L14: Bandpass Modulation - York University · L14: Bandpass Modulation SEC. 2.6 THE PUBLIC SWITCHED TELEPHONE NETWORK 149 Power Voice Upstream Downstream 256 4-kHz Channels 025 1100

EECS 4214, F15 1

L14: Bandpass Modulation

Sebastian Magierowski York University


LE/EECS 4214 Fall 2015

EECS 4214, F15 2

•  14.1 Basics •  14.2 Bandpass Signaling •  14.3 PSK: Phase Shift Keying •  14.4 FSK: Frequency Shift Keying •  14.5 ASK: Amplitude Shift Keying •  14.6 APK: Amplitude Phase Keying •  14.7 Signal Space Concepts •  14.8 Signal Space Examples

Outline


EECS 4214, F15 3

•  So far we have discussed communication pulses with spectra centred at DC

–  Basband signalling

Basics: Baseband


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EECS 4214, F15 4

•  But there are channels not conducive to this –  Wireless

Bandpass Channels


SEC. 2.3 WIRELESS TRANSMISSION 113

The location of these bands varies somewhat from country to country. In theUnited States, for example, the bands that networking devices use in practicewithout requiring a FCC license are shown in Fig. 2-13. The 900-MHz band wasused for early versions of 802.11, but it is crowded. The 2.4-GHz band is avail-able in most countries and widely used for 802.11b/g and Bluetooth, though it issubject to interference from microwave ovens and radar installations. The 5-GHzpart of the spectrum includes U-NII (Unlicensed National InformationInfrastructure) bands. The 5-GHz bands are relatively undeveloped but, sincethey have the most bandwidth and are used by 802.11a, they are quickly gainingin popularity.

26MHz

902MHz

928MHz

2.4GHz

5.25GHz

5.35GHz

5.47GHz

5.725GHz

U-NII bands

5.825GHz

2.4835GHz

ISM band

83.5MHz

100MHz

255MHz

ISM band

100MHz

ISM band

Figure 2-13. ISM and U-NII bands used in the United States by wireless devices.

The unlicensed bands have been a roaring success over the past decade. Theability to use the spectrum freely has unleashed a huge amount of innovation inwireless LANs and PANs, evidenced by the widespread deployment of technolo-gies such as 802.11 and Bluetooth. To continue this innovation, more spectrum isneeded. One exciting development in the U.S. is the FCC decision in 2009 toallow unlicensed use of white spaces around 700 MHz. White spaces are fre-quency bands that have been allocated but are not being used locally. The tran-sition from analog to all-digital television broadcasts in the U.S. in 2010 freed upwhite spaces around 700 MHz. The only difficulty is that, to use the whitespaces, unlicensed devices must be able to detect any nearby licensed trans-mitters, including wireless microphones, that have first rights to use the frequencyband.

Another flurry of activity is happening around the 60-GHz band. The FCCopened 57 GHz to 64 GHz for unlicensed operation in 2001. This range is anenormous portion of spectrum, more than all the other ISM bands combined, so itcan support the kind of high-speed networks that would be needed to streamhigh-definition TV through the air across your living room. At 60 GHz, radio

EECS 4214, F15 5

•  Wired can also be bandpass –  ADSL

–  Cable

Bandpass Channels


SEC. 2.6 THE PUBLIC SWITCHED TELEPHONE NETWORK 149

Pow

er

Voice Upstream Downstream

256 4-kHz Channels

0 25 1100 kHz

Figure 2-34. Operation of ADSL using discrete multitone modulation.

below the theoretical limit. It is up to the provider to determine how many chan-nels are used for upstream and how many for downstream. A 50/50 mix ofupstream and downstream is technically possible, but most providers allocatesomething like 80–90% of the bandwidth to the downstream channel since mostusers download more data than they upload. This choice gives rise to the ‘‘A’’ inADSL. A common split is 32 channels for upstream and the rest downstream. Itis also possible to have a few of the highest upstream channels be bidirectional forincreased bandwidth, although making this optimization requires adding a specialcircuit to cancel echoes.

The international ADSL standard, known as G.dmt, was approved in 1999. Itallows speeds of as much as 8 Mbps downstream and 1 Mbps upstream. It wassuperseded by a second generation in 2002, called ADSL2, with various im-provements to allow speeds of as much as 12 Mbps downstream and 1 Mbps up-stream. Now we have ADSL2+, which doubles the downstream speed to 24Mbps by doubling the bandwidth to use 2.2 MHz over the twisted pair.

However, the numbers quoted here are best-case speeds for good lines close(within 1 to 2 km) to the exchange. Few lines support these rates, and few pro-viders offer these speeds. Typically, providers offer something like 1 Mbpsdownstream and 256 kbps upstream (standard service), 4 Mbps downstream and 1Mbps upstream (improved service), and 8 Mbps downstream and 2 Mbpsupstream (premium service).

Within each channel, QAM modulation is used at a rate of roughly 4000symbols/sec. The line quality in each channel is constantly monitored and thedata rate is adjusted by using a larger or smaller constellation, like those inFig. 2-23. Different channels may have different data rates, with up to 15 bits persymbol sent on a channel with a high SNR, and down to 2, 1, or no bits per sym-bol sent on a channel with a low SNR depending on the standard.

A typical ADSL arrangement is shown in Fig. 2-35. In this scheme, a tele-phone company technician must install a NID (Network Interface Device) on thecustomer’s premises. This small plastic box marks the end of the telephone com-pany’s property and the start of the customer’s property. Close to the NID (orsometimes combined with it) is a splitter, an analog filter that separates the

182 THE PHYSICAL LAYER CHAP. 2

cables nowadays have 500–2000 houses, but as more and more people subscribeto Internet over cable, the load may become too great, requiring more splitting andmore fiber nodes.

2.8.3 Spectrum Allocation

Throwing off all the TV channels and using the cable infrastructure strictlyfor Internet access would probably generate a fair number of irate customers, socable companies are hesitant to do this. Furthermore, most cities heavily regulatewhat is on the cable, so the cable operators would not be allowed to do this even ifthey really wanted to. As a consequence, they needed to find a way to have tele-vision and Internet peacefully coexist on the same cable.

The solution is to build on frequency division multiplexing. Cable televisionchannels in North America occupy the 54–550 MHz region (except for FM radio,from 88 to 108 MHz). These channels are 6-MHz wide, including guard bands,and can carry one traditional analog television channel or several digital televisionchannels. In Europe the low end is usually 65 MHz and the channels are 6–8MHz wide for the higher resolution required by PAL and SECAM, but otherwisethe allocation scheme is similar. The low part of the band is not used. Moderncables can also operate well above 550 MHz, often at up to 750 MHz or more.The solution chosen was to introduce upstream channels in the 5–42 MHz band(slightly higher in Europe) and use the frequencies at the high end for the down-stream signals. The cable spectrum is illustrated in Fig. 2-52.

0 108

TV TV Downstream data

Downstream frequencies

Ups

trea

mda

taU

pstr

eam

freq

uenc

ies

FM

550 750 MHz

5 42 54 88

Figure 2-52. Frequency allocation in a typical cable TV system used for Inter-net access.

Note that since the television signals are all downstream, it is possible to useupstream amplifiers that work only in the 5–42 MHz region and downstreamamplifiers that work only at 54 MHz and up, as shown in the figure. Thus, we getan asymmetry in the upstream and downstream bandwidths because more spec-trum is available above television than below it. On the other hand, most userswant more downstream traffic, so cable operators are not unhappy with this fact

EECS 4214, F15 6

•  Instead of pulse you send modulated sine waves

•  Which result in bandpass channels (in frequency) –  With less spectral efficiency!

•  At first look

Bandpass Signalling


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s(t) = A(t) cos[!o

t+ �(t)]

EECS 4214, F15 7

•  Three main parameters to control

–  amplitude:

–  frequency:

–  phase:

•  These extra “levers” allow you to squeeze in more data per symbol

–  We’ll see this more clearly soon

Bandpass Signal Parameters


s(t) = A(t) cos[!o

t+ �(t)]

EECS 4214, F15 8

•  Common to designate amplitude in terms of signal energy per symbol

–  P = avg. power per symbol

A Note on Signal Settings


A =p2Arms =

p2A2

rms =p2P =

r2E

T

Arms

=

(1

T

ZT

0A2

cos

2(!

o

t)dt

) 12

=

Ap2

EECS 4214, F15 9

A Peak at Demodulation

•  Coherent Demodulation –  RX has to know the phase of

your sine wave to pick out the data •  Carrier recover needed

•  Non-Coherent Demodulation –  RX does not need to know the

phase of sine wave


EECS 4214, F15 10

•  Symbols have different phase

•  BPSK: Binary Phase Shift Keying

14.3 PSK: Phase Shift Keying


si

(t) =

r2E

Tcos[!

o

t+ �i

(t)], 0 t T, i = 1, . . . ,M

�i(t) =2⇡i

M

+1

-1

Binary Phase Shift Keying (BPSK) 0 T 2T 3T 4T 5T 6T t

1 1 1 1 0 0

�1(t) = ⇡ �2(t) = 0

EECS 4214, F15 11

•  Symbols have different frequency –  𝜔i has M discrete values i has M discrete values

–  Binary example:

14.4 FSK: Frequency Shift Keying


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-1 0 T 2T 3T 4T 5T 6T

t

Information 1 1 1 1 0 0

Frequency Shift Keying

EECS 4214, F15 12

•  Amplitude takes on different value for each symbol

•  Binary example –  OOK (On-Off Keying)

14.5 ASK: Amplitude Shift Keying


si

(t) =

r2E

i

Tcos[!

o

t+ �], 0 t T

Information 1 1 1 1 0 0

+1

-1 0 T 2T 3T 4T 5T 6T

Amplitude Shift Keying

t

EECS 4214, F15 13

•  A combination of ASK & PSK

•  QAM: Quadrature Amplitude Modulation

14.6 APK: Amplitude-Phase Keying


si

(t) =

r2E

i

Tcos[!

o

t+ �i

], 0 t T

EECS 4214, F15 14

•  A more general description of signals is possible –  Not just bandpass signals –  But also signal sequences!

•  The description consists of… –  …sequence of orthogonal signals

•  These descriptions allow more efficient organization of receiver structures for sophisticated signals

14.7 Signal-Space Concepts


EECS 4214, F15 15

•  Your signal can be represented as a point in an abstract space

–  Rather a vector to a point

•  The axes represent orthogonal signals –  Which constitute any signal you may wish to define in this space

Signal Space


si = (ai1, ai2, . . . , aiN )

sj = (aj1, aj2, . . . , ajN )

m(t) m = 0, 1, . . .

1

2

EECS 4214, F15 16

•  Basis Functions –  The functions that define our signal space coordinates

–  Defined over some time interval Ti to Tf •  Ti: initial time, Tf: final time

–  Make sure the functions are PERPENDICULAR to each other •  Orthonormal

Basis Functions


m(t) m = 0, 1, . . .

Z Tf

Ti

i(t) j(t)dt = �ij =

(0, if i 6= j

1, if i = j

EECS 4214, F15 17

•  Fourier series coefficients

Basis Function Example


m

(t) =

(✓1

T

◆ 12

,

✓2

T

◆ 12

cos(m!o

t),

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T

◆ 12

sin(m!o

t), . . .

)m = 1, 2, . . .

T = Tf � Ti

T =2⇡

!o

EECS 4214, F15 18

•  Finite set of N non-overlapping pulses

Basis Function Example


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EECS 4214, F15 19

•  Any of M symbols: •  Can be expressed as a N-term series expansion

–  of orthonormal basis functions

–  expansion coefficients aij

•  projection of ith symbol on jth basis function

Signals in Space


si(t), i = 1, 2, . . . ,M

si(t) =NX

j=1

aij j(t)

aij =

Z Tf

Ti

si(t) j(t)dt

EECS 4214, F15 20

•  Collection of M points in N-space –  Signal Constellation

Signal Space Pictorially


si = (ai1, ai2, . . . , aiN )

sj = (aj1, aj2, . . . , ajN )

1

2 N

aiN

ai1si(t) =NX

j=1

aij j(t)

aij =

Z Tf

Ti

si(t) j(t)dt

EECS 4214, F15 21

Signal Generation & Recovery

•  Generation •  Recovery


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si(t) =NX

j=1

aij j(t) aij =

Z Tf

Ti

si(t) j(t)dt

EECS 4214, F15 22

•  Draw the signal constellation

14.8 Signal Space Examples


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EECS 4214, F15 23

•  What value of A is needed to make these basis functions orthonormal?

Signal Space Example


LE/EECS 4214 E — Final Examination Name:

5. (4 points) Sequences and spaces. Only one part in this section.

(a) (4 points) What is the value of A in Volts needed to make the basis functions shown below orthonormal?

!1

t= 5ms

A!2

t = 5ms

A

t = 10ms

-A

t = 0ms

v v

t t

Page 7 of 10

EECS 4214, F15 24

•  For the basis functions shown, sketch the signal waveforms corresponding to the indicted constellation points

Signal Space Example


LE/EECS 4214 E — Final Examination Name:

1. (10 points) Signal basics.

(a) (5 points) Prove whether s1(t), and s2(t) functions are orthogonal or not.

s1(t)

0 1

s2(t)

t

t

(b) (5 points) For the basis functions �1(t) and �2(t) shown below sketch the waveforms corresponding to theconstellation points shown (the constellation points are alphabetically labelled so make sure to clearly indicatewhich label your waveform corresponds to).

0.51

!2(t)

t0.5 1

!1(t)

t

1

-1

1

-1

A•√2

!1

!2ab

c d

Page 2 of 9

EECS 4214, F15 25

•  The M-ary PSK

•  Just a harmonic signal with some phase •  Fourier series only needs 2 coefficients for this

–  Hence our basis functions can be…

M-ary PSK in Signal Space


si

(t) =

r2E

Tcos

✓!o

t+2⇡i

M

◆, i = 0, . . . ,M � 1

1(t) =

r2

Tcos(!

o

t) 2(t) =

r2

Tsin(!

o

t)

EECS 4214, F15 26

•  The resulting constellation is

M-ary PSK Constellation


aij =

Z Tf

Ti

si(t) j(t)dt

si

(t) =

r2E

Tcos

✓!o

t+2⇡i

M

◆, i = 0, . . . ,M � 1

pE

(ai1, ai2) = si =

✓pE cos

2⇡i

M,pE sin

2⇡i

M

◆

1

2

EECS 4214, F15 27

•  Constellation is…

2-ary PSK = BPSK


pE�

pE

1

s1(t) =

r2E

Tcos(!

o

t)�r

2E

Tcos(!

o

t) = s2(t)

�1(t) = 0 �2(t) = ⇡

1(t) =

r2

Tcos(!

o

t)

EECS 4214, F15 28

•  Quadrature Amplitude Modulation

•  Signal space coordinates: –  Typically chosen from points on a 2D square grid –  4-QAM example

QAM


si

(t) = ai1

r2

Tcos(!

o

t) + ai2

r2

Tsin(!

o

t)

(ai, bi)

si

(t) = ai

r2

Tcos(!

o

t) + bi

r2

Tsin(!

o

t)

(a, b)

(a,-b) (-a,-b)

(-a,b)

1

2

EECS 4214, F15 29

•  With 4-QAM –  bits represented per symbol: N = 2

–  Constellation points: M = 2N = 4

•  Many other possibilities (rectangular constellation) –  N = 3,4,5,... –  Even N preferred (easier coder)

•  Increasing N requires more power

Larger QAM Constellations


Quadrature Amplitude Modulation

QPSK 2 bits /symbolSymbol Rate = 1/2 bit 64 QAMSymbol Rate 1/2 bit rate 64 QAM

6 bits /symbolSymbol Rate = 1/6 bit rate

16 QAM 4 bits /symbol

Symbol Rate = 1/4 bit rate


128 QAM 7 bits /symbolSymbol Rate = 1/7 bit



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X-Series Spectrum Analyzers Seminar 2010










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M = 4 M = 16 M = 32 M = 64 M = 128 L = 2 L = 4 L = 6 L = 8 L = 12

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L14: Bandpass Modulation - York University · L14: Bandpass Modulation SEC. 2.6 THE PUBLIC SWITCHED TELEPHONE NETWORK 149 Power Voice Upstream Downstream 256 4-kHz Channels 025 1100