Introduction to Data Communication Lecture o2 Advanced Computer Networks (ACN) 545

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Mr. Thilak d e Silva. BSc . Eng., MSc , CEng , FIE(SL), FIET(UK), CITP(UK), MBCS(UK), MIEEE (USA). Introduction to Data Communication Lecture o2 Advanced Computer Networks (ACN) 545 . Today's Agenda. Analog and Digital Signals. Periodic and Non-periodic signals. - PowerPoint PPT Presentation

Transcript of Introduction to Data Communication Lecture o2 Advanced Computer Networks (ACN) 545

Introduction to Data CommunicationLecture o2

Advanced Computer Networks (ACN) 545

Mr. Thilak de Silva.BSc. Eng., MSc, CEng, FIE(SL), FIET(UK), CITP(UK), MBCS(UK), MIEEE (USA)

M.Sc. in IT - Year 1 Semester II - 2012

Today's Agenda

Analog and Digital Signals. Periodic and Non-periodic signals. Time domain and Frequency domain

representation. Fourier Analysis Nyquest theorem.

M.Sc. in IT - Year 1 Semester II - 2012

Session Outcomes

At the end of this session you will have a broad understanding of Analog and Digital signals, Fourier Analysis and Nyquest theorem.

M.Sc. in IT - Year 1 Semester II - 2012

Data and Signals

Data is in memory. It is converted in to Signals When transmitting Need a transmission media to transmit signals.

Signals can be divided as,▪ Analog Signals, Digital Signals▪ Periodic Signals, Non Periodic Signals

M.Sc. in IT - Year 1 Semester II - 2012

Analog Signals and Digital Signals Analog signals are continuous and

has infinitely many levels of intensity over a period of time.

Digital signals are discrete and has limited number of levels of intensity over a period of time.

M.Sc. in IT - Year 1 Semester II - 2012

Periodic and Non periodic Signals Can be analog or digital. Periodic signals – has a pattern

which repeats over identical periods. (Cycle)

Practically we do not have periodic signals.

Non periodic signals – changes without exhibiting a pattern or cycle that repeats over time.

M.Sc. in IT - Year 1 Semester II - 2012

Periodic analog signal

Non Periodic analog signal

Periodic Analog Signal

Has 3 parameters, Amplitude Frequency Phase

Time domain and Frequency domin representation Time Domain Representation shows changes in signal amplitude with respect to time

Frequency domain representationshow the relationship between amplitude and frequency 

M.Sc. in IT - Year 1 Semester II - 2012

Composite signals

A composite signal is made of many sine waves.

Fourier showed that any composite signal is actually a combination of simple sine waves with different frequencies, amplitudes and phases.

These are known as harmonics

M.Sc. in IT - Year 1 Semester II - 2012

Composite signals

A composite periodic Signal

Source - http://train-srv.manipalu.com/wpress/wp-content/uploads/2010/01/clip-image01617.jpg M.Sc. in IT - Year 1 Semester II - 2012

Fourier Analysis

Use to transform a time domain signal in to frequency components.

Only applicable for periodic signals. According to Fourier analysis any

signal is composed with several frequencies called harmonics.

M.Sc. in IT - Year 1 Semester II - 2012

Fundamental and harmonics

Fundamental frequency – f(first harmonic)Third harmonic – 3fFifth harmonic – 5f … Source -

http://train-srv.manipalu.com/wpress/wp-content/uploads/2010/01/clip-image01814.jpg M.Sc. in IT - Year 1 Semester II - 2012

Bandwidth

Range of frequencies / Difference between the highest and lowest frequencies

Source - http://train-srv.manipalu.com/wpress/wp-content/uploads/2010/01/clip-image02014.jpg

M.Sc. in IT - Year 1 Semester II - 2012

Signal strength

When adding two signals the strength of the resulting signal depends on the phase differences, amplitudes, frequencies etc.

Eg:-

In phase (add voltages)

Out phase (deduct voltages)

M.Sc. in IT - Year 1 Semester II - 2012

Bandwidth of a composite signal A digital signal has infinite number of

frequency components.(Bit rate)Speed = 1kb per

second

2 ms

F=1/TF=1/2*10F=5ooHzF=0.5Khz

-3

Required Bandwidth (Fundamental frequency) = ½*Bit Rate

T

Bit Pattern = 101010

M.Sc. in IT - Year 1 Semester II - 2012

If the bit pattern changes,Bit rate = 1kb per secondBit Pattern =

11001100

F=1/TF=1/4*10F=25oHzF=0.25Khz

-3

Required Bandwidth = 0.25Khz

M.Sc. in IT - Year 1 Semester II - 2012

T

4 ms

Regenerating the signal

At the receiving end the signal is regenerated by looking at the amplitude, IF amplitude is high 1 is generated IF amplitude is low 0 is generated

Therefore we must at least send the fundamental frequency.

That’s why we say that the bandwidth should be at least half of the bit rate. M.Sc. in IT - Year 1 Semester II - 2012

Regenerating the signal - impacts When frequency getting high the

amplitude gets low. Sending more and more harmonics

makes the signal regeneration easy. But this is expensive due to high

bandwidth. Deciding the number of harmonics

we send should be done based on the characteristics of the media.

M.Sc. in IT - Year 1 Semester II - 2012

Fourier Analysis and Transform Periodic Signal

Fourier Analysis

Non Periodic Signal Fourier Transform

A Discrete Frequency Spectrum

A Continuous Frequency SpectrumM.Sc. in IT - Year 1 Semester II - 2012

Digital signals Can have two or more discrete level.

Bit Rate – number of bits sent per second.

Baud rate – signal changing rate per second

Required bandwidth depends on the baud rate.

M.Sc. in IT - Year 1 Semester II - 2012

Digital signal having two levels and four levels

Two Levels One signal element

represents one bit Therefore bit rate=

baud rate

Four Levels One signal representation

has two bits Therefore bit rate=

baud rate

M.Sc. in IT - Year 1 Semester II - 2012

Source : http://train-srv.manipalu.com/wpress/wp-content/uploads/2010/01/clip-image02214.jpg

Increasing Levels,

We can increase the Bit Rate without increasing the Bandwidth.

But, The error probability is high, Circuit component cost is high, Effect of transmission impairments is

high,

There for we do not use 4 levels practically. M.Sc. in IT - Year 1 Semester II - 2012

Digital Impairments

Atténuation Delay Noise Jitter

M.Sc. in IT - Year 1 Semester II - 2012

Nyquist TheoremBitRate = 2 x bandwidth x log2L

Assumptions One signal element carry only one bit No noise, attanuation etc in the

transmission media

If there are noise and attanuation (Shannon Capacity)Capacity = bandwidth x log2(1 +

SNR) M.Sc. in IT - Year 1 Semester II - 2012

Number of Bits…? If Capacity = 20 Kbp/s & Bit Rate = 3 Kbp/s

Can represent maximum 6 bits per element.

2 = 64 combinations of amplitude or phase differences. (64 QAM)

M.Sc. in IT - Year 1 Semester II - 2012

2 bits per element

6 Kbp/s

3 bits per element

9 Kbp/s

4 bits per element

12 Kbp/s

5 bits per element

15 Kbp/s

6 bits per element

18 Kbp/s

7 bits per element

21 Kbp/s 6

References

Data Communications and Networking, Forouzan, Chapter 03, 4th Edition

M.Sc. in IT - Year 1 Semester II - 2012

Thank You…

M.Sc. in IT - Year 1 Semester II - 2012