UNIT - 2

1. UNIT - 2 www.bookspar.com | Website for students | VTU NOTES

2. Unit 2 Title: Data, Signals & Digital Transmission Syllabus: Analog & Digital Signals, Transmission impairments, Data rate limits, Performance, Digital to analog conversion, Analog to digital conversion, Transmission modes. www.bookspar.com | Website for students | VTU NOTES

3. Analog & Digital data Data can be in analog or digital. i) Analog Data – refers to information that is continuous and takes continuous values. Example: Human voice. ii) Digital data – refers to information that has discrete states and take discrete values. Example: Data stored in a computer memory. www.bookspar.com | Website for students | VTU NOTES

4. Analog & Digital signals Signals can be Analog or Digital. Before transmitting the data over a medium, the data must be converted in to electromagnetic signals. www.bookspar.com | Website for students | VTU NOTES

5. Analog signal An analog signal is a continuous signal. It has an infinite number of values in a range. Example - Human Voice. www.bookspar.com | Website for students | VTU NOTES

6. Digital signal A digital signal is a discrete signal. It has a limited number of values. www.bookspar.com | Website for students | VTU NOTES

7. A sine wave Mathematically, a Sine wave can be represented by S(t) = A sin(2Π f t + Φ) Where, S – instantaneous amplitude, A – peak amplitude, f – frequency and Φ - phase www.bookspar.com | Website for students | VTU NOTES

8. Amplitude Highest intensity of a signal represent the peak amplitude of the signal. The intensity is proportional to the energy it carries. Amplitude is measured in volts. www.bookspar.com | Website for students | VTU NOTES

9. Period & Frequency Period refers to the time taken by a signal to complete one cycle & expressed in seconds. It is denoted by ‘ T ‘ . Frequency refers to numbers of signals produced in one second & and expressed in hertz ( Hz ). It is denoted by ‘ f ‘. The relation between period and frequency is given by T x f = 1. Frequency and period are inverses of each other. www.bookspar.com | Website for students | VTU NOTES

10. Period and frequency www.bookspar.com | Website for students | VTU NOTES

11. Units of periods and frequencies www.bookspar.com | Website for students | VTU NOTES

12. Example 1 Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz. Solution We Know that, 1 ms = 10 3µs. 100 ms = 100  103µs = 105ms Now we use the inverse relationship to find the frequency, changing hertz to kilohertz 100 ms = 100  10-3 s = 10-1 s = T f = 1/10-1 Hz = 10  10-3 KHz = 10-2 KHz www.bookspar.com | Website for students | VTU NOTES

13. Note: • Frequency is the rate of change of the • signal with respect to time. • Change in a short span of time means • high frequency. • Change over a long span of time means • low frequency. www.bookspar.com | Website for students | VTU NOTES

14. Note: • If a signal does not change at all, its frequency is zero. • If a signal changes instantaneously (it jumps from one level to another in no time), its frequency is infinite, because its period is zero. www.bookspar.com | Website for students | VTU NOTES

15. Phase Phase describes the position of the waveform relative to time zero. It is measured in degrees or radians. www.bookspar.com | Website for students | VTU NOTES

16. Signals with different phases www.bookspar.com | Website for students | VTU NOTES

17. Example A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians? Solution We know that one complete cycle is 360 degrees. Therefore, 1/6 cycle is (1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 rad www.bookspar.com | Website for students | VTU NOTES

18. Sine wave examples www.bookspar.com | Website for students | VTU NOTES

19. Sine wave examples (continued) www.bookspar.com | Website for students | VTU NOTES

20. Sine wave examples (continued) www.bookspar.com | Website for students | VTU NOTES

21. Time and frequency domains Time-domain plot shows changes in signal amplitude with respect to time. Frequency-domain plot show a relations between amplitude and frequency . A signal with peak amplitude= 5 volts, and frequency =0 www.bookspar.com | Website for students | VTU NOTES

22. Time and frequency domains (continued) A signal with peak amplitude= 5 volts, and frequency = 8 www.bookspar.com | Website for students | VTU NOTES

23. Time and frequency domains (continued) A signal with peak amplitude= 5 volts, and frequency = 16 www.bookspar.com | Website for students | VTU NOTES

24. Note: A single-frequency sine wave is not useful in data communications; we need to change one or more of its characteristics to make it useful. www.bookspar.com | Website for students | VTU NOTES

25. Note: When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made of many frequencies. www.bookspar.com | Website for students | VTU NOTES

26. Note: According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes. Any composite signal is a sum of set sine waves of different frequencies, phases and amplitudes. Mathematically it is represented by S(t) = A1 sin(2Π f1 t + Φ1) + A2 sin(2Π f2 t + Φ2) + A3 sin(2Π f 3 t + Φ3 )+….. www.bookspar.com | Website for students | VTU NOTES

27. Three harmonics A graph with three harmonic waves www.bookspar.com | Website for students | VTU NOTES

28. Adding first three harmonics Fig. Adding first three harmonics To create a complete square wave sum up all the odd harmonics up to infinity. www.bookspar.com | Website for students | VTU NOTES

29. Signal corruption & Bandwidth A signal has to pass through a medium. One of the characteristics of the medium is frequency. The medium needs to pass every frequency and also preserve the amplitude and phase. No medium is perfect. A medium passes some frequencies and blocks some others. www.bookspar.com | Website for students | VTU NOTES

30. Bandwidth Bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. For example, Voice normally has a spectrum of 300 – 3300 Hz. Thus, requires a bandwidth of 3000 Hz. The bandwidth is a property of a medium. Bandwidth is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass. www.bookspar.com | Website for students | VTU NOTES

31. Bandwidth (continued) The medium can pass some frequencies above 5000 and below 1000, but the amplitude of those frequencies are less than those in the middle. www.bookspar.com | Website for students | VTU NOTES

32. Example: If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Solution B = fh- fl = 900 - 100 = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900 www.bookspar.com | Website for students | VTU NOTES

33. Example A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude. Solution B = fh- fl →20 = 60 - fl →fl = 60 - 20 = 40 Hz www.bookspar.com | Website for students | VTU NOTES

34. Example A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium? Solution The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz. The signal with frequency 1000 & 2000 Hz is totally lost. www.bookspar.com | Website for students | VTU NOTES

35. Digital Signals Data can be represented by a digital signal. Bit 1 can be encoded by positive voltage and bit 0 can be encoded by zero voltage. www.bookspar.com | Website for students | VTU NOTES

36. Bit rate & Bit interval Bit interval - is the time required to send one bit. Bit rate - is the number of bits sent in 1 second. It is expressed in bps. www.bookspar.com | Website for students | VTU NOTES

37. Example A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) Solution The bit interval is the inverse of the bit rate. Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106ms = 500 ms www.bookspar.com | Website for students | VTU NOTES

38. Example Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel?. Answer A page is an average of 24 lines with 80 characters in each line. If we assume that one character require 8 bits. The bit rate of the channel = 100 x 24 x 80 x 8 = 1,636,000 bps = 1.636 Mbps www.bookspar.com | Website for students | VTU NOTES

39. Bit Length It is the distance one bit occupies on the transmission medium. Bit Length = Propagation speed x Bit interval. Transmission of Digital Signals A digital signal can transmit by using either baseband transmission or Broadband transmission. Baseband transmission -means sending a digital signal with out changing it to an analog signal. Baseband transmission requires a low-pass channel. Digital transmission use a low-pass channel. www.bookspar.com | Website for students | VTU NOTES

40. Broadband transmission (modulation)- means sending a digital signal after changing it to an analog signal. Broadband transmission requires a bandpass channel. Analog transmission use a band-pass channel. www.bookspar.com | Website for students | VTU NOTES

41. Low-pass & Band-pass A channel is either low pass or band pass. A low – pass channel has a B/W With frequencies between 0 & f. A band pass has B/W with frequencies between f1 & f2. www.bookspar.com | Website for students | VTU NOTES

42. Data Rate Limit The data rate over a channel depends on 3 factors. i) The Band width available. ii) The levels of signals that can use for data transmission. iii) The quality of the channel. Two theoretical formulas were developed to calculate the data rate. 1) Nyquist Bit Rate - Noiseless Channel 2) Shannon Capacity - Noisy Channel Nyquist Bit Rate Maximum Bit rate = 2 x Bandwidth x log 2L where L is number of levels used to represent data. www.bookspar.com | Website for students | VTU NOTES

43. Example Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. Calculate the bit rate. Answer: BitRate = 2  3000  log2 2 = 6000 bps www.bookspar.com | Website for students | VTU NOTES

44. Example Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). Calculate the bit rate. Answer: Bit Rate = 2 x 3000 x log2 4 = 12,000 bps www.bookspar.com | Website for students | VTU NOTES

45. Shannon capacity of Bit rate Maximum data rate of a noisy channel, C=Band width x log 2 (1+SNR) Where SNR is the Signal to Noise Ratio. www.bookspar.com | Website for students | VTU NOTES

46. Signal – to – Noise Ratio (SNR) SNR is defined as SNR = Average signal power / Average noise power. It is described in decibel units, SNR dB. SNR dB = 10 log 10 SNR www.bookspar.com | Website for students | VTU NOTES

47. Example: Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. Calculate the channel capacity. Answer: C = B log2 (1 + SNR) = B log2 (1 + 0)= B log2 (1) = B  0 = 0 www.bookspar.com | Website for students | VTU NOTES

48. Example: A telephone line has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is 3162. Find the channel capacity. Answer: C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163) = 3000  11.62 = 34,860 bps www.bookspar.com | Website for students | VTU NOTES

49. Example We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63. what is the appropriate bit rate and signal level? Solution First, we use the Shannon formula to find our upper limit. C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 6 Mbps = 2  1 MHz  log2L L = 8 www.bookspar.com | Website for students | VTU NOTES

50. Transmission Impairments www.bookspar.com | Website for students | VTU NOTES