1
Chapter 3
Baseband Pusle and
Digital Signaling
2
Chapter Objectives
? Analog-to-digital signaling(PCM and delta
modulation)
? Binary and multilevel digitals signals
? Spectra and bandwidths of digital signals
? Prevention of intersymbol interference
? Time division multiplexing
? Packet transmission
3
3.1 Introduction(main goals)
? To study how analog waveforms can be
converted to digital waveforms,The most
popular technique is called PCM
? To learn how to compute the spectrum for
digital signals
? To examine how to filtering of pulse signals
affects out ability to recover the digital
information at the receiver(ISI)
? To study how we can multiplex data from
several digital bit streams into on high-speed
digital stream fro transmission over a digital
system.(TDM)
4
3.2 Pulse Amplitude Modulation
? PAM is an engineering term that is used to
describe the conversion of the analog signal to
a pulse-type signal in which the amplitude of
the pulse denotes the analog information
? The sampling theorem gives a way to
reproduce an analog waveform by using
sample values of that waveform and sin(x)/x
orthogonal functions
? PAM signaling is to provide another waveform
that looks like pulses,yet contains the
information that was present in tha analog
waveform,
5
3.2 Pulse Amplitude Modulation
? The pulse rate,fs,for PAM is the same as that
required by the sampling theorem,namely,
fs>=2B,where B is the highest frequency in the
analog waveform and 2B is called the Nyquist
rate,
? Classification,
Nature sampling(gating)
Instantaneous sampling(Flat-top type)
6
3.2 Pulse Amplitude Modulation(Nature
sampling or gating)
? Definition,if ω(t) is an analog waveform
bandlimited to B hertz,the PAM signal that
uses natural sampling is
ωs(t)=ω(t)s(t)
Where s(t) is a rectangular wave switching
waveform and fs=1/Ts>=2B
? ?
∑ ∏ )
1
(∑ ∏ )(
∑ ∏ /]-[s (t )
∞
∞-
∞
∞-
∞
∞-
??
?
????
?
kk
s
k
s
d
k
tT
k
t
kTt
???
?
7
3.2 Pulse Amplitude Modulation(PAM signal
with natural sampling)
(a) Baseband Analog Waveform
S(t)
t
(b) Switching waveform with Duty Cycle d=τ /T =1/3
t
W(t)
W (t)
s
Figure 3-1 PAM Signal with natural sampling
≈
T
τ
8
3.2 Pulse Amplitude Modulation(Generation of
PAM with natural sampling)
clock
S(f)
Analog bilateral switch
Ws(t)=w(t)*s(t)
Figure 3-2 Generation of PAM with natural
sampling (gating)
9
3.2 Pulse Amplitude Modulation(spectrum
of a naturally sampled PAM)
? Theorem,the spectrum for a naturally
sampled PAM signal is,
Where fs=1/Ts,ωs=2πfs,the duty cycle of s(t) is d=τ/Ts,
and W(f)=F[ω(t)] is the spectrum of the original
unsampled waveform,
∑ ∞
∞-=
)n-(ndπ ndπsin =)](ω[=)(
n sss
ffWdtfW F
( )
∑∑ ∏∑ ∏
∑ ∏
∞
∞-=
ω
∞
∞-=
∞
∞-=
∞
∞-=
)
1
-
τ
(=)
τ
-
τ
(=
τ/]-[=s (t )
k
tjn
n
kk
s
k
s
sec
d
k
tT
k
t
kTt
＝
10
3.2 Pulse Amplitude Modulation(spectrum
of a naturally sampled PAM)
? The spectrum of the PAM with
natural sampling is a function of the
spectrum of the analog input
waveform
? The spectrum of the input analog
waveform is repeated at harmonics of
the sampling frequency
? the PAM spectrum is zero for ± 3fs,
± 6fs,and so on,Because of d=1/3,and
the spectrum in these harmonic bands
is nulled out by the (sin(x)/x) function
? The bandwidth of the PAM signal is
much larger than the bandwidth of
the original ananlog signal,The null
bandwidth for the envelope of the
PAM signal is 3fs=12B,that is,the null
bandwidth of this PAM signal is 12
times the bandwidth of the analog
signal,
? Fig,3-3 Spectrum of a PAM
waveform with Natural
sampling(p132)
11
3.2 Pulse Amplitude Modulation(receiver)
? Fig.3-4 Demodulation
of a PAM signal
? Low-pass filter
B<fcutoff<fs-B
fs>=2B
If the analog
signal is under
sampled(fs<2B),
spectral overlapping
(aliasing) is raised,
? Product detection
noise due to power supply hum or noise due to
mechanical circuit vibration might fall in the band
corresponding to low frequency band and other
bands might be relatively noise free
Low Pass
Filter H(f)
Oscillator
Wo=nws
fco -fco
f
H(f)
B〈 fco〈 fs-B
Analog Multiplier
(four-quadrants multiplier) W(s) PAM (natural
sampling Cw(f)
12
3.2 Pulse Amplitude Modulation( instantaneous
sampled or Flat-top PAM)
? Definition,if ω(t) is an analog waveform
bandlimited to B hertz,the PAM signal
that uses natural sampling is
Where h(t) denotes the sampling-pulse
shape and,for flat-top sampling,the pulse
shape is,
Where τ<=Ts=1/fs and fs>=2B
∑ ∞
∞-= sss
)kT-) h ( t( k Tω=( t )ω
k
? ? /2|t| 0 a n d / 2,|t| 1∏ /h ( t ) ??? ??? f o rt ＝
13
3.2 Pulse Amplitude Modulation
? Fig 3-5 PAM signal with flat-top
sampling ∑
∞
∞-= sss
)kT-) h ( t( k Tω=( t )ω
k
? PAM with flat-top sampling
is called instantaneous
samples,since ω(t) is
sampled at t=kTs and
sample value ω(kTs)
determine the amplitude of
the flat-top rectangular
pulse
? The flat-top signal could be
generated by using a
sample-and-hold type of
electronic circuit
? Note that if h(t)=sinx/x with
overlapping pulses,then
becomes identical to the
sampling theorem
t
Ws(t)
0
(a) Baseband Analog waveform
t
(b) Impulse Train Sampling waveform
τ
→ ← Ts
Ws(t)
t
(c) Resulting PAM Signal
( flat-top sampling,d=τ/Ts=1/3 )
14
3.2 Pulse Amplitude Modulation—the spectrum
of flat-top PAM signal
? Theorem,the spectrum for a flat-top PAM
signal is,
where
Proof,
∑ ∞
∞-=
)k-()()/1(=)(
n sss
ffWfHTfW
)s i n())(()( f fthfH ?? ????F＝
∑ ∞
∞-=
sss )kT-) h ( t( k Tω=( t )ω
k
15
3.2 Pulse Amplitude Modulation—the spectrum
of flat-top PAM signal
16
3.2 Pulse Amplitude Modulation—the spectrum
of flat-top PAM signal
? Low-pass filter,
There is some high-frequency loss in the recovered
analog waveform due to the filtering effect,this loss
can be reduced by decreasing τ or by using some
additional gain the high frequency in the low-pass
filter transfer function,This filter is called
equalization filter,
The pulse width τ is also called the aperture since τ/Ts
determines the gain of the recovered analog signal,
which is small if τ is small relative to Ts
17
3.2 Pulse Amplitude Modulation( key results)
? The transmission of either naturally or instantaneously
sampled PAM over a channel requires a very wide
frequency response because of the narrow pulse width,
which imposes stringent requirements on the magnitude
and phase response of the channel,
? The bandwidth required is much larger than that of the
original analog signal,and the noise performance of the
PAM system can never be better than that achieved by
tansmitting the analog signal directly;
? It provide a means for converting an analog signal to a
PCM signal; and provide a means for breaking a signal
into time slots that multiple PAM signals carrying
information from different sources can be interleaved to
transmit all of the information over a single channel,
18
3.3 Pulse code modulation
? Definition,Pulse code modulation(PCM) is essentially
analog-to-digital conversion of a special type where
the information contained in the instantaneous
samples of an analog signal is represented by digital
words in a serial bit stream,
? Quantizing,instead of using the exact sampled value
of the analog waveform ω(kTs),the sample is
replaced by the closest allowed value where there are
M=2n allowed value,each corresponding to one of the
code words,
? Other popular types of AD conversion,including delta
modulation (DM) and differential pulse code
modulation (DPCM) will be discussed,
19
3.3 Pulse code modulation—the advantages of
PCM
? Relatively inexpensive digital circuitry may be used
extensively in the system
? PCM signals derived from all types of analog source
(audio,video) may be merged with data signals and
transmitted over a common high-speed digital
communication system,this merging is called time-
division multiplexing
? In long-distance digital telephone systems requiring
repeaters,a clean PCM waveform can be regenerated
at the output of each repeater,where the input
consists of a noisy PCM waveform,
? The noise performance of a digital system can be
superior to that of an analog system,
20
3.3 Pulse code modulation
(Sampling,Quantizing,and Encoding)
? PCM signal is generated by carrying out three basic operations,
sampling,Quantizing and Encoding;
? Sampling operation generates a flat-top PAM signal
? Fig.3-7 PCM transmission system
Signal in
Analog Low-pass
filter
Bandwidth=B
Band limited
analog Instantaneous
sampler
and hold
Flat-top
PAM Quantized No,
of levels=M
Quantized
PAM Encoder PCM
Signal
PCM transmitter (analog-to-digital conversion)
Telephone
line Regenerative
repeater
Telephone
line Regenerative
repeater
Regenerative
repeater
Telephone line
Channel (transmission path)
Regenerative
circuit
PCM Decoder Quantized PAM Low-pass (reconstruction)
filter
PCM receiver (digital-to-analog conversion)
Analog signal
out
21
3.3 Pulse code modulation
? Quantizing
If all of the steps are of equal size,quantizer is said to be uniform;
Quantizing effect introduced error into the recovered output analog
signal;
Quantizing error → Quantizing noise (a round -off error)
? Encoding
The PCM signal is obtained from the quantized PAM signal by
encoding each quantized sample value into a digital word
(Fig,3-8)
2 4 6 8 -2 -4 -6 -8
0 2
4
6 8
-4
-6
-8
M=8
Output
voltage
↑
V=8
Input voltage=x
↑
-V=8
(a) Quantize Output –Input Characteristics
22
3.3 Pulse code modulation
2
4
6
8
-2
-4
-6
-8
Sampling times T
s
t
Quantized PAM signals
Analog signal,a(t)
(b) Analog
Signal,Flat
– top PAM
Signal,and
Quantized
PAM Signal
1
2
-1
-2
Difference between analog signal
and quantized PAM signal
t
(c) Error Signal
1 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 1
→ ← PCM word
(d) PAM Signal
Fig 3-8 illustration of waveforms in a PCM system
23
3.3 Pulse code modulation
? Encoding
The PCM signal
is obtained from
the quantized
PAM signal by
encoding each
quantized sample
value into a
digital word
? Table3.1 Three-bit Gray code for M=8 levels
Quantized
sample Voltage
Gray Code Word
(PCM output)
+7
+5
+3
+1
1 10
1 11
1 01
1 00
-7
-5
-3
-1
0 00
0 01
0 11
0 10 each quantized sample ==> a digital word (n
binary digits) M= 2n
unique code words
24
3.3 Pulse code modulation
? Notes,
? Only one bit change for each step change
in the quantized level,
? Single errors in the received PCM code
word will cause minimal errors in the
recovered analog level,provided that the
sign bit is not in error
25
3.3 Pulse code modulation(bandwidth of PCM
signal)
Bit rate R=nfs
fs≥2B B—Bandwidth of analog signal
the bandwidth of the PCM waveform
BPCM≥1/2R= 1/2nfs
BPCM min = 1/2R = 1/2nfs
(A is （ sinx/x） type of pulse shape )
26
3.3 Pulse code modulation（ Effect of noise ）
? the main effects produce noise or distortion
? Quantizing noise that is caused by the M-step
quantized at PCM transmitter
? Bit errors in the recovered PCM signal,(channel
noise,improper channel filtering )
? Not strictly bandlimilted input analog signal,
(Aliasing noise )
27
3.3 Pulse code modulation
? Ratio of the recovered analog peak signal power to
the total average noise power is given by,
? Ratio of the average signal power to the average
noise power is
1 6 a )-(3 )1-(41 3/ 2
2
-
e
o u tpk PM
MNS
??）（
e
o u t PM
MNS
)1-(41/ 2
2
??）（
where M is the number of quantized levels,and Pe is
the probability of bit error
If Pe=0 (No ISI),the peak SNR and average SNR
resulting form only quantizing errors is respectively
2- 3=/ MNS o u tpk）（ 2=/ MNS o u t）（
28
3.3 Pulse code modulation--- Four types of the
quantizing noise
Overload noise ( flat-tops near the peak value )
the peak level exceed the design peak of V volts
Random noise
random quantization errors ---- white hissing sound
Granular noise unequal quantization errors form sample
to sample solution ; a nonuniform quantization,e.g,u-law
quantized
hunting noise when the input analog waveform is
nearly constant,the output samples oscillate between
two adjacent quantization levels,Solution,① filter out
the tone; ② designing the quantize so that there is no
vertical step at the constant valve step,
29
3.3 Pulse code modulation(6dB -rule)
? 6dB-rule
α+02.6=/ nNS dB）（
? where,n- the number of bits in the PCM word
α=4.77 ( the peak SNR ) α= 0 (the average SNR )
An additional 6-dB improvements in SNR is
obtained for each bit added to the PCM word,
Assumptions,① No bit errors
② the input signal level is large
enough to range over a significant number of
quantizing levels
30
3.3 Pulse code modulation
? Example 3-1 Design of a PCM signal for audio
voice-frequency (VF) Telephone Systems,
band, 300H ~ 3400 H
? Sampling frequency,(fSmin) = 3.4kHz * 2 = 6.8 kHz
? bit rate R= fs (samples/s)*n(bits/sample)
= 8 k sample/s * 8 bits/sample = 64 kbps
Is called Ds-0(digital signal,type zero)
? the minimum absolute bandwidth
(B)min = 1/2R = 32kHz ( for a sinx/x pulse shape)
BPCM =R = 64kH ( for rectangular pulse shape)
? peak SNR is
dBMNS outpk 9.52=2*33=/ 282- ）（＝）（
31
3.3 Pulse code modulation
Note,
? The inclusion of a parity bit does not affect
the quantizing noise,
? The performance of a PCM system (i.e,no
bit errors and no ISI ) is determined by M,the
number of quantizing steps used,
32
3.3 Pulse code modulation （ Nonumiform
Quantizing:μ-Law and A-Low compounding）
? No uniform Quantizing – a variable step size is
used
Analog
signal
A Compression
(nonlinear)
Amplifier
PCM
(uniform quantized)
Nonuniform
Quantizing
signal
Compression
quantizer
characteristic
Uniform quantizer
characteristic
0
0
0.2 0.4 0.6 0.8 1.0
V=1.0
0.2
0.4
0.6
0.8
1.0
(a) M=8
Quantizer
Characteristi
c
33
3.3 Pulse code modulation
0 0.2 0.4 0.6 0.8 1.0
0.2
0.4
0.6
0.8
1.0
μ=0 μ=1 μ=5
μ=100
μ=225
(b) u – law Characteristic (c) A-law Characteristic
0 0.2 0.4 0.6 0.8 1.0
0 0.2 0.4 0.6 0.8 1.0
0
0.2
0.4
0.6
0.8
1.0
A=1 A=2
A=5
A=87.6
A=100
Fig.3-9 Compression characteristics (first quadrant shown)
34
3.3 Pulse code modulation
35
3.3 Pulse code modulation
? U-Low compression characteristic,wher |w1(t)|≤ 1,
? u is a positive parameter (constant )
if u=0,it is linear amplification (Uniform
Quantization)
)μ+l n ( 1
|)( t )ω|μ+l n ( 1=|( t )ω| 1
2
? A-Low compression characteristic,wher |w1(t)|≤ 1,
? A is a positive parameter (constant )
if A=1,it is linear amplification (Uniform
Quantization)
)l n (+1
|)( t )ω|l n (+1=|( t )ω| 1
2 A
A
36
3.3 Pulse code modulation
? When compression is used at the transmitter,
expansion(i.e,decompression) must be used at
the receiver output to restore signal levels to
their correct relative values,The expandor
characteristic is the inverse of the
compression characteristic,and the
combination of a compressor and an expandor
is called a compandor
37
3.3 Pulse code modulation
? Output SNR α+02.6=/ nNS
dB）（
where α= 4.77-20log(V/xrms) (uniform quantizing)
or α ≈4.77-20log[ln(1+μ)] (μ-law companding)
or α ≈4.77-20log[1+lnA] ( A -law companding)
? Notice,1) the output SNR follows the 6-dB law;
? 2)the output SNR is a function of the input level for
uniform quantizing,but is relatively insensitive to
the input level for μ-law and A –law companding;
? 3) loading factor,V/xrms,,when V/xrms ＝ 4(12dB)
overloading Quantizing noise will be negligible
? 4) all of this results give a 6-dB increase in the
signal-to-quantizing noise ratio for each bit added to
the PCM code word,
38
3.3 Pulse code modulation
Fig,3-10 Out SNR of 8-bit PCM systems with an without
companding
-50 -40 -30 -20 -10
10
20
30
40
50
60
-4.77dB
Uniform quantization
(on companding)
μ-law companding
μ=255 (dB)
Relative input level 20 log(xrms/V)
39
3.4 Digital Signal
In this section,we will answer the following
questions,
How do we mathematically represent the
waveform for a digital signal?
How do we estimate the bandwidth of the
waveform?
40
3.4 Digital Signal
Waveform for digital signals can be expressed as an
orthogonal series with a finite number of terms N,
0
1=
T<t<0 )(φω=)(ω ∑
N
k
kk tt
ωk represent the digital data,N is the number of
dimensions required to describe the waveform,
When ωks have binary value,ω(t) is binary signal;
When ωks have more than two values,ω(t) is multilevel
signal;
For example,the letter X is assigned to the code word
0001101,ω1=0,ω2=0,ω3=0,ω4=1,ω5=1,ω6=0,ω7=1,
41
3.4 Digital Signal
? Baud(symbol) rate,
D=N/T0
Where N is the number of dimensions used in T0s
? bit rate,
R=n/T0
Where n is the number of data bit sent in T0s
? Notice,
1)For binary signal,the bit rate and the baud are equal,
i.e R = n∕T0=D=N∕T0,because n=N,
2) For multilevel signal,the bit rate and the baud are not
equal,i.e R = n∕T0 = l﹡ D = l﹡ N∕T0
because n=l﹡ N=(log2 L)﹡ N
42
3.4 Digital Signal
01= T<t<0 )(φω=)(ω ∑
N
k kk
tt Corresponds to the
orthogonal vector space represented by,
φω= ∑
1=
N
j jj
w
w is an N-dimensional vector in Euclidean vector space,
and {φk} is an orthogonal set of N-directional vectors
)ω,...,ω,ω(= 21 Nw
? Detecting using matched filter at receiver,
∫ 00 * )(φ)(ω)/1(=ω T kkk dtttK
Is also called correlation processing
43
3.4 Digital Signal (Bandwidth estimation)
The bandwidth of the
waveform ω(t) is
B≥N∕2T0 = D∕2 (Hz)
why? (dimension
theorem N=2BT0)
If the φk(t) are of the
sin(x)∕x type,then
B=D∕2 (Hz),otherwise B
＞ D ∕ 2 (Hz),
44
3.4 Digital Signal
Binary input (l=2bit) Output level (V)
11
10
00
01
+3
+1
-1
-3
? Table 3-4 a 2-bit digital-to-analog converter
45
3.4 Digital Signal
46
3.5 Line Codes and Spectra
? Binary 1’s and 0’s,such as PCM signaling,may be represented in
various serial-bit signaling formats called line codes,
? Two main Categories,Return-to-zero(RZ) and Nonreturn-to-
zero(NRZ) Return-to-zero(RZ),the waveforms returns to a
zero-volt level for a portion of the bit interval;
Line codes classifying further,
Unipolar Signaling (on-off keying);
Polar Signaling,1’s=>+A 0’s=>-A
Bipolar (pseudoternary) Signaling,Binary 1’s are represented
by alternately positive or negative values,The 0’s is represented
by a zero level,
Manchester Signaling,each binary 1 is represented by a positive
half-bit period pulse followed by a negative half-bit period pulse,
A binary 0 is represented by negative half-bit period pulse
followed by a positive half-bit period pulse,It is also called split-
phase encoding,
47
3.5 Line Codes and Spectra
48
3.5 Line Codes and Spectra
? Unipolar NRZ=> unipolar
? Polar NRZ=> polar
? Bipolar RZ=> bipolar
? bipolar has tow different conflicting
definitions,
1)in space communication industry,polar
NRZ is sometimes called bipolar NRZ,or
simply Bipolar
2) in telephone industry Bipolar denotes
pseudoternary signaling (used in this book)
49
3.5 Line Codes and Spectra
? Each of the line code has advantages and disadvantages,
advantages disadvantages
Unipolar
NRZ
Circuit only need a
+5V power supply
Dc coupled channel is
requires
Polar NRZ Dc coupled channel is
not required
Circuit need two power
supply
Mancherter
NRZ
Dc coupled channel is
not required
Double bandwidth is
required compared with
unipolar NRZ and polar
NRZ
50
3.5 Line Codes and Spectra
? Some of the desirable properties of a line code,
? Self-synchronization
? Low probability of bit error
? A spectrum that is suitable for the channel
? Transmission bandwidth,this should be as small as
possible
? Error detection capability
? Transparency,the data protocol and line code are
designed so that every possible sequence of data is
faithfully and transparently received,
51
3.5 Line Codes and Spectra (Power
Spectra for Binary line codes)
? approachs include deterministic and stochastic
? A line code can be represented by,
∑
∞
∞-=
s )nT-(=)(
n
n tfats
? Where f(t) is the symbol pulse shape and Ts is the
duration of one symbol,for binary signaling Ts=Tb,for
multilevel signaling Ts=lTb,{an} is the set of random
data
? Example,for the unipolar NRZ line code,f(t)=Π(t/Tb),
and an=+AV when a binary 1 is sent and an=0V when a
binary 0 is sent,
52
3.5 Line Codes and Spectra (Power
Spectra for Binary line codes)
? the general expression for the PSD of a digital signal is
∑ ∞
∞-=
k f Tπj2
2
sR (k )e
|)(|=)(
ksT
fFf
sP? Where F(f) is the Fourier transform of the pulse shape
f(t) and R(k) is the autocorrelation of the data,which is
given by,
∑ N
1= +
)(=)(
i iiknn
PaakR
? Where an and an+1 are the levels of the data pulse at the
nth and (n+k)th symbol positions,respectively,and Pi is
the probability of having the ith an an+1 product,
? Note that the spectra of the digital signal depends on
two things,1) the pulse shape used and 2) statistical
properties of the data,
53
3.5 Line Codes and Spectra(examples)
? Example 1 Unipolar NRZ signaling
? Example 2 Polar NRZ Signaling
? Example 3 unipolar RZ
? Example 4 Bipolar RZ Signaling
? Example 5 Manchester NRZ Signaling
54
3.5 Line Codes and Spectra (comparison)
55
3.5 Line Codes and Spectra
(Power Spectra for multilevel polar NRZ signal)
? multilevel signaling provide reduced bandwidth
compared with binary signaling,
digital code output (an)i
000 +7
001 +5 010 +3
011 +1
100 -1
101 -3
110 -5
111 -7
Table,3-5 3 bit DAC coding
Example,Let ?=3 then L=8
56
3.5 Line Codes and Spectra
t bit D/A transfer R bit/s D sign/s=R/?,
R bit/s
Single polar NRZ Binary
input ω1(t)
Multilevel polar NRZL carry
system output ω2(t)
Tb
Ts
t 0
ω1(t)
ω2(t)
Ts
t
7 6
5 4
3
-2
2 1
-1
-3 -4
-5 -6
-7
output L=8
Figure 3-32 Binary-to-multilevel polar NRZ signal conversion
57
3.5 Line Codes and Spectra
an=
+7
+5
+3
+1
-1
-3
-5
-7
probability of every level is equal
for k=0,R(0)= Σ (anan)iPi=21,where Pi= 8 1 i=1 8
for k≠0,R(k)=0
f(t)=π( ),where Ts=3Tb Ts t
F(f)=Ts( ) sinπf Ts πf Ts
58
3.5 Line Codes and Spectra
? The PSD for ω2(t) is
)0+21(|)(|=)(
2
ω 2
sT
fFfP
? and,
2
ω )π3
π3s i n(63(
2 b
b
b fT
fTTf ）＝P
? In general,for the case of L=2l levels,the PSD of a
multilevel polar NRZ signal with rectangular pulse
shapes is,
2)
π
πs i n((
b
b
N R Zm u l t i l e v e l fTl
fTlKf ）＝P
? Where K is a constant and the null bandwidth is,
B=R/l
59
3.5 Line Codes and Spectra
? Spectral Efficiency,the spectral efficiency η is,
Hzsb i tB )/(Rη ＝
? Where R is bit rate,B is transmission bandwidth。
The maximum spectral efficiency ηmax is,
Hzsb i tNSB )/)(+1(l o g=Cη 2m a x ＝
? Shannon’s theory does not tell us how to achieve a
system with the maximum theoretical spectral efficiency;
however,practical systems that approach this spectral
efficiency usually incorporate error correction coding
and multilevel signaling,
60
3.5 Line Codes and Spectra
Spectral efficiency for multilevel polar NRZ signal
Hzsb i tllRRB )/(=)//(Rη ＝＝
Code type First zero-point Bandwidth (Hz) Frequency efficiency η=R/B [(b/s)/Hz]
Single polarity NRZ R 1
Double polarity NRZ R 1
Single polarity RZ 2R 1/2
Double polarity RZ R 1
Manchester NRZ 2R 1/2
Multilevel polarity NRZ R/ ? ?
Table 3-6,spectral efficiencies of line code
61
3.5 Line Codes and Spectra
(Differential Coding)
? Background, when serial data are passed through many
circuits along a communication channel,the waveform is
often unintentionally inverted,For example the result
can occure in a twisted-pair transmission line channel
when a polar line code is used,
? In order to ameliorate the problem,differential coding is
often employed,The encoded differential data are
generated by,
en=dn⊕ en-1
Where ⊕ is a modulo 2 adder or an exclusive –OR gat
operator, The received encoded data are decoded by,
dn= en⊕ en-1
62
3.5 Line Codes and Spectra
? Fig.3-17 differential coding system
63
3.5 Line Codes and Spectra
? Table 3-4 example of differential coding
Encoding/decoding ref
Input sequence dn
Encoded sequence en
Received sequence (correct polarity) en
Decoded sequence dn
Received sequence (inverted polarity) en
Decoded sequence dn
1
1
0
1 1 0 1 0 0 1
0 1 1 0 0 0 1
0 1 1 0 0 0 1
1 1 0 1 0 0 1
1 0 0 1 1 1 1 0
1 1 0 1 0 0 1
64
3.5 Line Codes and Spectra
(Eye Patterns)
? The effect of
channel
filtering and
channel noise
can be seen by
observing the
received line
code on an
analog
oscilloscope
65
3.5 Line Codes and Spectra
? The eye pattern provides the following information,
? The timing error allowed on the sampler at the
receiver is given by the width inside the eye,called the
eye opening, Of course,the preferred time for
sampling is at the point where the vertical operning of
the eye is the largest;
? The sensitivity to timing error is given by the slope of
the open eye (evaluated at,or near,the zero-crossing
point)
? the noise margin of the system is given by the height
of the eye opening
66
3.5 Line Codes and Spectra
(Regenerative Repeaters)
? Background,when a line code digital signal is
transmitted over a hardwire channel,it is attenuated,
filtered,and corrupted by noise,Consequently,for
long lines,the data cannot be recovered at the
receiving end unless repeaters are placed in cascade
along the line and at the receiver,
? these repeaters amplify and ―clean up‖ the digital
periodically,
? For analog signal,in-band distortion would
accumulate from linear repeater to linear repeater;
but for a digital signal,nonlinear processing can be
used to regenerate a ―noise-free‖ digital signal,This
type of nonlinear processing is called a regenerative
repeater,
67
3.5 Line Codes and Spectra
Signal
input Regenerative
repeater Regenerative repeater
Signal
output ….
,
Regenerative
repeater
Long distance transmitting system
68
3.5 Line Codes and Spectra
(Bit synchronization)
? Synchronization signals are clock-type signals that are
necessary within a receiver (or repeater) for detection
(or regeneration) of the data from the corrupted input
signal,They have a precise frequency and phase
relationship with respect to the received input signal,
? Digital communications usually need at least threee
type of synchronization signals,
? Bit sync,to distinguish one bit interval from another;
? Frame sync,to distinguish groups of data;
? Carrier sync,for bandpass signaling with coherent
detection,
69
3.5 Line Codes and Spectra
? Systems are designed so that the sync is derived
? 1) directly from the corrupted signal;
? 2) from a separate channel that is used only to
transmit the sync information,which is often not
economically feasible to send sync over a separate
channel,
? 3) using the symmetry property of the line code itself
? The complexity of the bit synchronizer circuit depends
on the sync properties of the line code,for example,for
a unipolar RZ code,the bit sync clock signal can be
obtained by passing the received unipolar RZ
waveform through a narrowband bandpass filter that
is tuned to f0=R=1/Tb; or PLL at f=R,
70
3.5 Line Codes and Spectra
Example,Square-law bit synchronizer
71
3.5 Line Codes and Spectra
Example,Early-late bit synchronizer for polar NRZ signaling
72
3.5 Line Codes and Spectra
? ω1(t) is the filtered polar NRZ line code,and
ω1(τ0+nTb) denote a sample value of the line code at
the maximun of the eye opening,Because the pulse
shape of the line of the line code is approximately
symmetrical about the optimum clocking time for
alternating data,
|)Δ+nT+τ(ω|=|)Δ-nT+τ(ω| b01b01
|)Δ+nT+τ(ω||)Δ-nT+τ(ω|=( t )ω b01b012 －
)(ω=)(ω 23 tt
73
3.6 Intersymbol Interference
? what is InterSymbol Interference (ISI)?
? The absolute bandwidth of rectangular multilevel
pulse is infinity,If these pulses are filtered improperly
as they pass through a communication system,They
will spread in time,and the pulse for each symbol may
be smeared into adjacent time slots and cause
intersymbol interference,
? What cause ISI?
? 1) filtering improperly,such as he channels property
is not plat,and is limited
? 2) the absolute bandwidth of rectangular multilevel is
infinity,
74
3.6 Intersymbol Interference
Fig 3-23 example of ISI on received pulses in a binary communication system
Sampling points
(transmitter clock)
1
1 1 1
0 0 0
0
InterSymbol
Interference
0
0 0
0 0
0 t→
t→ t→ t→
t→ t→
Ts Ts
Ts
Input waveform ωin(t) individual pulse response
sampling points sampling points
receive waveform,
ωout(t)（ pulse response sum)
receive clock receive clock
75
3.6 Intersymbol Interference
? How can we eliminate ISI? Consider a digital
signaling system in which the flat-topped multilevel
signal at the input is,
58)-(3 )nT-(=)(ω ∑ sin n n that
? Where h(t)=Π(t/Ts) and a n may take on any of the
allowed L multilevels,The symbol rate is D=1/Ts
pulse/s,then:-
5 9 )-(3 h (t )*])nT-(δ[=)nT-(δ*)(=)(ω ∑∑ ssin n nn n tatthat
ωin(t) ωc(t) ωout(t)
Fig,3-24 baseband pulse-transmission system
send filter
HT(f)
channel(filter)
HC(f)
receive filter
HR(f) recovered rounded pulse (to sampling and
decoding circuit)
Flat-top
pulse
76
3.6 Intersymbol Interference
? The output of the linear system would be just the
input impulse train convolved with the equivalent
impulse response of the overall system; that is,
ωout(t)=[Σanδ( t-nTs)]*he(t) (3-60)
? Where the equivalent impulse response is,
he(f)=h(f)*hT(t)*hC(f)*hR(f) (3-61)
? Note that he(t) is also the pulse shape that will appear
at the output of the receive filter when a single flat-top
pulse is fed into the transmitting filter,
? The equivalent system transfer function is,
He(f)=H(f) HT(t) HC(f) HR(f) (3-62)
77
3.6 Intersymbol Interference
? The equation is used so that flat-top pulses will be
present at the input to the transmitting filter,The
receiving filter is given by,
? Where
H(f)=F[Π(t/Ts)]=Ts(sinπTs?/(πTs?)) (3-63)
6 4 )-(3 )()()( )(=)( fHfHfH fHfH
cT
eR
? When He(f) is close to minimize the ISI,HR(f) is called
an equalizing filter,The rounded pulse train at the
output of the receiving filter is,
ωout(t)=Σanhe( t-nTs) (3-63)
78
3.6 Intersymbol Interference
(Nyquist’s First Method (zero ISI))
? If the equivalent transfer function of system is He(f),
its impulse response satisfies the condition,
he(KTs+τ)= (3-66) C,k=0 0,k≠0
? Where k is an integer,Ts is the symbol (sample) clocking
period,τis the offset in the receiver sampling clock times
compared with the clock times of the input symbols,C is
a nonzero constant,That is,for a signal flat-top pulse of
level ―a‖ present at the input to the transmitting filter at
t=0,the received pulse would be a he (t), It would have
a value of aC at t=τ,but would not cause interference
at other sampling times,Because he(KTs+τ) = 0,for k
≠0
79
3.6 Intersymbol Interference
(Nyquist’s First Method (zero ISI))
? Minimum bandwidth of system without ISI
? If choose a (sinx/x) function for he(t) and τ=0,then,
6 7 )-(3 π πs i n=)( tf tfth
s
se
? Where fs=1/Ts,this satisfies Nyquist’s first criterion for
zero ISI,if the transmit and receive filters are designed
so that the overall transfer function is,
6 8 )-(3 1=)( ∏ ）（
sse f
f
ffH
? There will be no ISI,
He( f )
f 0 -f
s /2 f s /2
80
3.6 Intersymbol Interference
(Nyquist’s First Method (zero ISI))
? However,the (sinx)/x type of overall pulse shape has
two practical difficulties,
? The overall amplitude transfer characteristic He(f) has
to be flat over –B<f<B and zero elsewhere,The steep
skirt in the filter transfer function at f = ± B make it be
physically unrealizable,
? The synchronization of the clock in the decoding
sampling circuit has to be almost perfect,The sinx/x
pulse decays only as 1/x(too slowly),thus inaccurate
synchronization clock will cause ISI
Absolute bandwidth, B=fs/2
Symbol (baud) rate, D= 1/Ts=2B pulses/s
Spectral efficiency, η= R/B= ? D/B =2 ? (bit/s)/?
81
3.6 Intersymbol Interference
? Because of the difficulties for sinx/x pulse,we are
force to consider other pulse shapes that satisfy the
conditions,
? 1)Having slightly wider bandwidth than sinx/x pulse;
? 2)Going through zero at adjacent sampling points and
having an envelope that decays much faster than 1/x
so that clock jitter in the sampling times does not
cause appreciable ISI,One solution for equivalent
transfer function is the raised cosine-rolloff Nyquist
filter,
82
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Definition,the raised consine-rolloff Nyquist filter
has the transfer function,
1,| f |< f 1
He( f )= {1+cos[ ] },f 1<| f |<B (3-69)
0,| f |>B
2
1
2 fΔ
π(| f |- f 1)
? Where B is the absolute bandwidth and the parameters
f Δ=B- f 0 (3-70)
f 1 = f 0- f Δ (3-71)
f
fΔ fΔ
0.5
1.0
|He(f)|
-B -f0 -f1 B f0 f1
f0(1-r） f0(1+r)
83
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? f0 is the 6-dB bandwidth of the filter,The rolloff
factor is defined to be,
r= f Δ/f0 (3-72)
? The corresponding impulse response is,
2
Δ
Δ
0
0
0
1-
)(4-1
π2c o s?
π2
π2sin2=)]([=)(
tf
tf
tf
tfffHth
ee F
1.0
0.5
-2f0 -f0 0 -0.5f0 f0 1.5f0 2f0 f
|He(f)|
r=0
r=0.5
r=1
(a) extent and frequency response
1/2f0
f0
2f0
he(t)
t
r=1.0
r=0.5
r=0
Ts
1
2f0
-1 2f
0
-3 f
0
3 f
0
-2
f
0
2 f
0
(b) Impulse response
84
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Summary,(comparing with sinx/x pulse)
? 1) The filtering requirements are relaxed;
? 2) The clock timing requirement are relaxed,(since
the envelope of the impulse response decays faster
than 1/t,the order is 1/| t |3 for large values of t)
Absolute bandwidth, B= f0(1+r)
Symbol (baud) rate, D= 2B/(1+r) pulses/s because
he(t)=0 when t=n/(2f0) (n≠ 0) and on ISI when fs=1/Ts =2f0,
Note,the raised cosine filter is also called a Nyquist filter,
It is one of a more general class of filters that satify
Nyquist’s first criterion,
85
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Theorem,A filter is said to be a Nyquist filter if the
effective transfer funtion is,
He(f)= (3-75) Π( )+Y(f),| f |<2 f 0
0,other
2 f 0
f
Where Y(f) is a real function that is even symmetric
about f=0,that is
Y(-f)=Y(f),| f |<2 f 0 (3-76a)
And Y(f) is odd symmetric about f= f 0 ; that is,
Y(-f+f 0)= -Y(f+ f 0),| f |< f 0 (3-76b)
Then there will be no intersymbol interference at the
system output if the symbol rate is,
D= fs =2f 0 (3-77)
Proof,p186/notebook p
86
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
0.5
-0.5
0 -f0 -2f0 2f0 f0 f f ’
f
Y(f)
1.0
-f 0 f 0
f
Π( ) f
0
f
0
He(f)=
Π( )+Y(f),| f |<2 f 0
0,other
2 f 0
f
f
2f 0 -2f 0 f 0 -f 0 0
1.0
Figure.(3-27) Nyquist filter trait
? Note,the filter is noncausal,We could use a filter with a linear
phase characteristic He(f)ejωTd,that is,it is delayed by Td sec,
87
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Summary:1) Because Y(f) can be any real function that satisfies
Eqs.(3-76),Thus,an infinite number of filter characteristics can
be used to produce zero ISI,
? 2) at the digital receiver,in addition to minimizing the ISI,we
would like to minimize the effect of channel noise by proper
filtering,The filter that minimizes the effect of channel noise is
the match filter,if a match filter is used for HR(f) at the receiver,
the overall filter characteristic,He(f) will usually not satify the
Nyquist characteristic for minimum ISI,The channel noise at
output of receiver is minimum if transmitter and receiver filters
are designed as,
|HT|(f)= (3-78a) α|H (f)| √ |H
e(f)|
√ |He(f)| [Pn(f)] 1 4
and |HT|(f)= (3-78b) √ α |He(f)| √ |H
c(f)| [Pn(f)] 1 4
Pn(f) is the PSD for the noise at the receiver input,
α is an arbitrary positive constant
88
3.7 Differential Pulse code Modulation
? Adjacent samples of sampled audio or video signal are
close to the same value,That is,there is a lot of
redundancy (or correlation) in the signal samples,
Consequently,the bandwidth and the dynamic range
of a PCM system are wasted when redundancy
sample values are transmitted,
? Differential pulse code modulation (DPCM) is one
way to minimize redundant transmission and reduce
the bandwidth,in which the PCM signals
corresponding to the difference in adjacent sample
values is transmitted,
89
3.7 Differential Pulse code Modulation
? The linear prediction filter,by which the present
value can be estimated from the past values,may be
used in a differential configuration to produce DPCM,y ( n T s ) D e l a y T
s
D e l a y
T s
D e l a y
T s
…
a 1 a 2 a l a K
z ( n T s )
? The optimum tap
gains of transversal
filter are a function
of the correlation
properties of the
input signal,The
output samples are,
)79-3( )-(=)( ∑
1=
alTnTyanTz
K
l ssls
90
3.7 Differential Pulse code Modulation
? Like PCM,DPCM follows the 6-dB rule,
(S/N)out=6.02n+α
Where -3<α<15 for DPCM speech(p189)
91
3.8 Delta Modulation
? Delta modulation (DM) is a special case of DPCM in
which there are two quantizing levels,so that the
output is +Vc or - Vc,
? Compared to DPAM,because there are only two
quantizing levels (m=2),the DM system owns some
special properties,
? Only one bit is transmitted per sample;
? ADC (analog-to-digital converter) and DAC (digital-
to-analog converter) are not needed;
? Encoder is not needed (the comparator has the
complete encoder function); encoder function,convert the
multilevel DPAM signal to binary code words,
? the DM signal is a polar signal,
92
3.8 Delta Modulation
Fig 3-31 A typical DM system
(t)w
Low-pass
filter Sampler Comparator
Clock
fs Integrator or accumulator
Channel
Integrator
Analog
input signal
DM transmitter
w(t)
Flat topped
PAM
y(t) DM
z(t)
Analog output
signal
DM receiver
+ _
93
3.8 Delta Modulation
? Fig 3-32 DM system waveforms
δ
V
t
initial state )(tw
hunting noise
Slope overload
noise
)(tw
z(t)
(a),Analog input and accumulator output
Vc
-Vc
t
(b),Delta modulation waveform
94
3.8 Delta Modulation
(Granular Noise and Slope Overload Noise)
? Note that the accumulator output signal does not
always track the analog input signal,
? DM quantizing noise error,
the difference between analog input signal and
accumulator output signal
? classification,
slope overload noise
granular noise
condition,sampling period,Ts=1/f s
step size,δ=Δv
accumulator initial state value,Δv
input DM signal,± Vc or ± 1
95
3.8 Delta Modulation
? Slope overload noise
occurs when the step
size δ is too small for
the accumulator to
follow quich changes
in the input waveform,
? Granular noise occurs
for any step size,but is
smaller for a small
step size,
? Thus we would like to
have δ as small as
possible to minimize
the granular noise,
Optimum value
of δ
Sig
na
l-to
-no
ise
ra
tio
(d
B)
Slope overload
dominates
Granular
noise
dominates
Step size δ
? Fig,3-33.signal-to-noise out of
a DM system as a function of
step size
96
3.8 Delta Modulation
Example 3-5,Design of a DM system,
problem,find the step size δrequired to prevent slope overload
noise for the case when the input signal is a sine wave,
solution,the maximum slope that can be generated by the
accumulator output is,
δ/Ts=δfs (3-82)
For the case of the sine-wave input,where ω(t)=Acosωat,the slope is,
principle,to avoid or prevent slope overload noise,the maximum slope
of accumulator output signal must larger than the maximum slope
of the analog input signal,
tAdt td aa ωc o sω=)(ωAnd the maximum slope of the input signal is Aω
a,thus,to avoid slope
overload,we require that δfs> Aωa,Or
s
a
f
Afπ2>δ
97
3.8 Delta Modulation
However,we do not want to make δtoo much larger than
this value,or the granular noise will become too large,
As stated the above formula,if we want to keep the
granular noise under a certain value,and to prevent
slope overload noise,we can do,
(1),Decrease the amplitude A;
(2),Increase the sampling frequency fs;
(3) making the step size vary as a function of time as
the input waveform changes,
Classification,
?Adaptive Delta Modulation (ADM)
-- Discrete accumulator (change)
?Continuously Variable Slope Delta modulation (CVSD)
-- Continuous integrator (change)
98
3.8 Delta Modulation
Data Sequence
Number of
Successive
Binary 1’s or 0’s
Step-size
Algorithm
f(d)
x x 0 1 1 δ
x 0 1 1 2 δ
0 1 1 1 3 2δ
1 1 1 1 4 4δ
Table 3-7 A typical step-size Algorithm
x– don’t care
99
3.8 Delta Modulation
Principle,
(1)When ADM signal consists of data with successive
binary 1’s and 0’s or when two successive binary 0’s
occur,the step size is set to a value δ,
(2) If three successive binary 1’s or 0’s occur,the step
size is increased to 2δ
(3) If four successive binary 1’s or 0’s occur,the step
size is increased to 4δ
100
3.8 Delta Modulation
? Which is better,PCM or DM?
? Depending on the criterion used for comparison and
the type of message,
? Simple,low-cost --------- DM
? High output SNR,compatibility--------PCM
Chapter 3
Baseband Pusle and
Digital Signaling
2
Chapter Objectives
? Analog-to-digital signaling(PCM and delta
modulation)
? Binary and multilevel digitals signals
? Spectra and bandwidths of digital signals
? Prevention of intersymbol interference
? Time division multiplexing
? Packet transmission
3
3.1 Introduction(main goals)
? To study how analog waveforms can be
converted to digital waveforms,The most
popular technique is called PCM
? To learn how to compute the spectrum for
digital signals
? To examine how to filtering of pulse signals
affects out ability to recover the digital
information at the receiver(ISI)
? To study how we can multiplex data from
several digital bit streams into on high-speed
digital stream fro transmission over a digital
system.(TDM)
4
3.2 Pulse Amplitude Modulation
? PAM is an engineering term that is used to
describe the conversion of the analog signal to
a pulse-type signal in which the amplitude of
the pulse denotes the analog information
? The sampling theorem gives a way to
reproduce an analog waveform by using
sample values of that waveform and sin(x)/x
orthogonal functions
? PAM signaling is to provide another waveform
that looks like pulses,yet contains the
information that was present in tha analog
waveform,
5
3.2 Pulse Amplitude Modulation
? The pulse rate,fs,for PAM is the same as that
required by the sampling theorem,namely,
fs>=2B,where B is the highest frequency in the
analog waveform and 2B is called the Nyquist
rate,
? Classification,
Nature sampling(gating)
Instantaneous sampling(Flat-top type)
6
3.2 Pulse Amplitude Modulation(Nature
sampling or gating)
? Definition,if ω(t) is an analog waveform
bandlimited to B hertz,the PAM signal that
uses natural sampling is
ωs(t)=ω(t)s(t)
Where s(t) is a rectangular wave switching
waveform and fs=1/Ts>=2B
? ?
∑ ∏ )
1
(∑ ∏ )(
∑ ∏ /]-[s (t )
∞
∞-
∞
∞-
∞
∞-
??
?
????
?
kk
s
k
s
d
k
tT
k
t
kTt
???
?
7
3.2 Pulse Amplitude Modulation(PAM signal
with natural sampling)
(a) Baseband Analog Waveform
S(t)
t
(b) Switching waveform with Duty Cycle d=τ /T =1/3
t
W(t)
W (t)
s
Figure 3-1 PAM Signal with natural sampling
≈
T
τ
8
3.2 Pulse Amplitude Modulation(Generation of
PAM with natural sampling)
clock
S(f)
Analog bilateral switch
Ws(t)=w(t)*s(t)
Figure 3-2 Generation of PAM with natural
sampling (gating)
9
3.2 Pulse Amplitude Modulation(spectrum
of a naturally sampled PAM)
? Theorem,the spectrum for a naturally
sampled PAM signal is,
Where fs=1/Ts,ωs=2πfs,the duty cycle of s(t) is d=τ/Ts,
and W(f)=F[ω(t)] is the spectrum of the original
unsampled waveform,
∑ ∞
∞-=
)n-(ndπ ndπsin =)](ω[=)(
n sss
ffWdtfW F
( )
∑∑ ∏∑ ∏
∑ ∏
∞
∞-=
ω
∞
∞-=
∞
∞-=
∞
∞-=
)
1
-
τ
(=)
τ
-
τ
(=
τ/]-[=s (t )
k
tjn
n
kk
s
k
s
sec
d
k
tT
k
t
kTt
＝
10
3.2 Pulse Amplitude Modulation(spectrum
of a naturally sampled PAM)
? The spectrum of the PAM with
natural sampling is a function of the
spectrum of the analog input
waveform
? The spectrum of the input analog
waveform is repeated at harmonics of
the sampling frequency
? the PAM spectrum is zero for ± 3fs,
± 6fs,and so on,Because of d=1/3,and
the spectrum in these harmonic bands
is nulled out by the (sin(x)/x) function
? The bandwidth of the PAM signal is
much larger than the bandwidth of
the original ananlog signal,The null
bandwidth for the envelope of the
PAM signal is 3fs=12B,that is,the null
bandwidth of this PAM signal is 12
times the bandwidth of the analog
signal,
? Fig,3-3 Spectrum of a PAM
waveform with Natural
sampling(p132)
11
3.2 Pulse Amplitude Modulation(receiver)
? Fig.3-4 Demodulation
of a PAM signal
? Low-pass filter
B<fcutoff<fs-B
fs>=2B
If the analog
signal is under
sampled(fs<2B),
spectral overlapping
(aliasing) is raised,
? Product detection
noise due to power supply hum or noise due to
mechanical circuit vibration might fall in the band
corresponding to low frequency band and other
bands might be relatively noise free
Low Pass
Filter H(f)
Oscillator
Wo=nws
fco -fco
f
H(f)
B〈 fco〈 fs-B
Analog Multiplier
(four-quadrants multiplier) W(s) PAM (natural
sampling Cw(f)
12
3.2 Pulse Amplitude Modulation( instantaneous
sampled or Flat-top PAM)
? Definition,if ω(t) is an analog waveform
bandlimited to B hertz,the PAM signal
that uses natural sampling is
Where h(t) denotes the sampling-pulse
shape and,for flat-top sampling,the pulse
shape is,
Where τ<=Ts=1/fs and fs>=2B
∑ ∞
∞-= sss
)kT-) h ( t( k Tω=( t )ω
k
? ? /2|t| 0 a n d / 2,|t| 1∏ /h ( t ) ??? ??? f o rt ＝
13
3.2 Pulse Amplitude Modulation
? Fig 3-5 PAM signal with flat-top
sampling ∑
∞
∞-= sss
)kT-) h ( t( k Tω=( t )ω
k
? PAM with flat-top sampling
is called instantaneous
samples,since ω(t) is
sampled at t=kTs and
sample value ω(kTs)
determine the amplitude of
the flat-top rectangular
pulse
? The flat-top signal could be
generated by using a
sample-and-hold type of
electronic circuit
? Note that if h(t)=sinx/x with
overlapping pulses,then
becomes identical to the
sampling theorem
t
Ws(t)
0
(a) Baseband Analog waveform
t
(b) Impulse Train Sampling waveform
τ
→ ← Ts
Ws(t)
t
(c) Resulting PAM Signal
( flat-top sampling,d=τ/Ts=1/3 )
14
3.2 Pulse Amplitude Modulation—the spectrum
of flat-top PAM signal
? Theorem,the spectrum for a flat-top PAM
signal is,
where
Proof,
∑ ∞
∞-=
)k-()()/1(=)(
n sss
ffWfHTfW
)s i n())(()( f fthfH ?? ????F＝
∑ ∞
∞-=
sss )kT-) h ( t( k Tω=( t )ω
k
15
3.2 Pulse Amplitude Modulation—the spectrum
of flat-top PAM signal
16
3.2 Pulse Amplitude Modulation—the spectrum
of flat-top PAM signal
? Low-pass filter,
There is some high-frequency loss in the recovered
analog waveform due to the filtering effect,this loss
can be reduced by decreasing τ or by using some
additional gain the high frequency in the low-pass
filter transfer function,This filter is called
equalization filter,
The pulse width τ is also called the aperture since τ/Ts
determines the gain of the recovered analog signal,
which is small if τ is small relative to Ts
17
3.2 Pulse Amplitude Modulation( key results)
? The transmission of either naturally or instantaneously
sampled PAM over a channel requires a very wide
frequency response because of the narrow pulse width,
which imposes stringent requirements on the magnitude
and phase response of the channel,
? The bandwidth required is much larger than that of the
original analog signal,and the noise performance of the
PAM system can never be better than that achieved by
tansmitting the analog signal directly;
? It provide a means for converting an analog signal to a
PCM signal; and provide a means for breaking a signal
into time slots that multiple PAM signals carrying
information from different sources can be interleaved to
transmit all of the information over a single channel,
18
3.3 Pulse code modulation
? Definition,Pulse code modulation(PCM) is essentially
analog-to-digital conversion of a special type where
the information contained in the instantaneous
samples of an analog signal is represented by digital
words in a serial bit stream,
? Quantizing,instead of using the exact sampled value
of the analog waveform ω(kTs),the sample is
replaced by the closest allowed value where there are
M=2n allowed value,each corresponding to one of the
code words,
? Other popular types of AD conversion,including delta
modulation (DM) and differential pulse code
modulation (DPCM) will be discussed,
19
3.3 Pulse code modulation—the advantages of
PCM
? Relatively inexpensive digital circuitry may be used
extensively in the system
? PCM signals derived from all types of analog source
(audio,video) may be merged with data signals and
transmitted over a common high-speed digital
communication system,this merging is called time-
division multiplexing
? In long-distance digital telephone systems requiring
repeaters,a clean PCM waveform can be regenerated
at the output of each repeater,where the input
consists of a noisy PCM waveform,
? The noise performance of a digital system can be
superior to that of an analog system,
20
3.3 Pulse code modulation
(Sampling,Quantizing,and Encoding)
? PCM signal is generated by carrying out three basic operations,
sampling,Quantizing and Encoding;
? Sampling operation generates a flat-top PAM signal
? Fig.3-7 PCM transmission system
Signal in
Analog Low-pass
filter
Bandwidth=B
Band limited
analog Instantaneous
sampler
and hold
Flat-top
PAM Quantized No,
of levels=M
Quantized
PAM Encoder PCM
Signal
PCM transmitter (analog-to-digital conversion)
Telephone
line Regenerative
repeater
Telephone
line Regenerative
repeater
Regenerative
repeater
Telephone line
Channel (transmission path)
Regenerative
circuit
PCM Decoder Quantized PAM Low-pass (reconstruction)
filter
PCM receiver (digital-to-analog conversion)
Analog signal
out
21
3.3 Pulse code modulation
? Quantizing
If all of the steps are of equal size,quantizer is said to be uniform;
Quantizing effect introduced error into the recovered output analog
signal;
Quantizing error → Quantizing noise (a round -off error)
? Encoding
The PCM signal is obtained from the quantized PAM signal by
encoding each quantized sample value into a digital word
(Fig,3-8)
2 4 6 8 -2 -4 -6 -8
0 2
4
6 8
-4
-6
-8
M=8
Output
voltage
↑
V=8
Input voltage=x
↑
-V=8
(a) Quantize Output –Input Characteristics
22
3.3 Pulse code modulation
2
4
6
8
-2
-4
-6
-8
Sampling times T
s
t
Quantized PAM signals
Analog signal,a(t)
(b) Analog
Signal,Flat
– top PAM
Signal,and
Quantized
PAM Signal
1
2
-1
-2
Difference between analog signal
and quantized PAM signal
t
(c) Error Signal
1 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 1
→ ← PCM word
(d) PAM Signal
Fig 3-8 illustration of waveforms in a PCM system
23
3.3 Pulse code modulation
? Encoding
The PCM signal
is obtained from
the quantized
PAM signal by
encoding each
quantized sample
value into a
digital word
? Table3.1 Three-bit Gray code for M=8 levels
Quantized
sample Voltage
Gray Code Word
(PCM output)
+7
+5
+3
+1
1 10
1 11
1 01
1 00
-7
-5
-3
-1
0 00
0 01
0 11
0 10 each quantized sample ==> a digital word (n
binary digits) M= 2n
unique code words
24
3.3 Pulse code modulation
? Notes,
? Only one bit change for each step change
in the quantized level,
? Single errors in the received PCM code
word will cause minimal errors in the
recovered analog level,provided that the
sign bit is not in error
25
3.3 Pulse code modulation(bandwidth of PCM
signal)
Bit rate R=nfs
fs≥2B B—Bandwidth of analog signal
the bandwidth of the PCM waveform
BPCM≥1/2R= 1/2nfs
BPCM min = 1/2R = 1/2nfs
(A is （ sinx/x） type of pulse shape )
26
3.3 Pulse code modulation（ Effect of noise ）
? the main effects produce noise or distortion
? Quantizing noise that is caused by the M-step
quantized at PCM transmitter
? Bit errors in the recovered PCM signal,(channel
noise,improper channel filtering )
? Not strictly bandlimilted input analog signal,
(Aliasing noise )
27
3.3 Pulse code modulation
? Ratio of the recovered analog peak signal power to
the total average noise power is given by,
? Ratio of the average signal power to the average
noise power is
1 6 a )-(3 )1-(41 3/ 2
2
-
e
o u tpk PM
MNS
??）（
e
o u t PM
MNS
)1-(41/ 2
2
??）（
where M is the number of quantized levels,and Pe is
the probability of bit error
If Pe=0 (No ISI),the peak SNR and average SNR
resulting form only quantizing errors is respectively
2- 3=/ MNS o u tpk）（ 2=/ MNS o u t）（
28
3.3 Pulse code modulation--- Four types of the
quantizing noise
Overload noise ( flat-tops near the peak value )
the peak level exceed the design peak of V volts
Random noise
random quantization errors ---- white hissing sound
Granular noise unequal quantization errors form sample
to sample solution ; a nonuniform quantization,e.g,u-law
quantized
hunting noise when the input analog waveform is
nearly constant,the output samples oscillate between
two adjacent quantization levels,Solution,① filter out
the tone; ② designing the quantize so that there is no
vertical step at the constant valve step,
29
3.3 Pulse code modulation(6dB -rule)
? 6dB-rule
α+02.6=/ nNS dB）（
? where,n- the number of bits in the PCM word
α=4.77 ( the peak SNR ) α= 0 (the average SNR )
An additional 6-dB improvements in SNR is
obtained for each bit added to the PCM word,
Assumptions,① No bit errors
② the input signal level is large
enough to range over a significant number of
quantizing levels
30
3.3 Pulse code modulation
? Example 3-1 Design of a PCM signal for audio
voice-frequency (VF) Telephone Systems,
band, 300H ~ 3400 H
? Sampling frequency,(fSmin) = 3.4kHz * 2 = 6.8 kHz
? bit rate R= fs (samples/s)*n(bits/sample)
= 8 k sample/s * 8 bits/sample = 64 kbps
Is called Ds-0(digital signal,type zero)
? the minimum absolute bandwidth
(B)min = 1/2R = 32kHz ( for a sinx/x pulse shape)
BPCM =R = 64kH ( for rectangular pulse shape)
? peak SNR is
dBMNS outpk 9.52=2*33=/ 282- ）（＝）（
31
3.3 Pulse code modulation
Note,
? The inclusion of a parity bit does not affect
the quantizing noise,
? The performance of a PCM system (i.e,no
bit errors and no ISI ) is determined by M,the
number of quantizing steps used,
32
3.3 Pulse code modulation （ Nonumiform
Quantizing:μ-Law and A-Low compounding）
? No uniform Quantizing – a variable step size is
used
Analog
signal
A Compression
(nonlinear)
Amplifier
PCM
(uniform quantized)
Nonuniform
Quantizing
signal
Compression
quantizer
characteristic
Uniform quantizer
characteristic
0
0
0.2 0.4 0.6 0.8 1.0
V=1.0
0.2
0.4
0.6
0.8
1.0
(a) M=8
Quantizer
Characteristi
c
33
3.3 Pulse code modulation
0 0.2 0.4 0.6 0.8 1.0
0.2
0.4
0.6
0.8
1.0
μ=0 μ=1 μ=5
μ=100
μ=225
(b) u – law Characteristic (c) A-law Characteristic
0 0.2 0.4 0.6 0.8 1.0
0 0.2 0.4 0.6 0.8 1.0
0
0.2
0.4
0.6
0.8
1.0
A=1 A=2
A=5
A=87.6
A=100
Fig.3-9 Compression characteristics (first quadrant shown)
34
3.3 Pulse code modulation
35
3.3 Pulse code modulation
? U-Low compression characteristic,wher |w1(t)|≤ 1,
? u is a positive parameter (constant )
if u=0,it is linear amplification (Uniform
Quantization)
)μ+l n ( 1
|)( t )ω|μ+l n ( 1=|( t )ω| 1
2
? A-Low compression characteristic,wher |w1(t)|≤ 1,
? A is a positive parameter (constant )
if A=1,it is linear amplification (Uniform
Quantization)
)l n (+1
|)( t )ω|l n (+1=|( t )ω| 1
2 A
A
36
3.3 Pulse code modulation
? When compression is used at the transmitter,
expansion(i.e,decompression) must be used at
the receiver output to restore signal levels to
their correct relative values,The expandor
characteristic is the inverse of the
compression characteristic,and the
combination of a compressor and an expandor
is called a compandor
37
3.3 Pulse code modulation
? Output SNR α+02.6=/ nNS
dB）（
where α= 4.77-20log(V/xrms) (uniform quantizing)
or α ≈4.77-20log[ln(1+μ)] (μ-law companding)
or α ≈4.77-20log[1+lnA] ( A -law companding)
? Notice,1) the output SNR follows the 6-dB law;
? 2)the output SNR is a function of the input level for
uniform quantizing,but is relatively insensitive to
the input level for μ-law and A –law companding;
? 3) loading factor,V/xrms,,when V/xrms ＝ 4(12dB)
overloading Quantizing noise will be negligible
? 4) all of this results give a 6-dB increase in the
signal-to-quantizing noise ratio for each bit added to
the PCM code word,
38
3.3 Pulse code modulation
Fig,3-10 Out SNR of 8-bit PCM systems with an without
companding
-50 -40 -30 -20 -10
10
20
30
40
50
60
-4.77dB
Uniform quantization
(on companding)
μ-law companding
μ=255 (dB)
Relative input level 20 log(xrms/V)
39
3.4 Digital Signal
In this section,we will answer the following
questions,
How do we mathematically represent the
waveform for a digital signal?
How do we estimate the bandwidth of the
waveform?
40
3.4 Digital Signal
Waveform for digital signals can be expressed as an
orthogonal series with a finite number of terms N,
0
1=
T<t<0 )(φω=)(ω ∑
N
k
kk tt
ωk represent the digital data,N is the number of
dimensions required to describe the waveform,
When ωks have binary value,ω(t) is binary signal;
When ωks have more than two values,ω(t) is multilevel
signal;
For example,the letter X is assigned to the code word
0001101,ω1=0,ω2=0,ω3=0,ω4=1,ω5=1,ω6=0,ω7=1,
41
3.4 Digital Signal
? Baud(symbol) rate,
D=N/T0
Where N is the number of dimensions used in T0s
? bit rate,
R=n/T0
Where n is the number of data bit sent in T0s
? Notice,
1)For binary signal,the bit rate and the baud are equal,
i.e R = n∕T0=D=N∕T0,because n=N,
2) For multilevel signal,the bit rate and the baud are not
equal,i.e R = n∕T0 = l﹡ D = l﹡ N∕T0
because n=l﹡ N=(log2 L)﹡ N
42
3.4 Digital Signal
01= T<t<0 )(φω=)(ω ∑
N
k kk
tt Corresponds to the
orthogonal vector space represented by,
φω= ∑
1=
N
j jj
w
w is an N-dimensional vector in Euclidean vector space,
and {φk} is an orthogonal set of N-directional vectors
)ω,...,ω,ω(= 21 Nw
? Detecting using matched filter at receiver,
∫ 00 * )(φ)(ω)/1(=ω T kkk dtttK
Is also called correlation processing
43
3.4 Digital Signal (Bandwidth estimation)
The bandwidth of the
waveform ω(t) is
B≥N∕2T0 = D∕2 (Hz)
why? (dimension
theorem N=2BT0)
If the φk(t) are of the
sin(x)∕x type,then
B=D∕2 (Hz),otherwise B
＞ D ∕ 2 (Hz),
44
3.4 Digital Signal
Binary input (l=2bit) Output level (V)
11
10
00
01
+3
+1
-1
-3
? Table 3-4 a 2-bit digital-to-analog converter
45
3.4 Digital Signal
46
3.5 Line Codes and Spectra
? Binary 1’s and 0’s,such as PCM signaling,may be represented in
various serial-bit signaling formats called line codes,
? Two main Categories,Return-to-zero(RZ) and Nonreturn-to-
zero(NRZ) Return-to-zero(RZ),the waveforms returns to a
zero-volt level for a portion of the bit interval;
Line codes classifying further,
Unipolar Signaling (on-off keying);
Polar Signaling,1’s=>+A 0’s=>-A
Bipolar (pseudoternary) Signaling,Binary 1’s are represented
by alternately positive or negative values,The 0’s is represented
by a zero level,
Manchester Signaling,each binary 1 is represented by a positive
half-bit period pulse followed by a negative half-bit period pulse,
A binary 0 is represented by negative half-bit period pulse
followed by a positive half-bit period pulse,It is also called split-
phase encoding,
47
3.5 Line Codes and Spectra
48
3.5 Line Codes and Spectra
? Unipolar NRZ=> unipolar
? Polar NRZ=> polar
? Bipolar RZ=> bipolar
? bipolar has tow different conflicting
definitions,
1)in space communication industry,polar
NRZ is sometimes called bipolar NRZ,or
simply Bipolar
2) in telephone industry Bipolar denotes
pseudoternary signaling (used in this book)
49
3.5 Line Codes and Spectra
? Each of the line code has advantages and disadvantages,
advantages disadvantages
Unipolar
NRZ
Circuit only need a
+5V power supply
Dc coupled channel is
requires
Polar NRZ Dc coupled channel is
not required
Circuit need two power
supply
Mancherter
NRZ
Dc coupled channel is
not required
Double bandwidth is
required compared with
unipolar NRZ and polar
NRZ
50
3.5 Line Codes and Spectra
? Some of the desirable properties of a line code,
? Self-synchronization
? Low probability of bit error
? A spectrum that is suitable for the channel
? Transmission bandwidth,this should be as small as
possible
? Error detection capability
? Transparency,the data protocol and line code are
designed so that every possible sequence of data is
faithfully and transparently received,
51
3.5 Line Codes and Spectra (Power
Spectra for Binary line codes)
? approachs include deterministic and stochastic
? A line code can be represented by,
∑
∞
∞-=
s )nT-(=)(
n
n tfats
? Where f(t) is the symbol pulse shape and Ts is the
duration of one symbol,for binary signaling Ts=Tb,for
multilevel signaling Ts=lTb,{an} is the set of random
data
? Example,for the unipolar NRZ line code,f(t)=Π(t/Tb),
and an=+AV when a binary 1 is sent and an=0V when a
binary 0 is sent,
52
3.5 Line Codes and Spectra (Power
Spectra for Binary line codes)
? the general expression for the PSD of a digital signal is
∑ ∞
∞-=
k f Tπj2
2
sR (k )e
|)(|=)(
ksT
fFf
sP? Where F(f) is the Fourier transform of the pulse shape
f(t) and R(k) is the autocorrelation of the data,which is
given by,
∑ N
1= +
)(=)(
i iiknn
PaakR
? Where an and an+1 are the levels of the data pulse at the
nth and (n+k)th symbol positions,respectively,and Pi is
the probability of having the ith an an+1 product,
? Note that the spectra of the digital signal depends on
two things,1) the pulse shape used and 2) statistical
properties of the data,
53
3.5 Line Codes and Spectra(examples)
? Example 1 Unipolar NRZ signaling
? Example 2 Polar NRZ Signaling
? Example 3 unipolar RZ
? Example 4 Bipolar RZ Signaling
? Example 5 Manchester NRZ Signaling
54
3.5 Line Codes and Spectra (comparison)
55
3.5 Line Codes and Spectra
(Power Spectra for multilevel polar NRZ signal)
? multilevel signaling provide reduced bandwidth
compared with binary signaling,
digital code output (an)i
000 +7
001 +5 010 +3
011 +1
100 -1
101 -3
110 -5
111 -7
Table,3-5 3 bit DAC coding
Example,Let ?=3 then L=8
56
3.5 Line Codes and Spectra
t bit D/A transfer R bit/s D sign/s=R/?,
R bit/s
Single polar NRZ Binary
input ω1(t)
Multilevel polar NRZL carry
system output ω2(t)
Tb
Ts
t 0
ω1(t)
ω2(t)
Ts
t
7 6
5 4
3
-2
2 1
-1
-3 -4
-5 -6
-7
output L=8
Figure 3-32 Binary-to-multilevel polar NRZ signal conversion
57
3.5 Line Codes and Spectra
an=
+7
+5
+3
+1
-1
-3
-5
-7
probability of every level is equal
for k=0,R(0)= Σ (anan)iPi=21,where Pi= 8 1 i=1 8
for k≠0,R(k)=0
f(t)=π( ),where Ts=3Tb Ts t
F(f)=Ts( ) sinπf Ts πf Ts
58
3.5 Line Codes and Spectra
? The PSD for ω2(t) is
)0+21(|)(|=)(
2
ω 2
sT
fFfP
? and,
2
ω )π3
π3s i n(63(
2 b
b
b fT
fTTf ）＝P
? In general,for the case of L=2l levels,the PSD of a
multilevel polar NRZ signal with rectangular pulse
shapes is,
2)
π
πs i n((
b
b
N R Zm u l t i l e v e l fTl
fTlKf ）＝P
? Where K is a constant and the null bandwidth is,
B=R/l
59
3.5 Line Codes and Spectra
? Spectral Efficiency,the spectral efficiency η is,
Hzsb i tB )/(Rη ＝
? Where R is bit rate,B is transmission bandwidth。
The maximum spectral efficiency ηmax is,
Hzsb i tNSB )/)(+1(l o g=Cη 2m a x ＝
? Shannon’s theory does not tell us how to achieve a
system with the maximum theoretical spectral efficiency;
however,practical systems that approach this spectral
efficiency usually incorporate error correction coding
and multilevel signaling,
60
3.5 Line Codes and Spectra
Spectral efficiency for multilevel polar NRZ signal
Hzsb i tllRRB )/(=)//(Rη ＝＝
Code type First zero-point Bandwidth (Hz) Frequency efficiency η=R/B [(b/s)/Hz]
Single polarity NRZ R 1
Double polarity NRZ R 1
Single polarity RZ 2R 1/2
Double polarity RZ R 1
Manchester NRZ 2R 1/2
Multilevel polarity NRZ R/ ? ?
Table 3-6,spectral efficiencies of line code
61
3.5 Line Codes and Spectra
(Differential Coding)
? Background, when serial data are passed through many
circuits along a communication channel,the waveform is
often unintentionally inverted,For example the result
can occure in a twisted-pair transmission line channel
when a polar line code is used,
? In order to ameliorate the problem,differential coding is
often employed,The encoded differential data are
generated by,
en=dn⊕ en-1
Where ⊕ is a modulo 2 adder or an exclusive –OR gat
operator, The received encoded data are decoded by,
dn= en⊕ en-1
62
3.5 Line Codes and Spectra
? Fig.3-17 differential coding system
63
3.5 Line Codes and Spectra
? Table 3-4 example of differential coding
Encoding/decoding ref
Input sequence dn
Encoded sequence en
Received sequence (correct polarity) en
Decoded sequence dn
Received sequence (inverted polarity) en
Decoded sequence dn
1
1
0
1 1 0 1 0 0 1
0 1 1 0 0 0 1
0 1 1 0 0 0 1
1 1 0 1 0 0 1
1 0 0 1 1 1 1 0
1 1 0 1 0 0 1
64
3.5 Line Codes and Spectra
(Eye Patterns)
? The effect of
channel
filtering and
channel noise
can be seen by
observing the
received line
code on an
analog
oscilloscope
65
3.5 Line Codes and Spectra
? The eye pattern provides the following information,
? The timing error allowed on the sampler at the
receiver is given by the width inside the eye,called the
eye opening, Of course,the preferred time for
sampling is at the point where the vertical operning of
the eye is the largest;
? The sensitivity to timing error is given by the slope of
the open eye (evaluated at,or near,the zero-crossing
point)
? the noise margin of the system is given by the height
of the eye opening
66
3.5 Line Codes and Spectra
(Regenerative Repeaters)
? Background,when a line code digital signal is
transmitted over a hardwire channel,it is attenuated,
filtered,and corrupted by noise,Consequently,for
long lines,the data cannot be recovered at the
receiving end unless repeaters are placed in cascade
along the line and at the receiver,
? these repeaters amplify and ―clean up‖ the digital
periodically,
? For analog signal,in-band distortion would
accumulate from linear repeater to linear repeater;
but for a digital signal,nonlinear processing can be
used to regenerate a ―noise-free‖ digital signal,This
type of nonlinear processing is called a regenerative
repeater,
67
3.5 Line Codes and Spectra
Signal
input Regenerative
repeater Regenerative repeater
Signal
output ….
,
Regenerative
repeater
Long distance transmitting system
68
3.5 Line Codes and Spectra
(Bit synchronization)
? Synchronization signals are clock-type signals that are
necessary within a receiver (or repeater) for detection
(or regeneration) of the data from the corrupted input
signal,They have a precise frequency and phase
relationship with respect to the received input signal,
? Digital communications usually need at least threee
type of synchronization signals,
? Bit sync,to distinguish one bit interval from another;
? Frame sync,to distinguish groups of data;
? Carrier sync,for bandpass signaling with coherent
detection,
69
3.5 Line Codes and Spectra
? Systems are designed so that the sync is derived
? 1) directly from the corrupted signal;
? 2) from a separate channel that is used only to
transmit the sync information,which is often not
economically feasible to send sync over a separate
channel,
? 3) using the symmetry property of the line code itself
? The complexity of the bit synchronizer circuit depends
on the sync properties of the line code,for example,for
a unipolar RZ code,the bit sync clock signal can be
obtained by passing the received unipolar RZ
waveform through a narrowband bandpass filter that
is tuned to f0=R=1/Tb; or PLL at f=R,
70
3.5 Line Codes and Spectra
Example,Square-law bit synchronizer
71
3.5 Line Codes and Spectra
Example,Early-late bit synchronizer for polar NRZ signaling
72
3.5 Line Codes and Spectra
? ω1(t) is the filtered polar NRZ line code,and
ω1(τ0+nTb) denote a sample value of the line code at
the maximun of the eye opening,Because the pulse
shape of the line of the line code is approximately
symmetrical about the optimum clocking time for
alternating data,
|)Δ+nT+τ(ω|=|)Δ-nT+τ(ω| b01b01
|)Δ+nT+τ(ω||)Δ-nT+τ(ω|=( t )ω b01b012 －
)(ω=)(ω 23 tt
73
3.6 Intersymbol Interference
? what is InterSymbol Interference (ISI)?
? The absolute bandwidth of rectangular multilevel
pulse is infinity,If these pulses are filtered improperly
as they pass through a communication system,They
will spread in time,and the pulse for each symbol may
be smeared into adjacent time slots and cause
intersymbol interference,
? What cause ISI?
? 1) filtering improperly,such as he channels property
is not plat,and is limited
? 2) the absolute bandwidth of rectangular multilevel is
infinity,
74
3.6 Intersymbol Interference
Fig 3-23 example of ISI on received pulses in a binary communication system
Sampling points
(transmitter clock)
1
1 1 1
0 0 0
0
InterSymbol
Interference
0
0 0
0 0
0 t→
t→ t→ t→
t→ t→
Ts Ts
Ts
Input waveform ωin(t) individual pulse response
sampling points sampling points
receive waveform,
ωout(t)（ pulse response sum)
receive clock receive clock
75
3.6 Intersymbol Interference
? How can we eliminate ISI? Consider a digital
signaling system in which the flat-topped multilevel
signal at the input is,
58)-(3 )nT-(=)(ω ∑ sin n n that
? Where h(t)=Π(t/Ts) and a n may take on any of the
allowed L multilevels,The symbol rate is D=1/Ts
pulse/s,then:-
5 9 )-(3 h (t )*])nT-(δ[=)nT-(δ*)(=)(ω ∑∑ ssin n nn n tatthat
ωin(t) ωc(t) ωout(t)
Fig,3-24 baseband pulse-transmission system
send filter
HT(f)
channel(filter)
HC(f)
receive filter
HR(f) recovered rounded pulse (to sampling and
decoding circuit)
Flat-top
pulse
76
3.6 Intersymbol Interference
? The output of the linear system would be just the
input impulse train convolved with the equivalent
impulse response of the overall system; that is,
ωout(t)=[Σanδ( t-nTs)]*he(t) (3-60)
? Where the equivalent impulse response is,
he(f)=h(f)*hT(t)*hC(f)*hR(f) (3-61)
? Note that he(t) is also the pulse shape that will appear
at the output of the receive filter when a single flat-top
pulse is fed into the transmitting filter,
? The equivalent system transfer function is,
He(f)=H(f) HT(t) HC(f) HR(f) (3-62)
77
3.6 Intersymbol Interference
? The equation is used so that flat-top pulses will be
present at the input to the transmitting filter,The
receiving filter is given by,
? Where
H(f)=F[Π(t/Ts)]=Ts(sinπTs?/(πTs?)) (3-63)
6 4 )-(3 )()()( )(=)( fHfHfH fHfH
cT
eR
? When He(f) is close to minimize the ISI,HR(f) is called
an equalizing filter,The rounded pulse train at the
output of the receiving filter is,
ωout(t)=Σanhe( t-nTs) (3-63)
78
3.6 Intersymbol Interference
(Nyquist’s First Method (zero ISI))
? If the equivalent transfer function of system is He(f),
its impulse response satisfies the condition,
he(KTs+τ)= (3-66) C,k=0 0,k≠0
? Where k is an integer,Ts is the symbol (sample) clocking
period,τis the offset in the receiver sampling clock times
compared with the clock times of the input symbols,C is
a nonzero constant,That is,for a signal flat-top pulse of
level ―a‖ present at the input to the transmitting filter at
t=0,the received pulse would be a he (t), It would have
a value of aC at t=τ,but would not cause interference
at other sampling times,Because he(KTs+τ) = 0,for k
≠0
79
3.6 Intersymbol Interference
(Nyquist’s First Method (zero ISI))
? Minimum bandwidth of system without ISI
? If choose a (sinx/x) function for he(t) and τ=0,then,
6 7 )-(3 π πs i n=)( tf tfth
s
se
? Where fs=1/Ts,this satisfies Nyquist’s first criterion for
zero ISI,if the transmit and receive filters are designed
so that the overall transfer function is,
6 8 )-(3 1=)( ∏ ）（
sse f
f
ffH
? There will be no ISI,
He( f )
f 0 -f
s /2 f s /2
80
3.6 Intersymbol Interference
(Nyquist’s First Method (zero ISI))
? However,the (sinx)/x type of overall pulse shape has
two practical difficulties,
? The overall amplitude transfer characteristic He(f) has
to be flat over –B<f<B and zero elsewhere,The steep
skirt in the filter transfer function at f = ± B make it be
physically unrealizable,
? The synchronization of the clock in the decoding
sampling circuit has to be almost perfect,The sinx/x
pulse decays only as 1/x(too slowly),thus inaccurate
synchronization clock will cause ISI
Absolute bandwidth, B=fs/2
Symbol (baud) rate, D= 1/Ts=2B pulses/s
Spectral efficiency, η= R/B= ? D/B =2 ? (bit/s)/?
81
3.6 Intersymbol Interference
? Because of the difficulties for sinx/x pulse,we are
force to consider other pulse shapes that satisfy the
conditions,
? 1)Having slightly wider bandwidth than sinx/x pulse;
? 2)Going through zero at adjacent sampling points and
having an envelope that decays much faster than 1/x
so that clock jitter in the sampling times does not
cause appreciable ISI,One solution for equivalent
transfer function is the raised cosine-rolloff Nyquist
filter,
82
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Definition,the raised consine-rolloff Nyquist filter
has the transfer function,
1,| f |< f 1
He( f )= {1+cos[ ] },f 1<| f |<B (3-69)
0,| f |>B
2
1
2 fΔ
π(| f |- f 1)
? Where B is the absolute bandwidth and the parameters
f Δ=B- f 0 (3-70)
f 1 = f 0- f Δ (3-71)
f
fΔ fΔ
0.5
1.0
|He(f)|
-B -f0 -f1 B f0 f1
f0(1-r） f0(1+r)
83
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? f0 is the 6-dB bandwidth of the filter,The rolloff
factor is defined to be,
r= f Δ/f0 (3-72)
? The corresponding impulse response is,
2
Δ
Δ
0
0
0
1-
)(4-1
π2c o s?
π2
π2sin2=)]([=)(
tf
tf
tf
tfffHth
ee F
1.0
0.5
-2f0 -f0 0 -0.5f0 f0 1.5f0 2f0 f
|He(f)|
r=0
r=0.5
r=1
(a) extent and frequency response
1/2f0
f0
2f0
he(t)
t
r=1.0
r=0.5
r=0
Ts
1
2f0
-1 2f
0
-3 f
0
3 f
0
-2
f
0
2 f
0
(b) Impulse response
84
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Summary,(comparing with sinx/x pulse)
? 1) The filtering requirements are relaxed;
? 2) The clock timing requirement are relaxed,(since
the envelope of the impulse response decays faster
than 1/t,the order is 1/| t |3 for large values of t)
Absolute bandwidth, B= f0(1+r)
Symbol (baud) rate, D= 2B/(1+r) pulses/s because
he(t)=0 when t=n/(2f0) (n≠ 0) and on ISI when fs=1/Ts =2f0,
Note,the raised cosine filter is also called a Nyquist filter,
It is one of a more general class of filters that satify
Nyquist’s first criterion,
85
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Theorem,A filter is said to be a Nyquist filter if the
effective transfer funtion is,
He(f)= (3-75) Π( )+Y(f),| f |<2 f 0
0,other
2 f 0
f
Where Y(f) is a real function that is even symmetric
about f=0,that is
Y(-f)=Y(f),| f |<2 f 0 (3-76a)
And Y(f) is odd symmetric about f= f 0 ; that is,
Y(-f+f 0)= -Y(f+ f 0),| f |< f 0 (3-76b)
Then there will be no intersymbol interference at the
system output if the symbol rate is,
D= fs =2f 0 (3-77)
Proof,p186/notebook p
86
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
0.5
-0.5
0 -f0 -2f0 2f0 f0 f f ’
f
Y(f)
1.0
-f 0 f 0
f
Π( ) f
0
f
0
He(f)=
Π( )+Y(f),| f |<2 f 0
0,other
2 f 0
f
f
2f 0 -2f 0 f 0 -f 0 0
1.0
Figure.(3-27) Nyquist filter trait
? Note,the filter is noncausal,We could use a filter with a linear
phase characteristic He(f)ejωTd,that is,it is delayed by Td sec,
87
3.6 Intersymbol Interference
(Raised Cosine-Rolloff Nyquist Filtering)
? Summary:1) Because Y(f) can be any real function that satisfies
Eqs.(3-76),Thus,an infinite number of filter characteristics can
be used to produce zero ISI,
? 2) at the digital receiver,in addition to minimizing the ISI,we
would like to minimize the effect of channel noise by proper
filtering,The filter that minimizes the effect of channel noise is
the match filter,if a match filter is used for HR(f) at the receiver,
the overall filter characteristic,He(f) will usually not satify the
Nyquist characteristic for minimum ISI,The channel noise at
output of receiver is minimum if transmitter and receiver filters
are designed as,
|HT|(f)= (3-78a) α|H (f)| √ |H
e(f)|
√ |He(f)| [Pn(f)] 1 4
and |HT|(f)= (3-78b) √ α |He(f)| √ |H
c(f)| [Pn(f)] 1 4
Pn(f) is the PSD for the noise at the receiver input,
α is an arbitrary positive constant
88
3.7 Differential Pulse code Modulation
? Adjacent samples of sampled audio or video signal are
close to the same value,That is,there is a lot of
redundancy (or correlation) in the signal samples,
Consequently,the bandwidth and the dynamic range
of a PCM system are wasted when redundancy
sample values are transmitted,
? Differential pulse code modulation (DPCM) is one
way to minimize redundant transmission and reduce
the bandwidth,in which the PCM signals
corresponding to the difference in adjacent sample
values is transmitted,
89
3.7 Differential Pulse code Modulation
? The linear prediction filter,by which the present
value can be estimated from the past values,may be
used in a differential configuration to produce DPCM,y ( n T s ) D e l a y T
s
D e l a y
T s
D e l a y
T s
…
a 1 a 2 a l a K
z ( n T s )
? The optimum tap
gains of transversal
filter are a function
of the correlation
properties of the
input signal,The
output samples are,
)79-3( )-(=)( ∑
1=
alTnTyanTz
K
l ssls
90
3.7 Differential Pulse code Modulation
? Like PCM,DPCM follows the 6-dB rule,
(S/N)out=6.02n+α
Where -3<α<15 for DPCM speech(p189)
91
3.8 Delta Modulation
? Delta modulation (DM) is a special case of DPCM in
which there are two quantizing levels,so that the
output is +Vc or - Vc,
? Compared to DPAM,because there are only two
quantizing levels (m=2),the DM system owns some
special properties,
? Only one bit is transmitted per sample;
? ADC (analog-to-digital converter) and DAC (digital-
to-analog converter) are not needed;
? Encoder is not needed (the comparator has the
complete encoder function); encoder function,convert the
multilevel DPAM signal to binary code words,
? the DM signal is a polar signal,
92
3.8 Delta Modulation
Fig 3-31 A typical DM system
(t)w
Low-pass
filter Sampler Comparator
Clock
fs Integrator or accumulator
Channel
Integrator
Analog
input signal
DM transmitter
w(t)
Flat topped
PAM
y(t) DM
z(t)
Analog output
signal
DM receiver
+ _
93
3.8 Delta Modulation
? Fig 3-32 DM system waveforms
δ
V
t
initial state )(tw
hunting noise
Slope overload
noise
)(tw
z(t)
(a),Analog input and accumulator output
Vc
-Vc
t
(b),Delta modulation waveform
94
3.8 Delta Modulation
(Granular Noise and Slope Overload Noise)
? Note that the accumulator output signal does not
always track the analog input signal,
? DM quantizing noise error,
the difference between analog input signal and
accumulator output signal
? classification,
slope overload noise
granular noise
condition,sampling period,Ts=1/f s
step size,δ=Δv
accumulator initial state value,Δv
input DM signal,± Vc or ± 1
95
3.8 Delta Modulation
? Slope overload noise
occurs when the step
size δ is too small for
the accumulator to
follow quich changes
in the input waveform,
? Granular noise occurs
for any step size,but is
smaller for a small
step size,
? Thus we would like to
have δ as small as
possible to minimize
the granular noise,
Optimum value
of δ
Sig
na
l-to
-no
ise
ra
tio
(d
B)
Slope overload
dominates
Granular
noise
dominates
Step size δ
? Fig,3-33.signal-to-noise out of
a DM system as a function of
step size
96
3.8 Delta Modulation
Example 3-5,Design of a DM system,
problem,find the step size δrequired to prevent slope overload
noise for the case when the input signal is a sine wave,
solution,the maximum slope that can be generated by the
accumulator output is,
δ/Ts=δfs (3-82)
For the case of the sine-wave input,where ω(t)=Acosωat,the slope is,
principle,to avoid or prevent slope overload noise,the maximum slope
of accumulator output signal must larger than the maximum slope
of the analog input signal,
tAdt td aa ωc o sω=)(ωAnd the maximum slope of the input signal is Aω
a,thus,to avoid slope
overload,we require that δfs> Aωa,Or
s
a
f
Afπ2>δ
97
3.8 Delta Modulation
However,we do not want to make δtoo much larger than
this value,or the granular noise will become too large,
As stated the above formula,if we want to keep the
granular noise under a certain value,and to prevent
slope overload noise,we can do,
(1),Decrease the amplitude A;
(2),Increase the sampling frequency fs;
(3) making the step size vary as a function of time as
the input waveform changes,
Classification,
?Adaptive Delta Modulation (ADM)
-- Discrete accumulator (change)
?Continuously Variable Slope Delta modulation (CVSD)
-- Continuous integrator (change)
98
3.8 Delta Modulation
Data Sequence
Number of
Successive
Binary 1’s or 0’s
Step-size
Algorithm
f(d)
x x 0 1 1 δ
x 0 1 1 2 δ
0 1 1 1 3 2δ
1 1 1 1 4 4δ
Table 3-7 A typical step-size Algorithm
x– don’t care
99
3.8 Delta Modulation
Principle,
(1)When ADM signal consists of data with successive
binary 1’s and 0’s or when two successive binary 0’s
occur,the step size is set to a value δ,
(2) If three successive binary 1’s or 0’s occur,the step
size is increased to 2δ
(3) If four successive binary 1’s or 0’s occur,the step
size is increased to 4δ
100
3.8 Delta Modulation
? Which is better,PCM or DM?
? Depending on the criterion used for comparison and
the type of message,
? Simple,low-cost --------- DM
? High output SNR,compatibility--------PCM
