T - New Jersey Institute of Technology
Principles of Digital Data
Transmission
Chapter 7
Lectured by Dr. Yun Q. Shi
Dept. of Electrical & Computer Engr.
New Jersey Institute of Technology
shi@njit.edu
Text used for the course: <Modern Digital and Analog
Communication Systems>, 4th Edition, Lathi and Ding, Oxford
Introduction
• A significant portion of communication in 1990s was in analog.
• It has been replaced rapidly by digital communication.
• Now most of the communications become digital, with analog
communication playing a minor role.
• This chapter addresses the framework (several major aspects) of
digital data transmission
• All we learnt in this course so far is utilized here.
• Hence, this chapter is naturally a conclusion of this course.
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1
Digital Communication Systems:
Source
• Digital Communication Systems: Figure 7.1
Source encoder, Baseband modulation (Line coding), Digital
carrier modulation, Multiplexer, Channel, Regenerative repeater,
Receiver (detection).
• Source:
The input
– A sequence of digits
– We discuss mainly: The binary case (two symbols)
– Later in this chapter: The M-ary case (M symbols)
more general case
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Figure 7.1 Fundamental building blocks of digital
communication systems
Figure 7.1 Fundamental building blocks of digital communication systems
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2
Line Coder (Transmission Coder)
• The output of a multiplexer is coded into
electrical pulses or waveforms for the purpose
of transmission over the channel.
• This process is called a line coding or
transmission coding
• Many possible ways of assigning waveforms
(pulses) to the digital data.
1
• On-off:
Dr. Shi
↔
a
0
↔
pulse
p (t)
no
pulse
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5
Line Coder …
• Polar: 1 ↔
↔
0
a
a
pulse
pulse
p (t)
− p (t)
• Bipolar (pseudoternary or alternate mark inversion
(AMI))
1 ↔
0 ↔
a pulse p (t) or - p (t) depending on whether the previous 1
is encoded by -p (t) or p (t)
no pulse
In short, pulses representing consecutive 1’s alternate
in sign.
• All the above three could use half-width pulses. It is
possible to select other width.
• Figure 7.1 (parts a, b and c).
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3
Line Codes
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Line Coder …
• Full width pulses are often used in some
applications.
– i.e., the pulse amplitude is held to a constant value
throughout the pulse interval (it does not have a
chance to go to zero before the next pulse begins).
– These schemes are called non return-to-zero (NRZ)
in contrast to return-to-zero (RZ)
– Figure 7.2 shows
• An On-off NRZ signal
• A Polar NRZ signal in d and e parts of the figure
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4
Multiplexer
• Usually, the capacity of a practical channel >>
the data rate of individual sources.
• To utilize this capacity effectively, we combine
several sources through a digital multiplexer
using the process of interleaving.
• One way: A channel is time-shared by several
messages simultaneously.
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Regenerative Repeater
• Used at regularly spaced intervals along a digital
transmission line to: 1) detect the incoming digital
signal and 2) regenerate new clean pulses for further
transmission along the line.
• This process periodically eliminates, and thereby
combats the accumulation of noise and signal distortion
along the transmission path.
• The periodic timing information (the clock signal at R b
Hz) is required to sample the incoming signal at a
repeater.
• R b (rate): pulses/sec
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5
Regenerative Repeater…
• The clock signal can be extracted from the
received signal.
– e.g., the polar signal
when rectified ⇒ a clock signal at Rb Hz
• The on-off signal = A periodic signal at Rb
+ a polar signal (Figure 7.)
• When the periodic signal is applied to a resonant
circuit tuned to Rb Hz, the output, a sinusoid of
Rb Hz, can be used for timing.
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On-off Signal Decomposition
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6
Regenerative Repeater…
rectified
• The bipolar signal
an on-off signal
⇒ the clock signal can also be extracted.
• The timing signal (the output of the resonant circuit) is
sensitive to the incoming pattern, sometimes.
– e.g. in on-off, or
0
bipolar
↔
no pulse
• If there are too many zeros in a sequence, will have
problem.
– no signal at the input of the resonant circuit for a while
– sinusoids output of the resonant circuit starts decaying
• Polar scheme has no such a problem.
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Transparent Line Code
• A line code in which the bit pattern does not
affect the accuracy of the timing information is
said to be a transparent line code.
– The polar scheme is transparent.
– The on-off and bipolar are not transparent (nontransparent).
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7
Desired Properties of Line Codes
1.
Transmission bandwidth: as small as possible
2.
Power efficiency: for a specific bandwidth and detection error
rate, transmitted power should be as small as possible.
3.
Error detection and correction capability: as strong as possible
4.
Favorable power spectral density: desirable to have zero PSD at
(dc), called DC null, because of ac coupling requires this. If at
dc, ω = 0 , PSD ≠ 0 , then dc wanders in the pulse stream,
which is not desired.
5.
Adequate timing content: should be possible to extract the
clock signal (timing information).
6.
Transparency
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PSD of Various Line Codes
• Consider a general PAM signal, as shown in
Figure 7.4 (b):
Tb = 1
t = kTb
Rb
Rb
– the k th pulse in the pulse train y (t ) = ak p (t )
• p (t): the basic pulse
• P (ω) [P(f)]: Fourier spectrum of p (t)
• a k: arbitrary and random
– The on-off, polar, bipolar line codes are all special
cases of the general pulse train y(t), a k = 0, +1, -1
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8
A Random PAM signal
All 3 line codes
can be represented
this way
New approach to
determine PSD of
y(t)
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PSD of Various Line Codes …
• How to determine the PSD?
– Figure 7.4
– An attractive general approach
Rx (τ ), S x (ω ) need to be studied once
Then for different p(t), ⇒ different PSD
– x(t): an impulse train
– xˆ (t ) : a rectangular pulse train ( ε ⋅ hk = a k )
when ε → 0, xˆ (t ) → x(t )
(Figure 7.5)
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9
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PSD of Various Line Codes …
R0 = ak2
⇒ time average of a k2
R1 = ak .ak +1 ,L L, Rn = ak .ak +n
1
RX (τ ) =
Tb
S X (ω ) =
1
Tb
∞
∑ R δ (τ − nT )
n = −∞
n
b
∞
∑ Rn e− jnωTb =
n = −∞
1
Tb
∞
[
]
R
+
2
R
cos(
n
ω
T
)
∑
0
n
b
n =1
S y (ω ) = P (ω ) S X (ω )
2
Different line codes
Dr. Shi
⇒ different P(ω ) ⇒ different S y (ω )
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10
Polar Signaling
1 ↔ p(t)
0 ↔ -p(t)
} ⋅ ak = ±1, equally
R0 = ak2 = 1
R1 = ak .ak +1 = 0
Rn = 0, n ≥ 1
likely, ak2 = 1
Q ak .ak +1 = 1 or − 1
equally likely
Similar reasoning
S y (ω ) = P (ω ) .S X (ω )
2 1
= P(ω ) .
Tb
2
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Polar Signaling…
• To be specific, assume p(t) is a rectangular
pulse of width T b (a half-width rectangular
2
pulse)
t
p (t ) = re c t
T b
2
T
ω Tb
P ( ω ) = b s in c
2
4
T
ω Tb
S y ( ω ) = b s in c 2
4
4
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1st zero − cros sin g :
ω⋅Tb
4
= π ⇒ω =
or ⇒ f =
4π
Tb
2
= 2Rb
Tb
22
11
PSD of
polar signal (half-width rectangle)
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Polar Signaling …
• Comment 1:
•
•
•
•
The essential bandwidth of the signal (main
lobe) = 2 Rb
For a full-width pulse, ⇒ Rb
“Polar Signaling is not bandwidth efficient.”
Comment 2:
“No error-detection or error-correction
capability.”
Comment 3:
“Non-zero PSD at dc( ω = 0)”
⇒ This will rule out the use of ac coupling in transmission.
Comment 4:
“Most efficient scheme from the power
requirement viewpoint” for a given
power, the detection-error probability for a
polar scheme is the smallest possible.
Comment 5:
“Transparent”
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12
Achieving DC Null in PSD by
Pulse Shaping
Q P (ω ) =
∫
∴ P (0) = ∫
∞
−∞
∞
−∞
p ( t ) e − j ω t dt
p ( t )dt
⇒ If the area under p(t) = 0, P(0) = 0 .
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Manchester
(split-phase, twinned-binary) Signal
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13
On-off Signaling
• 1 ↔ p(t)
0 ↔ no pulse
R
0
R
n
S
X
=
1
2
1
=
4
=
(ω ) =
∞
1
1
+
2Tb
4Tb
1
1
+
4Tb
4Tb
Dr. Shi
n ≥ 1
for all
∞
∑
e
∑
e
− jn ω T b
n = −∞
n ≠0
− jn ω T b
n = −∞
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On-off Signaling
S X (ω ) =
1
2π
+
4Tb
4 T b2
∞
∑
n = −∞
δ ω −
2π n
Tb
P (ω )
2π ∞
2π n
S y (ω ) =
1
+
δ ω −
∑
4Tb
Tb n = −∞
Tb
For a half-width rectangular pulse,
2
S y (ω ) =
Tb
2π
ω Tb
sinc 2
1 + T
16
4
b
∞
∑
n = −∞
δ ω −
2π n
Tb
Note: 1. For some derivation, refer to the next four pages.
2. PSD is shown in Figure 7.8.
3. PSD consists of both a discrete part and a
continuous part.
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14
Some Derivation
(Lathi’s book, pp. 58-59, Example 2.12)
• A unit impulse train:
FS
g (t) =
∞
∑
n = −∞
D ne
jn ω b t
δ T (t ) = CombT (t ) = g (t )
0
=
=
0
1
Tb
∞
∑
e
n = −∞
1
2
+
Tb
Tb
jn ω b t
∞
∑
n = −∞
cos( n ω b t )
1 ∞
∑ δ ( f − nf 0 )
T 0 n = −∞
2π ∞
2π
G ( ω ) = FT { g ( t )} =
)
δ (ω − n
∑
T 0 n = −∞
T0
G ( f ) = FT { g ( t )} =
≅ ω 0 COMB
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ω0
(ω )
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Example 2.12
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15
Some Derivation…
• Fact 1 (Lathi’s book, p.83, Eq. (3.20a, 3.20b):
1↔ δ(f
)
e j 2πf bt ↔ δ ( f − f b )
• Fact 2:
[e
jω bt
]
↔ δ (ω − ωb )
Also,
(t ) ↔ 1
δ (t − nT b ) ↔
δ
Dr. Shi
e
− jn ω T
b
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Some Derivation …
• Fact 3:
(
)
∞
∑ δ t − nT
b
n1=4−∞
42443
FS
=
∞
jnω t
1
2π
b
ω =
∑ e
b
T n = −∞
T
b
1b4
44442444443
b FT
644744
8
∞ − jnω t
b
∑e
n = −∞
Dr. Shi
b FT
=
ω
(
∞
∑ δ ω − nω
b
b
n = −∞
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)
32
16
PSD of An On-off Signal
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On-off Signaling
• This is reasonable since, as shown in Fig. 7.2,
an on-off signal can be expressed as the sum of
a polar and a periodic component
• Comment 1: For a given transmitted power, it
is less immune to noise interference.
– Q Noise immunity ∝ difference of amplitudes
representing binary 0 and 1.
– If a pulse of amplitude 1 or –1 has energy E, then a
pulse of amplitude 2 has energy (2)2E = 4E.
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17
On-off Signaling …
• For polar:Q
1
Tb
∴
• For on-off:
digits are transmitted per second
polar signal power
1 2E
=
power = 4 E
2
T
b Tb
1 E
= E =
Tb Tb
Power is now as large
as twice of above.
• Comment 2: Not transparent
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Bipolar Signaling
(Pseudoternary or Alternate Mark Inverted (AMI))
A. 0 ↔ no pulse
1 ↔ p(t) or –p(t) depending on whether the
previous 1 was transmitted by –p(t) or p(t)
B. [p(t), 0 , -p(t)]: In reality, it is ternary
signaling.
C. Merit: a dc null in PSD
demerit: not transparent
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18
Bipolar Signaling…
1 N
N
1
( ± 1) 2 +
(0) =
N →∞ N
2
2
2
k
1 N
3N
1
R (1) = lim
( − 1) +
(0) = −
N →∞ N
4
4
4
111, 101, 110, 011, 010, 001, 000
ak ak + n
R (2)
D. R ( 0 ) = Nlim
→∞
R ( 2 ) = lim
N →∞
R n = lim
N →∞
1
N
1
N
∑a
2
k
= lim
1 N
N
5N
(1) +
( − 1) +
(0) = 0
N 8
8
8
∑a
k
a k + n = 0, n > 1
k
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Bipolar Signaling…
E.
S y (ω ) = P(ω ) S X (ω )
2
P (ω ) ∞
R n e − jnω Tb
=
∑
Tb
n = −∞
2
P (ω )
=
Tb
2
P (ω )
=
Tb
P (ω )
=
2 ⋅ Tb
Dr. Shi
2
∞
R
+
2
R n cos nω Tb
∑
0
n =1
2
(7.10b)
(7.10c)
1
1
+ 2 ⋅ ( − ) cos ω T b
4
2
P (ω )
(1 − co s ω Tb ) =
Tb
2
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ω Tb
sin 2
2
38
19
Bipolar Sampling …
⇒
As ω=0 (dc) regardless of P(ω)
A dc null (desirable for ac coupling)
S y (ω ) = 0
sin 2 (
ω Tb
2
)= 0
at
ω =
⇒ b a n d w id th = R b H z ,
2π
1
,∴ f =
= Rb
Tb
Tb
reg a rd less
For a half-width rectangular pulse,
Dr. Shi
⇒
Zero:
of
P (ω )
t
Tb
, P (t ) = rect
2
T b
2
ω
Tb
P (ω ) =
s in c
2
4
T
b
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S y (ω ) =
39
Tb
ωT
ωT
sinc 2 b sin 2 b
4
4
2
ɷ1Tb=4π
4π
ω1 =
Tb
f1 = 2 Rb
ɷ2Tb=2π
2π
ω2 =
Tb
f 2 = Rb
⇒
1
Rb =
Tb
essential ⋅ bandwidth
Half that of polar or on-off signaling.
Twice that of theoretical minimum bandwidth (channel b/w).
Table 3.1:
Dr. Shi
ω
t
ωτ
rect ↔ τ sin c
= τ sin c 2
τ
2
τ
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20
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Merits of Bipolar Signaling
1.Spectrum has a dc dull.
2.Bandwidth is not excessive.
3.It has single-error-detection capability.
Since, a single detection error ⇒ a violation of
the alternation pulse rule.
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21
Demerits of Bipolar Signaling
1. Requires twice as much power as a polar
signaling.
Distinction between A, -A, 0
vs.
Distinction between A/2, -A/2
2. Not transparent
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Pulse Shaping
S y (ω ) = P (ω ) S X (ω )
2
=
P (ω )
Tb
2
( R0 + 2
∞
∑R
n
cos nωTb )
[Slide 19]
n =1
• Different signaling (line coding, SX (ω) ) ⇒ diff.S y (ω )
• Different pulse shaping (P(ω)) ⇒ diff. S y (ω )
An additional factor.
• Intersymbol Interference (ISI) (refer to the next slide)
– Time-limited pulses (i.e., truncation in time domain) not band-limited
in frequency domain
– Not time-limited, causes problem band-limited signal (i.e.,
truncation in frequency domain)
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Intersymbol Interference (ISI)
in Detection
B. Sklar’s Digital Communications 1988
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Back to slide 44
45
Phase Shaping…
Intersymbol Interference (ISI)
• Whether we begin with time-limited pulses or
band-limited pulses, it appears that ISI cannot
be avoided.
An inherent problem in the finite bandwidth
transmission.
• Fortunately, there is a way to get away:
be able to detect pulse amplitudes correctly.
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23
Nyquist Criterion for Zero ISI
• The first method proposed by Nyquist.
1, t = 0
1
p (t ) =
0, t = ± nTb Tb =
Rb
(7.22)
T b : separation between successive transmitted pulses
• Example 6.1 (p.256)
p(t ) = sinc(2π Bt ),
B=
1
1
= Rb
2Tb 2
Since 2πBTb=π,
Tb is the 1st zero
crossing.
• Minimum bandwidth pulse that satisfies Nyquist
Criterion Fig. 7.10 (b and c)
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Minimum Bandwidth Pulse
(Nyquist Criterion)
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24
Nyquist Criterion for Zero ISI…
1
p (t ) = sinc (π Rb t ) =
0
P (ω ) =
ω
1
rect
Rb
2π R b
Table 3.1:
t=0
t = ± nTb
1
Tb =
Rb
bandwidth =
Rb
2
FT
ω
Sinc (Wt ) ↔ rect
π
2W
W
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Problems of This Method
•
Impractical
1. − ∞ < t < ∞
2. It decays too slowly at a rate 1/t.
⇒ Any small timing problem (deviation) ⇒ ISI
•
Solution: Find a pulse satisfying Nyquist criterion
(7.22), but decays faster than 1/t.
•
Nyquist: Such a pulse requires a bandwidth
k⋅
Dr. Shi
Rb
,
2
1≤ k ≤ 2
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25
p (t ) ↔ P (ω )
• Let
– The bandwidth of P(ω) is in (Rb /2, Rb)
– p(t) satisfies Nyquist criterion Equation (7.22)
• Sampling p(t) every Tb seconds by multiplying
p(t) by an impulse train δ T (t ).
b
⇒
b
⇒
p (t )δ
p (t) =
T
b
(t) = δ (t)
FT
1
T b
∞
∑
P (ω
− n ω
b
P (ω
− n ω
n = −∞
Equation
G (ω ) =
) = 1
( 6 .4 )
1
T s
∞
∑
n = −∞
s
)
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26
Derivation
• |P(ω)| is odd symmetric in y1-o-y2 system.
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Derivation…
ω
b
• The bandwidth of P(ω) is 2 + ω x
– ω x : the excess bandwidth
–
γ
roll − offfactor
=
=
excess bandwidth
theoretica l min imum bandwidth
ωx
ω
= 2⋅ x
ωb
ωb
0 ≤ r ≤1
2
R
rR
R
• The bandwidth of P(ω) is BT = 2b + 2b = ( 1 + γ ) 2b
• P(ω), thus derived, is called a Vestigial Spectrum.
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27
Realizability
• A physically realizable system, h(t), must be causal,
i.e., h(t) = 0 , for t < 0. (The n.c. & s.c.)
• In the frequency domain, the n.c. & s.c. is known as
Paley-Wiener criterion
∞
ln H (ω )
−∞
1+ω 2
∫
dω < ∞
• Note that, for a physically realizable system, H(ω) may
be zero at some discrete frequencies.
But, it cannot be zero over any finite band
• So, ideal filters are clearly unrealizable.
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• From this point of view, the vestigial spectrum P(ω) is
unrealizable.
• However, since the vestigial roll-off characteristic is
gradual, it can be more closely approximated by a
practical filter.
• One family of spectra that satisfies the Nyquist
criterion is
ωb
1 / 2
P (ω ) = 0
1
Dr. Shi
)
π (ω −
2
1 − sin
2ω x
Lathi & Ding-Digital Commun
ω−
ω >
ω <
ωb
2
ωb
2
ωb
2
< ωx
+ωx
−ωx
56
28
Realizability …
ωx = 0
ω
ωx = b
4
ωx =
ωb
2
(r = 0)
( r = 0 .5 )
( r = 1)
Shown in the figure on next slide
• Comment 1: Increasing ω x (or r) improves
p(t), i.e., more gradual cutoff reduces the
oscillatory nature of p(t), p(t) decays more
rapidly.
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Pulses satisfying Nyquist Criterion
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29
Realizability…
• Comment 2: As
γ
= 1, i.e., ω x
=
ωb
2
ω
ω
1
rect
P (ω ) = 1 + cos
2
2 Rb
4πRb
ω
ω
rect
= cos 2
4 Rb
4πRb
• This characteristic is known as:
The raised-cosine characteristic
The full-cosine roll-off characteristic
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Realizability…
p (t ) = Rb
A.
B.
C.
D.
co s(π R b t )
S in c (π R b t )
1 − 4 R 2bt 2
Bandwidth is Rb (γ =1)
p(0) = Rb
p(t) = 0 at all the signaling instants
&
at points midway between all the signaling instants
p(t) decays rapidly as 1/t3 ⇒ relatively insensitive
to derivations of Rb , sampling rate, timing jitter
and so on.
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30
Realizability…
E. Closely realizable
F. Can also be used as a duobinary pulse.
G. H c (ω ) : channel transfer factor
Pi ( k ) : transmitted pulse
Pi (ω ) ⋅ H c (ω ) = P (ω )
Received pulse at the detector input should be
P(ω) [p(t)].
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Signaling with Controlled ISI:
Partial Response Signals
• The second method proposed by Nyquist to
overcome ISI.
n = 0,1
1
• Duobinary pulse: p(nTb ) = 0 for all other n
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31
Signaling with Controlled ISI…
• Use polar signaling
1 ↔ p (t )
0 ↔ − p (t )
• The received signal is sampled at t = nTb
p(t) = 0 for all n except n= 0,1.
• Clearly, such a pulse causes zero ISI with all the
pulses except the succeeding pulses.
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Signaling with Controlled ISI…
• Consider two such successive pulses located at
0 and Tb.
– If both pulses are positive, the sample value at t = Tb
⇒ 2.
– If both pulses are negative, ⇒ -2
– If pulses are of opposite polarity, ⇒ 0
• Decision Rule: If the sample value at t = Tb is
• Positive ⇒ 1, 1
• Negative ⇒ 0, 0
• Zero ⇒
0, 1 or 1, 0
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32
Signaling with Controlled ISI…
• Table 7.1
Transmit ed Seq.
Sample of x(Tb)
Detected seq.
Dr. Shi
1 1 0 1 1 0 0 0 1 0 1 1 1
1 2 0 0 2 0 -2 -2 0 0 0 2 2
1 1 0 1 1 0 0 0 1 0 1 1 1
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Signaling with Controlled ISI:
Duobinary Pulses
• If the pulse bandwidth is restricted to be
it can be shown that the p(t) must be:
p (t) =
Rb
2
sin( π R b t )
π R b t (1 − R b t )
ω
2
P (ω ) =
cos
Rb
2 Rb
rect
ω
2π R b
−
e
j
ω
2 Rb
ωb
• Note: The raised-cosine pulse with ω = 2 , ( r = 1),
satisfies the condition (7.3b) to be signaling
with controlled ISI – refer to Figure 7.13 (slides 57).
x
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33
Signaling with Controlled ISI
Duobinary Pulses…
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Use of Differential Coding
• For the controlled ISI method,
0-valued sample ⇒ 0 to 1 or 1 to 0 transition.
• An error may be propagated.
• Differential coding helps.
• In differential coding, “1” is transmitted by a pulse
identical to that used for the previous bit. “0” is
transmitted by a pulse negative to that used for the
previous bit.
• Useful in systems that have no sense of absolute
polarity. Fig 7.17
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34
Differential Code
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Scrambling
• Purpose: A scrambler tends to make the data more
random by
– Removing long strings of 1’s or 0’s
– Removing periodic data strings.
– Used for preventing unauthorized access to the data.
• Example. (Structure)
Scrambler: S ⊕ D 3T ⊕ D 5T = T
D nT :
⊕:
T delayed by n units
Modulo 2 Sum
1⊕1 = 0
0⊕0=0
Dr. Shi
→ Incomplete version
1⊕ 0 = 1
0 ⊕1 = 1
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35
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Scrambling…
• Descrambler:
Incomplete version
R = T ⊕ (D
3
⊕ D 5 )T
= [1 ⊕ ( D 3 ⊕ D 5 )] T
= (1 ⊕ F ) T
∆
F = D
Dr. Shi
3
⊕ D5
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36
Example 7.2
• The data stream 101010100000111 is fed to the
scrambler in Fig. 7.19a. Find the scrambler output T,
assuming the initial content of the registers to be zero.
– From Fig. 7.19a we observer that initially T = S, and
the sequence S enters the register and is returned as
through the feedback path. This
( D 3 ⊕ D 5 ) S = FS
new sequence FS again enters the register and is
returned as F 2 S , and so on. Hence,
(7.41)
T = S ⊕ FS ⊕ F 2 S ⊕ F 3 S ⊕ ...
= (1 ⊕ F ⊕ F 2 ⊕ F 3 ⊕ ...) S
Complete version
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Example 7.2 …
• Recognizing that
We have
(
)(
F = D3 ⊕ D5
)
F 2 = D 3 ⊕ D 5 D 3 ⊕ D 5 = D 6 ⊕ D10 ⊕ D 8 ⊕ D8
• Because modulo-2 addition of any sequence
with itself is zero, D 8 ⊕ D 8 = 0 and
F 2 = D 6 ⊕ D10
• Similarly,
(
)(
)
F 3 = D 6 ⊕ D10 D 3 ⊕ D 5 = D 9 ⊕ D11 ⊕ D13 ⊕ D15
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37
Example 7.2 …
…… and so on ……
Hence,
T = (1⊕ D3 ⊕ D5 ⊕ D6 ⊕ D9 ⊕ D10 ⊕ D11 ⊕ D12 ⊕ D13 ⊕ D15 ⊕...)S
Because DnS is simply the sequence S delayed
by n bits, various terms in the preceding
equation correspond to the following sequences:
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Example 7.2 …
S = 1010101000 00111
D 3 S = 0001010101 00000111
D 5 S = 0000010101 0100000111
D 6 S = 0000001010 1010000011 1
D 9 S = 0000000001 0101010000 0111
D 10 S = 0000000000 1010101000 00111
D 11 S = 0000000000 0101010100 000111
D 12 S = 0000000000 0010101010 0000111
D 13 S = 0000000000 0001010101 00000111
D 15 S = 0000000000 000001010101000 00111
T = 1011100011 01001
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38
Example 7.2 …
• Note that the input sequence contains the periodic
sequence 10101010…, as well as a long string of 0’s.
• The scrambler output effectively removes the periodic
component as well as the long strings of 0’s.
• The input sequence has 15 digits. The scrambler output
up to the 15th digit only is shown, because all the
output digits beyond 15 depend on the input digits
beyond 15, which are not given.
• We can verify that the descrambler output is indeed S
when this sequence T is applied at its input.
• Homework: Learn to be able to determine where no
terms needs to be considered.
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Regenerative Repeater
•
Three functions:
1. Reshaping incoming pulses using an equalizer.
2. Extracting timing information
Required to sample incoming pulses at optimum
instants.
3. Making decision based on the pulse samples
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39
Regenerative Repeater…
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Preamplifier and Equalizer
• A pulse train is attenuated and distorted by
transmission medium, say, dispersion caused by an
attenuation of high-frequency components.
• Restoration of high frequency components ⇒ increase
of channel noise
• Fortunately, digital signals are more robust.
– Considerable pulse dispersion can be tolerated
• Main concern:
Pulse dispersion ⇒ ISI
⇒ increase error probability in detection
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40
Zero-Forcing Equalizer
• Detection decision is based solely on sample
values.
⇒ No need to eliminate ISI for all t.
All that is needed is to eliminate or minimize ISI
at their respective sampling instants only.
• This could be done by using the transversal-filter
equalizer, which forces the equalizer output pulse
to have zero values at the sampling (decisionmaking ) instant. (refer to the diagram in next slide)
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41
Zero-Forcing Equalizer …
c0 =1
ck = 0 ∀k ≠ 0
• Let
tap setting
⇒ P0 (t ) = Pr (t − NTb )
P0 (t ) = Pr (t )
If ignore delay.
• Fig 7.22 (b) indicates a problem :
a1, a-1, a2, a-2, …are not negligible due to dispersion.
• Now, want to force a1= a-1= a2= a-2 =…= 0
• Consider ck’s assume other values (other tap setting).
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Zero-Forcing Equalizer …
• Consider ck’s assume other values (other tap setting).
N
p0 ( t ) =
∑c
n pr ( t
− nTb )
n =− N
N
p0 ( kTb ) =
∑c
n pr [(k
− n )Tb ],
k = 0, ± 1, ± 2 ,...
n =− N
• For simplicity of notation:
p0 ( k ) =
∞
∑c
n =−∞
n
pr ( k − n)
• Nyquist Criterion:
Dr. Shi
p0 ( k ) = 0
∀k ≠ 0
p0 ( k ) = 1
for k = 0
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42
Zero-Forcing Equalizer …
•
⇒ A set of infinitely many simultaneous
equations:
2N+1: cn’s
Impossible to solve this set of equations.
• If, however, we specify the values of p0(k) only
at 2N+1 points:
k =0
1
p0 ( k ) =
0 k = ±1,±2,...,± N
then a unique solution exists.
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Zero-Forcing Equalizer …
• Meaning: Zero ISI at the sampling instants of N
preceding and N succeeding pulses.
• Since pulse amplitude decays rapidly, ISI
beyond the Nth pulse is not significant for N>2
in general.
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43
Zero-Forcing Equalizer …
• Page 7.37
Think about when n = 2, 1, 0, -1, -2; an example in next slide
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Zero-Forcing Equalizer …
• Page 7.37
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44
Eye Diagram
• A convenient way to study ISI on an oscilloscope
• Figure
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89
Timing Extraction
•
•
•
The received signal needs to be sampled at precise
instants.
Timing is necessary.
Three ways for synchronization:
1.
2.
3.
Dr. Shi
Derivation from a primary or a secondary standard (a
master timing source exists, both transmitter and receiver
follow the master).
Transmitting a separate synchronizing signal (pilot clock)
Self synchronization (timing information is extracted from
the received signal itself).
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45
Timing Extraction …
• Way 1:
Suitable for large volumes of data
high speed comm. Systems.
High cost.
• Way 2: Part of channel capacity is used to transmit
timing information.
Suitable for a large available capacity.
• Way 3: Very efficient.
• Examples:
– On-off signaling (decomposition: Fig. 7.2)
– Bipolar signaling rectification
on-off signaling
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Timing Jitter
• Small random deviation of the incoming pulses
from their ideal locations.
• Always present, even in the most sophisticated
communication systems.
• Jitter reduction is necessary about every 200
miles in a long digital link to keep the maximum
jitter within reasonable limits.
• Figure 7.23
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46
Figure 7.23
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93
Detection –Error Probability
• Received signal = desired + AWGN
Example: Polar signaling and transmission (Fig. 7.24,
the next slide)
Ap : Peak signal value
• Polar: Instead of ± Ap , received signal = ± Ap + n
• Because of symmetry, the detection threshold = 0.
– If sample value > 0, ⇒ “1”
– If sample value < 0, ⇒ “0”
– Error in detection may take place
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47
Figure 7.24
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Error Probability for Polar Signal
• With respect to Figure 7.24
P (∈ / 0 ) = probability that n > Ap
P (∈ / 1) = probability that n < -Ap
p (n) =
n: AWGN
⇒ P(∈ 0) =
=
Dr. Shi
1
2π
1
∞
Ap
∫σ
n
e
1
σ n 2π
−
1 2
( x)
2
∫
∞
Ap
2π σ n
e
1 n
−
2 σn
e
1 n
−
2 σ n
2
2
dn
x=
n
σn
Ap
dx = Q
σn
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48
Error Probability for Polar Signal…
1
Q( y ) =
2π
where
• Similarly,
∫
∞
y
e
−
x2
2
dx
Ap
P(∈ 1) = Q
σn
− Ap to − ∞
Symmetric
• Summary:
P (∈) = P (∈,0) + P (∈,1)
Here, assume P(0)= P(1)=1/2.
= P (∈ 0) P(0) + P (∈ 1) P(1)
=
A
1
[P(∈ 0) + P(∈ 1)] = Q p
2
σn
• Q(x):
Complementary error function: erfc (x)
1
Q ( x) ≅
x 2π
Dr. Shi
x2
0 .7 −
1 − 2 e 2 , x > 2
x
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Error Probability for Polar Signal…
• If
Ap
σn
= k , P(∈ 0) = P(∈ 1) = P(∈) = Q(k )
k
Q(k)=P(ε)
1
0.1587
e.g. 3.16 ×10−5 ⇒ one of
wrongly
Dr. Shi
2
3
4
5
6
0.0227 0.00135 3.16 ×10−5 2.87 ×10 −7 9 . 9 × 10 − 9
3.16 × 105
pulses will be possibly detected
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49
Error Probability for On-off Signals
• In this case, we have to distinguish between Ap
and 0.
1
Ap
• Threshold:
2
• Error Probabilities:
A
1
P(∈ 0) = prob. of
P(∈ 1) = prob. of
P(∈) =
n>
p
Ap = Q
σ
2
n
Ap
−1
n<
Ap = Q
2
2σ n
A
1
[P(∈ 0) + P(∈ 1)] = Q p
2
2σ n
2
Also assume 0 and 1
equally probable
P(∈) > P(ε ) (in the case of polar signaling)
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50
Error Probability for Bipolar Signals
1 ↔ Ap or − Ap
0 ↔ 0v
⇒
A
A
p
p
If the detected sample value: ∈ − 2 , 2 ↔ 0
Ap
P(∈ 0) = prob. n >
2
Ap
A
+ prob. n < − p
= prob. n >
2
2
o.w. ↔ 1
Ap
A
= 2Q p
= 2 prob. n >
2
2σ n
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101
Error Prob. for Bipolar Signals…
Ap
when
P (∈ 1) = prob . n < −
positive
pulse used
2
Ap
when negative
or
prob . n >
pulse used
2
Ap
= Q
2σ n
1
P (∈) = P (∈ 0 ) P ( 0 ) + P (∈ 1) P (1) = [P (∈ 0 ) + P (∈ 1) ]
2
Ap
> P (ε )
= 1 . 5Q
for on − off
signaling
2σ n
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51
Detection Error Probability…
Ap
Q
σn
A
on-off is in middle, Q p
2σ n
A
bipolar is the worst, 1.5Q p
2σ n
• Summary, polar is the best,
Assumption:
“0” and “1”
Equally likely
• Another factor:
P(∈) decreases exponentially with the signal power.
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Comparison among three line codes
• To obtain:
P(∈) = 0.286 × 10−6
Ap
• We need:
σn
Ap
σn
Ap
σn
=5
for polar case
= 10
Q Q(5) = 0.286 × 10−6
Ap
Q P (∈) = Q
2
σ
n
for bipolar case Q P (∈) = 1.5Q Ap
2σ
n
−6
= 0.286 × 10
for on-off case
= 10 . 16
c
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Ap
2σ n
= 5.08
104
52
Comparison among three line codes…
Ap
• For the same error rate, required SNR σ n ,
Polar < On-off < Bipolar
• For the same
A
SNR σ p
n
(from previous example)
, the caused error rate
Polar < On-off < Bipolar
(from next example)
• Polar is most efficient in terms of SNR vs. error rate
• Between, on-off and bipolar:
Only 16% improvement of on-off over bipolar
• Performance of bipolar case ≈ performance of on-off
case in terms of error rate.
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Example
a) Polar binary pulses are received with peak amplitude
Ap = 1 mV. The channel noise rms amplitude is 192.3
µV. Threshold detection is used, and 1 and 0 are
equally likely.
b) Find the error probability for
(i) the polar case, and the on-off case
(ii) the bipolar case if pulses of the same shape as
in part (a) are used, but their amplitudes are
adjusted so that the transmitted power is the same
as in part (a)
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53
Example …
a) For the polar case
Ap
σn
=
10 −3
= 5 .2
192.3(10 −6 )
From Table 8.2 (4th ed), we find
P (∈) = Q ( 5 . 2 ) = 0 . 9964 × 10 − 7
b) Because half the bits are transmitted by nopulse, there are, on the average, only half as
many pulses in the on-off case (compared to
the polar). Now, doubling the pulse energy is
accomplished by multiplying the pulse by 2
(in order to keep the power dissipation same)
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Example…
Thus, for on-off Ap is 2 times the Ap in the polar case.
•Therefore, from Equation (7.53)
A
P(∈) = Q p = Q (3.68) = 1.66 × 10 − 4
2σ n
•As seen earlier, for a given power, the Ap for both
the on-off and the bipolar cases are identical.
Hence, from equation (7.54)
Ap
P (∈) = 1.5Q
2σ n
Dr. Shi
= 1.749 × 10 −4
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54
M-ary Communication
• Digital communications use only a finite number of
symbols
• Information transmitted by each symbol increases with
M.
IM = log2 M bits
IM
: information transmitted
by an M-ary symbol
• Transmitted power increases as M2 ,
i.e., to increase the rate of communication by a factor
of log2 M , the power required increases as M2.
(see an example below)
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M-ary Communication…
• Most of the terrestrial digital telephone network:
Binary
• The subscriber loop portion of the integrated
services digital network (ISDN) uses the
quarternary code 2BIQ shown in Figure. 7.28
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55
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Pulse Shaping in
Multi-amplitude Case
• Nyquist Criterion can be used for M-ary case
Controlled ISI
• Figure 7.28: One possible M-ary Scheme
• Another Scheme: Use M orthogonal pulses:
ϕ1 (t ), ϕ 2 (t ),..., ϕ M (t )
• Definition:
Tb
c i = j
ϕ i ( t )ϕ j ( t ) =
0
0 i ≠ j
∫
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56
Pulse Shaping in
Multi-amplitude Case
• The figure in next slide: One example in M orthogonal
pulses:
2π ⋅ k ⋅ t
, 0 < t < Tb k = 1, 2,..., M
sin
Tb
ϕk (t ) =
0, otherwise
• In the set, pulse frequency:
k
1 1 2
M
: , ,...,
Tb Tb Tb
Tb
M times that of the binary
scheme
(see an example below)
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57
Pulse Shaping in
Multi-amplitude Case
• In general, it can be shown that the bandwidth
of an orthogonal M-ary scheme is M times that
of the binary scheme
– In an M-ary orthogonal scheme, the rate of
communication is increased by a factor of log 2 M at
the cost of an increase in transmission bandwidth by
a factor of M.
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115
Digital Carrier Systems
• So far: Baseband digital systems
– Signals are transmitted directly without any shift in
frequency.
– Suitable for transmission over wires, cables, optical
fibers.
• Baseband signals cannot be transmitted over a
radio link or satellites.
– Since it needs impractically large size for antennas
– Modulation (Shifting signal spectrum to higher
frequencies is needed)
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58
Digital Carrier Systems…
• A spectrum shift to higher frequencies is also
required when transmitting several messages
simultaneously by sharing the large bandwidth
of the transmission medium,
• FDM: Frequency-division Multiplexing.
(FDMA: Frequency-division Multiplexing
Access)
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Several Types of Modulation
• Amplitude-Shift Keying (ASK), also known as onoff keying (OOK)
m(t ) cos(ω c t ) : m(t ) : on-off baseband signal (modulating signal)
cos(ω c t )
Dr. Shi
: carrier
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59
Modulation Types…
• Phase-Shift Keying (PSK)
m(t ) : polar Signal
1 ↔ p(t) cos(ωct),0 ↔ − p(t) cos(ωct) = p(t) cos(π −ωct) = p(t) cos(ωct −π )
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119
Modulation Types…
• Frequency-Shift Keying (FSK)
– Frequency is modulation by the base band signal.
1 ↔ ω c1 ,0 ↔ ω c0
– Information bit resides in the carrier frequency.
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60
PSD of ASK, PSK & FSK
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FSK
• FSK signal may be viewed as a sum of two
interleaved ASK signals, one with a modulating
frequency ω c , the other, ω c
0
1
– Spectrum of FSK = Sum of the two ASK
– Bandwidth of FSK is higher than that of ASK or
PSK
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61
Demodulation
A. Review of Analog Demodulation
• Baseband signal and bandpass signal
The end of the 2nd part of the slides of Ch. 2 and Ch. 3
• Double-sideband Suppressed Carrier (DSB-SC)
Modulation
m(t) ↔ M(ω)
message signal
1
cos(ωct) ↔ [δ (ω +ωc ) +δ (ω −ωc )]
carrier
2
1
m(t)cos(ωct) ↔ [ M(ω +ωc ) + M(ω −ωc )] modulated signal
2
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Figure 4.1 (a), (b), (c) DSB-SC Modulation
124
62
DSB-SC Demodulation
• Demodulation:
Synchronization Detection
(Coherent Detection)
• Figure 4.1 (e) and (d)
Using a carrier of
exactly the same
frequency & pulse
Dr. Shi
125
DSB-SC Demodulation …
[m(t ) cos(ωct )]cos(ωct ) = e(t ) = 1 [m(t ) + m(t ) cos(2ωct )]
2
1
1
M (ω ) + [M (ω + 2ωc ) + M (ω − 2ωc )]
2
4
1
Aftera
LPF ⇒ M (ω )
2
E (ω ) =
1
m(t )
2
Dr. Shi
126
63
Figure 4.1 (d), (e) DSB-SC demodulation
127
AM Signal
• Motivation:
– Requirements of a carrier in frequency and phase
synchronism with carrier at the transmitter is too
sophisticated and quite costly
– Sending the carrier ⇒ a. eases the situation
b. requires more power in
transmission
c. suitable for broadcasting
Dr. Shi
128
64
Amplitude Modulation (AM)
ϕ AM (t ) = A cos(ω c t ) + m(t ) cos(ω c t )
= [A + m(t )]cos(ω c t )
ϕ AM (t ) ⇔
1
[M (ω + ω c ) + M (ω − ω c )]
2
+ π ⋅ A[δ (ω + ω c ) + δ (ω − ω c )]
Dr. Shi
129
• AM Signal and its Envelope
130
65
• Demodulation of AM signals
– Rectifier Detection:
Figure 4.11
Output:
– Envelope Detector:
1
π
m(t )
Figure 4.12
Output: A + m(t )
Dr. Shi
131
132
66
133
B. Back to Digital Communication System:
Demodulation
•
Demodulation of Digital-Modulated Signals is
similar to Demodulation of Analog-Modulated
Signals
•
Demodulation of ASK
Synchronous (Coherent) Detection
Non-coherent Detection: Say, envelope detection
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67
Demodulation of PSK
• Cannot be demodulated non-coherently
(envelope detection).
– Since, envelope is the same for 0 and 1
• Can be demodulated coherently.
• PSK may be demodulated non-coherently if use:
Differential coherent PSK (DPSK)
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Differential Encoding
• Differential coding:
– “1” encoded by the same pulse used to encode
the previous data bit no transition
– “0” encoded by the negative of pulse used to
encode the previous data bit transition
• Figure 7.30 (a)
• Modulated signal consists of pulses ± A cos(ω ct )
with a possible sign ambiguity
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68
Differential Encoding
• Figure
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Differential Encoding…
• Tb : one-bit interval
• If the received pulse is identical to the previous
pulse 1,
A2
Y (t ) = A2 cos 2 (ω c t ) =
[1 + cos(2ω ct )]
2
z (t ) =
2
A
2
• If 0 is transmitted
Y (t ) = − A2 cos 2 (ω c t ) = −
z (t ) = −
Dr. Shi
A2
[1 + cos(2ω ct )]
2
A2
2
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69
Demodulation of FSK
• FSK can be viewed as two interleaved ASK
signals with carrier frequency ω c and ω c ,
respectively.
• Hence, FSK can be detected by non-coherent
detection (envelope) or coherent detection
technique
0
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Demodulation of FSK
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70
Demodulation of FSK
• In (a), non-coherent detection (envelope),
ωc
– H0(ω)
0
Turned to
– H1(ω)
respectively
ωc
1
• Comparison: above or bottom
Whose output is large 0 or 1
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71
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