Introduction to Data Communication Lecture 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

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

representation.Fourier AnalysisNyquest theorem.


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


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


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.


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.


Periodic analog signal

Non Periodic analog signal

Has 3 parameters, Amplitude Frequency Phase

Time Domain Representation shows changes in signal amplitude with respect to time

Frequency domain representationshow the relationship between amplitude and frequency


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


A composite periodic Signal

Source - http://train-srv.manipalu.com/wpress/wp-content/uploads/2010/01/clip-image01617.jpg


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.


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

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

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)


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


Bit rate = 1kb per secondBit Pattern = 11001100

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

-3

Required Bandwidth = 0.25Khz


T

4 ms

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.


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.


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

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.



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

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.


Atténuation DelayNoise Jitter


BitRate = 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

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)


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

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


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

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