dsp slides module9 0

8/12/2019 Dsp Slides Module9 0

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Module Overview:

Module 9.1: The analog channel Module 9.2: Meeting the bandwidth constraint

Module 9.3: Meeting the power constraint

Module 9.4: Modulation and demodulation

Module 9.5: Receiver design

Module 9.6: ADSL

9


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Digital Sig

Module 9.1: Digital Commu


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Overview:

The many incarnations of a signal

Analog channel constraints

Satisfying the constraints

9.1


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Overview:




9.1


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Overview:




9.1


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Digital data throughputs

Transatlantic cable:

1866: 8 words per minute (

5 bps)

1956: AT&T, coax, 48 voice channels (3Mbps) 2005: Alcatel Tera10, ber, 8.4 Tbps (8 .4 1012 bps) 2012: ber, 60 Tbps

Voiceband modems 1950s: Bell 202, 1200 bps 1990s: V90, 56Kbps 2008: ADSL2+, 24Mbps

9.1


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Digital data throughputs

Transatlantic cable:

1866: 8 words per minute (

5 bps)

1956: AT&T, coax, 48 voice channels (3Mbps) 2005: Alcatel Tera10, ber, 8.4 Tbps (8 .4 1012 bps) 2012: ber, 60 Tbps

Voiceband modems 1950s: Bell 202, 1200 bps 1990s: V90, 56Kbps 2008: ADSL2+, 24Mbps

9.1


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Success factors for digital communications

1) power of the DSP paradigm: integers are easy to regenerate

good phase control

adaptive algorithms

9.1


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Regenerating signals

5

0

5

x (t )

9.1


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5

0

5

x (t )/ G + (t )

9.1


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5

0

5

G [x (t )/ G + (t )] = x (t ) + G (t )

9.1


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5

0

5

x 1(t ) = G sgn[x (t ) + (t )]

9.1


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2) algorithmic nature of DSP is a perfect match with information theory: JPEGs entropy coding

CDs and DVDs error correction

trellis-coded modulation and Viterbi decoding

9.1


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3) hardware advancement miniaturization

general-purpose platforms

power efficiency

9.1


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The many incarnations of a conversation

Base Station Switch

N

Switch CO

air copper

coaxcopper

ber

9.1


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The analog channel

unescapable limits of physical channels: bandwidth constraint

power constraint

both constraints will affect the nal capacity of the channel

9.1


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The analog channel


power constraint


9.1


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The analog channel


power constraint


9.1

h l h l


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The analog channels capacity

maximum amount of information that can be reliably delivered over (bits per second)

9.1

B d id h i


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Bandwidth vs capacity

simple thought experiment: we want to transmit information encoded as a sequence of digital sampl

continuous-time channel

we interpolate the sequence of samples with a period T s

if we make T s small we can send more info per unit of time...

... but the bandwidth of the signal will grow as 1/ T s

9.1

B d id h i


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9.1

B d idth it


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9.1

B d idth it


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9.1

Power and capacity


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Power and capacity

another thought experiment: all channels introduce noise; at the receiver we have to guess what wa

suppose noise variance is 1

suppose we are transmitting integers between 1 and 10: lots of guessing

transmit only odd numbers: fewer errors but less information

9.1

Power and capacity


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Power and capacity





9.1

Power and capacity


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Power and capacity





9.1

Power and capacity


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Power and capacity





9.1

Example: the AM radio channel


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cos(c t )

x (t )

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from 530kHz to 1.7MHz each channel is 8KHz

power limited by law:

daytime/nighttime

interference health hazards

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daytime/nighttime


9.1



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daytime/nighttime


9.1



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daytime/nighttime

interference

health hazards

9.1



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daytime/nighttime

interference

health hazards

9.1


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Example: the telephone channel


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p p

CO Network CO

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p p

one channel from around 300Hz to around 3000Hz

power limited by law to 0.2-0.7V rms

noise is rather low: SNR usually 30dB or more

9.1



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p p




9.1



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p p




9.1

The all-digital paradigm


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keep everything digital until we hit the physical channel

..01100

01010... TX D/A s

F s = 1 / T s

s [n]

9.1

Lets look at the channel constraints


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0 F min F max

bandwidth constraint

power constraint

9.1

Converting the specs to a digital design


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0 F min F max

9.1



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0 F min F max F s / 2

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min max0

9.1

Transmitter design


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some working hypotheses: convert the bitstream into a sequence of symbols a[n] via a mapper

model a[n] as a white random sequence (add a scrambler on the bitstrea

now we need to convert a[n] into a continuous-time signal within the co

9.1

Transmitter design


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9.1

Transmitter design


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9.1

Transmitter design


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..01100

01010... Scrambler Mapper ?

a[n]

9.1

First problem: the bandwidth constraint


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P a (e j ) = 2a

min max0

2a

0

9.1

First problem: the bandwidth constraint


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P a (e j ) = 2a

min max0

2a

0

9.1


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END OF MODULE 9.1


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Digital Sig

Module 9.2: Controlli

Overview:


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Upsampling

Fitting the transmitters spectrum

9.2

Overview:


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Upsampling

Fitting the transmitters spectrum

9.2

Shaping the bandwidth


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Our problem: bandwidth constraint requires us to control the spectral support of a sign

we need to be able to shrink the support of a full-band signal

the answer is multirate techniques

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9.2



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9.2

Multirate signal processing


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In a nutshell: increase or decrease the number of samples in a discrete-time signal

equivalent to going to continuous time and resampling

staying in the digital world is cleaner

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9.2



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9 2

Upsampling via continuous time


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x [n] Int x [n]

T s T s / K

x c (t )

9 2

Upsampling (K = 3)


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x [n ]

00

1

9 2

Upsampling (K = 3)


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x c (t )

00

1

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Upsampling (K = 3)


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00

1

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Upsampling (K = 3)


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x [n ]

00

1

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Upsampling


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As per usual, we can choose T s = 1...

x c (t ) =

m = x [m] sinc(t m)

x [n] = x c (n/ K )

=

m = x [m]sinc n

K m

9 2

Upsampling


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As per usual, we can choose T s = 1...

x c (t ) =

m = x [m] sinc(t m)

x [n] = x c (n/ K )

=

m = x [m]sinc n

K m

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Upsampling (frequency domain)


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3/ 4 / 2 0 / 2

0

1

X ( e j )

9 2



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3/ 4 / 2 0 / 2

0

1

X ( e j )

0 N N

N = / T s X ( j )

9 2



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3/ 4 / 2 0 / 2

0

1

X ( e j )

0 N N

N = / (T s / K ) X ( j )

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3/ 4 / 2 0 / 2

0

1

X ( e j )

0 N N

N = / (T s / K ) X ( j )

/ 4 / 2 0 / 2 0

1

X

( e j )

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Upsampling in the digital domain


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what can we do purely digitally? we need to increase the number of samples by K

obviously x U [m] = x [n] when m multiple of K

for lack of a better strategy, put zeros elsewhere

example for K = 3:

x U [m] = . . . x [0], 0, 0, x [1], 0, 0, x [2], 0, 0, . . .

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example for K = 3:

x U [m] = . . . x [0], 0, 0, x [1], 0, 0, x [2], 0, 0, . . .

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example for K = 3:

x U [m] = . . . x [0], 0, 0, x [1], 0, 0, x [2], 0, 0, . . .

9 2

Upsampling in the Time Domain


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00

1

9.2

Upsampling in the Time Domain


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00

1

9.2



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in the frequency domain

X U (e j ) =

m = x U [m]e j m

=

n= x [n]e j nK

= X (e j K

)

9.2



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X U (e j ) =

m = x U [m]e j m

=

n= x [n]e j nK

= X (e j K

)

9.2



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X U (e j ) =

m = x U [m]e j m

=

n= x [n]e j nK

= X (e j K

)

9.2


1)


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3/ 4 / 2 0 / 2

0

1

X ( e j )

9.2


1)


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3/ 4 / 2 0 / 2

0

1

X ( e j )

5 4 3 2 0 2 3 4 0

1

X ( e j )

9.2


1)


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3/ 4 / 2 0 / 2

0

1

X ( e j )

5 4 3 2 0 2 3 4 0

1

X ( e j )

/ 2 0 / 2 0

1

X U

( e j )

9.2


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1 )


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3/ 4 / 2 0 / 2

0

1

X ( e j

)

5 4 3 2 0 2 3 4 0

1

X ( e j )

/ 2 0 / 2 0

1

X U

( e j )

9.2


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back in time domain...


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insert K 1 zeros after every sample ideal lowpass ltering with c = / K

x [n] = x U (n)sinc(n/ K )

=

i = x U [i ]sinc

n i K

=

m = x [m]sinc

nK m

9.2


back in time domain...


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insert K 1 zeros after every sample ideal lowpass ltering with c = / K

x [n] = x U (n)sinc(n/ K )

=

i = x U [i ]sinc

n i K

=

m = x [m]sinc

nK m

9.2

Downsampling


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given an upsampled signal we can always recover the original:

x [n] = x U [nK ]

downsampling of generic signals more complicated (aliasing)

9.2

Downsampling


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given an upsampled signal we can always recover the original:

x [n] = x U [nK ]

downsampling of generic signals more complicated (aliasing)

9.2

Remember the bandwidth constraint?


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0 F min F max F s / 2

9.2

Heres a neat trick


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let W = F max F min ; pick F s so that: F s > 2F max (obviously)

F s = KW , K N

max min = 2W F s

= 2K

we can simply upsample by K

9.2

Heres a neat trick


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F s = KW , K N

max min = 2W F s

= 2K


9.2

Heres a neat trick


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F s = KW , K N

max min = 2W F s

= 2K


9.2

Heres a neat trick


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F s = KW , K N

max min = 2W F s

= 2K


9.2

Data rates


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upsampling does not change the data rate we produce (and transmit) W symbols per second

W is sometimes called the Baud rate of the system and is equal to the abandwidth

9.2

Data rates


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9.2

Data rates


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9.2

Transmitter design, continued


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..01100

01010... Scrambler Mapper K

a[n]

D/A s (t )

cos c n

b [n] s [n]

9.2

Meeting the bandwidth constraint


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min max 0

0

1

9.2



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min max 0

0

1

9.2



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min max 0

0

1

9.2



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min max 0

0

1

9.2

Raised Cosine


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/ 2 0 / 2 0

1

9.2


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Raised Cosine


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/ 2 0 / 2 0

1

9.2

Raised Cosine


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/ 2 0 / 2 0

1

9.2


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END OF MODULE 9.2


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Digital Sig

Module 9.3: Cont

Overview:


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Noise and probability of error Signaling alphabet and power

QAM signaling

9.3

Overview:


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QAM signaling

9.3

Overview:


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QAM signaling

9.3

Transmission reliability


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transmitter sends a sequence of symbols a[n]

receiver obtaines a sequence a[n]

even if no distortion we cant avoid noise: a[n] = a[n] + [n]

when noise is large, we make an error

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9.3



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9.3



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9.3

Probability of error


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depends on: power of the noise wrt power of the signal

decoding strategy

alphabet of transmission symbols

9.3



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decoding strategy


9.3



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decoding strategy


9.3

Signaling alphabets


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we have a (randomized) bitstream coming in we want to send some upsampled and interpolated samples over the cha

how do we go from bitstream to samples?

9.3

Signaling alphabets


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9.3

Signaling alphabets


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9.3

Mappers and slicers

mapper: lit i i g bit t i t h k


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split incoming bitstream into chunks

assign a symbol a[n] from a nite alphabet A to each chunk

slicer: receive a value a[n]

decide which symbol from A is closest to a[n] piece back together the corresponding bitstream

9.3

Example: two-level signaling

mapper:


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pp split incoming bitstream into single bits

a[n] = G if the bit is 1, a[n] = G if the bit is 0

slicer:

n-th bit =1 if a[n] > 0

0 otherwise

9.3


2


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0 5 10 15 20 25 30 2

1

0

1

9.3


2


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0 5 10 15 20 25 30 2

1

0

1

9.3


2


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0 5 10 15 20 25 30 2

1

0

1

9.3


l l k h b b l f f k h h


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lets look at the probability of error after making some hypotheses:

a[n] = a[n] + [n] bits in bitstream are equiprobable

noise and signal are independent

noise is additive white Gaussian noise with zero mean and variance 0

9.3


P err = P [ [n] < G | n-th bit is 1 ]P [ n-th bit is 1 ]+


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|P [ [n] > G

| n-th bit is 0 ]P [ n-th bit is 0 ]

= ( P [ [n] < G ] + P [ [n] > G ])/ 2= P [ [n] > G ]

=

G

1

2 20

e

2

2 20 d

= Q (G / 0) = 12

erfc((G / 0)/ 2)

9.3


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|P [ [n] > G


= ( P [ [n] < G ] + P [ [n] > G ])/ 2= P [ [n] > G ]

=

G

1

2 20

e

2

2 20 d

= Q (G / 0) = 12

erfc((G / 0)/ 2)

9.3




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|P [ [n] > G


= ( P [ [n] < G ] + P [ [n] > G ])/ 2= P [ [n] > G ]

=

G

1

2 20

e

2

2 20 d

= Q (G / 0) = 12

erfc((G / 0)/ 2)

9.3




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|P [ [n] > G


= ( P [ [n] < G ] + P [ [n] > G ])/ 2= P [ [n] > G ]

=

G

1

2 20

e

2

2 20 d

= Q (G / 0) = 12

erfc((G / 0)/ 2)

9.3


transmitted power


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2s = G 2 P [n-th bit is 1] + G 2 P [n-th bit is 0]

= G 2

P err = Q (s / 0) = Q ( SNR)

9.3


transmitted power


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= G 2

P err = Q (s / 0) = Q ( SNR)

9.3


transmitted power


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= G 2

P err = Q (s / 0) = Q ( SNR)

9.3


100


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10 10

10 20

0 10 20SNR (dB)

P e r r

9.3

Lesson learned:


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to reduce the probability of error increase G increasing G increases the power

we cant go above the channels power constraint!

9.3

Lesson learned:


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9.3

Lesson learned:


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9.3

Multilevel signaling


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binary signaling is not very efficient (one bit at a time) to increase the throughput we can use multilevel signaling

many ways to do so, we will just scratch the surface

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9.3



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9.3

PAM

mapper: split incoming bitstream into chunks of M bits


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p g

chunks dene a sequence of integers k [n] {0, 1, . . . , 2M 1} a[n] = G ((2M + 1) + 2 k [n]) (odd integers around zero)

slicer:

a

[n] = arg minaA[|a[n]a|]

9.3

PAM, M = 2 , G = 1


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3 1 1 30

distance between points is 2G

using odd integers creates a zero-mean sequence

9.3

PAM, M = 2 , G = 1


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3 1 1 30



9.3

PAM, M = 2 , G = 1


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3 1 1 30



9.3


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From PAM to QAM


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error analysis for PAM along the lines of binary signaling

can we increase the throughput even further?

heres a wild idea, lets use complex numbers

9.3

From PAM to QAM


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error analysis for PAM along the lines of binary signaling

can we increase the throughput even further?

heres a wild idea, lets use complex numbers

9.3

QAM

mapper: split incoming bitstream into chunks of M bits, M even

M/ 2 bi d PAM [ ]


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use M / 2 bits to dene a PAM sequence ar [n]

use the remaining M / 2 bits to dene an independent PAM sequence ai [

a[n] = G (ar [n] + jai [n])

slicer: a[n] = arg min

aA[|a[n]a|]

9.3

QAM, M = 2 , G = 1

Im


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Re

1 1

1

1

9.3

QAM, M = 4 , G = 1

Im

3


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Re

3 1 1 3

3

1

1

9.3

QAM, M = 8 , G = 1

Im


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Re

9.3

QAM

Im


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Re

9.3

QAM

Im


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Re

9.3

QAM

Im


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Re

9.3

QAM

Im


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Re

9.3

QAM

Im


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Re

9.3

QAM

Im


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Re

9.3

QAM, probability of error


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P err = P [|Re([n])| > G ] + P [|Im([n])| > G ]= 1 P [|Re([n])| < G |Im([n])| < G ]= 1 D f (z ) dz

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9.3



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9.3



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G

9.3



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P err e G

2 2

0

9.3


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transmitted power (all symbols equiprobable and independent):

2s = G 2 12M

aA|a|2


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= G 2 23

(2M 1)

P err e G

2 2

0 e 32 (M +1)

SNR

9.3


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100

10


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1010

10 204-point QAM16-point QAM64-point QAM

0 10 20 30SNR (dB)

P e r r

9.3


100

10


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1010


0 10 20 30SNR (dB)

P e r r

9.3


100

10


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100


0 10 20 30SNR (dB)

P e r r

9.3

QAM, the recipe

pick a probability of error you can live with (e.g. 10 6)


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nd out the SNR imposed by the channels power constraint

M = log 2 1 32

SNRln(p e )

nal throughput will be MW

9.3

QAM, the recipe



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nd out the SNR imposed by the channels power constraint

M = log 2 1 32

SNRln(p e )


9.3

QAM, the recipe



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nd out the SNR imposed by the channels power constraint M = log 2 1

32

SNRln(p e )


9.3

QAM, the recipe



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nd out the SNR imposed by the channels power constraint M = log 2 1

32

SNRln(p e )


9.3

QAM

where we stand: we know how to t the bandwidth constraint


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with QAM, we know how many bits per symbol we can use given the p

we know the theoretical throughput of the transmitter

but how do we transmit complex symbols over a real channe

9.3

QAM



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9.3

QAM



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9.3

QAM



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9.3

END OF MODULE 9.3


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Digital Sig


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Module 9.4: Modulation a

Overview:

Trasmitting and recovering the complex passband signal


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Design example

Channel capacity

9.4

Overview:



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Design example

Channel capacity

9.4

Overview:



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Design example

Channel capacity

9.4

QAM transmitter design

..01100


a[n]


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. . .b [n]

b [n] = b r [n] + jb i [n] is a complex-valued baseband signal

9.4

QAM transmitter design

..01100


a[n]


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. . .b [n]

b [n] = b r [n] + jb i [n] is a complex-valued baseband signal

9.4

Complex baseband signal

1


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min max 0 0

9.4

The passband signal

s [n] = Re

{b [n]e j c n

}


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{ }= Re{(b r [n] + jb i [n])(cos c n + j sinc n)}= b r [n]cos c n b i [n]sinc n

9.4

The passband signal

s [n] = Re

{b [n]e j c n

}


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9.4

The passband signal

s [n] = Re

{b [n]e j c n

}


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9.4


1


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DTFT {b r [n ]}

0

1

0 R e

9.4


1


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DTFT {b r [n ]}DTFT {b i [n ]}

0

1

0 R e

9.4


1

e


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DTFT {b i [n ]cos c n }DTFT {b i [n ]}

0

1

0 R e

9.4


1

e


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DTFT {b i [n ]cos c n }DTFT { b i [n ]sinc n }

0

1

0 R e

9.4

Recovering the baseband signal

lets try the usual method (multiplying by the carrier, see Module

s [n]cosc n = b r [n]cos2 c n b i [n]sinc n cos c n


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= b r [n]

1 + cos 2c n2 b i [n]

sin2c n2

= 12

b r [n] + 12

(b r [n]cos2c n b i [n]sin2c n)

9.4





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= b r [n]1 + cos 2c n2 b i [n]sin2c n

2

= 12

b r [n] + 12


9.4


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1 2 i 2


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= b r [n]1 + cos 2c n2 b i [n]sin2c n

2

= 12

b r [n] + 12


9.4


DTFT {b r [n]cos c n b i [n]sinc n}1


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0

1

0 R e

9.4


DTFT {(b r [n]cos c n b i [n]sinc n)cos c n}1


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0

1

0 R e

9.4


DTFT {(b r [n]cos c n b i [n]sinc n)cos c n}1


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0

1

0 R e

9.4


DTFT {b r [n]}1


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0

1

0 R e

9.4


as a lowpass lter, you can use the same lter used in upsampling

h d l h


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matched lter technique

9.4


as a lowpass lter, you can use the same lter used in upsampling

h d l h i


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matched lter technique

9.4


similarly:

s [n]sinc n = b r [n]cos c n sin c n b i [n]sin2 c n


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= 12

b i [n] + 12

(b r [n]sin2c n b i [n]cos2c n)

9.4


similarly:



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= 12

b i [n] + 12


9.4


similarly:



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= 12

b i [n] + 12


9.4


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QAM receiver, idealized design

s (t ) cosc

nsin c n

+

b r [n]

j b i [n]

s [n]


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K Slicer Descrambler ..0110001010.a[n]

9.4

Example: the V.32 voiceband modem

analog telephone channel: F min = 450Hz, F max = 2850Hz

usable bandwidth: W = 2400Hz, center frequency F c = 1650Hzk h


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pick F s = 3 2400 = 7200Hz, so that K = 3 c = 0 .458

9.4



usable bandwidth: W = 2400Hz, center frequency F c = 1650Hzi k F 3 2400 7200H h K 3


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9.4



usable bandwidth: W = 2400Hz, center frequency F c = 1650Hzi k F 3 2400 7200H h K 3


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9.4



usable bandwidth: W

= 2400Hz, center frequency F

c = 1650Hz i k F 3 2400 7200H th t K 3


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9.4


maximum SNR: 22dB

pick P err = 10 6

using QAM, we ndM l 1

3 1022/ 104 1865


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M = log2 1 2 ln(10 6) 4.1865so we pick M = 4 and use a 16-point constellation

nal data rate is WM = 9600 bits per second

9.4


maximum SNR: 22dB

pick P err = 10 6

using QAM, we ndM l 1

3 1022/ 104 1865


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9.4


maximum SNR: 22dB

pick P err = 10 6

using QAM, we ndM l g 1

3 1022/ 104 1865


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9.4


maximum SNR: 22dB

pick P err = 10 6

using QAM, we ndM = log 1

3 1022/ 104 1865


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9.4

Theoretical channel capacity

we used very specic design choices to derive the throughput

what is the best one can do?

Shannons capacity formula is the upper boundC W l (1 SNR )


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C = W log2 (1 + SNR )

for instance, for the previous example C 17500 bps the gap can be narrowed by more advanced coding techniques

9.4




Shannons capacity formula is the upper boundC W l (1 SNR )


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C = W log2 (1 + SNR )


9.4




Shannons capacity formula is the upper boundC W l (1 + SNR )


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C = W log2 (1 + SNR )


9.4




Shannons capacity formula is the upper boundC W log (1 + SNR )


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C = W log2 (1 + SNR )


9.4




Shannons capacity formula is the upper boundC = W log (1 + SNR )


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C = W log2 (1 + SNR )


9.4

END OF MODULE 9.4


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Digital Sig

Module 9.5


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Overview:

Adaptive equalization

Timing recovery


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9.5

Overview:


Timing recovery


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9.5

A blast from the past


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9.5

A blast from the past

a sound familiar to anyone whos used a modem or a fax machine

whats going on here?


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9.5

Graphically

s (t ) cosc

nsin c n

+

b r [n]

j b i [n]

s [n]


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K Slicer Descrambler ..01100

01010.

a[n]

9.5

Pilot tones

if s [n] = cos(( c + 0)n):

b [n] = H{cos((c + 0)n)cos(c n) j cos((c + 0)n)sin(c n)}= H{cos(0n) + cos((2 c + 0)n) j sin((2c + 0)n) + j sin= cos( 0n) + j sin(0n)


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cos( 0n) j sin(0n)

= e j 0n

9.5

Pilot tones

if s [n] = cos(( c + 0)n):



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cos( 0 ) j sin(0 )

= e j 0n

9.5

Pilot tones

if s [n] = cos(( c + 0)n):



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( 0 ) j ( 0 )

= e j 0n

9.5

Pilot tones

if s [n] = cos(( c + 0)n):



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( 0 ) j ( 0 )

= e j 0n

9.5

In slow motion


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9.5

Its a dirty job...

but a receiver has to do it: interference

propagation delay linear distortion


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clock drifts

9.5

Its a dirty job...




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clock drifts

9.5

Its a dirty job...




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clock drifts

9.5


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Its a dirty job...

but a receiver has to do it: interference handshake and line probing



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clock drifts

9.5

Its a dirty job...


propagation delay delay estimation linear distortion


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clock drifts

9.5


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Its a dirty job...


propagation delay delay estimation linear distortion adaptive equalization


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clock drifts timing recovery

9.5

The two main problems

s [n] D/A D ( j ) A/D

T s T s

s (t ) s (t )

h l di t ti D( j)


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channel distortion D ( j )

(time-varying) discrepancies in clocks T s = T s

9.5


s [n] D/A D ( j ) A/D

T s T s

s (t ) s (t )

channel distortion D( j)


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9.5


s [n] D/A D ( j ) A/D

T s T s

s (t ) s (t )

channel distortion D( j)


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9.5

Delay compensation

Assume the channel is a simple delay: s (t ) = s (t d ) D ( j ) = channel introduces a delay of d seconds

we can write d = ( b + )T s with b N

and | | < 1/ 2 b is called the bulk delay


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is the fractional delay

9.5


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Delay compensation





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9.5

Delay compensation





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9.5

Offsetting the bulk delay (T s = 1)

s [n ]

00

1


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9.5


s ( t )

00

1


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9.5


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b

s [n ]

00

1


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b +

9.5

Estimating the fractional delay

transmit b [n] = e j 0n (i.e. s [n] = cos(( c + 0)n))

receive s [n] = cos(( c + 0)(n b )) after demodulation and bulk delay offset:

b [n] = e j 0(n )

multiply by known frequency


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multiply by known frequencyb [n]e j 0n = e j 0

9.5




b [n] = e j 0(n )



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9.5




b [n] = e j 0(n )



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9.5




b [n] = e j 0(n )



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9.5

Compensating for the fractional delay

s [n] = s (n )T s (after offsetting bulk delay) we need to compute subsample values

in theory, compensate with a sinc fractional delay h[n] = sinc(n + )

in practice, use local Lagrange approximation


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p g g pp

9.5






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9.5






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9.5






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9.5


n 1 n n + 1 1

0

1

2


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1

9.5


n 1 n n + 1

n + 1

0

1

2


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1

9.5

Lagrange approximation (see Module 6.2)

as per usual, choose T s = 1 we want to compute x (n + ), with | | < 1/ 2 local Lagrange approximation around n

x L(

n;t ) =

N

k = N

x [n

k

]L(N )

k (t )

L(N ) (t ) =N t i k = N N


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Lk (t ) =i = N

i = n

k i

k = N , . . . , N

x (n + ) x L(n; )

9.5



x L(n; t ) =

N

k = N

x [n

k ]L(N )

k (t )

L(N ) (t ) =N t i k = N N


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Lk (t ) =i = N

i = n

k i

k = N , . . . , N

x (n + ) x L(n; )

9.5



x L(n; t ) =

N

k = N

x [n

k ]L(N )

k (t )

L(N ) (t ) =N t i k = N N


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Lk (t ) i = N

i = n

k i

k N , . . . , N

x (n + ) x L(n; )

9.5

Lagrange interpolation (N = 1)

n 1 n n + 1 1

0

1

2


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9.5


n 1 n n + 1

n + 1

0

1

2


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9.5


n 1 n n + 1n + 1

0

1

2


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9.5


n 1 n n + 1n + 1

0

1

2


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9.5


n 1 n n + 1n + 1

0

1

2


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9.5


n 1 n n + 1n + 1

0

1

2


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9.5


n 1 n n + 1n + 1

0

1

2


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9.5

Lagrange interpolation as an FIR

x (n + ) x L(n; ) dene d [k ] = L

(N )k ( ), k = N , . . . , N

d [k ] form a (2N + 1)-tap FIR

x L(n; ) = ( x d )[n]


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9.5


x (n + ) x L(n; ) dene d [k ] = L

(N )k ( ), k = N , . . . , N


x L(n; ) = ( x d )[n]


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9.5


x (n + ) x L(n; ) dene d [k ] = L

(N )k ( ), k = N , . . . , N


x L(n; ) = ( x d )[n]


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9.5


x (n + ) x L(n; ) dene d [k ] = L

(N )k ( ), k = N , . . . , N


x L(n; ) = ( x d )[n]


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9.5

Example (N = 1, second order approximation)

L(1) 1(t ) = t t 1

2

L(1)0 (t ) = (1

t )(1 + t )

L(1)1 (t ) = t t + 1

2


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9.5

Example (N = 1, second order approximation)

d 0.2[n] =0.08 n = 10.96 n = 0

0.12 n = 10 otherwise


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9.5

Delay compensation algorithm

estimate the delay

compute the 2N + 1 Lagrangian coefficients

lter with the resulting FIR


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9.5


estimate the delay




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9.5


estimate the delay




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9.5

Compensating for the distortion

s [n] D/A D ( j ) A/D s (t ) s (t )


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9.5

Compensating for the distortion

s [n] D (z ) s [n]


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9.5

Example: adaptive equalization

s [n] D (z ) E (z ) s e [n] = ss [n]


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9.5


in theory, E (z ) = 1 / D (z )

but we dont know D (z ) in advance

D (z ) may change over time


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9.5






289/354

9.5






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9.5


s [n] E (z )

- s [n]

s e [n]

e [n]


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9.5

Adaptive equalization: bootstrapping via a training sequenc

a t [n] TX . . .s [n]


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9.5

Adaptive equalization: bootstrapping via a training sequenc

a t [n] TX . . .s [n]

s [n] E (z )

s e [n]

s [n]e [n]


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- Modulator at [n

9.5

Adaptive equalization: online mode

s [n] E (z ) Demod Slicer

- Modulator

s e [n]

s [n]e [n]


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9.5

So much more to do...

how do we perform the adaptation of the coefficients?

how do we compensate for differences in clocks?

how do we recover from interference?

how do we improve resilience to noise?

adaptive signal processing


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9.5








296/354


9.5








297/354


9.5








298/354


9.5








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9.5


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END OF MODULE 9.5


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Digital Sig

M


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Overview:

Channel

Signaling strategy

Discrete Multitone Modulation (DMT)


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9.6


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Overview:

Channel

Signaling strategy

Discrete Multitone Modulation (DMT)


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9.6

The telephone network today

voicenetwork CO

CO

DSLAM


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internet

9.6

The telephone network today

voicenetwork CO

CO

DSLAMlast mile


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internet

9.6

The last mile

copper wire (twisted pair) between home and nearest CO

very large bandwidth (well over 1MHz)

very uneven spectrum: noise, attenuation, interference, etc.


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9.6

The last mile





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9.6

The last mile





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9.6


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The ADSL channel

0 1MHz


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9.6

Idea: split the band into independent subchannels

0 1MHz


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9.6

Subchannel structure

allocate N subchannels over the total positive bandwidth

equal subchannel bandwidth F max / N

equally spaced subchannels with center frequency kF max / N , k = 0 , . . . , N


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9.6






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9.6






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9.6

The digital design

pick F s = 2F max (F max is high now!)

center frequency for each subchannel k = 2kF max / N

F s =

22N

k

bandwidth of each subchannel 22N

to send symbols over a subchannel: upsampling factor K 2N


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9.6

The digital design



F s =

22N

k

bandwidth of each subchannel 22N to send symbols over a subchannel: upsampling factor K 2N


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9.6

The digital design



F s =

22N

k



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9.6

The digital design



F s =

22N

k



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9.6

The digital design (N = 3)

0


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9.6


0 = 0 1 = 2 / 6 2 = 4 / 6 1 2


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9.6


0 = 0 1 = 2 / 6 2 = 4 / 6 1 2


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9.6

The digital design

put a QAM modem on each channel

decide on constellation size independently

noisy or forbidden subchannels send zeros


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9.6

The digital design





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9.6

The digital design





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9.6

The subchannel modem

e j (2/ 2N )kn

ak [m] 2N

b k [n] c k [n]


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9.6

The bank of modems

e j (2/ 2N )(0 n )

a0[m] 2N b 0[n] c 0[n]

e j (2/ 2N )(1 n )

a1[m] 2N b 1[n] c 1[n]

e j (2/ 2N )(2 n )

a2[m] 2N

b 2[n] c 2[n] + Re

. . .e j (2/ 2N )( N 2)n

aN 2[m] 2N b N 2[n] c N 2[n]

e j (2/ 2N )( N 1)n

c [n]


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aN 1[m] 2N b N 1[n] c N 1[n]

9.6

If it looks familiar...

check back Module 4.3, the DFT reconstruction formula:

A 0 0

0

A 1 1

1

A 2 2

2 + x [n]

. . .

AN 2N 2

N 2

AN 1N 1

N 1


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N 1

9.6

DMT via IFFT

we will show that transmission can be implemented efficiently via an IF

Discrete Multitone Modulation


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9.6

DMT via IFFT

we will show that transmission can be implemented efficiently via an IF

Discrete Multitone Modulation


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9.6

The great ADSL trick

instead of using a good lowpass lter, use the 2N -tap interval indi

h[n] =1 for 0 n < 2N 0 otherwise


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9.6

Interval indicator signal (Module 4.7)

0 2N 10

1


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9.6

DTFT of interval signal (Module 4.7)

/ N 0 0

1

| H ( e j ) |


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9.6

DTFT of interval signal (Module 4.7)

/ N 0 0

1

| H ( e j ) |


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9.6

Back to the subchannel modem

e j (2/ 2N )kn

ak [m] 2N b k [n] c k [n]

rate: B symbols/sec 2NB samples/sec


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9.6


e j (2/ 2N )kn

ak [m] 2N b k [n] c k [n]

rate: B symbols/sec 2NB samples/sec


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9.6


by using the indicator function as a lowpass:

e j (2/ 2N )nk

ak [n/ 2N ] c k [n]


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9.6

The bank of modems, revisited

e j (2/ 2N )(0 n)

a0[n/ 2N ] e j (2/ 2N )(1 n)

a1[n/ 2N ] e j (2/ 2N )(2 n)

a2[n/ 2N ] + Re s [n]. . .

e j (2/ 2N )(N 2)n

aN 2[n/ 2N ] e j (2/ 2N )(N 1)n

aN 1[n/ 2N ]

c [n]


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9.6

The complex output signal

c [n] =N 1

k =0

ak [n/ 2N ]e j 22N nk

= 2 N IDFT2N a0[m] a1[m] . . . aN 1[m] 0 0 . . . 0(m = n/ 2N )


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9.6

The complex output signal

c [n] =N 1

k =0

ak [n/ 2N ]e j 22N nk

= 2 N IDFT2N a0[m] a1[m] . . . aN 1[m] 0 0 . . . 0(m = n/ 2N )


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9.6

We can do even better!

we are interested in s

[n

] = Re{c

[n

]} = (c

[n

] + c

[n

])/ 2 it is easy to prove (exercise) that:

IDFT x 0 x 1 x 2 . . . x N 2 x N 1= IDFT x 0 x N 1 x N 2

c [n] = 2N

IDFT a0[m] a1[m] . . . aN 1[m] 0 0 . . . 0 [n]

therefore

s [n] = N IDFT 2a0[m] a1[m] . . . aN 1[m] aN 1[m] aN 2[m]


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9.6


we are interested in s [n] = Re

{c [n]

} = ( c [n] + c [n])/ 2

it is easy to prove (exercise) that:


c [n] = 2N

IDFT a0[m] a1[m] . . . aN 1[m] 0 0 . . . 0 [n]

therefore



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9.6



{c [n]

} = ( c [n] + c [n])/ 2



c [n] = 2N

IDFT a0[m] a1[m] . . . aN 1[m] 0 0 . . . 0 [n]

therefore



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9.6



{c [n]

} = ( c [n] + c [n])/ 2



c [n] = 2N

IDFT a0[m] a1[m] . . . aN 1[m] 0 0 . . . 0 [n]

therefore



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9.6

ADSL transmitter

2a 0[m ] a 1[m ] a 2[m ]

a N 1[m ] a N 2[m ]

I F F T

s [2Nm ]s [2Nm + 1]

s [2Nm + 2]s [2Nm + 3]s [2Nm + 4]s [2Nm + 5]

s [2Nm + N 1]s [2Nm + N 2]s [2Nm + N 3]s [2Nm + N 4]

p a r a

l l e

l t o

s e r

i a l

s


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9.6

ADSL specs

F max = 1104KHz

N = 256

each QAM can send from 0 to 15 bits per symbol

forbidden channels: 0 to 7 (voice)

channels 7 to 31: upstream data

max theoretical throughput: 14.9Mbps (downstream)


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9.6

ADSL specs

F max = 1104KHz

N = 256






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9.6

ADSL specs

F max = 1104KHz

N = 256






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9.6

ADSL specs

F max = 1104KHz

N = 256






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9.6

ADSL specs

F max = 1104KHz

N = 256






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9.6

ADSL specs

F max = 1104KHz

N = 256






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9.6

END OF MODULE 9.6


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END OF MODULE 9


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