~yao Polytechnic University, Brooklyn, NY11201 Yao Wang ...©Yao Wang, 2006 EE3414: Final Review 3 Color Principle • How a human perceives color – Three types of cones sensitive

Yao WangPolytechnic University, Brooklyn, NY11201

http://eeweb.poly.edu/~yao

Final Term Review

©Yao Wang, 2006 EE3414: Final Review 2

Topics Covered After First Exam

• Image processing basics– Color perception and specification– Color quantization– Image noise removal, sharpening, edge detection

• Image coding– Transform– Prediction – JPEG and JPEG2000

• Analog communications– Modulation and multiplexing

• Digital communications– Digital modulation– Error detection and correction

• Communication networks• Wireless communications


Color Principle

• How a human perceives color– Three types of cones sensitive to red, green, and blue respectively

• How to generate different colors in display– By mixing three primary colors, Red, Green and Blue– The mixing following additive rules (frequency components add)– Know some simple mixing rule:

• R+G+B light in equal proportion = white• How to generate different colors in print

– By mixing three primary colors, Cyan, Magenta, Yellow– The mixing follows subtractive rule – Know some simple mixing rule:

• C+M+M paint in equal proportion = black• How is a color image stored

– Consists of three separate component images: ex. R,G,B– Convert to CMY+K for printing


Color Specification

• Color representation– Primary colors: RGB or CMY– Hue, saturation, intensity (luminance): HSI, YUV, YIQ– 8 bits for each color component, or 24 bits total for each pixel– Total of 16 million colors

• Color coordination conversion

−

=

BGR

YMC

255255255


Color Quantization

• Uniform scalar quantization– Each color component quantized independently using a uniform

quantizer• Vector quantization (adaptive quantization, adaptive palette

– Three color components quantized jointly– Finding the best set of colors in an image to represent all colors in

an image• Color dithering (perceptual quantization)

– Randomly perturb quantized color pixels to reduce the contour effect

• Know the advantages of different approaches. But actual computation for vector quantization and dithering is not required


Noise Removal

• How do averaging and weighted averaging filters work?– How to apply a filter mask

• How does the median filter work?• What method is better for additive Gaussian noise?• What method is better for salt-and-pepper noise?• What is the challenge in noise removal?

– Trade-off between noise removal and detail preserving


Example: Weighted Average

100100100100100

100195205200100

100200200195100

100203205200100

100100100100100

100100100100100

100156175156100

100175201174100

100158176156100

100100100100100

121

242

121

×161

100100100100100

100144166144100

100168200167100

100145167144100

100100100100100

111

111

111

×91


Example: 3x3 Median

100100100100100

100195205200100

100200200195100

100203205200100

100100100100100

100100100100100

100100195100100

100200200200100

100100200100100

100100100100100

Matlab command: medfilt2(A,[3 3])


Sharpening and Deblurring

• Sharpening:– Enhance edges– Use a high-emphasis filter– Implementation similar to noise removal with weighted

averaging filter, but filter coefficients differ

• Debluring– Model the bluring process as a low-pass filter– Apply inverse filter– Not required for the exam


Frequency Domain Characterization

• Know the desired frequency response for different filters– Noise removal: low pass– Edge detection: high pass– Sharpening: high emphasis

• Determination of the Fourier transform (2D DTFT) of given simple filters NOT required

• Should be able to perform linear filtering and median filters with given filters on small images


Image Coding

• What is linear transform– Basis images– Computation

• How to use transform for image coding– Transform, quantization, runlength coding

• JPEG and JPEG2000 standards


What is Linear Transform?

• Represent an image as the linear combination of some basis images and specify the linear coefficients.

+t1 t2 t3 t4


Basis Images of 8x8 DCT

Low-Low

Low-High

High-Low

High-High


Approximation by DCT Basis

Original

With 8/64Coefficients




Example: 4x4 DCT

( )

......,

111111111111

1111

41,

111111111111

1111

41,

1111111111111111

41,

1111111111111111

41

:yields using

3827.09239.09239.0

3827.0

21

821cos

815cos

89cos

83cos

21;

111

1

21

47cos

45cos

43cos

4cos

21;

0.9239-0.3827-0.38270.9239

21

87cos

85cos

83cos

8cos

21;

1111

21: are basis DCT 1D

0,k,21

42(k),

21

41(0),)12(

4*2kcos (k) Using

2,20,22,00,0

3210

,

−−−−−−

−−

=

−−−−−−−−

=

−−−−−−−−

=

=

=

−

−=

=

−−

=

=

=

=

=

≠====

+=

UUUU

uuU

uuuu

Tlkk,l

nk nu

π

π

π

π

π

π

π

π

π

π

π

π

ααπα

( )

( )

tscoefficienfrequency high some offcut webecause original, n theslower tha variesimage tedreconstruc The

0.050.310.680.610.951.631.381.110.951.631.91.440.861.381.541.09

S

:yieldsonly tscoefficien 2x2 topfromtion Reconstruc5.04619.001913.0

3034.05.00391.2001913.05.04619.0

1152.130793.05.4:)dct2(S)"" (using yields Matlab usingn calculatio Completing

5.0421221121131221

41

1221121013100221

,

111111111111

1111

41),(

5.44

18122112101310022141

1221121013100221

,

1111111111111111

41),(

e.g, yields, ),( Using

compute ,

1221121013100221

For

2,32,2

0,00,0

,,

,

=

−−−−

−−−

=

−=−=−−−+−++−++−−=

−

−−−−−−

−−

==

==−++++++++++++++=

−

==

=

−

=

T

SU

SU

SU

S

T

T

T

T

lklk

lk


Quantization of DCT Coefficients

• Use uniform quantizer on each coefficient• Different coefficient is quantized with different step-size (Q):

– Human eye is more sensitive to low frequency components– Low frequency coefficients with a smaller Q– High frequency coefficients with a larger Q– Specified in a normalization matrix– Normalization matrix can then be scaled by a scale factor (QP)


Default Quantization Table in JPEG Coder

9999999999999999

9999999999999999

9999999999999999

9999999999999999

9999999999996647

9999999999562624

9999999966262118

9999999947241817

For luminance For chrominance

The encoder can specify the quantization tables different from the default ones as part of the header information

lena256_gray.tif QP=1

QP=0.5 QP=2


What Should You Know

• How to perform 2D transform: forward and inverse transform– Manual calculation for small sizes, using inner product notation– For the exam: the basis vector/image will be given– Know how to reconstruct from selected coefficients specified through

“masks”• Why DCT is good for image coding

– Real transform, easier than DFT– Most high frequency coefficients are nearly zero and can be ignored– Different coefficients can be quantized with different accuracy based

on human sensitivity• How to quantize DCT coefficients

– A quantization matrix specifies the default quantization stepsize for each coefficient

– The matrix can be scaled using a user chosen parameter (QP)– Know how to determine quantized index and quantized value

for each coefficient given a quantization matrix


JPEG Standard

• The baseline JPEG is a DCT-based coder– Divide each image into 8x8 blocks– Each block undergoes DCT– DCT coefficients are quantized according to a quantization

matrix, scaled by a quantization parameter– Quantized coefficient indices are ordered into 1D array using

zig-zag ordering– The ordered coefficients are represented in symbols of (run-

length of zeros, non-zero value)– Each symbol is coded using Huffman coding


A Typical Transform Coder

Forw

ard

Tran

sfor

m

Qua

ntiz

er

Run

-Len

gth

Cod

er

Inve

rse

Tran

sfor

m

Inve

rse

Qua

ntiz

er

Run

-Len

gth

Dec

oder

InputSamples

CoefficientIndices

TransformCoefficients

OutputSamples

CodedBitstream

QuantizedCoefficients

Channel


Coding of Quantized DCT Coefficients

• DC coefficient: Predictive coding– The DC value of the current block is predicted from that of the previous

block, and the error is coded using Huffman coding• AC Coefficients: Runlength coding

– Many high frequency AC coefficients are zero after first few low-frequency coefficients

– Runlength Representation:• Ordering coefficients in the zig-zag order• Specify how many zeros before a non-zero value• Each symbol=(length-of-zero, non-zero-value)

– Code all possible symbols using Huffman coding• More frequently appearing symbols are given shorter codewords• For more details on the actual coding table, see Handout (Sec.8.5.3 in

[Gonzalez02]• One can use default Huffman tables or specify its own tables.• Instead of Huffman coding, arithmetic coding can be used to achieve

higher coding efficiency at an added complexity.• Specific mapping from symbol to Huffman codes is not required in the

exam


Zig-Zag Ordering of DCT Coefficients

Zig-Zag ordering: converting a 2D matrix into a 1D array, so that the frequency (horizontal+vertical) increases in this order, and the coefficient variance (average of magnitude square) decreases in this order.


Example

qdct =

2 5 0 -2 0 -1 0 0

9 1 -1 2 0 1 0 0

14 1 -1 0 -1 0 0 0

3 -1 -1 -1 0 0 0 0

2 -1 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Run-length symbol representation:{2,(0,5),(0,9),(0,14),(0,1),(1,-2),(0,-1),(0,1),(0,3),(0,2),(0,-1),(0,-1),(0,2),(1,-1),(2,-1), (0,-1), (4,-1),(0,-1),(0,1),EOB}

EOB: End of block, one of the symbol that is assigned a short Huffman codeword


JPEG for Color Images

• Color images are typically stored in (R,G,B) format• JPEG standard can be applied to each component separately

– Does not make use of the correlation between color components– Does not make use of the lower sensitivity of the human eye to

chrominance samples• Alternate approach

– Convert (R,G,B) representation to a YCbCr representation• Y: luminance, Cb, Cr: chrominance

– Down-sample the two chrominance components• Because the peak response of the eye to the luminance component

occurs at a higher frequency (3-10 cpd) than to the chrominance components (0.1-0.5 cpd).

• JPEG standard can handle an image consisting of many (up to 100) components



• JPEG Processing for a gray scale image:– Divide the image into data units of 8x8 blocks– Each 8x8 block go through 3 steps:

• Transform (DCT), quantization, run-length+Huffman coding– Quality vs. bit rate trade-off can be controlled by setting a quality factor, which

scales a predefined quantization matrix.• How does JPEG handle color images

– The RGB coordinate is converted to YCbCr• How to do the conversion?

– The Cb and Cr components are typically down sampled by a factor of 2 in both horizontal and vertical directions

• Why should we down-sample chrominance components?– Each component is divided into 8x8 blocks and coded as above

• Performance of JPEG– Very good quality at 1bpp for 24 bpp color images, acceptable quality at 0.5

bpp, poor quality at 0.25 bpp or below.• JPEG2000 not required for the exam


Communications

• Analog modulation and multiplexing• Digital modulation• Error detection and correction• Channel capacity


A Typical Analog Communication System

Modulator Transmitter

Demodulator Receiver

Signal to be transmitted(analog or digital)

Received signal


Modulation = Frequency Shifting

0 fc

Basebandsignal

Modulated signal

Frequency


• By multiplying with a sinusoid signal !

How do we shift the frequency of a signal? (Amplitude Modulation)

)(tx )cos()()( ttxty cω=

frequencycarrier :signalcarrier )cos(

c

ct

ω

ω


Frequency Domain Interpretation of Amplitude Modulation

From Figure 7.5 in Signals/Systems

)(tx

)cos( tcω

)cos()()( ttxty cω=


How to get back to the baseband? (Demodulation)

• By multiplying with the same sinusoid + low pass filtering!

)(ty)(tw

)cos( tcωmω−

mω−

)(ωH

2)(tx

LPF


Frequency Domain Interpretation of Demodulation

Figure 7.7 in Signals and Systems


Frequency Division Multiplexing


)cos()()( ttxty aaa ω=



)()()()( tytytytw cba ++=


FDM Transmitter



FDM Receiver

)cos( taω


Demultiplexing Demodulation


Quadrature Amplitude Modulation

• With amplitude modulation: a signal with bandwidth B needs 2B channel bandwidth – This is called double sideband (DSB) AM– Other techniques can reduce the bandwidth requirement

• Single sideband (SSB)• Vestigial sideband (VSB)

• By using QAM, we can send 2 signals each with bandwidth B over a channel bandwidth of 2B– Equivalent to each signal with bandwidth B


Quadrature Amplitude Modulation (QAM)

• A method to modulate two signals onto the same carrier frequency, but with 90o phase shift

)2cos( 1tfπ

)2sin( 1tfπ

)(1 ts

)(2 ts

)(tmLPF

LPF

)2cos( 1tfπ

)2sin( 1tfπ

)(1 ts

)(2 ts

)(tm

QAM modulator QAM demodulator



• Understand the principle of amplitude modulation – Know how to modulate a signal to a certain frequency– Know how to demodulate a signal back to the baseband– Can write the equation and draw block diagram for both modulation

and demodulation– Can plot the signal spectrum after modulation and demodulation– Bandwidth requirement: need 2B channel bandwidth to send a

signal bandwidth B– By using QAM, we can send 2 signals each with bandwidth B over

a channel bandwidth of 2B• Understand the principle of frequency division multiplexing

– Can write the equation and draw block diagram for both multiplexing and demultiplexing

– Can determine the stopband/pass band frequencies for any filters in the block diagram

• SSB, VSB not required for the exam• FM and PM not required for the exam


Digital Communication

• Digital modulation:– Convert digital bits to analog waveforms– Amplitude Shift Keying– QAM– Parameters of modulation

• Energy efficiency

• Channel error detection and correction• Channel capacity• Advantages of digital communication


Amplitude Shift Keying (ASK)

M-ary ASK: each group of log2M bits generates a symbol. The number corresponding to the symbol controls the amplitude of a sinusoid waveform. The number of cycles in the sinusoid waveform depends on the carrier frequency.(Also known as Pulse Amplitude Modulation or PAM)

4-ASK: 2 bits/symbol (00=-3, 01=-1, 11=1, 10=3)

Example: Given a sequence: 01001011…, what is the analog form resulting from 4-ASK?

Symbol representation: “-1”,”-3”,”3”,”1”

Waveform:

“00”(-3A) “01”(-A) “11” (A) “10”(3A)


Quadrature Amplitude Modulation (QAM)

M-ary QAM uses symbols corresponding to sinusoids with different amplitude as well as phase, arranged in the two-dimensional plane.

Ex. 4-QAM (only phase change):

sin(ωct)

00=cos(ωct-π/4)01=cos(ωct-3π/4)

11=cos(ωct-5π/4) 10=cos(ωct-7π/4)

cos(ωct)

Note this is equivalent to analog QAM if we interpret the first bit and second bit coming from two pulse sequences!


Example of 4-QAM

01 00 10 11

Example: Given a sequence: 01001011…, what is the analog form resulting from 4-ASK?

Using the previous mapping, the analog waveform for the above sequence is


Vector Representation for ASK

• M-ary ASK (M=4, send two bits at a time)

• Labeling is done in such a way that adjacent points only differ in one bit (called Gray mapping)

A-Ax x xx

-3A 3A

11 10 00 01

Ex: A digital sequence 100110 is sent as {-A 3A –A…}


Two Dimensional Modulation

• Modulated signal – 4-QAM (Quadrature Amplitude Modulation)

– 8-PSK (Phase Shift Keying)

)2sin()2cos()( tfAtfAts cscci ππ +=

x

x x

x10 00

0111

xx

xx

x x

xx

A

AA


Parameters of Modulation

• Three important parameters of a modulation scheme:– Minimum distance dmin: The smallest distance among points

in vector representation, which affects error detection capability

– Average energy Eav– Number of bits/symbol= log2(M)

• Example:– 4-ASK: dmin=2A, Eav=2(A2+9A2)/4=5A2, log2(M)=2– 4-QAM: dmin=2A, Eav=2A2, log2(M)=2– For the same Eav and M, we want to maximize dmin (minimize

effect of transmission noise)– For the same dmin and M, we want to minimize Eav (minimize

power consumption)


Channel Capacity (Noiseless Case)

• To send a digital sequence, we send a sequence of pulses– level A representing “1”, and level –A for “0”

• If the interval of each pulse is T, what is the maximum frequency?– The maximum frequency occurs when we have alternating 1 and –

1, spanning 2T time per cycle– This signal has a maximum frequency of fmax=1/2T– With a channel bandwidth B, we can send at most C=2B

bits/second• B>= fmax =1/2T ---> T>=1/2B, C=1/T<=2B

• Instead of two-level pulse, we can use N-level pulse (m=log2Nbits/pulse), we can send C=2Bm bits/second (Nyquist channel capacity)

• By increasing N (hence m), we can reach infinite bit rate!– What is wrong?– Here we assume the channel is noiseless, every level can be

distinguished correctly


Channel Capacity (Noisy Case)

• When there are more levels in a pulse, the signal difference between two adjacent levels is smaller (for the same total dynamic range). Noise in communication channel is more likely to make the received/detected level differ from the actual level.

• The channel capacity depends on the signal to noise ratio (SNR)– SNR = signal energy / noise energy

• Shannel Channel Capacity– C=B log2 (1+SNR) bits/second

2 level 4 level


Advantages of Digital Communication

• More tolerant to channel noise. – With amplitude shift keying: as long as the noise does not change

the amplitude from one level to another level, the original bits can be inferred

– With QAM: as long as the received signal is more close to the original symbol than its neighboring symbols

– Each repeater can regenerate the analog modulated signals from demodulated bits (noise do not accumulate)

• Can insert parity bits before sending data to allow detection/correction of errors

• Can apply digital compression techniques to reduce data rate subject to distortion criterion

• Different signals can be multiplexed more easily – Internet packets can contain any types of signals, cell phones can

send different types of data



• Basic modulation scheme: M-ASK and M-QAM • Parameters of digital modulation schemes

– Can calculate basic parameters for a given modulation scheme– Understand the design objective and know how to compare

different modulation schemes – Relation of bit error probability with SNR: Not required in the exam

• Channel error detection and correction– Should understand how simple parity check works and how

repetition coding works– Not required in the exam

• Channel capacity– Understand the role of channel bandwidth and SNR in determining

channel capacity – Can determine channel capacity based on given SNR and

bandwidth


Communication Networks

• Functions of a network• Packet switching vs. circuit switching• Multiple access methods

– FDMA, TDMA, WDM

• How wireless and wired phone system work


� A communication network provides a general solution to the problem of connecting many devices:

– Connect each device to a network node– Network nodes (switches) exchange information and carry the

information from a source device to a destination device

General Communication Network

CommunicationNetwork

Network Node

host


Network Functions

• Addressing– Identify which network input is to be connected to which network output

• Traffic control– To ensure the smooth flow of information through the network– Reroute or prevent information traffic during congestion

• Network management– Monitor the network performance– Detect and recover from faults– Configure the network resources

• Multiplexing– connect multiple information flows into shared connection lines

• Routing– determine the path across the network

• Switching – Switch traffic from one link to another


Switching Approaches

• Circuit-switched (e.g., Telephone network) • Packet-switched

– Connectionless• packet headers carry destination addresses and routing through the network

is based on the destination address • e.g., Internet - IP protocol based

– Connection-oriented• a "connection" is set up (i.e., a route is selected) prior to data transfer and

released after data transfer.• e.g., ATM: Asynchronous Transfer Mode

Switched Communication Networks

Circuit-Switched Networks Packet-Switched Networks

Datagram Networks Virtual Circuit Networks


� In a circuit-switched network, a dedicated communication path is established between two stations through the nodes of the network

� The dedicated path is called a circuit-switched connection or circuit

� A circuit occupies a fixed capacity of each link for the entire lifetime of the connection. Capacity unused by the circuit cannot be used by other circuits

� Data is not delayed at the switches� Most important circuit-switching networks

– Telephone networks– ISDN (Integrated Services Digital Networks)

Circuit Switching


Packet Switching

� Example: Internet � Operates much like the postal system, where an envelope is sent in the

mail with a destination address� Each intermediate post office determines the next hop post office based on

this address� Intermediate post offices are comparable to IP routers

� Data are sent as formatted bit-sequences, so-called packets� Header and Trailer carry control information� Each packet is passed through the network from node to node along some

path (Routing)� At each node the entire packet is received, stored briefly, and then

forwarded to the next node (Store-and-Forward Networks)� No capacity is allocated for packets

Header Data Trailer


Datagram Packet Switching

� Packets are called datagrams� The network nodes process each packet independently

� If Host A sends two packets back-to-back to Host B over a datagram packet network, the network cannot tell that the packets belong together.

� In fact, the two packets can take different routes.� Implications of processing packets independently

– A sequence of packets can be received in a different order than it was sent

– Each packet header must contain the full address of the destination


Virtual-Circuit Packet Switching

� Virtual-circuit packet switching is a hybrid of circuit switching and packet switching

� All data is transmitted as packets� All packets from one packet stream are sent along a pre-established path

(virtual circuit)� Guarantees in-sequence delivery of packets� However, packets from different virtual circuits may be interleaved� Examples

– X.25, since the 1970s, used in many public packet switching networks– ATM (Asynchronous Transfer Mode), developed in the 1980s, for

transmission of voice, video, and data in a single network


Comparison

� Dedicated transmission path

� Continuous transmission

� Path stays fixed for entire connection

� Call setup delay� Negligible transmission

delay� No queueing delay� Busy signal overloaded

network� Fixed bandwidth for

each circuit � No overhead after call

setup

Circuit Switching� No dedicated

transmission path� Transmission of packets� Route of each packet is

independent� No setup delay� Transmission delay for

each packet� Queueing delays at

switches� Delays increase in

overloaded networks� Bandwidth is shared by

all packets� Overhead in each

packet

Datagram Packet Switching

� No dedicated transmission path

� Transmission of packets� Path stays fixed for

entire connection� Call setup delay� Transmission delay for

each packet� Queueing delays at

switches� Delays increase in

overloaded networks� Bandwidth is shared by

all packets� Overhead in each

packet

VC Packet Switching


What should we use for multimedia applications?

• Which is better for telephone ?– Circuit switching vs. packet switching (voice over IP)?

• Challenges in delivering real-time multimedia traffic– Delay sensitive– High data rate– Should we use circuit switching or packet switching?

Connectionless or connection oriented?• Open question


B B

C C

A A

B

C

A

B

C

A

MUXMUX

(a) (b)Trunkgroup

Figure 4.1

Multiplexing of Signals

• Multiplexers are used to multiplex data bits from different communication sessions on to one link. – Time Division Multiplexing (TDM)– Frequency Division Multiplexing (FDM)– Wavelength Division Multiplexing (WDM).


(a) Each signal transmits 1 unit every 3T seconds

(b) Combined signal transmits 1 unit every T seconds

tA1 A2

tB1 B2

tC1 C2

3T0T 6T

3T0T 6T

3T0T 6T

tB1 C1 A2 C2B2A1

0T 1T 2T 3T 4T 5T 6TFigure 4.3

Time Division Multiplexing


A CBf

Cf

Bf

Af

W

W

W

0

0

0

(a) Individual signals occupy W Hz

(b) Combined signal fits into channel bandwidth

Figure 4.2

Frequency Division Multiplexing


λ1

λ2

λm

OpticalMUX

λ1

λ2

λm

OpticaldeMUX

λ1 λ2. λm

Opticalfiber

Figure 4.18

Wavelength Division Multiplexing


Wireless Communications

• Wireless channel model– Multiple path + fading

• Cellular architecture• Network elements• Wireless systems


Wireless Channel

• Reflection from buildings, cars etc: Multipath• Scattering: Fading


Multipath Fading Model

Transmitted signal s Received signal r

t t

1 A1

A2A3

T1 T2 T3

• Amplitudes Ai and delays Ti randomly vary in time

Delay T1

Delay T2s

A2

nA1

r

A is randomly varying, bad if A is <<1


Effect of Fading

0 5 10 15 20 25 30 3510-4

10-3

10-2

10-1

100

SNR (dB)

PR

OB

AB

ILIT

Y O

F E

RR

OR

FADING GAUSSIAN

Ex: to obtain Pe=10^-4, need 34-8=26 dB higher PSNR, about 1000 times more transmission energy


Cellular Concept

• Break the coverage area into cells (AT&T, 1960’s)


Advantages/disadvantages of cellular

• Advantages– Received power (and hence SNR) decreases with distance

• Path loss: – Mobile user in one cell is far away from other cells

→ Causes very small interference in other cells. Hence we can reuse the same spectrum.Previous figure illustrates reuse factor K=7.

– Limited coverage area means low power base stations.• Disadvantages

– Need a large number of cells in metropolitan areas• Large number of base stations (expensive)

– Need to contact another base-station (handoff) when a user moves from one cell to another

• Large number of handoffs.

α−= dPP transrec


Wireless Networks

TERMINALS (MS)

BASE STATION (BS)

SWITCH (MSC)

DATABASES (HLR, VLR)


Wireless Network Architecture

HCAUm

E Ai Di

MSC PSTN ISDN

HLRMSC ACBSMS

EIR VLR VLRG

F B D


Multiple Access

• Allow multiple mobile users to send data to the same base station– FDMA (same as before)– TDMA (same as before)– CDMA


Time Division Multiple Access (TDMA)

Time

Frequency

User 1 User 2 User 3 User 4

• Each user transmits at a separate time slot in the assigned frequency band. • Needs for buffering in transmitter and receiver for voice communications

T sec

W Hz


Frequency Division Multiple Access (FDMA)

Time

FrequencyW Hz

T secUser 1

User 2

User 3User 4

• Each user transmits at a separate frequency slot


Code 2

Code 4

Code 3

Code 1

Transmitted bit

• Each user transmits all the time over all the frequency band, but has a different “ spreading code.”• Code i belongs to user i.• The base station differentiates users based on their codes.

Code Division Multiple Access (CDMA)


Spreading Codes

Transmitted bit

S(t)

bit

chip

b=1 b=-1

t

t

Transmitted signal

• User transmits signal bS(t) where

=−=

=0 bit if ,11bit if ,1

b


Orthogonal Codes

• Spreading codes belonging to different users are orthogonal

S1(t)

t1

t1

S2(t)

-1

• Note S1=(1,1), S2=(1,-1) and

•The base station can easily identify users (HW10)• Problem: Mobile users asynchronous, and hence they transmit at different times. So it is hard for s1(t) and s2(t) to be orthogonal.

0)()(, 2121 =>==< ∫ dttstsssρ


Process involved in Making a Phone Call

• Cellphone:– Initiate phone – Search for the phone– Respond to the page

message– Assign radio channel– Conversation– Handoff– Terminate

• Wired phone (PSTN)• No initiation stage (a

wirephone is always connected to its designated switching center)

• Search for the phone (could be a cell phone)

• Establish connection• Conversation• Terminate


Wireless Systems

• Cellular phone:– First Generation: analog– Second Generation: digital, TDMA and CDMA– Third Generation: Higher data rates (up to 2MBits/sec)

• All based on CDMA, adaptive modulation and error correction• Wireless LAN’s (Wi-Fi)

– Provide high data rates in limited geographical areas• WLAN complements cellular:

– Cellular provides seamless coverage and mobility– WLAN provides high speed in select areas

• Bluetooth:– Low power, short range wireless communications– Eliminate wires connecting keyboard, computer, printer etc

• WiMAX– Wireless alternative to cable modems, digital subscriber lines (DSL) and

fiber optic links


What You Should Know

• Communication networks– terminals, switches (routers) and links (wired or wireless)– Circuit switching (telephone) vs. packet switching (Internet)– Connection oriented vs connectionless

• Wireless channel properties– Fading and multiple path– Effect of fading on transmission reliability

• Cellular network architecture– Terminals, base stations, switches (switching between base stations

when the terminal moves from one cell to another)– Advantages/disadvantages of using small cells

• Main advantage: reduce transmission power by mobile– Multiple access methods

• TDMA, FDMA, CDMA• Process involved in making a phone call (wired and wireless)

– Details NOT required• Technical details of different wireless systems NOT required.


Final Exam

• Time– Thursday 4/27 11-12:50

• Closed book– 1 sheet of notes (one sided) allowed, can use a calculator – No peak into neighbors– Cheating will result in a failing grade!

• What you should know: – Should be able to do all the HW problems (HW6-13) and quizzes

(quiz5-9) without checking into the solutions!

• Office hours:– Tuesday (4/25): Yao Wang 1:30-3:30 PM (LC256)– Wed (4/28): Zhengye Liu 2-6 PM (LC220)

~yao Polytechnic University, Brooklyn, NY11201 Yao Wang ...©Yao Wang, 2006 EE3414: Final Review 3 Color Principle • How a human perceives color – Three types of cones sensitive

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