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University of Delaware CPEG 419 1 Today Finish cable modem Cell phones Start link layer!

Today Finish cable modem Cell phones Start link layer!

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Today Finish cable modem Cell phones Start link layer!. TDM vs FDM. FDM. TDM. Time Division Multiplexing (TDM). channel 1. channel 2. channel n. FDM. Channel 1. Channel 2. Channel n. very high bandwidth (fiber). Head end. coaxial. fiber. Internet over Cable. - PowerPoint PPT Presentation

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Page 1: Today Finish cable modem Cell phones Start link layer!

University of Delaware CPEG 419 1

Today Finish cable modem Cell phones

Start link layer!

Page 2: Today Finish cable modem Cell phones Start link layer!

University of Delaware CPEG 419 2

TDM vs FDM

TDM FDM

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University of Delaware CPEG 419 3

Time Division Multiplexing (TDM)

channel 1

channel 2 channel n

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University of Delaware CPEG 419 4

FDM

Channel 1 Channel 2 Channel n

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Internet over Cable

Headend

very high bandwidth

(fiber)

fiber

coaxial

A group of homes share the cable. (As oppose to DSL.) The number of homes per group is around 500-2000.

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Cable Spectrum

Upstream5 – 42 MHZ

TVFM

54 MHz88MHz

108MHz

TVHBOshoppingDownstream

data

550MHz 750MHz

Each TV channel gets its own frequency.This is called frequency multiplexing

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Cable Modem

Why put downstream data at highend of spectrum? When they made the system they put in amplifiers that

could work at these high frequencies (why?) So the downstream was all set to go. But upstream

amplifiers had to be installed. Downstream data also uses FDM: 6MHz or 8MHz

channel width with QAM-64 (what is the data rate?) (This should be 6Mbaud – 8Mbaud)

But with overhead you only get 27Mbps. The upstream also uses FDM, but with QPSK (2

bits per symbol).

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How fast is V32?The phone system transmits 300 to 3400 Hz

So what bandwidth can we use. How fast can we send symbols?We can send signals at 3100*2 = 6200 baud.We need to send signal two ways, so we can at most send 3100 baud. But there needs to be a guard band (so they don’t overlap). So 2400 baud is used.

Use 2400 sample each way - duplex

Definition: a duplex connection means that we can send data in both directions at the same time.A simplex or half-duplex connection only sends data in one direction at a time.

So 2400 * 6 = 14400 bps

What is the baud rate?

V.34 2400 baud - with 12 data bits/symbolV.34 bis 2400 baud – with 14 data bits/symbolThat’s the fastest there is!

To get 56K you send at 4000 baud (if the phone system can handle it)

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DSL

VoicePOTS (stands for ?)(channel 0)

empty

A total of 256 4kHz channels so 8 kBaud per channel

25kHz (channel 6)

Upstream

downstream

250 parallel channels: Each data channel uses QAM 16 (with 1 parity bit).

What is the maximum data rate?

The quality of each channel is monitored and adjusted. So channels may transmit at different speeds

channel 6+32

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Cable modems

On modem boot-up, the head-end tells the modem which channel (time and frequency) to use for upstream and downstream transmission.

Many users may share the same channel.

To share the same channel, cable uses time division multiplexing (TDM) and media access control (MAC).

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Back to Cable

Downstream is easy. The head-end just transmits packets as it wants. Each packet has a label, so the modem can detect that the packet belongs to it.

Upstream is hard. The upstream channel is shared. TDM is used, but still each TDM slot is shared

by many users. What happens if two users try to send

upstream data on the same channel?

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Cable: Upstream Media Access (link layer) Time is divided into minislot. It is possible to transmit 8 bytes in one

minislot. The modem asks the headend if it can transmit a packet. The headend responds with an acknowledgment and tells the modem

which minislots it can use. Problem: How can the modem sent the request without permission to use

minislots? Solution: on boot-up, the headend tells the modem which minislots it can

use for requesting minislots and the headend never allocates these minislots for upstream data.

Problem: These special control minislots are shared by many users (why?), so what happens if two users make a request at the same time?

Solution: If two users transmit at the same time, the signal cannot be understood by the headend and is ignored. Thus, no acknowledgement is made. The competing modems then wait a random amount of time and try again. It if fails again, then they wait a random amount of time again, but the maximum time they might wait is doubled.

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Cable MAC (media access control )modem headend

modem sends a request for upstream minislots

Headend gets request.Thinks about it.And sends acknowledgement with which minislots to use.

modem sends data

upstream request for bandwidth minislots

data does not overlap with request minislot

time

} minislot

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Cable MAC: ContentionTwo modems send request for bandwidth at the same time. The headend can figure out what

was transmitted and ignores itThe modems wait for the acknowledgement that will never arrive. The amount of time they wait is random.

This time the red modem gets through.

The blue modem tries again, but the green modem also sends a request.

Green gets throughNow the blue doubles the maximum random amount of time it waits before sending another request.

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Cable vs. DSL Cable could give higher bandwidth, but it might give

less. It depends on the number of users. If there are too many users in a group, the cable

operator has to put in a fiber and headend. That cost money, so they try not to do it.

DSL can promise 1Mbps down and 256kbps up, and you will likely get it.

Note that sharing bandwidth is much more efficient than assigning each user a fixed chunk.

The telephone system is very reliable. When was the last time you picked up the phone and there was no dial tone (major earthquake or huge storm). Cable will go down when the power goes out.

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Mobile Phone

FE

GA

D

CB

F

GE

FD

E

C

A

•One transmitter/receiver in each cell.•The base station transmit at very high power and has sensitive receivers. This way the cell phone can broadcast at low power and have a simply receiver.•The size of the cell depends on the transmitting power of the base station.•In first generation phones, there were 832 channels. Each channel was 30kHz. (which kind of multiplexing?)•Multipath reflections are a problem.•Each cell uses a different set of channels (why?)•What to do if there are more users in a cell?

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2nd Generation Mobile Phones Types of 2nd generation phone systems: D-AMPS, GSM, CDMA and

PDC. (PCS is second generation, but not a standard like these, just mean second generation)

All Digital Phone is recorded by phone, converted to digital and compressed.

The compressed signal (numbers) are sent to the base station. Now a single 30KHz channel can be shared by 3 users using TDM.

1 2 3 4 5 6from mobile to base

40ms

1 2 3 4 5 6from base to mobile

1850.01MHz

1930.05MHz

1 slot: 324 bits64 bits control101 bits of error correction159 bits of data

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GSM

Widely used outside the US (but some here as well).

Similar to the D-AMPS, but much higher bandwidth, 200kHz vs. 30kHz.

There are 992 time slots. Max data rate is 33.854 kbps. After

overhead and error correction, there is 13kbs of speech data.

D-AMPS is 8kbs of speech data.

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CDMA – Code Division Multiple Access

Verizon and Sprint uses CDMA. AT&T uses D-AMPS. D-AMPS and GSM use FDM and TDM. CDMA is

totally different. And is the best. All users transmit over the same frequencies at the

same time!?! Idea: each phone uses a different code. If you listen on the wrong code, it sounds like

noise. If you listen on the right code it sounds like the right signal (with the other transmissions sounding like noise).

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CDMA

Each bit is transmitted into m short time intervals called chips.

Each user has a unique m-bit chip sequence. To send a 1, the user sends the sequence. To

send a 0, the user sends the complement.

A (one): -1 -1 -1 +1 +1 -1 +1 +1A (zero): +1 +1 +1 -1 -1 +1 -1 -1

A: -1 -1 -1 +1 +1 -1 +1 +1

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CDMA

A: -1 -1 -1 +1 +1 -1 +1 +1B: -1 -1 +1 -1 +1 +1+1 -1C: -1 +1 -1 +1 +1 +1 -1 -1D: -1 +1 -1 -1 -1 -1 +1 -1

Send: A=1, B=1, C=0, D=1

codes

A(1): -1 -1 -1 +1 +1 -1 +1 +1B(1): -1 -1 +1 -1 +1 +1+1 -1C(0): +1 -1 +1 -1 -1 -1 +1 +1D(1): -1 +1 -1 -1 -1 -1 +1 –1Total:-2 –2 0 -2 0 -2 +4 0

received signal

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CDMA

Total: S = [-2 –2 0 -2 0 -2 +4 0]

(S A)/8 =(-2*-1 + -2*-1 + 0*-1 + -2*+1 + 0*+1 + -2*-1 + 4*+1 + 0*+1)/8=(2+2-2+2+4)/8=1

Idea: A, B, C and D are orthonormal set of vectors:

(A B)/8 = 0, (A C)/8 = 0, (A D)/8 = 0, (B C)/8 = 0, (B D)/8 = 0,(C d)/8 = 0,

orthogonal

(A A)/8 = 1, (B B)/8 = 1, (C C)/8 = 1, (D D)/8 = 1normalized

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FDM

Code is sin(wi t)

Send code*Value. (e.g., 14.967384 sin (wi t))

i

ii twV sinTotal sent signal =

How to decode?

The Fourier expansion is an orthogonal decomposition.Each channel sends one (or a few) elements in the decomposition.The Fourier transform can decompose the signal to get the components (channels).

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3G (third generation phones)

They expect more data than voice (we’ll see) Upper maximum data rates: 2Mbps while

stationary, 384kbps while walking, and 144kbps while driving. (we’ll see).

High-quality voice transmission (never! – blade runner)

Messaging (no sweat) Web surfing (ok) Multimedia (at 144kbps?)

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3G ideas

Seamlessly move from cell to cell. Currently, when you leave a cell, you are dropped but the transmitter and quickly find a new one (maybe).

GPRS (general packet radio service) - Packet network over the network.

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Packet Switched Network vs Circuit Switched Network

Circuit switched – each connection gets bandwidth (or time slot) and uses this at will (or not). The connection is maintained for the duration of the connection.

Packet switched – Every bit of data is clumped together (packet) and sent into a large network. The packet has an address that tells the network where the packet goes.

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Circuit switched

That piece of wire is yours for the call

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Packet Switched

More efficient because bandwidth is not wasted.

But each packet has to have address (so less data).

Each node has to compete to get bandwidth. This can lead to inefficiencies.

Suited to data. But gets tricky for multimedia or real-time transmissions.

So 3G will be mixed

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Circuit Switched

Inefficient Channel capacity dedicated for duration of

connection If no data, capacity wasted

Once a call starts, it will not be dropped.Set up (connection) takes timeOnce connected, transfer is transparentDeveloped for voice traffic (phone)

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Data Link Layer

So far, sending signals over transmission medium. Suppose I can send data, then what?

Data link layer: responsible for error-free (reliable) communication between adjacent nodes.

Functions: framing, error control, flow control, addressing (in multipoint medium).

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Flow Control

What is it? Ensures that transmitter does not

overrun receiver: limited receiver buffer space.

Receiver buffers data to process before passing it up.

If no flow control, receiver buffers may fill up and data may get dropped.

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Stop-and-Wait

Simplest form of flow control. Transmitter sends frame and waits. Receiver receives frame and sends ACK. Transmitter gets ACK, sends other frame, and

waits, until no more frames to send.

Good when few frames. Problem: inefficient link utilization.

In the case of high data rates or long propagation delays.

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Sliding Window 1

Allows multiple frames to be in transit at the same time.

Receiver allocates buffer space for n frames.

Transmitter is allowed to send n (window size) frames without receiving ACK.

Frame sequence number: labels frames.

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Sliding Window 2

Receiver ack’s frame by including sequence number of next expected frame.

Cumulative ACK: ack’s multiple frames.Example: if receiver receives frames

2,3, and 4, it sends an ACK with sequence number 5, which ack’s receipt of 2, 3, and 4.

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Sliding Window 3

Sender maintains sequence numbers it’s allowed to send; receiver maintains sequence number it can receive. These lists are sender and receiver windows.

Sequence numbers are bounded; if frame reserves k-bit field for sequence numbers, then they can range from 0 … 2k -1 and are modulo 2k.

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Sliding Window 4

Transmission window shrinks each time frame is sent, and grows each time an ACK is received.

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Example: 3-bit sequence number and window size 7

A B0 1 2 3 4 5 6 7 0 1 2 3 4... 0 1 2 3 4 5 6 7 0 1

2 3 40

1

20 1 2 3 4 5 6 7 0 1 2 3 4

0 1 2 3 4 5 6 7 0 1 2 3 4RR3

0 1 2 3 4 5 6 7 0 1 2 3 4

3456RR40 1 2 3 4 5 6 7 0 1 2 3 4

0 1 2 3 4 5 6 7 0 1 2 3 4

0 1 2 3 4 5 6 7 0 1 2 3 4

0 1 2 3 4 5 6 7 0 1 2 3 4 0 1 2 3 4 5 6 7 0 1 2 3 4

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Sliding Window (cont’d)

RR n acknowledges up to frame n-1.There is also RNR n, which ack’s up to

frame n-1 but no longer accepts more frames.

RNR shuts down the receive window and consequently the transmission window.

Need subsequent RR to re-open window.

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Piggybacking

When both endpoints transmit, each keeps 2 windows, transmitter and receiver windows.

Each send data and need to send ACKs.

When sending data, transmitter can “piggyback” the acknowledgment information.

When no data, send just the ACK.

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Duplicate ACKs

When no data, must re-send last ACK.

Duplicate ACKs: report potential errors.

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Error Detection

Transmission impairments lead to transmission errors: change of 1 or more bits in transmitted frame.

Transmission errors defined using probabilities: transmission medium modeled as a statistical system.

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Error Probabilities 1Definitions:

Pb probability of single bit error (bit error rate); constant and independent for each bit.

P1 probability frame received with no errors.

P2 probability frame received with 1 or more undetected errors.

P3 probability frame received with 1 or more detected bit errors, but no undetected ones.

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Error Probabilities 2

If no error detection mechanism, P3 = 0.

P1 = (1 - Pb)F and P2 = (1- P1), where F is size of frame in bits.

P1 decreases as Pb increases.

P1 decreases as F increases.

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Example64-kbps ISDN channel’s bit error rate is

less than 10-6. User requirement of at most 1 frame with undetected bit error per day. Frame is 1000 bits. In a day, 5.529 x 106 frames transmitted. Required frame error rate of 1/ 5.529 x 106,

or P2 = 0.18 x 10-6.

But Pb = 10-6, so P1 = (1-Pb)F = 0.999 and P2 = 1 - P1 = 10-3, which is >>> required P2

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Error Detection Schemes Transmitter adds additional bits for error

detection.Transmitter computes error detection bits

as function of original data.Receiver performs same calculation and

compares results. If mismatch, then error.P3 probability error detection scheme

detects error; P2 residual error rate or probability error goes undetected.

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Parity

Simplest error detection scheme.Append parity bit to data block.Example: ASCII transmission

1 parity bit appended to each 7-bit ASCII character.

Even parity: 8-bit code has even number of 1’s.

Odd parity: 8-bit code has odd number of 1’s.

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Parity Check

Example: transmitting ASCII “G” (1110001) using odd parity. Code transmitted is 11100011. Receiver checks received code and if odd

number of 1’s, assumes no error. Suppose it receives 11000011, then

detects error. NOTE: If more than 2 bits in error, may

not be detected.

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Cyclic Redundancy CheckCRC is one of the most effective and

common error detecting schemes.Let M be m-bit message, G (r+1)-bit

pattern. Transmitter appends r 0’s to M, 2r*M. Divide 2r*M by G and add remainder to 2r*M

forming T (m+r bits), which is transmitted. Receiver computes T/G; if remainder, then

error.

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CRC Example

Frame M 1010001101 = x9+x7+x3+x2+x0.

Pattern G 110101.Dividing (frame*25) by pattern results

in 01110.Thus T 101000110101110.Receiver can detect errors unless

received message Tr is divisible by G.

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CRC

Patterns are expressed as polynomials G(x).

Example: CRC-16 = x16+x15+x2+1 CRC-CCITT = X16+x12+x5+1

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CRC-Based Detection

If suitably selected polynomials, CRC can detect: All single-bit errors. All double-bit errors, as long as P(X) has

at least three 1’s. Any odd number of errors as long as P(X)

contains factor (X+1). Any burst error whose length is <=

sizeof(FCS).

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Error ControlMechanisms to detect and correct

transmission errors.Consider 2 types of errors:

Lost frame: frame is sent but never arrives. Damaged frame: frame arrives but in error.

Error control: combination of error detection, feedback (ACK or NACK) from receiver, and retransmission by source.

Coupled with flow control feedback.

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ARQ

ARQ: automatic repeat request.Works by creating a reliable data link

from an unreliable one.3 versions:

Stop-and-wait ARQ. Go-back-N ARQ. Selective-reject ARQ.

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Stop-and-Wait ARQSingle outstanding frame at any time.Simple but inefficient.Use of timers to trigger retransmission of

data or ACKs.2 types of errors:

Damaged or lost frame. Damaged or lost ACK.

Sequence numbers alternate between 0 and 1.

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Stop-and-Wait ARQ: ExampleSender ReceiverFrame 0

ACK1

Frame 1ACK 0

Frame 0

Timeout

Frame 0

ACK 1

Timeout

Frame 0

ACK 1 B discards duplicate.

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Go-Back-N ARQVariation of sliding window for error

control.Allows a window’s worth of frames to be in

transit at any time.RR: ack’s receipt of frame.REJ: negative acknowledgment indicating

the frame in error.Destination discards frame in error plus

subsequent frames.

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Go-Back-N ARQ ExampleS R S Rf0

f1f2

rr3f3

f4

f5rr4

Errorf6

rej5f7

f5f6

rr6 f7

5, 6, 7rexm.

f7

f0

f1rr0

rr(P bit=1)

rr2

f2

Timeout

Discarded

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Go-Back-N ARQ Issues

For k-bit sequence number, maximum window size is (2k-1). If window size is too large, ACKs may be

ambiguous: not clear if ACK is a duplicate ACK (errors occurred).

Example: 3-bit sequence number and 8 -frame window.Source transmits f0, gets back rr1, then

sends f1--f0, and gets back another rr1. ???

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Selective-Reject ARQ

Only frames transmitted are the ones that are NACK’ed (SREJ) or that timeout.

More efficient than Go-Back-N regarding amount of reXmissions.

But, receiver must buffer out-of-order frames.

More restriction on maximum window size; for k-bit sequence #’s, 2k-1 window.

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Example Data Link Layer Protocol

High-Level Data Link Control (HDLC) Widely-used (ISO standard). Single frame format. Synchronous transmission.

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HDLC: Frame Format

Flag: frame delimiters (01111110). Address field for multipoint links. 16-bit or 32-bit CRC. Refer to book (pages 176-185) for more

details.

8bits

8ext.

8 or16

variable 16 or32

8

flag address control data FCS flag

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Other DLL Protocols 1

LAPB: Link Access Procedure, Balanced. Part of the X.25 standard. Subset of HDLC. Link between user system and switch. Same frame format as HDLC.

LAPD: Link Access Procedure, D-Channel. Part of the ISDN standard.

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Other DLL Protocols 2

LLC: Logical Link Control. Part of the 802 protocol family for LANs. Link control functions divided between

the MAC layer and the LLC layer. LLC layer operates on top of MAC layer.

Dst.MACaddr

Src.MACaddr

FCSDst.LLCaddr

Src.LLCaddr

LLCctl. DataMAC

control

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Other DLL Protocols 3

SLIP: Serial Line IP Dial-up protocol. No error control. Not standardized.

PPP: Point-to-Point Protocol Internet standard for dial-up

connections. Provides framing similar to HDLC.

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Multiplexing

Sharing a link/channel among multiple source-destination pairs.

Example: high-capacity long-distance trunks (fiber, microwave links) carry multiple connections at the same time.

MU

X

...

DE

MU

X ...

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Multiplexing Techniques

3 basic types: Frequency-Division Multiplexing (FDM). Time-Division Multiplexing (TDM). Statistical Time-Division Multiplexing

(STDM).

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FDM 1

High bandwidth medium when compared to signals to be transmitted.

Widely used (e.g., TV, radio).Various signals carried simultaneously

where each one modulated onto different carrier frequency, or channel.

Channels separated by guard bands (unused) to prevent interference.

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FDM 2

Time

Frequency

1 2 N

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TDM 1

TDM or synchronous TDM.High data rate medium when

compared to signals to be transmitted.

Time

Frequency

12

N

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TDM 2

Time divided into time slots.Frame consists of cycle of time slots.In each frame, 1 or more slots

assigned to a data source.

1 2 N... 1 2 ... N

frame Time

U1 U2 ... UN

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TDM 3

No control info at this level.Flow and error control?

To be provided on a per-channel basis. Use DLL protocol such as HDLC.

Examples: SONET (Synchronous Optical Network) for optical fiber.

+’s: simple, fair.-’s: inefficient.

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Statistical TDM 1Or asynchronous TDM.Dynamically allocates time slots on

demand.N input lines in statistical multiplexer, but

only k slots on TDM frame, where k < n.Multiplexer scans input lines collecting

data until frame is filled.Demultiplexer receives frame and

distributes data accordingly.

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STDM 2

Data rate on mux’ed line < sum of data rates from all input lines.

Can support more devices than TDM using same link.

Problem: peak periods. Solution: multiplexers have some buffering

capacity to hold excess data. Tradeoff data rate and buffer size (response

time).