Wireless PHY: Digital Demodulation and Wireless Channels

Wireless PHY: Digital Demodulation and

Wireless Channels

Y. Richard Yang

09/13/2012

2

Outline Admin and recap Digital demodulation Wireless channels

3

Admin Assignment 1 posted

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Demodulation Low pass filter and FIR Convolution Theorem

Digital modulation/demodulation ASK, FSK, PSK General representation

Recap

Recap: gi() for BPSK

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1: g1(t) = cos(2πfct) t in [0, T]

0: g0(t) = -cos(2πfct) t in [0, T]

Note: g1(t) = -g0(t)

cos(2πfct)[0, T]1-1g1(t)g0(t)

Recap: Signaling Functions gi() for QPSK

6

11: cos(2πfct + π/4) t in [0, T]

10: cos(2πfct + 3π/4) t in [0, T]

00: cos(2πfct - 3π/4) t in [0, T]

01: cos(2πfct - π/4) t in [0, T]

Q

I

11

01

10

00

Recap: QPSK Signaling Functions as Sum of cos(2πfct), sin(2πfct)

7

11: cos(π/4 + 2πfct) t in [0, T]-> cos(π/4) cos(2πfct) + -sin(π/4) sin(2πfct)

10: cos(3π/4 + 2πfct) t in [0, T]-> cos(3π/4) cos(2πfct) + -sin(3π/4) sin(2πfct)

00: cos(- 3π/4 + 2πfct) t in [0, T]-> cos(3π/4) cos(2πfct) + sin(3π/4) sin(2πfct)

01: cos(- π/4 + 2πfct) t in [0, T]-> cos(π/4) cos(2πfct) + sin(π/4) sin(2πfct)

sin(2πfct)

11

00

10

cos(2πfct)

[cos(π/4), sin(π/4)]

01[cos(3π/4), sin(3π/4)]

[cos(3π/4), -sin(3π/4)]

[-sin(π/4), cos(π/4)]

We call sin(2πfct) and cos(2πfct) the bases.

Recap: Demodulation/Decoding

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Considered a simple setting: sender uses a single signaling function g(), and can have two actions send g() or nothing (send 0)

How does receiver use the received sequence x(t) in [0, T] to detect if sends g() or nothing?

Recap: Design

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Streaming algorithm: use all data points in [0, T] As each sample xi comes in, multiply it by a factor hT-i-1

and accumulate to a sum y

At time T, makes a decision based on the accumulated sum at time T: y[T]

xTx2x1x0

h0h1h2hT

****

Determining the Best h

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where w is noise,

Design objective: maximize peak pulse signal-to-noise ratio


11

Assume Gaussian noise, one can derive

Using Fourier Analysis and Convolution Theorem:


12

Apply Schwartz inequality

By considering


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Determining Best h to Use

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xTx2x1x0

gTg2g1g0

****

xTx2x1x0

h0h1h2hT

****

Summary of Progress

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After this “complex” math, the implementation/interpretation is actually the following quite simple alg: precompute auto correlation: <g, g>

compute the correlation between received x and signaling function g, denoted as <x, g>

if <x, g> is closer to <g, g> • output sends g

else• output sends nothing

Applying Scheme to BPSK

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since g0 = -g1 <x, g0> = - <x, g1> <g0, g0> = - <g0, g1>

rewrite as if |<x, g1> - <g1, g1>| < |<x, g1> - <g0, g1>|

• pick 1 else

• pick 0

cos(2πfct)[0, T]1-1g1(t)g0(t)

Interpretation

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For any signal s, <s, g1> computes the coordinate of s when using g1 as a base cleaner if g1 is normalized, but we do not

worry about it yet

g1=cos(2πfct)[0, T]

<g1(t), g1(t)><g0(t), g1(t)>=-<g1(t), g1(t)>

<x, g1(t)>

Applying Scheme to QPSK: Attempt 1

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Consider g00 alone, compute <x, g00> … Consider g01 alone, compute <x, g01> … Consider g10 alone, compute <x, g10> … Consider g11 alone, compute <x, g11> … Issues

Complexity:• Need to compute M correlation, where M is number of

signaling functions• Think of 64-QAM

Objective• the previous scheme is defined for a single signaling

function, does it work for M?

Decoding for QPSK using bases

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4 signaling functions g00(), g01(), g10(), g11() For each signaling function, computes

correlation with the bases (cos(), sin()), e.g., g00: [a00, b00] What is the meaning of a00, b00?

For received signal x, computes ax=<x, cos> and bx=<x, sin> (how many correlation do we do now?)

QPSK Demodulation/Decoding

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sin(2πfct)

cos(2πfct)

[a01,b01]

[a10,b10]

[a00,b00]

[a11,b11]

[ax,bx]

Q: how to decode?

Look into Noise

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Assume sender sends gm(t) [0, T] Receiver receives x(t) [0, T]

Consider one sample

where w[i] is noise Assume white noise, i.e., prob w[i] = z is

2

2

2

21)(

z

ezf

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Likelihood What is the likelihood (prob.) of observing

x[i]? it is the prob. of noise being w[i] = x[i] – g[i]

What is the likelihood (prob.) of observing the whole sequence x? the product of the probabilities

Likelihood Detection

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Suppose we know

Maxim likelihood detection picks the m with the highest P{x|gm}.

From the expression

We pick m with the lowest ||x-gm||2

Back to QPSK

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QPSK Demodulation/Decoding

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sin(2πfct)

cos(2πfct)

[a01,b01]

[a10,b10]

[a00,b00]

[a11,b11]

[ax,bx]

Q: what does maximum likelihood det pick?

General Matched Filter Detection: Implementation for Multiple Sig Func.

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Basic idea consider each gm[0,T] as a point (with

coordinates) in a space

compute the coordinate of the received signal x[0,T]

check the distance between gm[0,T] and the received signal x[0,T]

pick m* that gives the lowest distance value

Computing Coordinates

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Pick orthogonal bases {f1(t), f2(t), …, fN(t)} for {g1(t), g2(t), …, gM(t)}

Compute the coordinate of gm[0,T] as cm = [cm1, cm2, …, cmN], where

Compute the coordinate of the received signal x[0,T] as x = [x1, x2, …, xN]

Compute the distance between r and cm every cm and pick m* that gives the lowest distance value

Example: Matched Filter => Correlation Detector

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receivedsignal x

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BPSK vs QPSK

BPSK

QPSK

fc: carrier freq.Rb: freq. of data10dB = 10; 20dB =100

11 10 00 01

A

t

BPSK vs QPSK A major metric of modulation performance is

spectral density (SD)

Q: what is the SD of BPSK vs that of QPSK? Q: Why would any one use BPSK, given higher

QAM?31

Spectral Density =

bit rate-------------------

width of spectrum used

Context Previous demodulation considers only

additive noise, and does not consider wireless channel’s effects

We next study its effects

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33


Signal Propagation

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Isotropic radiator: a single point equal radiation in all directions (three dimensional) only a theoretical reference antenna

Radiation pattern: measurement of radiation around an antenna

zy

x

z

y x idealisotropicradiator

Antennas: Isotropic Radiator

Q: how does power level decrease as a function of d, the distancefrom the transmitter to the receiver?

36

Free-Space Isotropic Signal Propagation

In free space, receiving power proportional to 1/d² (d = distance between transmitter and receiver)

Suppose transmitted signal is cos(2ft), the received signal is

Pr: received power Pt: transmitted power Gr, Gt: receiver and

transmitter antenna gain (=c/f): wave length

Sometime we write path loss in log scale: Lp = 10 log(Pt) – 10log(Pr)

dcdtftfEd)]/(2cos[),(

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Log Scale for Large SpandB = 10 log(times)

Slim/Gates

~100B

Obama

~10M

~10K

1000 times

40 dB

10,000 times

30 dB

10,000 x 1,000

40 + 30 = 70 dB

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Path Loss in dBdB = 10 log(times)

source

10 W

d1

1 mW

1 uW

1000 times

40 dB

10,000 times

30 dB

10,000 x 1,000

40 + 30 = 70 dBpower

d2

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dBm (Absolute Measure of Power)dBm = 10 log (P/1mW)

source

10 W

d1

1 mW

1 uW

1000 times

40 dB

10,000 times

30 dB

10,000 x 1,000

40 + 30 = 70 dBpower

d2

40 dBm

-30 dBm

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Number in Perspective (Typical #)

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Exercise: 915MHz WLAN (free space) Transmit power (Pt) = 24.5 dBm Receive sensitivity = -64.5 dBm

Receiving distance (Pr) =

Gt=Gr=1

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Two-ray Ground Reflection Model

Single line-of-sight is not typical. Two paths (direct and reflect) cancel each other and reduce signal strength

Pr: received power Pt: transmitted power Gr, Gt: receiver and

transmitter antenna gain hr, ht: receiver and

transmitter height

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Exercise: 915MHz WLAN (Two-ray ground reflect) Transmit power (Pt) = 24.5 dBm Receive sensitivity = -64.5 dBm

Receiving distance (Pr) =

Gt=Gr=hr=ht=1

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Real Antennas Real antennas are not isotropic radiators Some simple antennas: quarter wave /4 on car roofs or

half wave dipole /2 size of antenna proportional to wavelength for better transmission/receiving

/4/2

Q: Assume frequency 1 Ghz, = ?

http://www.amazon.com/exec/obidos/tg/detail/-/B00007KDVJ/104-4604240-8024759?v=glance&s=electronics&n=507846&me=ATVPDKIKX0DER&vi=pictures&img=14

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Figure for Thought: Real Measurements

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Receiving power additionally influenced by shadowing (e.g., through a wall or a door) refraction depending on the density of a medium reflection at large obstacles scattering at small obstacles diffraction at edges

reflectionscattering

diffraction

shadow fadingrefraction

Signal Propagation: Complexity

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Signal Propagation: Complexity

Details of signal propagation are very complicated

We want to understand the key characteristics that are important to our understanding

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Intro shadowing

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Shadowing Signal strength loss after passing

through obstacles

Same distance, but different levels of shadowing: It is a random, large-scale effect depending on the environment

Example Shadowing Effects

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i.e. reduces to ¼ of signal10 log(1/4) = -6.02

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JTC Indoor Model for PCS: Path Loss

)(10 nLdBLogAL fA: an environment dependent fixed loss factor

(dB)B: the distance dependent loss coefficient,d : separation distance between the base station

and mobile terminal, in metersLf : a floor penetration loss factor (dB)n: the number of floors between base station

and mobile terminal

Shadowing path loss follows a log-normal distribution (i.e. L is normal distribution) with mean:

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JTC Model at 1.8 GHz

)(10 nLdBLogAL f

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Intro Shadowing Multipath

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Signal can take many different paths between sender and receiver due to reflection, scattering, diffraction

Multipath

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Example: reflection from the ground or building

Multipath Example: Outdoor

ground

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Multipath Effect (A Simple Example)

d1 d2

1

11 ][2cos

dtf cd

ft2cos

2121 22)(2 21dd

cddfff c

dcd

2

22 ][2cos

dtf cd

phase difference:

Assume transmitter sends out signal cos(2 fc t)

Multipath Effect (A Simple Example) Where do the two waves totally

destruct?

Q: where do the two waves construct?

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integer2121

dd

cddf

Option 1: Change Location If receiver moves to the right by /4:

d1’ = d1 + /4; d2’ = d2 - /4;

->

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21

21

21

2

)4/(4/22

''2

dd

dd

dd

By moving a quarter of wavelength, destructiveturns into constructive.Assume f = 1G, how far do we move?

Option 2: Change Frequency

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Change frequency:

2121'

ddcff

2121 22 ddcddf

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Multipath Delay SpreadRMS: root-mean-square

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Multipath Effect(moving receiver)

d1 d2

1

11 ][2cos

dtf cd

ft2cos

example

2

22 ][2cos

dtf cd

Suppose d1=r0+vt

d2=2d-r0-vtd1d2

d

Derivation

62

])[sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

)sin()sin(2

])[2cos(])[2cos(

0

0

0

0000

020020

00

2

2)2(

22

2][2][2

2][2][2

2

cvrd

cvf

cd

cdvtr

cd

cdvtr

cd

cvtrdvtr

cvtrdvtr

tftftftf

cvtrd

cvtr

ttf

ftf

ftf

ftf

tftf

cvtrd

cvtr

cvtrd

cvtr

See http://www.sosmath.com/trig/Trig5/trig5/trig5.html for cos(u)-cos(v)

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Derivation

63

])[sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

)sin()sin(2

])[2cos(])[2cos(

0

0

0

0000

020020

00

2

2)2(

22

2][2][2

2][2][2

2

cvrd

cvf

cd

cdvtr

cd

cdvtr

cd

cvtrdvtr

cvtrdvtr

tftftftf

cvtrd

cvtr

ttf

ftf

ftf

ftf

tftf

cvtrd

cvtr

cvtrd

cvtr



Derivation

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])[sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

)sin()sin(2

])[2cos(])[2cos(

0

0

0

0000

020020

00

2

2)2(

22

2][2][2

2][2][2

2

cvrd

cvf

cd

cdvtr

cd

cdvtr

cd

cvtrdvtr

cvtrdvtr

tftftftf

cvtrd

cvtr

ttf

ftf

ftf

ftf

tftf

cvtrd

cvtr

cvtrd

cvtr



Derivation

65

])[sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

)sin()sin(2

])[2cos(])[2cos(

0

0

0

0000

020020

00

2

2)2(

22

2][2][2

2][2][2

2

cvrd

cvf

cd

cdvtr

cd

cdvtr

cd

cvtrdvtr

cvtrdvtr

tftftftf

cvtrd

cvtr

ttf

ftf

ftf

ftf

tftf

cvtrd

cvtr

cvtrd

cvtr



Derivation

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])[sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

)sin()sin(2

])[2cos(])[2cos(

0

0

0

0000

020020

00

2

2)2(

22

2][2][2

2][2][2

2

cvrd

cvf

cd

cdvtr

cd

cdvtr

cd

cvtrdvtr

cvtrdvtr

tftftftf

cvtrd

cvtr

ttf

ftf

ftf

ftf

tftf

cvtrd

cvtr

cvtrd

cvtr



Derivation

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])[sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

])[2sin(])[2sin(2

)sin()sin(2

])[2cos(])[2cos(

0

0

0

0000

020020

00

2

2)2(

22

2][2][2

2][2][2

2

cvrd

cvf

cd

cdvtr

cd

cdvtr

cd

cvtrdvtr

cvtrdvtr

tftftftf

cvtrd

cvtr

ttf

ftf

ftf

ftf

tftf

cvtrd

cvtr

cvtrd

cvtr



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Waveformv = 65 miles/h, fc = 1 GHz: fc v/c =

10 ms

deep fade

Q: How far does a car drive in ½ of a cycle?

])[sin(])[2sin(2 02cvrd

cvf

cd ttf

109 * 30 / 3x108 = 100 Hz

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Multipath with Mobility

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Effect of Small-Scale Fading

no small-scalefading

small-scalefading

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signal at sender

Multipath Can Spread Delay

signal at receiver

LOS pulsemultipathpulses

LOS: Line Of Sight

Time dispersion: signal is dispersed over time

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JTC Model: Delay SpreadResidential Buildings

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signal at sender

Multipath Can Cause ISI

signal at receiver


LOS: Line Of Sight

Dispersed signal can cause interference between “neighbor” symbols, Inter Symbol Interference (ISI)

Assume 300 meters delay spread, the arrival time difference is 300/3x108 = 1 ns if symbol rate > 1 Ms/sec, we will have serious ISI

In practice, fractional ISI can already substantially increase loss rate

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Channel characteristics change over location, time, and frequency

small-scale fading

Large-scalefading

time

power

Summary: Wireless Channels

path loss

log (distance)

Received Signal Power (dB)

frequency

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Preview: Challenges and Techniques of Wireless Design

Performance affected

Mitigation techniques

Shadow fading(large-scale fading)

Fast fading(small-scale, flat fading)Delay spread (small-scale fading)

received signal

strength

bit/packet error rate at deep fade

ISI

use fade margin—increase power or reduce distance

diversity

equalization; spread-spectrum; OFDM;

directional antenna

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Representation of Wireless Channels

Received signal at time m is y[m], hl[m] is the strength of the l-th tap, w[m] is the background noise:

When inter-symbol interference is small:

(also called flat fading channel)

Backup Slides

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Received Signal

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2

2

1

1 )]/(2cos[)]/(2cos[),(d

cdtfd

cdtftfEd

cfd

cfd 12 22diff phase

d2

d1 receiver

cddf )(2 12

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Multipath Fading with Mobility: A Simple Two-path Example

r(t) = r0 + v t, assume transmitter sends out signal cos(2 fc t)

r0

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Received Waveform

v = 65 miles/h, fc = 1 GHz: fc v/c = 109 * 30 / 3x108 = 100 Hz

10 ms

Why is fast multipath fading bad?

deep fade

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Small-Scale Fading

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signal at sender

Multipath Can Spread Delay

signal at receiver


LOS: Line Of Sight

Time dispersion: signal is dispersed over time

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Delay Spread RMS: root-mean-square

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signal at sender

Multipath Can Cause ISI

signal at receiver


LOS: Line Of Sight

dispersed signal can cause interference between “neighbor” symbols, Inter Symbol Interference (ISI)

Assume 300 meters delay spread, the arrival time difference is 300/3x108 = 1 msif symbol rate > 1 Ms/sec, we will have serious ISI

In practice, fractional ISI can already substantially increase loss rate

85

Channel characteristics change over location, time, and frequency

small-scale fading

Large-scalefading

time

power

Summary: Wireless Channels

path loss

log (distance)

Received Signal Power (dB)

frequency

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Dipole: Radiation Pattern of a Dipole

http://www.tpub.com/content/neets/14182/index.htmhttp://en.wikipedia.org/wiki/Dipole_antenna

Free Space Signal Propagation

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1 0 1

t

1 0 1

t

1 0 1

t

at distance d?

Why Not Digital Signal (revisited) Not good for spectrum usage/sharing The wavelength can be extremely large

to build portal devices e.g., T = 1 us -> f=1/T = 1MHz ->

wavelength = 3x108/106 = 300m

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Exercise Suppose fc = 1 GHz

(fc1 = 1 GHz, fc0 = 900 GHzfor FSK)

Bit rate is 1 Mbps Encode one bit at a time Bit seq: 1 0 0 1 0 1

Q: How does the wave look like for?

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11 10 00 01

Q

I

11

01

10

00

A

t

Wireless PHY: Digital Demodulation and Wireless Channels

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