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Digital
Communication
The Digital Communications module builds on the previous course by introducing methods of
using analog modulation schemes to transmit digital information. This module uses a block
diagram approach to generate digital modulation scenarios using analog blocks learned in the first
semester (mixers, filters, etc.). The modulation schemes start with simple ASK and then proceed
through FSK and PSK, finishing up with QAM, QPSK, and higher. Emphasis is put on the
bandwidth utilization of each technique and on throughput.
1.
Given the four major digital modulation schemes (ASK, PSK, FSK, QAM), compare and
contrast each with respect to bandwidth requirements and ease of demodulation.
2.
Given a PSK or QAM modulation system, draw the constellation diagram and discuss
bit-error rate as it relates to the constellation diagram.
3.
Given each of the digital modulation schemes, illustrate methods of demodulation so that
the original data stream is successfully recovered.
4.
Given a communication application with specific power constraints (Eb/No), select a
digital modulation method and provide justification.
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TABLE OF CONTENTS
1.
2.
3.
Classification of Communication Systems ............................................................. 3
Bandwidth Considerations ...................................................................................... 4
Information Capacity .............................................................................................. 5
4.
Digital Modulation .................................................................................................. 9
ASK................................................................................................................... 9
FSK ................................................................................................................. 10
PSK ................................................................................................................. 12
QPSK .............................................................................................................. 12
MPSK.............................................................................................................. 14
Noise and Error Rate Performance ....................................................................... 17
5.
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CLASSIFICATION OF COMMUNICATION SYSTEMS
Analog modulating
Waveform
AM, FM, PM
Analog modulating
Waveform
Digital modulating
Waveform
Analog Modulated
Waveform
Digital Modulated
Waveform
PCM
ASK, FSK, PSK
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Analog modulated
Waveform
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BANDWIDTH CONSIDERATIONS
Rectangular (NRZ)





Rectangular
bipolar(RZ)
 

 
Raised cosine
Transmission systems are designed to use pulses whose shapes maximize the percentage
of total signal power within the first 'lobe' of the spectrum. For a channel with a
bandwidth whide enough to include the first lobe, transmitting raised cosine pulses result
in the reception of more power with less pulse distortion than for rectangular pulses.
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CHANNEL CAPACITY OF A WHITE ADDITIVE GAUSSIAN NOISE
CHANNEL
Hartley Law
C  B T
where
C=information capacity
B=bandwidth
T=transmission time
Shannon's limit for Information Capacity (also known as Shannon-Hartley law):

S 
S

C  2 B log 2 M   2 B log 2  1    B log 2 1  
N
 N

S

 3.32 B log 10 1  
 N
Cc = Capacity in bps
B = Channel Bandwidth in Hz
S
= Ratio of Signal Power to Noise power.
N
The above equation is also referred to as the Shannon-Hartley law.
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Example
Determine the channel capacity if the SNR is 20 dB, and the bandwidth available is 4
kHz (Appropriate for telephone communications)
Solution

10 
C  4 log2 1  10 


 4 log2 101
20
 26.63kbits / sec
value of 100 is appropriate for an SNR of 20 dB.
Example
If it is required to transmit at 50 kbit/s, and a bandwidth of 1 MHz is used, determine the
minimum SNR required.
Solution
50 = 1000 log 2(1+S/N)
C
S
 
 2 W   1
N
 0.035
 14.5dB
This shows that in some sense it may be possible to transmit using signals below the
noise level, using wide bandwidth communication, as in “spread spectrum”
communications.
Let
Eb = Bit energy = STb = S/Rb
N0 = Noise density = N/B
Where Tb = Bit time
And Rb = Bit rate
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Hence:
E b  SB
N 0 N Rb

S E b Rb E b C c


N N0 B
N0 B
If bit rate Rb = CC, then

S Eb Cc

N
N0 B
The Shannon-Hartley law can then be expressed as:
C c  log 1  Eb C c 
2

B
N0 B 

B   C  
E
b

  2 B   1
N 0 Cc 

c
Now for B>>Cc


 c
 B 


2
C




e
C c ln 2 
B


 1  C c ln 2
B
and hence for B>>Cc
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E
N
b
 ln 2  1.6dB
0
This value is referred to as the Shannon Limit
Example
Telephony circuit using 3 kHz bandwidth and S/N=30 db.
Using Shannons Law
CC = 3103 log2(1+103) = 29.92 kb/s since 30 db = 103
For binary information, ie M=2,
Shann-Hartley law gives for a noiseless channel
CC = 23103 log2(2) = 23103 b/s = 6 kb/s
The number of states required to approach Shannon limit
2B log2(M) = B log2(1+S/N) = 29.92 kb/s
log2M = 29.92kbps/6k
M = 31.6  32 states or 32 level code system can be used.
Such a system can use PAM where each bit is assigned 1 of 32 possible levels
or
Using binary PCM will required 5 bits (25 = 32) but will take 5 times longer to get the
information transferred through the 3kHz telephone line.
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DIGITAL MODULATION
v(t)=V sin(2 f t +  )
ASK
FSK
PSK
QAM
ASK
v(t)=s(t)sin(2fct)
s(t) is baseband signal
fc is the carrier frequency
Frequency Spectrum
ASK is rarely used in wireless applications, since noise and multipath effects can
adversely affect the amplitude, causing great errors on decoding.
FSK and PSK are more robuse in the presence of multipath.
Hybrid forms of ASK PSK or FSK exist.
For ASK, bandwidth = baud rate
Digital Baseband
modulating signal
fm
ASK Modulated Signal
fc
2fm
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LPF
Sample &
Hold
Threshold
Comparison
cos(  t)
Receiver Block Diagram
FSK
v(t)=sin(2f1t) for bit 1
v(t)=sin(2f2t) for bit 0
The carrier frequencies are usually picked to be orthogonal to each other (ie The
minimum frequency spacing f=1/Tb where Tb is the bit period.
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Frequency
Modulator
BPF
Envelope
Detector
BPF
Envelope
Detector
Data Output
BPF
noise
Transmitter --- channel --- Receiver Block Diagram
FSK Bandwidth = baud rate + frequency shift
for BFSK, baud rate = bit rate
Bandwidth of FSK Signal
BW=(f2-f1)+2fm
where BW is the total signal bandwidth
f2 and f1 are the mark and space frequencies
fm is the bandwidth of base signal
This is illustrated with the diagram as follow:
Digital Baseband
modulating signal
fm
f1
2fm
f2
FSK Modulated Signal
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The minimum bandwidth occurs when the two "humps" touch each other.
BWmin = 4fm
PSK
v(t)=sin(2ft) for bit 1
v(t)=sin(2ft+) for bit 0
This is called BPSK as there are two phases 0 and  representing 0 and 1.
See illustration for the waveform.
QPSK
s1(t)=sin(2ft - ) for bits 01
s2(t)=sin(2ft - 3) for bits 00
s3(t)=sin(2ft + 3) for bits 10
s4(t)=sin(2ft + ) for bits 11
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See illustration for the waveform.
Constellation Diagram
Block Diagram of Transmitter
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Block Diagram of Receiver
LPF
BP
Filter
Power
Splitter
Carrier
Recovery
series to parallel
conversion
90 deg
LPF
MPSK
for symbols s1-sM
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QAM
s1(t) = 1.848sin(2ft - 3/4) for bits 001
s2(t) = 0.765sin(2ft - 3) for bits 000
s3(t) = 0.765sin(2ft - ) for bits 010
s4(t) = 1.848sin(2ft - ) for bits 011
s5(t) = 1.848sin(2ft + 3) for bits 101
s6(t) = 0.765sin(2ft + 3) for bits 100
s7(t) = 0.765sin(2ft + ) for bits 110
s8(t) = 1.848sin(2ft + ) for bits 111
Waveform
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Constellations
8 QAM Encoder Block Diagram
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Noise and Error Rate Performance
The Basic SNR Parameter for Digital Communication Systems.
In Digital System Eb/No, a normalized version of SNR is used.
Eb = Bit energy = STb =S/Rb where Rb is bit rate, S is signal power, Tb is bit time
No = noise power spectral density = N/W where W is bandwidth, N is noise power
Eb/No=STb/N/W=S/Rb/NW=S/N(W/R)
One important metric is plot of bit error probability PB vs Eb/No. This is a 'water-falllike' shape.
PB
Eb/No
For a given probability of error, the smaller the required Eb/No, the more efficient is the
detection process.
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