Digital Data Transmission
ECE 457
Spring 2005
Analog vs. Digital

Analog signals

x(t)
Value varies continuously
t

Digital signals


Value limited to a finite set
Binary signals




x(t)
Has at most 2 values
Used to represent bit values
Bit time T needed to send 1 bit
Data rate R=1/T bits per second
t
x(t)
1
0
T
1
0 0
1
0
t
Information Representation
• Communication systems convert information into
a form suitable for transmission
• Analog systemsAnalog signals are modulated
(AM, FM radio)
• Digital system generate bits and transmit digital
signals (Computers)
• Analog signals can be converted to digital signals.
Digital Data System
Figure 7-1
Block diagram of a digital data system. (a) Transmitter.
(b) Receiver.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
Components of Digital
Communication
• Sampling: If the message is analog, it’s converted
to discrete time by sampling.
(What should the sampling rate be ?)
• Quantization: Quantized in amplitude.
Discrete in time and amplitude
• Encoder:
– Convert message or signals in accordance with a set of
rules
– Translate the discrete set of sample values to a signal.
• Decoder: Decodes received signals back into
original message
Different Codes
0
1
1
0
1
0
0
1
Performance Metrics
• In analog communications we want, mˆ (t )  m(t )
• Digital communication systems:
–
–
–
–
Data rate (R bps) (Limited) Channel Capacity
Probability of error Pe
Without noise, we don’t make bit errors
Bit Error Rate (BER): Number of bit errors that occur
for a given number of bits transmitted.
• What’s BER if Pe=10-6 and 107 bits are
transmitted?
Advantages
• Stability of components: Analog hardware
change due to component aging, heat, etc.
• Flexibility:
– Perform encryption
– Compression
– Error correction/detection
• Reliable reproduction
Applications
• Digital Audio
Transmission
• Telephone channels
• Lowpass
filter,sample,quantize
• 32kbps-64kbps
(depending on the
encoder)
• Digital Audio
Recording
• LP vs. CD
• Improve fidelity
(How?)
• More durable and
don’t deteriorate with
time
Baseband Data Transmission
Figure 7-2
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.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
• Each T-second pulse is a bit.
• Receiver has to decide whether it’s a 1 or 0
( A or –A)
• Integrate-and-dump detector
• Possible different signaling schemes?
Receiver Structure
Figure 7-3
Receiver structure and integrator output. (a) Integrate-anddump receiver. (b) Output from the integrator.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
Receiver Preformance
• The output of the integrator:
V
t 0 T
 [s(t )  n(t )]dt
t0
 AT  N

 AT  N
t 0 T
A
is
sent
A
is
sent
• N   n(t )dt is a random variable.
• N is Gaussian. Why?
t0
Analysis
E[ N ]  E[
t 0 T

n(t )dt ] 
t0
t 0 T
 E[n(t )]dt  0
t0
Var[ N ]  E[ N 2 ]  E 2 [ N ]
 E[ N 2 ]
Why ?
2
  t 0 T




 E   n(t )dt  

 

 t0


t 0 T t 0 T
  E[n(t )n(s)]dtds
t0

t 0 T t 0 T
 
t0

t0
t0
N0
 (t  s )dtds
2
Why ?(White
N 0T
2
• Key Point
– White noise is uncorrelated
noise
is
uncorrelat ed !)
Error Analysis
• Therefore, the pdf of N is:
f N ( n) 
e
 n 2 /( N 0T )
N 0T
• In how many different ways, can an error
occur?
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)
Figure 7-4
Illustration of error probabilities for binary signaling.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
 AT
•
P( Error | A) 

e

• Similarly,
P( Error |  A) 


 2 A2T
dn  Q
 N0
N 0T






dn  Q

N 0T





 n 2 / N 0T
e
 n 2 / N 0T
AT
2 A2T
N0
• The average probability of error:
PE  P( E | A) P( A)  P( E |  A) P( A)
 2 A2T
 Q

N0





• Energy per bit:
Eb 
t 0 T
2
2
A
dt

A
T

t0
• Therefore, the error can be written in terms
of the energy.
• Define
A2T Eb
z

N0
N0
• Recall: Rectangular pulse of duration T
seconds has magnitude spectrum
ATsinc (Tf )
• Effective Bandwidth:
• Therefore,
Bp  1/ T
A2
z
N0 Bp
• What’s the physical meaning of this
quantity?
Probability of Error vs. SNR
Figure 7-5
PE for antipodal baseband
digital signaling.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
Error Approximation
• Use the approximation
u 2 / 2
e
Q(u ) 
, u  1
u 2
 2 A2T
PE  Q
 N0

z

e

, z  1
 2 z

Example
• Digital data is transmitted through a
baseband system with N0  107W / Hz , the
received pulse amplitude A=20mV.
a)If 1 kbps is the transmission rate, what is
probability of error?
1
1
 3  103
T 10
A2
400 10 6
SNR  z 
 7
 400 10  2  4
3
N 0 B p 10 10
Bp 
e z
PE 
 2.58 10 3
2 z
b) If 10 kbps are transmitted, what must be the
value of A to attain the same probability of
error?
A2
A2
2
3
z
 7

4

A

4

10
 A  63.2mV
4
N 0 B p 10 10
• Conclusion:
Transmission power vs. Bit rate
Binary Signaling Techniques
Figure 7-13
Waveforms for ASK, PSK, and
FSK modulation.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
ASK, PSK, and FSK

Amplitude Shift Keying (ASK)
 A cos( 2f c t )
s (t )  m(t ) Ac cos( 2f c t )   c
0

1
0
1
1
m(t)
m( nTb )  1
m( nTb )  0
AM Modulation

Phase Shift Keying (PSK)
 A cos( 2f c t )
s (t )  Ac m(t ) cos( 2f c t )   c
 Ac cos( 2f c t   )

m( nTb )  1
0
1
1
m(t)
m( nTb )  1
Frequency Shift Keying
 A cos(2f1t )
s (t )   c
 Ac cos(2f 2t )
1
m( nTb )  1
PM Modulation
1
0
1
1
m( nTb )  1
FM Modulation
Amplitude Shift Keying (ASK)
• 00
• 1Acos(wct)
• What is the structure of the optimum
receiver?
Receiver for binary signals in
noise
Figure 7-6
A possible receiver structure for detecting binary signals in
white Gaussian noise.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
Error Analysis
• 0s1(t), 1s2(t) in general.
• The received signal:
y (t )  s1 (t )  n(t ), t0  t  t0  T
OR
y (t )  s2 (t )  n(t ), t0  t  t0  T
• Noise is white and Gaussian.
• Find PE
• In how many different ways can an error occur?
Error Analysis (general case)
• Two ways for error:
» Receive 1 Send 0
» Receive 0Send 1
• Decision:
» The received signal is filtered. (How does this
compare to baseband transmission?)
» Filter output is sampled every T seconds
» Threshold k
» Error occurs when:
v(T )  s01 (T )  n0 (T )  k
OR
v(T )  s02 (T )  n0 (T )  k
• s01, s02 , n0 are filtered signal and noise terms.
• Noise term: n 0 (t ) is the filtered white Gaussian
noise.
• Therefore, it’s Gaussian (why?)
• Has PSD:
N0
2
S n0 ( f ) 
2
H( f )
• Mean zero, variance?
• Recall: Variance is equal to average power of the
noise process
 
2

N0
2
H
(
f
)
df
 2
• The pdf of noise term is:
f N ( n) 
e
 n 2 / 2 2 0
2 2
• Note that we still don’t know what the filter is.
• Will any filter work? Or is there an optimal one?
• Recall that in baseband case (no modulation), we
had the integrator which is equivalent to filtering
with
1
H( f ) 
j 2f
• The input to the thresholder is:
V  v(T )  s01 (T )  N
OR
V  v(T )  s02 (T )  N
• These are also Gaussian random variables; why?
• Mean: s01(T ) OR s02 (T )
• Variance: Same as the variance of N
Distribution of V
• The distribution of V, the input to the
threshold device is:
Figure 7-7
Conditional probability density functions of the filter output
at time t = T.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
Probability of Error
• Two types of errors:

P( E | s1 (t ))  
e
[ v  s0 1 (T )]2 / 2 2
2 2
k
k
P( E | s2 (t )) 


e
 k  s01 (T ) 
dv  Q




[ v  s0 2 (T )]2 / 2 2
2 2
 k  s02 (T ) 
dv  1  Q




• The average probability of error:
PE 
1
1
P[ E | s1 (t )]  P[ E | s2 (t )]
2
2
• Goal: Minimize the average probability of
errror
• Choose the optimal threshold
• What should the optimal threshold, kopt be?
• Kopt=0.5[s01(T)+s02(T)]
•
 s02 (T )  s01 (T ) 
P Q
E


2


Observations
• PE is a function of the difference between the two
signals.
• Recall: Q-function decreases with increasing
argument. (Why?)
• Therefore, PE will decrease with increasing
distance between the two output signals
• Should choose the filter h(t) such that PE is a
minimummaximize the difference between the
two signals at the output of the filter
Matched Filter
• Goal: Given s1 (t ), s2 (t ) , choose H(f) such
s (T )  s (T )
d

that
is maximized.

• The solution to this problem is known as the
matched filter and is given by:
02
01
h0 (t )  s2 (T  t )  s1 (T  t )
• Therefore, the optimum filter depends on
the input signals.
Matched filter receiver
Figure 7-9
Matched filter receiver for binary signaling in white
Gaussian noise.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
Error Probability for Matched
Filter Receiver
• Recall P  Q d2 
 
• The maximum value of the distance,
E
d max 
2
2
( E1  E2  2 E1 E2 12 )
N0
• E1 is the energy of the first signal.
• E2 is the energy of the second signal.
E1 
t 0 T

2
s1 (t ) dt
t0
E2 
t 0 T

t0
12 
1
E1 E2
2
s2 (t ) dt

 s (t )s
1

2
(t )dt
• Therefore,
 E  E  2 E E  1/ 2 
2
1 2 12 

PE  Q  1


2N0

 

• Probability of error depends on the signal energies
(just as in 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
(   1 )
12
ASK
s1 (t )  0, s2 (t )  A cos( 2f ct )
•
•
•
•
The matched filter: A cos(2f ct )
1
Optimum Threshold: 4 A T
Similarity between signals?
Therefore, P  Q 4ANT   Q z 
2
2
E

0

• 3dB worse than baseband.
PSK
s1 (t )  A sin( 2f ct  cos 1 m), s2 (t )  A sin( 2f c t  cos 1 m)
• Modulation index: m (determines the phase
jump)
• Matched Filter:  2 A 1  m cos(2f t )
• Threshold: 0
• Therefore, P  Q( 2(1  m ) z )
• For m=0, 3dB better than ASK.
2
c
2
E
Matched Filter for PSK
Figure 7-14
Correlator realization of optimum receiver for PSK.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
FSK
• s1 (t )  A cos( 2f ct ), s2 (t )  A cos(2 ( f c  f )t )
• f  m
T
• Probability of Error: Q( z )
• Same as ASK
Applications
• Modems: FSK
• RF based security and access control
systems
• Cellular phones