Digital data transmission

Digital Data Transmission
ECE 457
Spring 2005

A n a l o g v s . D i g i t a l
 Analog signals
Value varies continuously
 Digital signals
Value limited to a finite set
 Binary signals
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)
t
x(t)
t
x(t) 1
0 0 0
1 1
T 0

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.
Principles of Communications, 5/E by Rodger Ziemer and William Tranter
Copyright © 2002 John Wiley & Sons. Inc. All rights reserved.
(b) Receiver.

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,
• Digital communication systems:
mˆ (t) @ m(t)
– Data rate (R bps) (Limited) Channel Capacity
– Probability of error
P
e – 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 P=10-6 and 107 bits are transmitted?
e

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.

• 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-and-dump
receiver. (b) Output from the integrator.

Receiver Preformance
• The output of the integrator:
t T
ò +
0
= + V s t n t dt
î í ì
t
0
[ ( ) ( )]
AT +
N A is sent
AT N A is sent
- + -
=
t T
ò + =
• 0
N n ( t )
dt
is a random variable.
• t
N is Gaussian. 0
Why?

Analysis
t T
0
t T
+ +
0
ò ò
E N E n t dt E n t dt
= = =
[ ] [ ( ) ] [ ( )] 0
2 2
0
0
t
Var N E N E N
t
= -
[ ] [ ] [ ]
2
E N Why
[ ] ?
é
t T
ì
ïî
ïí
+
ò
E n t dt
t T
+ +
ò ò
t
t T
=
=
=
ü
ïþ
ïý
ù
ú úû
ê êë
+ +
ò ò
= -
t
t
t T
t
t T
N T
0
0
0
2
0
0
0
0
0
0
0
0
E n t n s dtds
t
=
• Key Point
( )
2
[ ( ) ( )]
N d
t s dtds Why White noise is uncorrelated
( ) ?( !)
0
2
– White noise is uncorrelated

Error Analysis
• Therefore, the pdf of N is:
2
n N T
/( 0 )
N T
-
=
f n e
N
0
( )
p
• 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.

•
AT n N T
/ 2 ( | )
• Similarly,
æ
-
2
dn Q A T
æ
¥ -
2
/ 2 ( | )
dn Q A T
• The average probability of error:
ö
÷ ÷
ø
ç ç
è
= ò =
-¥
-
0
0
0
2
N
N T
P Error A e
p
ö
÷ ÷
ø
ç ç
è
- = ò =
0
0
0
2
N
N T
P Error A e
AT
n N T
p
P = P E A P A + P E - A P -
A E
( | ) ( ) ( | ) ( )
ö
÷ ÷
ø
æ
Q A T
ç ç
è
=
2 2
0
N

• Energy per bit:
t T
= = ò +
2 2
0
E A dt A T
t
b
0
• Therefore, the error can be written in terms
of the energy.
• Define
E
z = A T = b
0 0
2
N
N

• Recall: Rectangular pulse of duration T
seconds has magnitude spectrum
ATsinc(Tf )
• Effective Bandwidth:
• Therefore,
B T p =1/
z A
2
N B
0
p =
• What’s the physical meaning of this
quantity?

Probability of Error vs. SNR
Figure 7-5
PE for antipodal baseband
digital signaling.

Error Approximation
• Use the approximation
, 1
2
Q u e
P Q 2
A T
, 1
2
( )
2
0
2 / 2
ö
>> @ ÷ ÷
ø
æ
ç ç
è
=
@ >>
-
-
z
z
e
N
u
u
z
E
u
p
p

Example
• Digital data is transmitted through a
baseband system with N = 10 -
7W / Hz
, the
0
received pulse amplitude A=20mV.
a)If 1 kbps is the transmission rate, what is
probability of error?
SNR z A
= = = ´
400 10
3
2
6
7 3
0
2
3
3
2.58 10
1 1
2
400 10 4
10 10
10
10
-
-
-
-
-
-
@ = ´
= ´ =
´
= = =
z
P e
N B
T
B
z
E
p
p
p

b) If 10 kbps are transmitted, what must be
the value of A to attain the same
probability of error?
2
A A A mV
z A
2
= = -
N B
p
2 3
= Þ = ´ Þ =
4 4 10 63.2
7 4
´
-
10 10
0
• Conclusion:
Transmission power vs. Bit rate

Binary Signaling Techniques
Figure 7-13
Waveforms for ASK, PSK, and
FSK modulation.

ASK, PSK, and FSK
 Amplitude Shift Keying (ASK)
A f t m nT
î í ì
cos(2 ) ( ) 1
c c b
c c m nT
 Phase Shift Keying (PSK)
A p
f t m nT
cos(2 ) ( ) 1
c c b
c c A f t m nT
 Frequency Shift Keying
=
=
= =
0 ( ) 0
( ) ( ) cos(2 )
b
s t m t A f t
p
p
î í ì
=
+ =-
= =
cos(2 ) ( ) 1
( ) ( ) cos(2 )
c c b
s t A m t f t
p p
p
A f t m nT
î í ì
=
p
cos(2 ) ( ) 1
=-
=
c 1
b
A f t m nT
cos(2 ) ( ) 1
( )
c 2
b
s t
p
1 0 1 1
m(t)
AM Modulation
1 0 1 1
m(t)
PM Modulation
1 0 1 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.

Error Analysis
• 0s1(t), 1s2(t) in general.
• The received signal:
y ( t ) s ( t ) n ( t ),
t t t T
OR
= + £ £ +
1 0 0
y ( t ) = s ( t ) + n ( t ),
t £ t £ t +
T
2 0 0
• 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 ) s ( T ) n ( T )
k
OR
= + >
01 0
v ( T ) = s ( T ) + n ( T )
<
k
02 0

• s , s , n
are filtered signal and noise terms.
01 02 0 • Noise term: n0(t)
is the filtered white Gaussian
noise.
• Therefore, it’s Gaussian (why?)
• Has PSD:
S ( f ) =
N 0 H ( f ) 2
n 0 2
• Mean zero, variance?
• Recall: Variance is equal to average power of the
noise process
¥
2 N0 H( f ) 2 df
ò 2
-¥
s =

• The pdf of noise term is:
0
2 2
ps
2
n s
/ 2
2
-
=
f ( n )
e
N
• 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
j f
H f
2p

• The input to the thresholder is:
V = v T = s T +
N
OR
( ) ( )
01
V = v ( T ) = s ( T )
+
N
02
• These are also Gaussian random variables; why?
• Mean:
s ( T ) OR s ( T
) 01 02 • 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.

Probability of Error
• Two types of errors:
v s T
P E s t e dv Q k s T
ö çè
s
[ ( )] / 2
ps s
1 ( )
ö çè
¥ - -
s
k v s T
[ ( )] / 2
k
• The average probability of error:
÷ø
= = - æ -
÷ø
= ò
= æ -
ò
-¥
- -
ps s
2
( | ( ))
( )
2
( | ( ))
02
2
2
01
2
1
2 2
02
2 2
01
P E s t e dv Q k s T
1 2 P P E s t P E s t E = +
[ | ( )]
[ | ( )] 1
2
1
2

÷ø
• Goal: çè
Minimize the average probability of
errror
• Choose the optimal threshold
• What should the optimal threshold, kbe?
opt • K=0.5[s(T)+s(T)]
opt0102• P = Q æ s ( T ) -
s ( T ) 02 01 ö E
2s

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
( ), ( ) 1 2 s t s t
• Goal: Given , choose H(f) such
that d = s ( T ) - s ( T
) 02 01 is maximized.
s
• The solution to this problem is known as the
matched filter and is given by:
( ) ( ) ( ) 0 2 1 h t = s T - t - s 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.

Error Probability for Matched
Filter Receiver
• Recall
P = Q æ
d E
çè
ö • The maximum value of the distance,
÷ø
2
2 ( 2 )
d = + -
max E E E E r
1 2 1 2 12
0
2
N
• E1 is the energy of the first signal.
• E2 is the energy of the second signal.
t +
T
ò
2
E s t dt
1 1
t
t +
T
ò
=
0
( )
E =
s t dt
t
0
0
0
( )
2
2 2
1 ( ) ( )
s t s t dt
12 ò
E E
1 2
1 2
¥
-¥
r =

• Therefore,
ù
ú ú
û
é
æ + -
ç ç
ê ê
è
ë
1/ 2
ö
÷ ÷
ø
P =
Q E
E E E E
2
N
r
1 2 1 2 12
0
2
• 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
( r = -
1 12 )

ASK
( ) 0, ( ) cos(2 ) 1 2 s t s t A f t c = = p
• The matched filter:
• Optimum Threshold:
• Similarity between signals?
• Therefore,
Acos(2 f t) c p
1
A2T
4
Q( z )
ö
æ
2
4
P Q A T E = ÷ ÷
N
ø
ç ç
è
=
0
• 3dB worse than baseband.

PSK
= p + - = p - -
( ) sin(2 cos ), ( ) sin(2 cos 1 )
s t A f t m s t A f t m 1 c 2
c
1
• Modulation index: m (determines the phase
jump)
• Matched Filter:
- 2A 1 - m2 cos(2 p
f t) c • Threshold: 0
• Therefore,
P = Q( 2(1 -
m2 )z ) E • For m=0, 3dB better than ASK.

Matched Filter for PSK
Figure 7-14 Correlator realization of optimum receiver for PSK.

FSK
( ) cos(2 ), ( ) cos(2 ( ) ) 1 2 s t A f t s t A f f t c c = p = p + D
•
•
Df =m
T
• Probability of Error:
• Same as ASK
Q( z )

Applications
• Modems: FSK
• RF based security and access control
systems
• Cellular phones

Digital data transmission

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