Digital Communication Principle

Flexible · Affordable · Accessible
the people’s university
TEL 213/05 Telecommunication Principle
Tutorial 2: Digital
Communication Principle
Semester January 2012

Basic Roadmap of Unit 2

Digital Versus Analog
Communication
Digital Communication Analogue Communication
More immune to noise as signal can
be regenerated if it is below the
threshold.
Less immune to noise
Has error detection and correction
techniques
No error detection and correction
Compatible with time division
multiplexing
Compatible with frequency division
multiplexing
Smaller ICs possible with greater
processing capability
Smaller ICs not possible
Can be processed using digital signal
processing techniques including
signal manipulation
Signal manipulation not possible with
DSP
More bandwidth required Less bandwidth required
More complex and requires more
circuitry
Less complex with less required
circuitry

Sampling
• The challenge is always to represent analog signals in digital form to ease
transmission.
• To do this, sampling of analog data needs to be done.
• Sampling is a process of approximation (estimation) of an analog quantity.
• After sampling, mathematical modeling can be done to represent the signal
in digital form.
• Sampling must be done at regular intervals and must cover most of the data
(at least twice the bandwidth frequency) to have an accurate depiction of the
whole data.
• This concept is called the Nyquist-Shannon Sampling Theory
• The phenomenon called “aliasing” (misrepresentation) will happen if not
enough samples are taken to represent the whole population.
Sampling Frequency = 2 * Bandwidth

Sampling an Analog Signal

Example
• If a 12 bits A/D converter were to be used,
how many voltage increments are there?
• 12 bits would produce 212
or 4096 voltage
increments.

Example
• Calculate the minimum voltage step increment for a 10 bit A/D converter
assuming that the input voltage is from 0V to 6V.
• Number of voltage levels =2 to the power of 10=1024 voltage levels
• Number of Increments = 1024-1=1023
• Minimum voltage step increment or maximum amount of error=

Example
• An information signal to be transmitted digitally is a rectangular wave with a
period of . It is given that the wave will be adequately passed if the
bandwidth includes the fourth harmonic. Calculate the signal frequency, the
frequency of the fourth harmonic and the minimum sampling frequency
(Nyquist rate).

Noise and S/N
• Noise is an electronic signal that is a mixture of
many random frequencies at different
amplitudes that gets added to a radio or
information signal as it is transmitted from one
place to another as it is processed
• The signal-noise (S/N) ratio is also called SNR,
and is an indication of the relative strengths of
the signal and noise in a communication system.
The stronger the signal and the weaker the
noise, the higher the S/N ratio

Formula for S/N calculation

Example

Bit Error Ratio in digital
Communication System
• Bit error ratio (BER) is defined as the
possibility of a bit being received in error in
a digital communication system

Solution – ERF Table

Bit Error Rate (BER)

Example
• Find the BER of a 100kbits/s assuming
unipolar transmission. The SNR is given
as 1.2dB.

Data Transmission
• All data need to be converted to ASCII
code first: http://www.ascii-code.com/

Value “M” being transmitted
serially
M = 010011101(first 0 is not taken into account, and total is 8 bits). t is the time
between each bit, known as bit interval. Bit time is the total time taken and can
be expressed in bps

Example 1
What is the bit time at 230.4kbps?
st µ34.410*34.4
230400
1 6
=== −

No PAM
• Compute the bit rate that it will take to transmit a
decimal number '201' using a bit interval of 1
microseconds using no modulation (serial
transmission)

With PAM
• Repeat the transmission with 2 bits of PAM transmission and calculate the
bit rate of this transmission.

PCM
• Digitizing = converting analogue signals to digital signals.
• Pulse code modulation (PCM) is commonly used
The resulting 4 bit PAM of a signal is found to be
10,9,8,11. Draw the resultant PCM.
Solution
10=1010
9=1001
8=1000
11=1011
10 9

FSK
• Frequency-shift keying (FSK) uses two
sine wave frequencies are used to
represent binary 0s and 1s
Binary Signal
FSK Signal

Problem with FSK

Binary Phase Shift Keying

DPSK

Line Encoding

Example
• Transmit the word “Data Com” in the
transmission line and calculate the
LRC/BCC and parity of VRC (odd)

Solution – Step 1
• D = 01000100
• A = 01000001
• T = 01010100
• A = 01000001
• <space> = 00100000
• C = 01000011
• O = 01001111
• M = 01001101

Solution – Step 2
A (input) B (input) Q (even parity) (odd parity)
0 0 0 1
0 1 1 0
1 0 1 0
1 1 0 1
Q
Character D A T A C O M LRC
or
BCC
(LSB) 0 1 0 1 0 1 1 1 1
0 0 0 0 0 1 1 0 0
1 0 1 0 0 0 1 1 0
0 0 0 0 0 0 1 1 0
0 0 1 0 0 0 0 0 1
0 0 0 0 1 0 0 0 1
(MSB) 1 1 1 1 0 1 1 1 1
Parity of VRC
(odd)
1 1 0 1 0 0 0 1 0

XOR Addition
•When the number of 1's is ODD, the
result of the XOR operation is '1'.
•When the number of 1's is EVEN (or
none present), the result of the XOR
operation is '0'.

Detecting Errors
Error Bit

Hamming Detection (FEC)
Example
• The data word is 01101010. Use
Hamming FEC Method to transmit this
data across the transmission line

Step 1 – Hamming – find the
value of n
8 bit data word to be transmitted
Experimented Value (2)
False
8 bit data word to be transmitted
Experimented Value (4)
True (therefore n=4)

Step 2: Initial Hamming Table
• Total bits required = 8+4=12 bits
• Insert the 4 required Hamming bits
between the transmitted data:
12 11 10 9 8 7 6 5 4 3 2 1
H 0 1 H 1 0 H 1 0 H 1 0

Step 3:
• Determine which bits are already ‘1’
12 11 10 9 8 7 6 5 4 3 2 1
H 0 1 H 1 0 H 1 0 H 1 0
Position 2 = 0010
Position 5 = 0101
Position 8 = 1000
Position 10 = 1010

Step 4: XOR Bits that are ‘1’

Step 5: Insert Final XOR into
Hamming Table
12 11 10 9 8 7 6 5 4 3 2 1
H 0 1 H 1 0 H 1 0 H 1 0
12 11 10 9 8 7 6 5 4 3 2 1
0 0 1 1 1 0 0 1 0 1 1 0
Final Transmitted Data

Hartley’s Law
C=2B
whereby C = channel capacity
expressed in bits per second and B
is the channel bandwidth
whereby the S/N is the signal to
noise ratio in power.
)1)(2log( 10
N
S
BC +=
With noise

Example – Shanon-Hartley’s
Law
Calculate the maximum channel capacity of a voice-graded telephone
line with a bandwidth of 3100hz and a S/N of 30dB.
10003log)
10
30
(log
)10/log(
log10
11
===
=
=
−−
P
dBantiP
PdB
bpsC
N
S
BC
31000)10(3100
1097.9)3(32.31001log32.31001log
1001log3100)10001(log3100)1(log
102
222
==
≈===
=+=+=
32
5log
5
6200
31000
)3100(2
31000
2
log
2
2
=
=
====
N
antiN
B
C
N
32 channels of multilayer encoding is required for a maximum channel capacity
Of 31kbps with S/N of 30dB

Summary
• A signal needs to be sampled as the first process in converting an analogue
signal to a digital signal
• To properly sample a signal and to ensure aliasing does not occur and the
whole signal can be adequately reconstructed from its digital representation,
the sampling frequency needs to be twice the value of the original signal.
• The process of quantization and mapping changes the digitally acquired
signals to become numerical values
• Pulse Code Modulation (PCM) uses multiplexing to convert the signals into
a stream of data.
• The encoding process is the last process of the A/D converter and serves to
change the numerical values for optimization prior to transmission.
• There are 3 main types of encoding available namely source encoding,
channel encoding and line encoding.

Thank you!

Digital Communication Principle

More Related Content

What's hot (20)

Viewers also liked (20)

Similar to Digital Communication Principle (20)

More from Dr. Ghanshyam Singh (15)

Recently uploaded (20)

Digital Communication Principle