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8/2/2019 Advance Communication System Lectures Part 4
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Learning Outcomes
29 April 20121
Interpret various type of PulseModulation (PAM, PWM,PPM,PCM).
Analyze Pulse Modulation anddiscuss the process of Sampling,
Quantization and Coding, whichforms the fundamental for digitaltransmission of any signal
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Pulse Modulation
29 April 20122
Pulse Modulation is a process of sampling analogsignal and then converting them into discretepulses and transporting the pulses from a sourceto a destination over a transmission medium.
A device to perform this is called ADC (Analog-to-Digital Converter) & DAC (Digital-to-AnalogConverter).
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Pulse Modulation
29 April 20123
1. PAM (Pulse Amplitude Modulation)2. PWM (Pulse Width Modulation)
3. PPM (Pulse Position Modulation)
4. PCM (Pulse Code Modulation)
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PAM (Pulse Amplitude Modulation)
29 April 20124
It is used to describe the conversion ofanalog signal to pulse-type signal in whichthe amplitude of the pulse denotes the
analog information.
In addition, it is a series of pulses in whichthe amplitude of each pulse represents the
amplitude of the information signal at agiven time.
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Pulse Modulation
29 April 20125
PWM (Pulse Width Modulation)
It is a pulse duration modulation (PDM) or pulselength modulation. The width of pulse is varied
proportional to the Amplitude of the analog signal atthe time signal is sampled.
PPM (Pulse Position Modulation)
It is a series of pulses in which the timing of each
pulse represents the amplitude of the informationsignal at a given time.
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PCM (Pulse Code Modulation)
29 April 20126
It is a series of pulse in which the amplitude of theinformation signal at a given time is coded as abinary number. The pulses are of fixed length and
fixed amplitude.
PCM is generated by 3 processes; Sampling,Quantization & Encoding.
An Integrated circuit that perform PCM encodingand decoding function is called CODER ORDECODER.
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29 April 20127
Pulse Modulation
Analog signal
Sample pulse
Pulse widthmodulation
Pulse position modulation
Pulse amplitudemodulation
Pulse codemodulation
8 bit
ts
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Pulse Modulation
Pulse Modulation consists of:
Easily effected bynoise
Less susceptible tonoise
Less susceptible tonoise compared toPAM
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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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Sampling An analog signal must be sampled at Nyquist rate to avoid aliasing
Quantization & Coding
Quantization : Process ofMapping samples of a continuousamplitude waveform to a finite set of amplitudes.
A fixed number of levels including the maximum and
minimum value of the analog signal
Number oflevels is determined by the number of bits usedfor coding
Coding : translate the quantized sample into a code
number.
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29 April 201211
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Quantization Interval (V)
Represent the voltage value for each quantized level
For example: For a sampled signal(mp) 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 torepresent 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
V102
L
mV
p
quantization
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13
M = L
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+mp
-mp
0
1 11
1 10
1 01
1 00
0 000 01
0 10
0 11
t
V
We assume that the amplitude of
m(t) is confined to the range (-mp,
mp)
L
mV
p2
nL 2
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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 of4.375 V.
The difference between the sampled value and the quantized value
results in quantization noise.
V375.4
2
V25.1V50 V
V
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+mp
-mp
0
1 11
1 10
1 01
1 00
0 000 01
0 10
0 11
valueSign bit
t
100 101 111 111 111 110 101 000 010 011 011 010 000 001 110 110 110 100
Quantization error
Qe
PCM code
t
The same code representing several
samples with different amplitudes
Step size
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The binary codes used for PCM are n-bit codes (sign-magnitudecode) where the MSB bit is the sign bit. If PCM is 3-bit codes,
then the sign and magnitude are shown below:
In terms of Voltage, the maximum signal voltages are 3 V or -3 V
and the minimum signal voltages are 1 V or -1 V.
Sign Magnitude Decimal value Quantization range
(V)
1 1 1 +3 +2.5 to +3.5
1 10 +2 +1.5 to +2.5
1 01 +1 +0.5 to 1.5
1 00 +0 0 to +0.5
0 00 -0 0 to -0.5
0 01 -1 -0.5 to -1.50 10 -2 -1.5 to -2.5
0 11 -3 -2.5 to -3.5
Folded binary code
0 V codes each
have an input
range equal to
only one half aquantum
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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
UNIFORM QUANTIZATION
Uniform quantization is a quantization process with a uniform (fixed) quantization
interval/step size.
Example :n = 3 ,L = 8 , signal +5 V ; => V = 1.25 V . Bit rate: sb nff
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Input analog signal
Sampling pulse
PCM code
Quantization
PAM signal
What is the PCM code for +2.6 V??
+2.6
V
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Input analog signal
Sampling pulse
PCM code
PAM signal
Question: What is the quantized interval and PCM code for +1.75 V??
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Quantization Error/ Noise
Folded PCM code = sample voltage
resolutionFor input at 2.6 V, the PCM code is therefore:
2.6/1 = 2.6
But since there is no code for +2.6, the magnitude is
rounded off to the nearest valid code, which is 111(+3V)
Thus there is difference of 0.4
QUANTIZATION ERROR (Qe)
or also known as quantization noise (Qn)
Qe =sample voltage - original analog signal
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Maximummagnitude Qe isequal to one-half
a quantum
Low Resolution
, more accuratethe quantizedsignal willresemble the
original analogsample
2229 April 2012
resolution
2eQ
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Linear input-output transfer curve
Linear
Quantization
Error
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Dynamic Range
max max
min2 1resolution
nV V
DR V
DR = dynamic range (unitless)
Vmin= the quantum value
Vmax= the maximum voltage magnitude of the DACs
n = number of bits in a PCM code (excl. sign bit)
20log 2 1n
dBDR
Ratio of the largest possible magnitude to the smallest
(other than 0) magnitude that can be decoded by the
digital-to-analog converter (DAC) in the receiver
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Number of bits used for a PCM code depends onthe dynamic range
DR = 2n-1
Thus 2n= DR + 1
And therefore, The minimum number of bit used:n = log ( DR + 1 )
log 2
2 1 2n nDR
For n > 4
20log 2 1 20 log2 6n
dBDR n n
D i R
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No of Bits No of Levels DR (dB)
1 2 6.02
2 4 12
3 6 18.1
4 16 24.1
5 32 30.1
6 62 36.1
7 128 42.1
8 256 48.2
9 512 54.2
10 1024 60.2
11 2048 66.2
12 4096 72.2
13 8192 78.3
14 16348 84.3
15 32768 90.3
16 65536 96.3
Dynamic Range
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Dynamic Range
Total dynamic range in (dB) = 6 x (number of bits)
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Coding Efficiency
minimum number of bitscoding efficiency= 100
actual number of bits
Coding efficiency is a numerical indication of how
efficiently a PCM code is utilized
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EXAMPLEA PCM systems has the following specification:
Maximum Analog Input Frequency = 4 kHz
Maximum decoded voltage at the receiver =2.55 V
The dynamic range = 46 dB
Determine the following :
(a) Minimum Sampling Rate
(b) Minimum number of bits used in PCM code
(c) Resolution
(d) Quantization Error(e) Coding Efficiency
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Solution
(a) The minimum sampling rate:
fs
= 2fa
= 2(4 kHz) = 8 kHz
(b) Calculate the Dynamic range :
46 = 20log(Vmax/ Vmin)
Vmax/ Vmin = antilog (46/20) = 199.5Thus, the minimum number of bit
used:
n = log (199.5 + 1) / Log 2 = 7.63
(c) Resolution is defined as:
Vmax / 2n - 1 = 0.01 V
(d) Quantization Error :Q = resolution / 2 = 0.01 V / 2 =
(e) Coding Efficiency
Coding efficiency= (8.63/9)(100)= 95.89%
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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:
(b) LSB voltage, V :(c) Voltage value for 101101 ;
(d) Voltage value for 011001 ;
(e) PCM Code for input +13.62 V
(g)PCM Code for input9.37 V
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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, V : 31.5/64 = 0.492 V(c) Voltage value for 101101 ; +(13 x 0.492) = +6.4 V
(d) Voltage value for 011001 ; (25 x 0.492) = -12.3 V
(e) Code for input +13.62 V
= 13.62/0.492 = 27.68 28 => 111100(g)Code for input9.37 V
= 9.37/0.492 = 19.04 19 => 010011