 Datacom 1 30
CRC Example
Fall 2000 Datacom 1 31
Internet Checksum Algorithm
• View message as a sequence of 16-bit integers; sum using 16-bit ones-complement arithmetic; take ones-complement of the result.
u_shortcksum(u_short *buf, int count){ register u_long sum = 0; while (count--) { sum += *buf++; if (sum & 0xFFFF0000) { /* carry occurred, so wrap around */ sum &= 0xFFFF; sum++; } } return ~(sum & 0xFFFF);}