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CSE 3213, W14 1

L12: Basic Baseband Digital Comms

Sebastian Magierowski York University

L12: Baseband Basics

SC/CSE 3213 Winter 2014

CSE 3213, W14 2

•  Basic digital communicator blocks •  Nyquist pulses •  Channel capacity

Outline

L12: Baseband Basics

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•  Attempting to send a sequence of digits through a continuous channel

•  Not easy… •  …modulerize the design

Underlying Idea

L12: Baseband Basics

channel 1101011 1101011

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•  Exchanging sequences of discrete symbols

•  source coding –  increase redundancy

The Information Theory Perspective

•  channel coding –  error correction (add

redundancy)

•  line coding –  map bits to symbols

L12: Baseband Basics

source coder

channel/line coder

channel channel/line decoder

source decoder

data symbols 0 ––– –1 1 ––– +1

00 ––– –3 01 ––– –1 10 ––– +1 11 ––– +3

00 ––– –1 01 ––– –j 10 ––– +1 11 ––– +j

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Line Coding Examples

L12: Baseband Basics

source coder

channel/line coder

channel channel/line decoder

source decoder

data symbols

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•  Transmit filter (TX filter) •  Map symbols into analog waveforms that match channel

response –  bandwidth –  location (baseband/passband)

Continuous-Time Communication

L12: Baseband Basics

channel coder tx filter channel rx filter channel

decoder

symbols waveforms

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•  Appreciate need to design your waveforms to channel

•  Prolonged rise and fall times corrupt energies of adjacent symbols: intersymbol interference (ISI)

Baseband Channel Problems

L12: Baseband Basics [Razavi12]

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•  Random bit patterns sinc2 spectrum

•  Need extremely wide LPF to pass this without distortion

•  Otherwise deal with ISI (intersymbol interference)

A Spectral Perspective

L12: Baseband Basics [Razavi12]

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•  What is the ideal bandlimited pulse shape for the tx filter to produce?

•  A sinc pulse…

•  …has spectrum that is neatly confined (to ±1/2Tb) –  And there’s another important advantage of this…

Pulse Shaping, Sinc Pulses

L12: Baseband Basics [Razavi12]

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•  Pulse goes through zero at equally spaced intervals of Tb –  …such pulses spaced by Tb do not interfere

•  Nyquist pulses

Simple Baseband Modulator (Pulse Shaper)

L12: Baseband Basics

coder tx filter channel rx filter decoder

coder sinc FIR channel sampler decoder

[Razavi12]

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Example of Composite Waveform

Three Nyquist pulses shown separately

•  + s(t) •  + s(t-T) •  - s(t-2T)

Composite waveform r(t) = s(t)+s(t-T)-s(t-2T)

Samples at kT r(0)=s(0)+s(-T)-s(-2T)=+1 r(T)=s(T)+s(0)-s(-T)=+1 r(2T)=s(2T)+s(T)-s(0)=-1

Zero ISI at sampling times kT

L12: Baseband Basics -2

-1

0

1

2

-2 -1 0 1 2 3 4 t

T T T T T T

-1

0

1

-2 -1 0 1 2 3 4 t

T T T T T T

r(t)

+s(t) +s(t-T)

-s(t-2T)

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•  For brickwall channel… –  Wc: stop frequency

•  For sinc pulses with 2Tb between zero-xing… –  spectrum limited to 1/2Tb

•  Therefore to prevent distortion (no ISI)… –  make: 1/2Tb = Wc

•  Therefore for Wc channel… –  can send: symbols per second = 2•Wc

•  E.g.: 1 MHz channel allows 2 MSps

Achievable Symbol Rate

L12: Baseband Basics

Wc

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•  Simple sinc has a lot of temporal energy away from main pulse

–  Sampling error can lead to lots of corruption

•  Raised cosine (RC) waveform trades bandwidth for this problem

•  excess-bandwidth/roll-off factor, α –  10-20%

Raised Cosine (RC)

L12: Baseband Basics

coder RC channel sampler decoder

[Razavi12]

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•  Very important for receiver to filter-out noise –  minimize corruption to sampler

•  Split up RC for simultaneous noise filtering and pulse shaping

Root Raised Cosine (RRC)

L12: Baseband Basics

coder RRC channel decoder sampler RRC

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•  What if channel not a perfect brickwall or RC filter? –  our previous assumptions would not match such a general channel

and hence be distorted

•  Compensate for channel with another filter: equalizer (EQ)

Equalization & Matched Filtering

L12: Baseband Basics

coder RRC channel decoder sampler RRC EQ

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•  Account for alphabet size: M –  binary, M = 2 –  multilevel M > 2 –  4-level:

•  R = 2Wc•log2(M) •  For example:

–  Wc = 1 MHz –  M = 4 –  R = 4 Mbps (2 MSps)

Achievable Bit Rate, R

L12: Baseband Basics

00 ––– –1 01 ––– –1/3 10 ––– +1/3 11 ––– +1

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•  What is the bit-rate relative to the spectrum used: •  Simple 2-level baseband modulation 2 bits/s•Hz •  or 2 symbols/s•Hz

Spectral Efficiency

L12: Baseband Basics

2Wc•log2(M) Wc

v =

bit-rate bandwidth

v =

bits s • Hz

[ ]

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•  Receiver makes decision based on transmitted pulse level + noise •  Error rate depends on relative value of noise and level spacing •  Large (positive or negative) noise values can cause wrong decision •  Noise level impacts 8-level signaling more than 4-level signaling

Noise Limits Accuracy

4-LEVEL 8-LEVEL

typical noise

+A

+A/3

-A/3

-A

+A

+5A/7

+3A/7

+A/7

-A/7

-3A/7

-5A/7

-A

L12: Baseband Basics

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•  Noise is characterized by probability density of amplitude samples

•  Likelihood that certain amplitude occurs

•  Thermal electronic noise is inevitable (due to vibrations of electrons)

•  Noise distribution is Gaussian (bell-shaped) as below

Noise Distribution

L12: Baseband Basics

222

21 σ

σπxe−

x 0

time

x

Pr[X(t)>x0 ] = ?

Pr[X(t)>x0 ] = Area under graph

x0

x0

σ2 = Avg Noise Power

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•  Error occurs if noise value exceeds certain magnitude

•  Prob. of large values drops quickly with Gaussian noise

•  Target probability of error achieved by designing system so separation between signal levels is appropriate relative to average noise power

Probability of Error

L12: Baseband Basics

1.00E-121.00E-111.00E-101.00E-091.00E-081.00E-071.00E-061.00E-051.00E-041.00E-031.00E-021.00E-011.00E+00

0 2 4 6 8 δ/2σ

Pr[X(t)>δ/2 ]

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Shannon Channel Capacity: •  The maximum reliable transmission rate over an ideal

channel with bandwidth Wc Hz, with Gaussian distributed noise, and with SNR S/N is

C = Wc log2(1 + S/N) bits per second •  Reliable means error rate can be made arbitrarily small

by proper coding

Shannon Channel Capacity

L12: Baseband Basics

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•  Consider a 3 kHz channel with 8-level signaling. Compare bit rate to channel capacity at 20 dB SNR

•  3 kHz telephone channel with 8 level signaling Bit rate = 2•3000 pulses/sec • 3 bits/pulse = 18 kbps

•  20 dB SNR means 10 log10(S/N) = 20 Implies S/N = 100

•  Shannon Channel Capacity is then C = 3000 log (1 + 100) = 19,963 bits/second

Example

L12: Baseband Basics