digicom 102

DIgital Communication ECE 422L

2013

Bandpass Modulation

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Transmission of Digital Signal

1. Digital transmission• Baseband data transmission• Data is directly transmitted without

carrier• Suitable for short distance

transmission

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Transmission of Digital Signal

2. 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 is the smallest quantity, a bit, 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 element vs. Signal element

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

• r as 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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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)

Amplitude Shift Keying

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Amplitude Shift Keying

modulation produces aperiodic composite signal, with continuous set of frequencies

bandwidth is proportional to the signal ( baud ) rate

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Amplitude Shift Keying

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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)

Example 3Example 3

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

SolutionSolution

In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.

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

In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 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 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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0 → regular frequency ; 1 → increased frequency

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

5.33

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?

The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means

Example 6Example 6

SolutionSolution

5.34

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

5.35

Bandwidth of MFSK

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

• Binary PSK (BPSK)implemented using two signal elementsone with phase 0o and other with 180o

Phase Shift Keying

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

(less bandwidth)

Binary PSK

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phase 0o → 0 bit ; phase 180o → 1 bit

Binary PSK

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the digital signal used here is polar NRZ

Binary PSK : implementation

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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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8 PSK: waveform

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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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Constellation diagram

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P = 0P = 180

• Binary PSK

Constellation diagram

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• 4-PSK characteristics

Constellation diagram

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• 8-PSK characteristics

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Constellation diagram

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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

QAM is a combination of ASK and PSK

Quadrature Amplitude Modulation

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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

QAM is a combination of ASK and PSK

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

Time domain for an 8-QAM signal

16-QAM constellations

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Bit and baud rate comparison

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 8N

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

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

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

References

• Data and Computer Communications Ninth Edition

– by William Stallings

• Advance Electronic Communication System– By Wayne Tomasi

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