Digital Communication 2

DIgital Communication ECE 421L

2013

II. Bandpass Modulation

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Types of Signal Transmission

Digital transmission•Baseband data transmission•Data is directly transmitted without carrier•Suitable for short distance transmission

Analog Transmission• Passband data

transmission• Data modulates high

frequency sinusoidal carrier

• Suitable for long distance transmission

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Analog Transmission

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Data element vs. Signal element

• a data element (bit) is the smallest quantity, that can represent a piece of information

• a signal element (vehicle / carrier) carries data elements (passengers) - can contain one or more bits

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bit rate : the number of data elements transmitted per second

baud rate : the number of signal elements transmitted per second

Bit and Baud

Bit and Baud

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Data (bit) rate vs. signal (baud) rate

• r is the number of data elements carried by each signal element

• N = bit rate and S = baud rate• S = N x (1 ÷ r) in bauds• r = log2 L where L is the type of signal elementin analog transmission, S ≤ r

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An analog signal carries 4 bits per signal element. If 1000 signal elements are transmitted per second, find the bit rate.

r = 4 S = 1000 N = S x r = 4000 bps

Example 1Example 1

SolutionSolution

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An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud.How many data elements are carried by each signal element ?

N = 8000S = 1000r = (N ÷ S) = 8

Example 1Example 1

SolutionSolution

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Analog Transmission

• three mechanisms of modulating digital data into an analog signal by altering any of the three characteristics of analog signal:

amplitude → ASK : amplitude shift keyingfrequency → FSK : frequency shift keyingphase → PSK : phase shift keying

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Types of Analog Transmission

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Amplitude Shift Keying

•amplitude of the carrier signal is varied to create signal elements frequency and phase remain constant•implemented using two levels

– Binary ASK (BASK)– also referred to as on-off-keying (OOK)


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• modulation produces aperiodic composite signal, with continuous

set of frequencies• bandwidth is proportional

to the signal ( baud ) rate

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• In data communications, it is normally to use full-duplex links with communication in both directions.

• bandwidth is divided into two with two carrier frequencies, as• The figure shows the positions of two carrier frequencies

and the bandwidths.• The available bandwidth for each direction is now 50 kHz,

which leaves us with a data rate of 25 kbps in each direction.


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4, 8,16 … amplitudes can be used for the signal

data can be modulated using 2, 3, 4 … bits at a time

in such cases, r = 2, r = 3, r = 4, ….

Multi-level ASK (MASK)

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Example 3Example 3

Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex.

SolutionSolution• In ASK: baud rate = bit rate Therefore baud

rate = 2000bps.• minimum bandwidth =baud rate. Therefore,

minimum bandwidth = 2000 Hz.

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Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate?

• baud rate = bandwidth, Therefore, baud rate = 5000 bps.

• baud rate = bit rate, Therefore, bit rate = 5000 bps.

Example 4Example 4

SolutionSolution

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• generate carrier using an oscillator• multiply the digital signal by the carrier

signal

Binary ASK : implementation

Merits and Demerits• Values represented by different amplitudes of

carrier• Usually, one amplitude is zero

– i.e. presence and absence of carrier is used• Susceptible to sudden gain changes• Inefficient• Typically used up to 1200bps on voice grade

lines• Used over optical fiber

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• frequency of the carrier signal is varied to represent data

• frequency of the modulated signal is constant for the duration of one signal element and changes for the next signal element if the data element changes amplitude

• Amplitude and Phase remain constant for all signal elements

Frequency Shift Keying

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• implemented using two carrier frequencies:• F1,(space frequency) data elements 0 • f2, (mark frequency) data elements 1

• both f1 and f2 are 2Δf apart

Binary FSK

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Binary FSK

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Modulation Index

∆f = frequency deviationfa = modulating frequency

fb = input bit rate

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• use of a voltage controlled oscillator (VCO)• VCO changes its frequency according to

input voltage

Binary FSK : implementation

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Minimum Shift Keying FSK

• Continuous-phase frequency shift keying• Mark and Space frequency are synchronized

with the input bit rate.• Mark and Space frequency are selected such

that they are separated from the center frequency by an exact odd multiple of one-half of the bit rate

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Minimum Shift Keying FSK

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• it requires synchronizing circuits and is therefore more expensive to implement.

Merits and Demerits

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We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?

Fc = 250 kHz.(midband) We choose 2Δf to be 50 kHz;

Example 6Example 6

SolutionSolution

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We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.

Example 7Example 7

SolutionSolution

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5.35

Bandwidth of Multi level FSK

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Merits and Demerits• Values represented by different frequencies (near

carrier)• Less susceptible to error than ASK• Typically used up to 1200bps on voice grade

lines• High frequency radio• Even higher frequency on LANs using co-ax• Used in cordless and paging system

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• Phase of the carrier signal is varied to represent two or more different signal elements

• amplitude and frequency remain constant

Phase Shift Keying

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• helps defining the amplitude and phase of a signal element

• signal element type is represented as a dot

• the bit or combination of bits it carries is written next to the dot

• diagram has two axesX-axis → related to the in-phase carrierY-axis → related to the quadrature carrier

Constellation diagram

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phase 0o → 1 bit ; phase 180o → 0 bitbandwidth requirement is the same as that of ASK

phase = 0o

phase =180o

Binary PSK

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P = 0P = 180

• Binary PSK


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Merits (a) less susceptible to noise (b) requires only one carrier

(less bandwidth)

Binary PSK

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Binary PSK : implementation

Balanced Modulator

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M-ary Encoding

• M represents the number of possible of condition

Ex. M= 4, 8

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• use of two bits at a time in each signal element → decrease of baud rate → reduction of required bandwidth

• uses two separate BPSK modulations : one in-phase and the other out-of-phase

(quadrature)

Quadrature PSK

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serial toparallel

converter

serial to parallel converter sends one bit to one modulator and the next bit to the other modulator

Quadrature PSK: implementation

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P = 90 P = 180 P = 180 P = 270 P = 0

Quadrature PSK

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• 4-PSK characteristics


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8 PSK: waveform

• 8-PSK characteristics

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ASK BPSK QPSKuses only an

in-phase carrier

A = 1P = 0

A = 1P = 180

A = √2P = +45

Comparison!!

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Differential PSK

– Phase shifted relative to previous transmission rather than some reference signal

– eliminates the need for the synchronous carrier in the demodulation process and this has the effect of simplifying the receiver.

– receiver only needs to detect– phase changes.

Differential PSK

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• small differences in phase are difficult to detect (PSK)

• QAM works on the basis of altering two characteristics of the carrier :

amplitude and phase• two carriers, one in-phase and another

quadrature with two different levels are used

Quadrature Amplitude Modulation

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Quadrature Amplitude Modulation

• Uses more phase shifts than amplitude shifts to reduce noise susceptibility

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(a) 4-QAM with four signal element types similar to ASK or OOK(b) 4-QAM similar to QPSK(c) 4-QAM with a signal with two positive levels(d) 16-QAM with 8 signal levels : 4 +ve & 4 -ve

Constellation diagrams

The 4-QAM and 8-QAM constellations

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Time domain for an 8-QAM signal

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16-QAM constellations

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ModulationModulation UnitsUnits Bits/BaudBits/Baud Baud rateBaud rate Bit Rate

ASK, FSK, 2-PSKASK, FSK, 2-PSK Bit 1 N N

4-PSK, 4-QAM4-PSK, 4-QAM Dibit 2 N 2N

8-PSK, 8-QAM8-PSK, 8-QAM Tribit 3 N 3N

16-QAM16-QAM Quadbit 4 N 4N

32-QAM32-QAM Pentabit 5 N 5N

64-QAM64-QAM Hexabit 6 N 6N

128-QAM128-QAM Septabit 7 N 7N

256-QAM256-QAM Octabit 8 N 8N62

Bit and baud rate comparison

Example 8Example 8

A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?

SolutionSolution

The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud

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Compute the bit rate for a 1000-baud 16-QAM signal.

SolutionSolution

A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus,

(1000)(4) = 4000 bps

Example 9Example 9

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Compute the baud rate for a 72,000-bps 64-QAM signal.

SolutionSolution

A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000 / 6 = 12,000 baud

Example 10Example 10

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Digital Transmission

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Digital Transmission

Digital to Digital Conversion

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Techniques:•Line coding•Block coding•Scrambling

• Also known as Line Encoding

• Baud rate determine the bandwidth

Digital Transmission Methods

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• Nonreturn to Zero• Unipolar• Bipolar

• Return to zero• Unipolar• Bipolar• Bipolar-AMI

• Manchester

Nonreturn to Zero

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• the signal remains at the binary level assigned to it for the entire bit time.

• the voltage does not return to zero during the binary 1 interval

• normally generated inside computers, at low speeds, when asynchronous transmission is being used.

Nonreturn to Zero - Bipolar

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• A bipolar NRZ signal has two polarities, positive and negative.

• The voltage levels are +12 and -12 V. • The popular RS-232 serial computer interface uses bipolar

NRZ, where a binary 1 is a negative voltage between -3 and -25 V and a binary 0 is a voltage between +3 and +25 V.

Return to Zero - Unipolar

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• The binary 1 level occurs for 50 percent of the bit interval, and the remaining bit interval is zero.

• Only one polarity level is used. • Pulses occur only when a binary 1 is transmitted; no pulse is

transmitted for a binary 0.

Return to Zero - Bipolar

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• A 50 percent bit interval 13-V pulse is transmitted during a binary 1, and a 23-V pulse is transmitted for a binary 0.

• Because there is one clearly discernible pulse per bit, it is extremely easy to derive the clock from the transmitted data.

• For that reason, bipolar RZ is preferred over unipolar RZ.

Return to Zero – Bipolar AMI

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• During the bit interval, binary 0s are transmitted as no pulse.• Binary 1s, also called marks, are transmitted as alternating

positive and negative pulses.• One binary 1 is sent as a positive pulse, the next binary 1 as

a negative pulse, the next as a positive pulse, and so on.

Manchester

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• Also referred to as biphase encoding, can be unipolar or bipolar.

• A binary 1 is transmitted first as a positive pulse, for one half of the bit interval, and then as a negative pulse, for the remaining part of the bit interval.

• A binary 0 is transmitted as a negative pulse for the first half of the bit interval and a positive pulse for the second half of the bit interval

Manchester

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• The fact that there is a transition at the center of each 0 or 1 bit makes clock recovery very easy.

• However, because of the transition in the center of each bit, the frequency of a Manchester-encoded signal is two times an NRZ signal, doubling the bandwidth requirement.

• It is widely used in LANs.

Reference

• Data Communication – by Forouzan

• Advance Electronic Communication – by Robert Tomasi

• Principles of Electronic Communication Systems– Louis E. Frenzel Jr.

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Engineering

Digital Communication 2