Introduction to Digital Communications
Aaron Gulliver
Dept. of Electrical and Computer Engineering
University of Victoria
Analog vs. Digital
• Analog signals– Value varies continuously
• Digital signals– Values limited to a finite set
• Binary signals– Two valued– Time T needed to send 1 bit– Data rate R=1/T bits per second
t
x(t)
t
x(t)
t
x(t) 1
0 0 0
1 1
0T
2Introduction to Digital Communications2008
Information Representation
• Communication systems must convert information into a form suitable for transmission
• Analog systemsAnalog signals are directly modulated
– AM, FM radio
• Digital systems Generate bits and transmit digital signals
– Computer communications, Cellular telephones
• Analog signals can be converted into digital signals
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Digital Communication System
(a) Transmitter. (b) Receiver.
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Digital Transmitter• Matches the message to the channel• If the message is analog, it must be sampled in time and
quantized in amplitude. – discrete signal in time and amplitude
• Encoder: – adds redundancy for error correction.
• Modulation encodes the message into the amplitude, phase or frequency of the carrier signal (PSK, FSK, QAM, OFDM, PAM, PPM)
• Advantages: – Reduces noise and interference– Multiplexing– Channel assignment
• Examples: television, radio, 802.11, cellphones, bluetooth, GPS, …
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Modulation
• Convert the digital information into waveforms suitable for the channel
• OOK
• PAM
0 1 1 0 1 0 0 1
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Receiver
7
• Extracts the message from the received signal
• Operations: Filtering, Amplification, Demodulation
• The ideal receiver output is a scaled, delayed version of the message signal
• Decoder:
• estimates the original message from the received signal.
2008 Introduction to Digital Communications
Channel
8
• Physical medium that that the signal is transmitted through
• Examples: Air, wires, coaxial cables, fiber optic cables
• Every channel introduces some amount of distortion, noise and interference
• The channel properties determine
• Data throughput of the system
• Quality of service (QoS) offered by the system
2008 Introduction to Digital Communications
Noise and Interference
9
• Internal Noise
– Generated by components within a communication system (thermal noise)
• External Noise and Interference
– Atmospheric noise (electrical discharges)
– Man-made noise (ignition noise)
– Multipath interference (multiple transmission paths)
– Multiple access interference (signals from other users)
2008 Introduction to Digital Communications
Advantages
• Many sources are digital in nature– Data, images, text, video, music
• Different sources can be treated the same• Flexibility
– Encryption– Compression (source coding)– Error correction/detection
• Reliable reproduction of signals - regeneration– Two states vs. infinite variety of shapes
• Greater immunity to noise and interference• Power efficient and spectral efficient
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Introduction to Digital Communications 11
Digital versus Analog• Advantages of digital communications
– Regenerator receiver
– Different kinds of digital signals can be treated identically.
Data
Voice
Media
Propagation distance
Originalpulse
Regeneratedpulse
bits are bits!
2008
Digital Advantages
• Source coding compression algorithms can dramatically reduce the bit rate required to represent signals without significant distortion.
• Signal processing and channel coding techniques have significantly increased the bit rate that can be supported by a physical channel.
• Integrated circuits make complex signal processing and coding functions cost effective.
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Disadvantages
• Complex signal processing• Synchronization problems• Non-graceful degradation in performance as
the SNR decreases
2008 Introduction to Digital Communications 13
Performance Metrics• In analog communications we want
• In digital communications– Data rate (R bps) (limited by the Channel Capacity)
– Resources consumed: bandwidth, power
– Quality of the communications link : typically measured in terms of the Bit Error Rate (BER) or probability of error, PE
• Number of bit errors that occur for a given number of bits transmitted.
• Optical channels: Pe = 10-9
• Wireless channels: voice Pe = 10-3 data Pe = 10-6
– Propagation and processing delay
– Timing jitter in the bitstream at the receiver
)()(ˆ tmtm
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Applications
• Internet (last mile, VOIP)
• Local and long distance telephone channels
• Fibre optics (backbone and fibre to the home)
• Satellite communications (HDTV)
• CDs and DVDs
• Digital audio (mp3)
• Wireless Communications
• Cellular Communications
– GSM, TDMA, FDMA, CDMA, 3G
• Wireless LANs (802.11)
• WiMAX
• Bluetooth (headsets)
• Cordless telephones
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Goals of Digital Communications Design
• Maximize bit rate R
• Minimize probability of error PE
• Minimize required signal-to-noise ratio (SNR)
• Minimize required bandwidth W
• Maximize system utilization Capacity
• Minimize system complexity
• Minimize cost $
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Baseband Data Transmission - PAM
System model and waveforms
for synchronous baseband
digital data transmission.
(a) Baseband digital data
communication system.
(b) Typical transmitted
sequence. (c) Received
sequence plus noise.
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• Each T second pulse represents a bit of data
• Receiver has to decide whether a 1 or 0 was received (A or –A)
• Integrate-and-dump detector
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Receiver Structure
Receiver structure and integrator output. (a) Integrate-and-dump receiver. (b) Output from the integrator.
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Receiver Preformance• The output of the integrator is
V is a random variable• N is Gaussian noise
0
0
[ ( ) ( )]
is sent
is sent
t T
t
V s t n t dt
AT N A
AT N A
0
0
( )
t T
t
N n t dt
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2005-01-21 Lecture 1 24
Noise in communication systems
Thermal noise is described by a zero-mean Gaussian random process, n(t).
Its PSD is flat, hence, it is called white noise
N0/2 W/Hz
Probability density function
Power spectral density
2 2/(2 )
2( )
2
n
N
ef n
Analysis
since AWGN is uncorrelated
2)(
2
)]()([
)(][
][][][
0)]([])([][
00
2
2
22
0
0
0
0
0
0
0
0
0
0
0
0
0
0
TNdtdsst
N
dtdssntnE
dttnENE
NENENVar
dttnEdttnENE
Tt
t
Tt
t
Tt
t
Tt
t
Tt
t
Tt
t
Tt
t
25Introduction to Digital Communications2008
Error Analysis
• The pdf of N is
• In how many different ways can an error occur?
TN
enf
TNn
N
0
)/( 02
)(
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Error Analysis• Two ways in which errors occur
– A is transmitted, AT+N<0 (0 received,1 sent)
– -A is transmitted, -AT+N>0 (1 received,0 sent)
Error probabilities for binary signaling.
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• Similarly
• The average probability of error is
20/ 2
00
2( | )
AT n N Te A T
P Error A dn QNN T
20/ 2
00
2( | )
n N T
AT
e A TP Error A dn Q
NN T
0
22
)()|()()|(
N
TAQ
APAEPAPAEPPE
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• Energy per bit
• Therefore, PE can be written in terms of the energy.
• Define
TAdtAE
Tt
t
b
220
0
00
2
N
E
N
TAz b
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• Recall: Rectangular pulse of duration Tseconds has magnitude spectrum
• Effective Bandwidth
• Therefore
TBp /1
pBN
Az
0
2
sinc( )AT Tf
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Probability of Error vs. SNR
Pe for antipodal
baseband digital
signaling.
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Probability of Error Approximation
• Use the approximation
1,2
2
1,2
)(
0
2
2/2
zz
e
N
TAQP
uu
euQ
z
E
u
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Example
• Digital data is transmitted through a baseband system with , the received pulse amplitude is A = 20mV.
a) If the transmission rate is 1kbps, what is the probability of error?
3
3
2 62
7 3
0
3
1 110
10
400 10400 10 4V
10 10
2.58 102
p
p
z
E
BT
ASNR z
N B
eP
z
7
0 10 W/HzN
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b) If 10 kbps are transmitted, what must the value of A be to attain the same probability of error?
• Conclusion: tradeoff is
Transmission power vs. Bit rate
2 22 3
7 4
0
4 4 10 63.2mV10 10p
A Az A A
N B
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Bandpass Modulation
• There are 3 parameters
– Amplitude A(t) ― Amplitude Modulation
– Frequency f(t) ― Frequency Modulation
– Phase φ(t) ― Phase Modulation
V(t) = A cos(2πfc t + Φ)
Binary Signaling Techniques
Waveforms forASK, PSK, and FSK modulation.
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ASK, PSK, and FSK
• Amplitude Shift Keying (ASK)
• Phase Shift Keying (PSK)
• Frequency Shift Keying
0)(0
1)()2cos()2cos()()(
b
bcc
ccnTm
nTmtfAtfAtmts
1)()2cos(
1)()2cos()2cos()()(
bcc
bcc
ccnTmtfA
nTmtfAtftmAts
1)()2cos(
1)()2cos()(
2
1
bc
bc
nTmtfA
nTmtfAts
1 0 1 1
1 0 1 1
1 0 1 1
AM Modulation
PM Modulation
FM Modulation
m(t)
m(t)
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Amplitude Shift Keying (ASK)
What is the structure of the optimum receiver?
1 cos(2 )cA f t
0 0
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Receiver for Binary Signals in Noise
Receiver structure for detecting binary signals in additive white Gaussian noise (AWGN)
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Error Analysis
• 0s1(t), 1s2(t)
• Received signal:
• Noise is white and Gaussian
• Find PE
– In how many different ways can an error occur?
1 0 0
2 0 0
( ) ( ) ( ),
or
( ) ( ) ( ),
y t s t n t t t t T
y t s t n t t t t T
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Error Analysis (General Case)• Two types of errors:
• Receive 1 Send 0
• Receive 0 Send 1
• Decision process: • The received signal is filtered
• Filter output is sampled every T seconds
• Threshold k
• An error occurs when:
kTnTsTv
kTnTsTv
)()()(
or
)()()(
002
001
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• are filtered signal and noise terms.
• Noise term: is filtered white Gaussian noise.– therefore it is Gaussian
• The PSD is
– mean zero
– variance is equal to the average power of the noise process
00201 ,, nss
20 )(2
)(0
fHN
fSn
dffHN 202 )(2
( )on t
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• The pdf of the noise term is
• Note that we still don’t know what the filter is.
• Will any filter work? Or is there an optimal one?
• Recall that in the baseband case (no modulation), we used an integrator
– equivalent to filtering with
2
2/
2)(
022
n
N
enf
fjfH
2
1)(
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• The input to the threshold device is
• These are also Gaussian random variables
– mean:
– variance: same as the variance of N
01
02
( ) ( )
or
( ) ( )
V v T s T N
V v T s T N
01 02( ) or ( )s T s T
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Distribution of V
• The distribution of V, the input to the threshold device is
Conditional probability density functions of the filter output at time t = T
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Probability of Error
• Two types of errors
• The average probability of error
)(1
2))(|(
)(
2))(|(
02
2
2/)]([
2
01
2
2/)]([
1
2202
2201
TskQdv
etsEP
TskQdv
etsEP
k Tsv
k
Tsv
)](|[2
1)](|[
2
121 tsEPtsEPPE
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• Goal: Minimize the average probability of error
– choose the optimal threshold
• What should the optimal threshold, kopt be?
– kopt=0.5[s01(T)+s02(T)]
2
)()( 0102 TsTsQPE
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Observations
• PE is a function of the difference between the two signals.
• Recall: Q-function decreases with increasing argument.
• Therefore, PE will decrease with increasing distance between the two output signals
• Choose the filter h(t) such that PE is a minimum– maximize the difference between the two signals at the
output of the filter
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Matched Filter
• Goal: Given , choose H(f) such that
is maximized.
• The solution to this problem is known as the matched filter and is given by
• Therefore, the optimum filter depends on the input signals.
)(),( 21 tsts
)()( 0102 TsTsd
)()()( 120 tTstTsth
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Matched Filter Receiver
Matched filter receiver for binary signaling in additive white Gaussian noise (AWGN).
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Error Probability for Matched Filter Receiver
• Recall
• The maximum value of the distance is
• E1 is the energy of the first signal
• E2 is the energy of the second signal
2
dQPE
)2(2
122121
0
2
max EEEEN
d
dttstsEE
)()(1
21
21
12
51Introduction to Digital Communications
0 0
0 0
2 2
1 1 2 2( ) ( )
t T t T
t t
E s t dt E s t dt
2008
• Therefore
• Probability of error depends on the signal energies (just as in the baseband case), noise power, and the similarity between the signals.
• If we make the transmitted signals as dissimilar as possible, then the probability of error will decrease.
• This is achieved with
2/1
0
122121
2
2
N
EEEEQPE
112
52Introduction to Digital Communications2008
ASK
• The matched filter:
• Optimum Threshold:
• Similarity between signals?
• Therefore
• 3dB worse than baseband.
)2cos()(,0)( 21 tfAtsts c
)2cos( tfA c
TA2
4
1
zQN
TAQPE
0
2
4
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PSK
• Modulation index: m (determines the phase shift)
• Matched Filter with threshold 0
• For m = 0, 3dB better than ASK
)cos2sin()(
)cos2sin()(
1
2
1
1
mtfAts
mtfAts
c
c
)2cos(12 2 tfmA c
))1(2( 2 zmQPE
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Matched Filter for PSK
Optimum correlation receiver for PSK.
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