COSC 3213: Communication Networks: Chapter 3 Handout #4 · 2008. 10. 14. · CSE3213 F08 M. Mandelbaum 4 Line Coding (2) 2. Polar NRZ: Bit 1 is represented by +A/2 Volts Bit 0 is

Instructor: Dr. Marvin MandelbaumDepartment of Computer Science

York UniversitySection E F08

Topics:1. Line Coding: Unipolar, Polar,and Inverted NRZ; Bipolar; Manchester and

Differential Manchester. 2. Digital Modulation Schemes: ASK, PSK, FSK, and QAM3. Modems4. Transmission Media

Garcia: Sections 3.5 – 3.8

COSC 3213: Communication Networks: Chapter 3 Handout #4

CSE3213 F08 M. Mandelbaum 2

Line Coding

Binary info to Digital signal

• How to choose method? Criteria?– Maximize Bit rate subject to bandwidth constraint– Ease of retrieving information– Error detection capability– Immune to noise– Complexity– cost

1 0 1 0 1 1 0 01

Signal (Digital)

Binary Info


Line Coding (1)

Line Coding:

Converts a binary sequence into a digital signalUnipolar NRZ: Bit 1 is represented by +A Volts

Bit 0 is represented by 0 Volts

Average transmitted power per pulse = 1/2 ×

(A2) + 1/2 ×

(0) = A2

/ 2Average value of signal = A / 2 VoltsUnipolar: since always positive or zero never negative since have the same signNon-return-to-zero: Uses two voltage levels to depict the bit value and keeps them constant for the whole

bit duration. (that is there is no return to zero or transition during the bit interval sometime called NRZ-Level

1 0 1 0 1 1 0 01Unipolar

NRZ


Line Coding (2)

2.

Polar NRZ:

Bit 1 is represented by +A/2 VoltsBit 0 is represented by –A/2 Volts

Average transmitted power per pulse = 1/2 ×

(A/2)2

+ 1/2 ×

(−A/2)2

= A2

/ 4Half the power used as compared to Unipolar

NRZ with same distance between levelsAverage value of signal = 0 VoltsProblem systematic error in polarity causing error in allPolar: uses positive one value (for bit 1) and negative for the otherNRZ: constant level of voltage for duration.

1 0 1 0 1 1 0 01Unipolar

NRZ

Polar NRZ


Line Coding (3)

3.

NRZ-Inverted:

First bit 1 is represented by +A/2 VoltsNo change for bit 0; Flip to the opposite voltage for next bit 1

Average transmitted power per pulse = A2

/ 4Average value of signal = 0 VoltsNRZ since stays level or constant voltage for bit duration. But Inverted because different appearance of a

specific bit value may have a new inverted polarity.Differential encoding because information in terms of change between successive data symbols

1 0 1 0 1 1 0 01Unipolar

NRZ

Polar NRZ

NRZ-Inverted(DifferentialEncoding)


Line Coding (4)4.

Bipolar:

Bit 0 is represented by 0 VoltsFirst bit 1 by +A/2 Volts; Consecutive 1’s by +A/2 and –A/2


/ 8Average value of signal = 0 VoltsBetter scheme for strings of 1s in data. Better for passing channels that block low frequencies.Long strings of zeros cause problem. Bipolar since one of the bit symbols is represented by opposite

polarity each time.

1 0 1 0 1 1 0 01Unipolar

NRZ

Polar NRZ

NRZ-Inverted

BipolarEncoding


A/2-A/2

A/2-A/2

Line Coding (5)

5.

Manchester:

Bit 1

Bit 0


/ 4; Average value of signal = 0 Volts

1 0 1 0 1 1 0 01Unipolar

NRZ

Polar NRZ

NRZ-Inverted

BipolarEncoding

Manchester


A/2-A/2

Line Coding (6)

6.

Differential Manchester:

Bit 1

Next 0: No change

Next 1: Flip to


/ 4; Average value of signal = 0 Volts

A/2-A/2

1 0 1 0 1 1 0 01UnipolarNRZ

Polar NRZ

NRZ-Inverted

BipolarEncoding

Manchester

DifferentialManchester


Line Coding (7)Power spectra of different line coding schemes:

NRZ is typically used for lowpass

channels.Bipolar is used for telephone transmission system that do not pass frequencies < 200 HzManchester used for LAN (Manchester for Ethernet & Differential for Token ring) where

bandwidth efficiency is relatively not important

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0

0.2

0.4

0.6

0.8 1

1.2

1.4

1.6

1.8 2

f T

pow

er d

ensi

tyNRZ: UnipolarBipolarManchester


Digital Modulation (1)

Why Modulation:1.

Modulation shifts the frequency content of message signal to a range that is passed by the channel.Example: Spectrum of the signal is represented in green while the channel is represented in blue

2.

Selecting a different carrier frequency for different signals lead to multiple access.3.

Effect of channel noise can be reduced by selecting an appropriate carrier frequency.

ff2f1 fc0

ff2f1 fc0

Output is zero

Modulation shifts the signal spectrum.Signal is transmitted without distortion.



Amplitude Shift Keying (ASK): Represent bit 1 with cos(2πfc t)Represent bit 0 with 0 Volts

Information 1 1 1 10 0

0 T 2T 3T 4T 5T 6TASK t



Frequency Shift Keying (FSK):

Represent bit 1 with cos(2πf1

t)Represent bit 0 with cos(2πf2

t)



0 T 2T 3T 4T 5T 6TFSK t


Digital Modulation (3) Phase Shift Keying (PSK):

Represent bit 1 with cos(2πfc

t)Represent bit 0 with cos(2πfc

t + π) = −cos(2pfc

t)


0 T 2T 3T 4T 5T 6TFSK t


0 T 2T 3T 4T 5T 6TPSK t


PSK - Modulation

1. Represent binary information with a polar NRZ

2. Multiply Xi (t) with sine wave generator


Baseband Signal Xi (t)0 2T 3T 6TT 4T 5T t

Ak x

cos(2πfc t)

Yi (t) = Ak cos(2πfc t)


1 / W


PSK - Demodulation

1. Multiply PSK signal with sine wave generator

Output can be expressed as2Yi (t) cos(2πfc t) = Ak ( 1 + cos(2π2fc t))

Transmission rate = 1 bit / pulse duration = W bps for a channel with BW = W Hz

0 T 2T 3T 4T 5T 6Tt

Yi (t) x

2 cos(2πfc t)

2Yi (t) cos(2πfc t)


1 / W


PSK – Demodulation (2)

2. Filter out the high frequency component

3. Ak is retrieved


Yi (t) x

2 cos(2πfc t)

2Yi (t) cos(2πfc t)Low-pass Filterwith cutoff fc Hz

Baseband Signal Xi (t)0 2T 3T 6TT 4T 5T t


Trig Identities

• 2cos2(x) = 1 + cos(2x)• 2sin2(x) = 1 – sin(2x)• 2cos(x)sin(x) = 0 + sin(2x)

cos(A +- B) = cos(A)cos(B) –+ sin(A)sin(B)

sin(A +- B) = sin(A)cos(B) +- cos(A)sin(B)

Show or check these identities below


Quadrature Amplitude Modulation QAM

• Signaling technique to speed up data rate from W to 2W• Put ASK and PSK together• Send two signals simultaneously on same carrier• f1

= f• f2

= f + π/2

shift 90 degree

M DM

f1

f2

f1

f2 Signal 2

Signal 1

Signal 1

Signal 2

f1

+ f2 + π


1.

Split information stream (rate 2W) into two sequences of odd and

even numbered bits each rate W

2.

Determine the bipolar NRZ Bk

for even and Ak

for odd3.

Multiply Ak

with cosine wave and Bk

with sin wave: Modulate

Ak x

cos(2πfc t)

Yi (t) = Ak cos(2πfc t)

Bk x

sin(2πfc t)

Yq (t) = Bk sin(2πfc t)

+ Y(t)

QAM – Modulation

to speed up data rate

1110 … (even #)

0010 … (odd #)Splitter01011100 …

Y(t) = Ak cos(2πfct) + Bk sin(2πfct)

In phaseQuadrature phase

θ

= 2π

fc

t

In band widthBand pass channel

f1 fc f2

Two phase system

Send over bandpass channel


QAM – Signal Constellation

3.

Signal Constellation: is a method of representing signal states in terms of inphase (cos(2πfc t))

and quadrature (sin(2πfc t)) components.How to draw signal constellation for Y(t)

= Ak cos(2πfc t) + Bk sin(2πfc t)⎯

Take cos(2πfc t) as the horizontal axis and sin(2πfc t) as the vertical axis⎯

Derive all possible combinations of Y(t) = Ak cos(2πfc t) + Bk sin(2πfc t) by selecting different combinations of Ak and Bk. They both have values 1 and -1

⎯

Represent each combination of Y(t) = Ak cos(2πfc t) + Bk sin(2πfc t) as a coordinate (Ak , Bk ) on the Cartesian axis

Activity I: Draw constellation for QAM system where Ak and Bk have values (−1,−2/3,1/3,1)?

sin(2πfc t)

cos(2πfc t)1

1

−1

−14 levels / pulse2 bits / pulse2W bps

1 1

1 -1

-1

1

-1

-1

voltage


QAM - Demodulation

Multiply Y(t) with sine and cosine wave followed by lowpass

filtering

Activity II:

Show that the above system restores Ak and Bk ?

Y(t) x

2cos(2πfc t)

x

2sin(2πfc t)

Low-pass Filterwith cutoff W/2 Hz

Low-pass Filterwith cutoff W/2 Hz

Ak

Bk

QAM is a 2-d system Effective signal rate of 2W pulses per second over a bandpass

channel of W HzNyquist

rate


Telephone Modems

Telephone Channels:⎯

Passband

range: 500 Hz to 2900 Hz (BW = ?)⎯

SNR = 40 dB⎯

Duration of pulse = ? (1 / BW) Why?⎯

Channel Capacity = ?

⎯

Activity III:

Derive the pulse rate and channel capacity of the telephone channel?

Pulse Rate Modulation V.32bis

2400 pulses/s Trellis 128 w/ 64 valid 6 x 2400 = 14,400 bps

2400 pulses/s Trellis 32 w/ 16 valid 4 x 2400 = 9,600 bps

2400 pulses/s QAM 4 (all valid) 2 x 2400 = 4,800 bps

V.64bis

2400 – 3429 pulses/s Trellis 960 2400 – 33,600 bps


Digital Media

1.

Twisted Pair2.

Coaxial Cable3.

Optical Fiber4.

Microwave

The topic will not be covered explicitly in the classReview section 3.7 for details

Slide Number 1Line CodingLine Coding (1)Line Coding (2)Line Coding (3)Line Coding (4)Line Coding (5)Line Coding (6)Line Coding (7)Digital Modulation (1)Digital Modulation (2)Digital Modulation (3)Digital Modulation (3) Phase Shift Keying (PSK): PSK - ModulationPSK - DemodulationPSK – Demodulation (2)Trig IdentitiesQuadrature Amplitude Modulation QAMQAM – Modulation �to speed up data rateQAM – Signal ConstellationQAM - DemodulationTelephone ModemsDigital Media

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COSC 3213: Communication Networks: Chapter 3 Handout #4 · 2008. 10. 14. · CSE3213 F08 M. Mandelbaum 4 Line Coding (2) 2. Polar NRZ: Bit 1 is represented by +A/2 Volts Bit 0 is