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Pemodulatan Digi
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In the early 90s, telecommunication networks is changing towards
digital world. With the rapid advancement in the fields of VLSI and
microprocessor, several telecommunication components such astransmission line and switching has been using digital signals in
their operation.
Therefore, information signals must be changed to digital form so
that it can be transmitted through this network.
Several techniques requiring full coding of the original signal will
be used:
4.0 Introduction4.0 Introduction
-
Pulse Code Modulation (PCM)
Differential PCM (DPCM)
Adaptive Differential PCM (ADPCM)
Delta Modulation (DM)
Adaptive Delta Modulation (ADM)
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Advantages :
Immunity to noise
Easy storage and processing:
Regeneration
Easy to measure
Enables encryption
Data from several sources can be integrated andtransmitted using the same digital communication
system Error correction detection can be utilized
Disadvantages :
Requires a bigger bandwidth
Analog signal need to be changed to digital first Not compatible to analog system
Need synchronization
Pemodulatan Digi
Voice : Analog : 4 kHz
Digit : 2 x 4 kHz x 8 bit = 64 kb/s
BWmin 32 kHz
MP, DSP, RAM, ROM, Computer
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4.2 TRANSMISSION METHOD FOR ANALOG &
DIGITAL SIGNALS
Analog
input
Analog channel
Baseband
Analog
output
Analog
inputModulator De
modulator
Analog
output
Analog
channel
Digital
input
encoder decoderDigital
channel
Digital
output
Digital
inputModem Modem
Analog
channel
Digital
output
Analoginput ADC &encoderDecoder& DAC
Analogoutput
Digitalchannel
Analog
input
Analog
outputAnalog
channel
ADC &
encoderModem
ADC &
decoderModem
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4.3 Pulse Modulation4.3 Pulse Modulation
PAM (Pulse Amplitude Modulation) => VPAM Vm PWM (Pulse Width Modulation) => Vm PPM (Pulse Position Modulation) => d (pulse delay) Vm PCM (Pulse Code Modulation)
Pulse Modulation consists of:
Easily effected by
noise
Less susceptible to
noise
Less susceptible to
noise compared to
PAM
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ms ff 2
Pemodulatan Digi
X
Digital signal
s(t)
ms(t)m(t)
m(t)
t
ms(t)
t
s(t)
t
Ts
Nyquist theorem
states that:
( ) [ ]
s
s
s
sss
s
fT
tttT
ts
22
.....3cos22cos2cos211
==
++++=
where
s
sf
T1
=
( ) ( ) ( )
( ) ( ) ( ) ( )[ ].....3cos22cos2cos21
++++=
=
ttmttmttmtmT
tstmtm
sss
s
s
Fourier series for impulse train :
Therefore :
8/14/2019 b4 Pulse Modulation -comm theorem
8/52Pemodulatan Digi
s s
ms +
m m0
ms ms +
ms
)(sM
sT
1
m m0
)(M
1
s s0
)(S
sT
2
0t
)(tm
sT6 sT60
)(ts
t
sT
sT6 sT60
)(tms
tsT
Time domain Frequency domain
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9/52Pemodulatan Digi
m m0
)(rM
1
m m0
)(M
1
X
Pulse signal
s(t)
ms(t)
m(t) h(t) mr(t)
TX RX
Low pass filter
s s
ms +
m m0
ms ms +
ms
)(sM
sT
1
n n0
)(H
sT
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Sampling process shown previously uses an ideal pulsesignal
However, it is quite difficult to generate an ideal pulse
signal practically The usual pulse signal generated is as shown below:
1
2 2( ) kos
di mana
sin
n
n s s s
sn
s
A A nt s t c
T T T
n
Tc
n
T
=
= +
=
Pemodulatan Digi
t
s(t)
Ts
A
-pulse widthTs pulse period
sT
nsinc=
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Natural Sampling Flat-top Sampling
Information signal
Pulse signal
Sampled signal (PAM)
t
m(t)
t
s(t)
Ts
t
ms(t)
Ts
t
ms(t)
Ts
( )sT
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In every sampling methods, the pulse amplitude is directlyproportional to the amplitude of the information signal
Practically, an ideal sampling is difficult to generate
However, by using an ideal and natural sampling, noisecan be eliminated, which is not the case for flat-topsampling
Pemodulatan Digi
Ideal Sampling Flat-top Sampling
ms(t)
t
Natural Sampling
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X
Pulse signal
s(t)
ms(t)m(t)
m(t)
t
ms(t)
t
s(t)
t
Mathematical analysis:
sn sss T
nt
T
n
TT
ts 2
cossinc2
)(
1
=
+=
Fourier series for pulse signal, s(t) :
)()()( tstmtms =
Therefore, the sampled signal:
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+=
=1
2cos
2).()(
n s
n
ss
sT
ntc
TTtmtm
=
+=1
2cos
2)()()(
n s
n
ss
sT
ntc
T
tm
T
tmtm
....6
cos2)(
4cos
2)(2cos
2)()()(
3
21
+
+++=
ss
sssss
s
T
tc
T
tm
T
tc
T
tm
T
tc
T
tm
T
tmtm
For n = 1, 2 , 3 ..
The above expression shows that the frequency components of the
sampled signal is at fs, 2f
sand 3f
s. Components 2f
sand 3f
sis a replica of
the spectrum of the sampled signal.
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....6cos2)(
4cos
2)(2cos
2)()()(
3
21
+
+++=
ss
sssss
s
Ttc
Ttm
T
tc
T
tm
T
tc
T
tm
T
tmtm
f
3f2s 2fs3fsfs0 fs-fm
fs+fm 2fs+fm 3fs+fm3fs-fm2fs-fm
ms(f)
Spectrum of the sampled
signal
The spectrum of the sampled signal has sidebands fs f
m, 2f
s f
m, 3f
s
fm
and so on.
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The choice of sampling frequency, fs
must follow the sampling theorem
to overcome the problem of aliasing and loss of information
(a) Sampling frequency=> fs1 < 2fm(max)
f2fs1 3fs1fs1fm
Aliasingms(f)
(b) Sampling frequency=> fs2 > 2fm(max)
f
2fs2 3fs2fs2fm
ms(f)
Shannon sampling
theorem=> fs 2fmNyquist frequency
fs = 2fm= fN
A bandlimited signal thathas a maximum
frequency, fmax can be
regenerated from the
sampled signal if it is
sampled at a rate of at
least 2fmax.
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4.4 Detection of Sampled Signal4.4 Detection of Sampled Signal
By using LPF to the sampled signal, ms(t)
LPFms(t) m(t)
Cut-off frequency , fo for LPF must be within the range: fm fo fs - fm Eventhough the sampled signal can be detected easily atfs = 2fm, but usually
fs > 2fm . The main reason is to have a guardband .
Therefore, the maximum frequency that can be processed by the sampleddata using sampling frequency, fs (without aliasing) is:
=> fm = fs/ 2 = 1 / 2Ts
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=+=
1
2cos2)()()(n s
n
ss
sTntc
Ttm
Ttmtm
From the sampling process, the sampled signal:
s
nT
nc
sinc=where :
sT
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( ) ( ) ( )
( ) ( )
( )
( )
( ) ( )
( ) ( )
+++++
+++++=
+++
+++=
+++++=
++=
+++=
+++=
=
=
=
.........2cos2cos2cos2
coscoscos2cos1
...cos2cos22cos2
coscos2cos2cos1
....3cos22cos2cos21cos1
cos21cos1
cos2cos1cos1
cos2
cos1cos1
1
1
1
ttt
tttt
T
ttt
tttt
T
tttT
t
tnT
t
tnT
tT
t
tnT
tT
ttm
msmss
msmssm
s
mss
mssm
s
sss
s
m
n
s
s
m
n
s
s
m
s
m
n
s
s
m
s
ms
replacing ( ) ( )ttm mcos1+=
=
+=1
2cos
2)()()(
n sss
sT
nt
T
tm
T
tmtm
inside
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It can be shown that the output sampled signal is the same as the output
PAM signal when :
sT
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4.5 Pulse Width Demodulation (PWD)4.5 Pulse Width Demodulation (PWD)
)(tvm
(pulse width) follows the instantaneous value of the informationsignal vm(t) :
tmoo cos+=
( )tmo cos1+=
o
represents the width that is
fixed according to the minimum
value of the information signal
The equation shows that the pulse
width, of the output signalPWM varies according to the
instantaneous value of the
information signal.
( ) { }...2cos2cos21 +++== ttT
v sss
PWMPWM
Replacing inside the general equation of the sampled signal:
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( )
( ){ }...2cos2cos21
cos1+++
+== tt
T
tv ss
s
moPWM PWM
++
++++= ...cos2cos2
coscos2cos2cos2cos21
tt
ttttt
Tv ms
msmss
s
oPWM
+++++
++++=
...)2cos()2cos()cos(
)cos(cos2cos2cos21
ttt
tttt
Tv
msmsms
msmss
s
oPWM
Generation of PWM signal is by changing the value of sample signal of the
PAM signal into a specific period
vPWM (t)vPAM (t)555 timer
(a) PWM generation using voltage to time converter
LPFvPWM (t) vm(t)
(b) PWM detection using LPF
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fs
> 2fm
fs = 2fm fs < 2fm
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Sampling Quantization Coding
A method used to represent an analog signal in terms of digital word
Constitutes 3 process:
1. Sampling the analog signal
2. Quantization of the amplitude of the sampled signal
3. Coding of the quantized sample into digital signal
LPF S/H ADC PCM
S/H : Sample and hold
circuit
Analogsignal
Anti aliasing
filter ADC : analog to digital converter
PCM process:
fs
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4.6.1 Sampling4.6.1 Sampling
An analog signal must be sampled at Nyquist rate to avoid
aliasing
4.6.2 Quantization & Coding
Process of estimating the sampled amplitude into a value suitable for
coding (ADC).
A fixed number of levels including the maximum and minimum value of
the analog signal
Number of levels is determined by the number of bits used for coding
3 t th t l d i th ti ti
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Quantization Interval
Represent the voltage value for each quantized level
For example: For a sampled signal that has 5V amplitude, Vpp
= 10 V
divide by the quantized level,L = 8 level,
Therefore, quantized interval ,
Quantization level,L= 2n
Quantization level depends on the number of binary bits, n used to
represent each sample.
For example:For= 3; Quantization level,L = 23 = 8 level.
In this example, first level (level 0) is represented by 000, whereas bit
111 represents the eigth level
V25.18
V10==V
3 terms that are commonly used in the quantization
process:
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Quantization value, Vk
The middle voltage for each quantized level
For example: forn = 3, quantized level,L = 8 and a sampled sinusoidalsignal with +5 V ,
The middle quantized value for level 0,
In this example, for a sample that is in level 0 segment will be
represented by bit 000 with a voltage value of 4.375 V. The difference
between the sampled value and the quantized value results in
quantization noise.
V375.4
2
V25.1V50 =+=V
For voice communication 256 levels
are commonly used (i.e n = 8)
4 6 3 UNIFORM QUANTIZATION
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t
Level 0 : 000
Level 1 : 001
Level 2 : 010
Level 3 : 011
Level 4 : 100
Level 5 : 101
Level 6 : 110
Leve l 7 : 111
1.9V
+5.0V
-5.0V
4.375V
3.125V
1.875V
0.625V
-0.625V
-1.875V
-3.125V
-4.375V
4.3V
1.9V
-3.2V
-4.5V
Quantization level &
binary representationQuantized
value
Sampled signal
4.6.3 UNIFORM QUANTIZATION
Uniform quantization is a quantization process with a uniform (fixed)
quantization interval.
Example : n = 3 ,L = 8 , signal +5 V ; => Vk = 1.25 V . Bit rate: sb nff =
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Pemodulatan Digi
+mp
-mp
0
0 11
0 10
0 01
0 00
1 001 01
1 10
1 11
valueSign but
t
000 001 011 011 011 010 001100 110 111 111 110 100 001 010 010 010 000
Quantization error
Qe
PCM code
t
The same code representing several
samples with different amplitudes
Step size
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Pemodulatan Digi
May add to or substract from the
actual signal
Quantization error (Qe) is also called Quantization noise (Qn) . And its
maximum magnitude is one half of the voltage of the minimum step
size .
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4.6.3.2 Quantization error4.6.3.2 Quantization error
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PCM
system
Example :
Vpp = 31.5 V
6 bit code (5 bits for
magnitude and 1 bit
for sign
(a) No of levels: 26 = 64
(b) LSB voltage, : 31.5/64 = 0.492 V(c) Maximum quantization level, /2 = 0.25 V(d) Voltage value for 101101 ; +(13 x 0.492) = +6.4 V
(e) Voltage value for 011001 ; (25 x 0.492) = -12.3 V
(f) Code for input +13.62 V
= 13.62/0.492 = 27.68 28 => 111100(g)Code for input 9.37 V
= 9.37/0.492 = 19.04 19 => 010011
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4.6.4 Non uniform quantization4.6.4 Non uniform quantization
nonuniform: to improve SNR (SQR)
More levels is available for low level amplitudes compared to highamplitude
Increase SNR for low level amplitude and decrease SNR for higheramplitudes
analog compression is done to the input signal before sampling and
quantization at the transmitter
Expansion is done at the receiver
COMPANDING (compression and expanding)
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example : Non-Linear
Quantization
Pemodulatan Digi
Companding => Compress - Expanding
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Companding Compress Expanding
A method used to produce a uniform SNR for all input signal range is
compression-expansion (Companding).
Input signal is compressed at the transmitter and expanded at the
receiver
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=> Analog Compression process is done on the input signal
before sampling and coding
=> Digital compression process is done after the signal is
sampled
Companding => Compress - Expanding
analog signal
(input)analog
compressorADC
DACAnalog
expander
Analog signal
(output)
PCM with analog compress-expand
To digital channel
analog signal
(input)ADC
DACDigital
expander
Analog signal
(output)
PCM with digital compress-expand
To digital channelDigital
compressor
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Pemodulatan Digi
2 Popular companding system (standardized by ITU)
EUROPE => A - Law
USA/NORTH AMERICA => - Law
Axfor
xA
for
A
AxA
Ax
y1
0
11
log1
log1
)log(1
+
+
+
=
A - compressor paramater. Usually
the value ofA is 87.6.
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Pemodulatan Digi
USA/NORTH AMERICA => - Law
Law is a standard compress-expand that is used in America
and Japan. The value of usedis 255 (8 bit).
( )
++=1log
)1log( xy
)(maki
i
E
E
x = )(makoo
E
E
y =
For both laws, the values ofxand
y refers to the equation below:
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Example 4.3 :
A compress-expand system using Law ( = 255) is used for a signalwith range 0 to 10V. Determine the output of the system if the input is 0 and
7.5V.Solution :
Given = 255 and Ei(mak)
= 10 V
ForEi= 0 V
)(maki
i
E
Ex = 0
10
0==x;
Output :
( )
++
= 1log)1log( x
y ; ( )2551log))0(2551log(
+
+=y
0=y
ForEi= 7.5 V
)(maki
i
E
E
x= 75.0
10
5.7==x;
( )
+
+=
1log
)1log( xy
Output :
( )2551log))75.0(2551log(
+
+=y
948.0=y
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4.6.5 Bit rate for PCM transmission
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Europe bit rate(Mb/s)
2.048
8.448
34.368
139.264
565.148
Telephone
channel
30
120
480
1920
7680
SDH 2.5Gb/s
Telephone
channel
North America bit
rate(Mb/s)
24 1.544
48 3.152
96 6.321
672 44.736
4032 274.176
European standard : A-Law
30 + 2 control channel = 32
Bit rate= 32 x 8 bit/sample x 8000 sample/s
= 2.048 Mb/s
North American standard (NAS) : -LawFor every 24 sample, 1 bit is added for
synchronization
For 24 sampel => 24 x 8 bit/sample+ 1 bit = 193 bits
Bit rate= 193 x 8000 = 1.544 Mb/sNeeds Multiplexing Process of transmitting two orNeeds Multiplexing Process of transmitting two or
more signals simultaneouslymore signals simultaneously
Example : PCM TDM CEPT SystemExample : PCM-TDM CEPT System
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Example : PCM-TDM CEPT SystemExample : PCM-TDM CEPT SystemFrame structure and Timing : European standard PCM system : E Line
(a) bits per time slot (b) time slots per frame (c) frames per multiframe
488 ns
3.9 s3.9 s
125 s125 s
2 ms
8 bits pertime slot
Bit duration
30 signal + 2 control = 32 channels = 1 frame
Signalling & synchronization
16 frames = 1 multiframeDuration of multiframe
CEPT system 32 channels (30 signals + 2 control)
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Frame structure and timing
Number of channel = 32
Number of bits in one time slot = 8
32 channels = 1 frame
Number of bits in a frame = 32 x 8 = 256 bits
This frame must be transmitted within the sampling period
and thus 8 x 103 frames are transmitted per second.
Therefore :
Transmission rate = 8 x 103 x 256 = 2.048 Mb/s
Bit duration = 1 / 2.048 x 106 = 488 ns
Duration of a time slot = 8 x 488 ns = 3.9 sDuration of a frame = 32 x 3.9 s = 125 s => (= 1 / 8 kHz = 125 s)Duration of a multi frame = 16 x 125 s = 2 ms
CEPT telephone system hierarchy
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MUX
1
MUX
2
MUX3
MUX4
30
Voice
channels
.
.
.
.
.
.
E1 line
2.048 Mbps
E2 line
8.448 MbpsE1
E1
E1
E2
E2
E2
E3
E3
E3
E3 line
34.368 Mbps
E4 line
139.264 Mbps
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There are 2 main components in the DM generator circuit,i.e comparator and integrator.
Pemodulatan Digi
+
-
X
-+
Pulse signal
s(t)
comparator
integrator
d(t)xDM
(t)m(t)e(t)
)(~ tm
Comparator will compare the error signal e(t), where)(~)()( ttt
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Output signal from comparator has the following function:
The output from the comparator will be sampled with apulse signal at a rate of 1/T
s.
Next, DM signal will be generated with the equation below:
The DM signal will be feed back, but before that this signalwill be integrated first
This signal will determine the error value e(t).
)()()( tmtmte =
=
=
=
=
n
ss
nsDM
nTtnTe
nTttetx
)()](sgn[
)()](sgn[)(
Pemodulatan Digi
+== )](sgn[)( tetd
0)(
0)(
te
te
=
=n
snTetm )](sgn[)(~
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Pemodulatan Digi
)(tm
t
Ts
)(~ tmEffects of steep
slope
0001010111111101100010000000
If e(t) < 0 or - , it will be coded as 0
If e(t) > 0 or +, it will be coded as 1
A steep slope results in noise in DM signal. To avoid this fromhappening, it has to follow the following condition:
dt
tdmmak
Ts
)(>
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4 8 1 Line code format4.8.1 Line code format A. NRZ (Non Return to Zero)
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4.8.1 Line code format4.8.1 Line code format
Digital Signal Encoding Formats
- Popular method
- easy
- Data does not return to 0 in one
clock interval
- No synchronization. Can use start
bit for synchronization purposes
1. NRZ-L (NRZ-Level)
1 => High level
0 => Low level
2. NRZ-M (NRZ-Mark)
1 => transition at the starting interval
0 => no transition
3. NRZ-S (NRZ-Space)
1 => no transition
0 => transition at the starting interval
B. RZ (Return to Zero)
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Digital Signal Encoding Formats
Return to 0 at the half bit interval
The same
advantages/disadvantages with
NRZ
Overcome by using bipolar signal
and alternating pulse for
synchronization
4. RZ (Unipolar)
1 => High level
0 => Low level
5. RZ (Bipolar)
1 => Alternately +ve
0 => Alternately ve
6. RZ (AMI Alternately Mark Inversion)
1 => Alternately +ve and -ve
0 => Low level
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