0% found this document useful (0 votes)
26 views63 pages

Chapter 5 - Digital Modulation

Uploaded by

mrunmay318
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPT, PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
26 views63 pages

Chapter 5 - Digital Modulation

Uploaded by

mrunmay318
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPT, PDF, TXT or read online on Scribd
You are on page 1/ 63

Digital Modulation Techniques

Contents
Definitions
Coherent binary modulation schemes
Non- coherent binary modulation schemes
M-ary modulation
Performance measure : Bandwidth, Power spectra,
advantages, disadvantages,
Comparison of binary and quaternary modulation
schemes
Hierarchy of Modulation Schemes
Digital Modulation can be classified as
1) According to receiver type :
- Coherent Technique
- Non- Coherent Technique
2) According to signaling schemes :
- Binary Signaling
- M-ary Signaling
3) Hybrid schemes:
Binary M-ary Hybrid
ASK QPSK QAM
FSK M-ary FSK (QASK)
PSK M-ary PSK
Performance measure characteristics of Modulation
schemes :
1) Probability of Error (Pe)

2) Power Spectra

3) Bandwidth Efficiency
ASK (Amplitude Shift Keying)

VASK(t) = d(t) sin(2πfct)


VASK(t) = sin(2πfct) ……. bit 1
VASK(t) = 0 ……… bit 0
ASK Modulator
ASK Demodulator
ASK Restoration
Bandwidth
BW = 2fb

Geometrical representation (Signal Space Diagram)


- Euclidean distance , d = √Eb
- Probability of Error (Pe)

Merits
Demerits
BPSK

VBPSK(t) = A.cos(ωct) ……………Bit 1


= A.cos(ωct + π)
= -A.cos(ωct) ………….Bit 0
BPSK Modulator
Signal Space Representation : Geometrical
representation
- d = 2√Eb
- Importance of Euclidean distance
- As d increases, Pe decreases

- Bandwidth = 2fb
QPSK
Need
Concept
Advantages
- Reliable performance
- Reduces Pe
- Efficient utilisation of BW
QPSK Modulator
Dibit Formation
QPSK Transmitter
QPSK Phasor Diagram
Symbol and corresponding
phase shifts
QPSK Demodulator
Bandwidth of QPSK Signal
BW of QPSK = 2 fb/N = 2 fb /2 = fb
Advantages:
- For the same bit error BW required will be half of
BPSK
- Good Noise Immunity
- Possible to increase bit rate without increasing BW.
M-ary Modulation
Review of Binary and Quad Modulation Schemes
Concept: Send one of the M possible signals, during
each signaling interval of T duration.
No. of possible signals, M = 2^N
Types
1. M-ary PSK
2. M-ary QAM
3. M-ary FSK
1.M-ary PSK
M-ary PSK signal generation-
1. Group N-bits together
2. N-bit symbols are extended over period of NTb.
3. Possible symbols, M.
4. M symbols are represented by sinusoids of
duration 2
M

Ts=N Tb, which is differing in phase by


Mathematical expression of M-ary PSK :
Geometrical Representation
(Signal Space Diagram)
Euclidean Distance
Distance Between Signal Points
- Euclidean Distance , d = 2√Es sin (π / M)
- Substitute different values of M

As d increases error probability decreases.


8-PSK System
Here M=8, therefore N=3
Uses 8 different phase shifts to transmit 8 symbols.
Baud Rate = (1/3) Bit Rate
Bandwidth of M- ary PSK
M-ary PSK Transmitter
Advantages-
- As N increases, BW reduces
- Information is transmitted through signal
phase.
Disadvantages-
- Tra. & Rec. is complex
- Pe increases, as N increases
Conserves BW at the cost of increased transmission
power and error probability.
In digital communication system, the bit rate of NRZ
data stream is 1 Mbps and carrier frequency of
transmission is 100 MHz. Find symbol rate of
transmission and BW requirement of channel for
following techniques –
i) BPSK
ii) QPSK
iii) M-ary PSK
Assume you are required to transmit fb = 90 Mbps
data in authorized BW of 20 MHz. Which modulation
technique you will use.
QASK/ QAM
Amplitude + Phase Shift Keying
BPSK, QPSK, M-ary PSK – Same amplitude
 Direct modulation of carriers in quadrature
is involved
QAM- Geometrical Representation
In this Signal space representation : Ts = 4 Tb
M = No.of possible symbols = 16
System must generate 16 distinguishable signals
Here. Each signal point is equallydistant from its
neighbouring point , d = 2a
Consider Ist quad.
- Calculate Avg. normalised energy of Ist quad , Es
- d = 2√0.1Es
Probability of error
BPSK : d = 2√Eb
QPSK : d = 2√Eb
M-ary PSK : 16 –ary PSK : d =
QASK : 16 – ary QASK : d = 2√0.4 Eb

QASK : Low error rate compared to 16 ary PSK


: Higher error rate compared to QPSK
Types of QAM
Scheme Bits/symbol Number of symbols
4-QAM 2 4
8-QAM 3 8
16-QAM 4 16
32-QAM 5 32
64-QAM 6 64
QASK Transmitter
Bit stream b(t) is applied to Serial to Parallel conv
(Clk period = Ts sec)
B(t) are stored by conv and represented in parallel
form.
 O/p of DACs Carriers
Ae(t)
Ao(t)
BW of QASK System
Comparison of QPSK & QASK
FSK

VFSK(t) = A.cos(ωct + d(t) Ωt)


VFSK(t) = A. cos(ωc + Ω)t -----------Bit 1
= A. cos(ωc – Ω)t-----------Bit 0
FSK Modulator
BFSK Transmitter
FSK Demodulator
Geometrical representation (Constellation Diagram)
=
- d = √2Eb
Consider following signal space diagram.
- Compare it’s performance with std. 8 level PSK
system
- Which system will give lower Pe?
M-ary FSK Transmitter
M-ary FSK System
PSD of M-ary FSK
BW of M-ary FSK
Advantages:
- Improved Noise Immunity
- Pe reduces as M increases
Disadvantages:
- Correctly tuned filters are required for reception
- Large BW requirements
Performance Comparison of
Digital Modulation Techniques
Parameters :
- Information is transmitted though
- Equation of system
- No. of bits/ symbol
- Symbol Rate
- No. of possible symbols (M)
- Euclidean distance
- Noise Immunity
- Bandwidth
- Probability of Error
Prepare a comparison table

You might also like