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

 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

 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

 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

 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

 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 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 representation show the relationship between amplitude and frequency M.Sc. in IT - Year 1 Semester II

 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

A composite periodic Signal Source - M.Sc. in IT - Year 1 Semester II

 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

Fundamental frequency – f (first harmonic) Third harmonic – 3f Fifth harmonic – 5f … Source - M.Sc. in IT - Year 1 Semester II

 Range of frequencies / Difference between the highest and lowest frequencies Source - M.Sc. in IT - Year 1 Semester II

 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

 A digital signal has infinite number of frequency components. (Bit rate)Speed = 1kb per second 2 ms F=1/T F=1/2*10 F=5ooHz F=0.5Khz -3 Required Bandwidth (Fundamental frequency) = ½*Bit Rate T Bit Pattern = M.Sc. in IT - Year 1 Semester II

Bit rate = 1kb per second Bit Pattern = F=1/T F=1/4*10 F=25oHz F=0.25Khz -3 Required Bandwidth = 0.25Khz M.Sc. in IT - Year 1 Semester II 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. M.Sc. in IT - Year 1 Semester II

 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

 Periodic Signal Fourier Analysis  Non Periodic Signal Fourier Transform A Discrete Frequency Spectrum A Continuous Frequency Spectrum M.Sc. in IT - Year 1 Semester II

 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

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 Source :

 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

 Atténuation  Delay  Noise  Jitter M.Sc. in IT - Year 1 Semester II

BitRate = 2 x bandwidth x log 2 L  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 log 2 (1 + SNR) M.Sc. in IT - Year 1 Semester II

 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 bits per element6 Kbp/s 3 bits per element9 Kbp/s 4 bits per element12 Kbp/s 5 bits per element15 Kbp/s 6 bits per element18 Kbp/s 7 bits per element21 Kbp/s 6

 Data Communications and Networking, Forouzan, Chapter 03, 4 th Edition M.Sc. in IT - Year 1 Semester II