Telecommunication Networks (ECEg-5311)
Chapter Two
Digital Telecommunication Transmission Principles
Outline
▪ Introduction
▪ Digital Representation of Information
▪ Digital Processing of Analog Signals
▪ Line Coding
▪ Digital Modulation Techniques
▪ Digital Hierarchy Technologies
2
Introduction
Fig. 1: Basics of a communication system.
3
Introduction …
▪ A transmission system makes use of a physical transmission
medium or channel that allows the propagation of energy in the
form of pulses or variations in voltage, current or light intensity.
▪ In analog communications, the objective is to transmit a
waveform which is a function that varies continuously with
time.
▪ This function of time must be reproduced exactly at the output
of the analog communication system.
▪ In practice, communication channels do not satisfy this
condition, so some degree of distortion is unavoidable.
4
Introduction …
▪ In digital transmission, the objective is to transmit a given
symbol that is selected from some finite set of possibilities.
▪ Specifically, in binary digital transmission the objective is to
transmit either a 0 or 1.
▪ This can be done by transmitting positive voltages for a certain
period of time to convey 1 and a negative voltage to convey 0.
▪ The task of the receiver is to determine the input symbol with
high probability.
▪ The system will operate correctly as long as the receiver can
determine whether the original voltage was positive or negative.
5
Digital Representation of Information
Fig. 2: Block diagram of digital transceiver.
6
Digital Representation of Information …
▪ Networks are driven by the applications they support and must
therefore be designed to accommodate the requirements
imposed by the information types in the applications.
▪ These information types include text, speech, audio, data,
images and video.
▪ These information types can be classified into two broad
categories.
i.
Block information:
✓
include files that contain text, numerical or graphical information
✓
can range from a few bytes to hundreds of kilobytes
✓
data compression is performed to reduce the file size
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Example: Data Compression
▪ An individual color image produces a huge number of bits .
▪ A pixel is defined as a single dot in a digitized information.
▪ For example, an 8x10-inch picture scanned at a resolution of
400x400-pixels per square inch yields 400x400x8x10=12.8
Megapixels.
▪ A color image is decomposed into red, green and blue subimages.
▪ Normally eight bits are used to represent each of the red, green
and blue color components resulting in 12.8 Megapixels x 3
bytes/pixel=38.4 Megabytes .
8
Example: Data Compression …
▪ At a speed of 28.8 kbps, this image would require about 3 hours
to transmit!
▪ Clearly, data compression methods are required to reduce the
transmission time.
▪ Some of the data compression standards include:
✓
The Graphics Interchange Format (GIF)
✓
The Joint Photographic Experts Group (JPEG)
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Digital Representation of Information ...
Table: Block-oriented Information
10
Digital Representation of Information …
ii.
Stream information:
✓
information that is produced continuously by the source
✓
include active music, voice and video information
✓
require relatively high file size compared to block information
▪ The voice signal in telephone systems is sampled at a rate of
8000samples/second.
▪ Each sample is then represented by 8 bits resulting in a bit rate
of 8000samples/sec x 8 bits/sample=64 kbps.
▪ Music signals vary much more rapidly than voice signals.
▪ Audio compact disc (CD) systems sample the music signals at
44000 samples/second at a resolution of 16 bits.
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Digital Representation of Information …
▪ For a stereo music system, the resulting bit rate is 44,000
samples/second x 16 bits/sample x 2 channel=1.4 Megabits/sec.
▪ One hour of music will then produce 317 Mbytes of
information.
▪ The sub-band coding technique used in the MPEG audio
standard can reduce this bit rate to 14 kbps to about 100 kbps.
▪ Video signals can be viewed as a succession of pictures that is
fast enough to give the human eye the appearance continuous
motion.
▪ Typical videoconferencing systems operate with frames of 176
x 144 pixels at 10 to 36 frames/second.
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Digital Representation of Information …
▪ Broadcast television requires greater resolution than
videoconferencing, i.e., 720 x 480 pixels/frame, and can contain
a high degree of motion.
▪ The MPEG-2 coding system can achieve a reduction from the
uncompressed bit rate of 249 Mbps to the range of 2 to 6 Mbps.
▪ The recently approved high-definition television system
operates with a resolution of 1920 x 1080 pixels/frame and 30
frames/second.
▪ The uncompressed bit rate is 1.6 Gigabits/second.
▪ The MPEG-2 coding can reduce this to 19 to 38 Mbps which
can be supported by digital transmission systems.
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Digital Representation of Information …
▪ The video image pixel rates for different systems is shown
below.
14
Digital Representation of Information …
15
Digital Processing of Analog Signals
▪ Most real-world signals are analog in nature.
✓ They are continuous in time and amplitude
✓ They are converted to voltage or current signals using sensors and
transducers
▪ The voltage or current signals are then converted into digital
signals for digital transmission.
✓ Sampling -> quantization -> coding
▪ Finally, the analog signals must be reconstructed at the receiver.
✓ Reconstruction or digital-to-analog conversion
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Pulse Code Modulation (PCM)
▪ PCM consists of three steps to digitize an analog signal:
1. Sampling
2. Quantization
3. Binary Encoding
▪ Before we sample, we have to filter the signal to limit the
maximum frequency of the signal as it affects the sampling rate.
▪ Filtering should ensure that we do not distort the signal, i.e.,
remove high frequency components that affect the signal shape.
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Pulse Code Modulation (PCM) …
➢ The Analog-to-digital Converter (ADC)
performs three functions:
Analog
Input
Signal
– Sampling
• Makes the signal discrete in time.
• If the analog input has a bandwidth
of W Hz, then the minimum sample
frequency such that the signal can be
reconstructed without distortion.
Sample
ADC
– Quantization
Quantize
Encode
111
110
101
100
011
010
001
000
• Makes the signal discrete in
amplitude.
• Round off to one of q discrete levels.
– Encode
• Maps the quantized values to digital
words that are n bits long.
Digital Output
Signal
111 111 001 010 011 111 011
Pulse Code Modulation (PCM) ...
Fig. Components of PCM Encoder
19
Sampling
▪ Analog signal is sampled at every Ts seconds.
▪ Ts is referred to as the sampling interval (sampling period).
▪ fs = 1/Ts is called the sampling rate or sampling frequency.
▪ There are 3 sampling methods:
i.
Ideal - an impulse at each sampling instant
ii. Natural - a pulse of short width with varying amplitude
iii. Flattop - sample and hold, like natural but with single
amplitude value
▪ The process is referred to as pulse amplitude modulation PAM
and the outcome is a signal with analog (non integer) values.
20
Sampling …
Fig. Three different sampling methods for PCM
21
Sampling …
▪ According to the Nyquist theorem, the sampling rate must be at
least 2 times the highest frequency contained in the signal.
Fig. Nyquist sampling rate for lowpass and bandpass signals
22
Quantization
▪ Sampling results in a series of pulses of varying amplitude
values ranging between two limits: min and max.
▪ The amplitude values are infinite between the two limits.
▪ We need to map the infinite amplitude values onto a finite set
of known values.
▪ This is achieved by dividing the distance between min and max
into L zones, each of height 
max − min 2 A
=
=
L
L
where A is the peak amplitudeof the message signal,
i.e., [- A, A]
23
Quantization Levels
▪ The midpoint of each zone is assigned a value from 0 to L-1
(resulting in L values)
▪ Each sample falling in a zone is then approximated to the value
of the midpoint.
▪ Assume we have a voltage signal with amplitudes Vmin=-20V
and Vmax=+20V.
▪ We want to use L=8 quantization levels.
▪ Zone width  = [20 -(-20]/8 = 5
▪ The 8 zones are: -20 to -15, -15 to -10, -10 to -5, -5 to 0, 0 to
+5, +5 to +10, +10 to +15, +15 to +20
▪ The midpoints are: -17.5, -12.5, -7.5, -2.5, 2.5, 7.5, 12.5, 17.5
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Quantizing Error
▪ The difference between the input and output signals of the
quantizer is known as the quantizing error or quantizing noise.
▪ It is apparent that with a random input signal, the quantizing
error qe varies randomly in the interval:
−

 qe 
2
2
▪ Assuming that the error is likely to lie anywhere in the range
(-Δ/2, Δ/2), the mean-square quantizing error is given by:
1 /2 2
2 A2
E (qe ) =  qe dqe =
= 2

/
2

12 3L
2
25
Output Signal to Quantizing Noise Ratio
▪ The output signal-to-quantizing-noise ratio in a PCM system is
defined as the ratio of average signal power to average
quantizing noise power.
▪ For a full-scale sinusoidal message signal with amplitude A, the
average signal and quantizing noise powers are given by:
A2
2 A2
2
S=
and N q = E (qe ) =
= 2
2
12 3L
▪ The output signal-to-quantizing-noise ratio of a PCM system is
then given by:
2
 S 
 S 
A
/2
3 2



( SNR ) o =
= 2
= L  ( SNR ) o ,dB = 
= 1.76 + 20 log L
2
N 


 q  o A /(3L ) 2
 N q  o ,dB
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Encoding
▪ After quantization, each zone is then assigned a binary code.
▪ The number of bits required to encode the zones, or the number
of bits per sample as it is commonly referred to, is obtained as
follows:
nb = log 2
L
▪ For the given example, nb = 3
▪ The 8 zone (or level) codes are therefore: 000, 001, 010, 011,
100, 101, 110, and 111
▪ Assigning codes to zones:
✓
000 will refer to zone -20 to -15, 001 to zone -15 to -10, etc….
27
Encoding…
Fig. Quantization and encoding of a sampled signal
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Bit Rate and Bandwidth Requirements of PCM
▪ The bit rate of a PCM signal can be calculated by multiplying
the number of bits per sample and the sampling rate, i.e.,
Bit Rate = nb x f s
▪ The bandwidth required to transmit this signal depends on the
type of line encoding used.
▪ A digitized signal will always need more bandwidth than the
original analog signal.
▪ Thus, higher bandwidth is the price we pay for robustness and
other features of digital transmission.
29
PCM Decoder
▪ To recover an analog signal from a digitized signal we follow
the following steps:
✓ We use a hold circuit that holds the amplitude value of a pulse till
the next pulse arrives.
✓ We pass this signal through a low pass filter with a cutoff
frequency that is equal to the highest frequency in the pre-sampled
signal.
▪ The higher the value of L, the less distorted a signal is
recovered.
30
PCM Decoder …
Fig. PCM decoder components
31
Delta Modulation
▪ This scheme sends only the difference between pulses, if the
pulse at time tn+1 is higher in amplitude value than the pulse at
time tn, then a single bit, say a “1”, is used to indicate the
positive value.
▪ If the pulse is lower in value, resulting in a negative value, a
“0” is used.
▪ This scheme works well for small changes in signal values
between samples.
▪ If changes in amplitude are large, this will result in large
errors.
32
Delta Modulation …
Fig. The process of delta modulation
33
Delta Modulation …
Fig. Delta modulation components
34
Delta Modulation …
Fig. Delta demodulation components
35
Examples on PCM
Example-1:
Find the Nyquist rate and Nyquist interval of an analog signal
given by:
x(t ) = 5 cos1000t cos 4000t
Solution:
x(t ) = 5 cos1000t cos 4000t  x(t ) = 2.5[cos 3000t + cos 5000t ]
5000
 fm =
= 2500 Hz = 2.5kHz
2
1
1
 f s = 2 f m = 2(2500) = 5000 Hz and Ts = =
= 0.2ms
f s 5000
 The Nyquist rate is 5kHz and the Nyquist interval is 0.2ms
36
Examples on PCM …
Example-2:
A binary channel with bit rate 36kbps is available for PCM
voice transmission. Find appropriate values of the sampling
rate fs, the quantizing level L and the binary digits nb assuming
fm=3.2kHz.
Solution:
f s  2 f m = 6400 and nb f s  Rb = 36000 , where Rb : channel bit rate
 nb 
Rb 36000

= 5.6
fs
6400
 nb = 5, L = 25 = 32
37
Examples on PCM …
Example-3:
In a binary PCM system, the output signal-to-quantizing-noise
ratio is to be held to a minimum of 40dB. Determine the
number of required levels and the corresponding output signalto-quantizing noise ratio.
Solution:
In a binanry PCM system, L = 2 n , n is the number of binary digits
 S 


= 1.76 + 20 log 2 n = 1.76 + 6.02n dB
N 
 q  o.dB
38
Examples on PCM …
Solution:
 S

N
 q

 S

= 40dB  

N
 o , dB
 q

 = 10,000

o
2  S
3  N q

20,000

L =
=
= 81.6  82

3
o
The number of binary digits n is :
n = log 2
82
= 6.36  7
The number of levels required is L = 2 7 = 128.
The corresponding output signal - to - quantizing - noise ratio is :
 S

N
 q


= 1.76 + 6.02 * 7 = 43.9dB

 o , dB
39