ADC 11 Digital Modulation

ANALOG & DIGITAL COMMUNICATION

By Engr. Hyder Bux MangrioInstitute of Information & Communication

TechnologiesMehran University of Engineering and Technology,

Jamshoro.

09TL-BATCH

Lecture#:39-43Digital Modulation

05/03/[email protected] 1

Transmitting digital data through the public telephone network

For the telephone network, modem are used that are produce signals in the voice frequency range

Public telephone system 300Hz to 3400Hz Use modem (modulator-demodulator)

Amplitude shift keying (ASK) Frequency shift keying (FSK) Phase shift keying (PK)


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 Up to 1200bps on voice grade lines Used over optical fiber


Most common form is binary FSK (BFSK) Two binary values represented by two

different frequencies (near carrier) Less susceptible to error than ASK Up to 1200bps on voice grade lines High frequency radio Even higher frequency on LANs using

coax


Phase of carrier signal is shifted to represent data

Binary PSKTwo phases represent two binary digits

Differential PSKPhase shifted relative to previous

transmission rather than some reference signal


Draw the output waveforms of the following binary bit stream 1101001 into;

1. ASK2. PSK3. FSK(2 cycles per unit for logic 1 and one

cycle per unit for logic 0)


Draw the output wave form of following data 010101001 into

1.ASK2.BPSK3.DPSK


Balanced Modulator acts as a phase reversing switch.

BM has two inputs: a carrier that is in phase with reference oscillator and the binary digital data.


Level converte

rBPFBM

Carrier oscillato

r

Modulated PSK output

Binary Data


Binary Data Output phase

Logic 0Logic 1

1800

00

sin wct(00)Logic 1

-sin wct(1800)Logic 0

cos wct(+900)

-cos wct(-900)

00 ReferenceLogic 1

±1800 Logic 0

cos wct

-cos wct

(a)Truth Table(b)Phasor diagram

(c) Constellation

The coherent carrier recovery circuit detects and regenerates a carrier signal in both frequency and phase coherent with the original transmitted carrier.

The Balanced modulator is product detector; the output is product of two inputs (the BPSK signal and recovered carrier).

LPF separates the recovered binary data from complex demodulated signal.



BPF

Clock recover

y

Coherent carrier

recovery

BM LPFLevel

converter

sinwct

For BPSK, the output rate of change(baud) is equal to the input rate of change (bps), and the widest output bandwidth occurs when the input binary data are alternating 1/0 sequence.

The fundamental frequency (fa) of an alternative 1/0 bit sequence is equal to one half of the bit rate (fb/2)

BPSK output=[sin(2π fa)t]x[sin(2π fc)t]


Where fa = maximum fundamental frequency of

binary input.fc = reference carrier frequency The minimum double-sided Nyquist

bandwidth is


b

b

a

aCaC

aCaC

ffB

fBffffB

tfftff

222

])(2cos[21])(2cos[

21

Quaternary phase shift keying QPSK is an M-ary encoding scheme. Four output phases (+450, +1350, -450, -

1350) are possible for a single carrier frequency.

The binary input data are combined into groups of two bits called dibits.

QPSK transmitter Two bits (a dibit) are clocked into the bit

splitter QPSK modulator is two BPSK modulator

combined in parallel.05/03/[email protected] 21


BPF

Bit splitter

Q

I

+2

Balanced

Modulator

Ref. Carrier oscillato

r

BPF

Linear Summe

r

900 phase shift

BM

BPF

Binary input data fb

Bit clock

I channel fb/2

Q channel fb/2

Logic 1= +1 VLogic 0= -1 V

Logic 1= +1 VLogic 0= -1 V

cosωct

±cosωct

sinωct

±sinωct


Binary input QPSK outputPhase

I Q

0 0 0 1 1 0 1 1

-135-45+135+45

-cos wct

cos wct

sin wct-sin wct

sin wct-sin wct

cos wct

-cos wct

Q Icos wct+sin wct1 1(sin wct+450)

Q Icos wct-sin wct1 0(sin wct+1350)

Q I-cos wct-sin wct0 0(sin wct-1350)

Q I-cos wct+sin wct0 1(sin wct-450)


+900

QPSKSignal

BPF IQ

LPF

Product detector

Product detector

LPF Clock recovery

Power splitter Carrier

I channel

Q channel

QPSK Receiver

Recovery binary value

The highest fundamental frequency at the input and the fastest rate of change at the output of the balance modulator is equal to the one fourth of the binary input rate.

BW= fb/2


Offset QPSK 8-PSK 16-PSK


To achieve high data rate with narrowband channel is to increase the number of bits per symbol

Use combination of amplitude and phase modulation known as Quadrature Amplitude Modulation (QAM)

QAM has finite number of allowable amplitude phase combinations

Constellation diagram shows the possibilities for hypothetical system with sixteen amplitude-phase combinations.


Each transmitted symbol represents four bits

Each dot represents a possible amplitude/phase combination or state

QAM is more efficient in term of bandwidth than either FSK or QPSK but it is also more susceptible to noise

Another disadvantage compared to FSK is that QAM signals like analog signals, vary in amplitude. This means that transmitter amplifiers must be linear.



Product Modulator

2 to 4 level converter

2 to 4 level converter

Product Modulator

Reference Oscillator

+900

Q I CLinear summer

BPF

I channel

Q channel

PAM

PAM

8-QAM output

I/Q C Output

0 0 -0.541V

0 1 -1.307

1 0 +0.541

1 1 +1.307


000

001

101

100

010

011

110

111

110

010

011

111101

001

000

100

Fig: Phasor diagramFig: Constellation diagram

32

All advanced modems use a combination of modulation techniques to transmit multiple bits per baud.

Multiple amplitude and multiple phase shifts are combined to transmit several bits per symbol.

QPSK (Quadrature Phase Shift Keying) uses multiple phase shifts per symbol.

Modems actually use Quadrature Amplitude Modulation (QAM).

These concepts are explained using constellation points where a point determines a specific amplitude and phase.

33

(a) QPSK. (b) QAM-16. (c) QAM-64.

The North American TDMA digital cell phone standard transmit at 24.3 kilo baud using DQPSK. What is channel data rate.


A modem uses sixteen different phase angles and four different amplitudes. How many bits does it transmit for each symbol.


How much bandwidth would be required to transmit a DS-1 signal (1.544Mbps) using a four level code

(a)Assuming a noiseless channel?(b)With a signal to noise ratio 15dB?


Consider a QPSK system that will transmit three bits of information per symbol

(a)How many phase angle are needed?(b)Draw a vector diagram for such a

system


A microwave radio system uses a 256-QAM, there are 256 possible amplitudes and phase combinations.

(a)How many bits per symbol does it uses?(b)If it has channel with 90MHz bandwidth,

what is its maximum data rate, ignoring noise?


ADC 11 Digital Modulation

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