1. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 UNIT - 3 www.bookspar.com | Website for students | VTU NOTES

2. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Analog Transmission & Multiplexing Digital – to – Analog conversion, Analog – to – analog conversion, Multiplexing, Spread Spectrum. www.bookspar.com | Website for students | VTU NOTES

3. Analog Transmission www.bookspar.com | Website for students | VTU NOTES

4. Modulation of Digital Data www.bookspar.com | Website for students | VTU NOTES

5. Digital-to-analog modulation Modulation Converting digital signals to analog signals is called modulation. Example Transmition of data from one computer to another computer using telephone line. The digital data is modulated on an analog signal. www.bookspar.com | Website for students | VTU NOTES

6. Types of digital-to-analog modulation Modulating techniques are 1)Amplitude Shift Keying (ASK) 2) Frequency Shift Keying (FSK) 3) Phase Shift Keying (PSK) 4) Quadrature Amplitude Modulation (QAM) www.bookspar.com | Website for students | VTU NOTES

7. Bit rate is the number of bits transmitted per second. Baud rate is the number of signal elements transmitted per second. A signal element carries one or more bits. Baud rate = The bit rate / Number of bits represented by each signal unit. Baud rate is less than or equal to the bit rate. Bit Rate & Baud Rate (Signal elements) www.bookspar.com | Website for students | VTU NOTES

8. Example An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate Solution Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps www.bookspar.com | Website for students | VTU NOTES

9. Example The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? Solution Baud rate = 3000 / 6 = 500 baud/s Baud rate = Bit rate / Number of bits represented by each signal unit. Baud rate = 3000 / 6 = 500 baud/s www.bookspar.com | Website for students | VTU NOTES

10. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example An analog signal has a bit rate of 8000 bps & a baud rate of 1000 baud. How many data elements are carried by each signal element?. How many signal elements do we need?. Solution Baud rate, S= 1000, bit rate, N = 8000, Data Elements, r=?, Signal Elements, L=? S = N/r r = N/S r= 8000/1000=8 bits /baud. r = log 2 L L = 2 r L = 2 8 L= 256 www.bookspar.com | Website for students | VTU NOTES

11. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Carrier Signal In analog transmission, the sending device produces a high frequency signal that acts as a basis for the information signal. This base signal is called as carrier signal or carrier frequency. The receiving device is tuned to the frequency of the carrier signal of the sender. The digital information then modulates the carrier signal by modifying one or more of its characteristics ( amplitude, frequency or phase). This modification is called modulation ( or Shift key). www.bookspar.com | Website for students | VTU NOTES

12. Amplitude Shift Keying (ASK / BASK) In ASK, the frequency & phase of the carrier signal remain constant and amplitude changes to represent the bit 1 or 0. A Zero amplitude represents bit 0 and max amplitude represent bit 1. A bit duration is the period of time that defines 1 bit. The amplitude of the bit during each duration is constant. In ASK baud rate and bit rate are same. ASK is susceptible for noise. www.bookspar.com | Website for students | VTU NOTES

13. Relationship between baud rate and bandwidth in ASK Bandwidth of a signal is the total range of frequencies occupied by that signal. The most significant frequencies are between f c – N baud /2 + f c +N baud /2, f c is the carrier frequency. In ASK, BW = (1+d) x N baud, where N baud is the baud rate, d is the factor related to the modulation process. www.bookspar.com | Website for students | VTU NOTES

14. Example Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex. Solution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz. www.bookspar.com | Website for students | VTU NOTES

15. Example Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? Solution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps. www.bookspar.com | Website for students | VTU NOTES

16. Example Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions. Solution For full-duplex ASK, the bandwidth allocated for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band, fc (forward) = 1000 + 5000/2 = 3500 Hz fc (backward) = 11000 – 5000/2 = 8500 Hz www.bookspar.com | Website for students | VTU NOTES

17. Solution to Example In data communication we use full duplex links with communication in both directions. Then we divide the bandwidth into two with two carrier frequencies. www.bookspar.com | Website for students | VTU NOTES

18. Frequency Shift Keying ( FSK /BFSK) In FSK, the Amplitude & phase of the carrier signal remain constant and frequency changes to represent the bit 1 or 0.The frequency of the signal during each bit duration is constant. www.bookspar.com | Website for students | VTU NOTES

19. Relationship between baud rate and bandwidth in FSK FSK spectra is a combination of two ASK spectra centered on f c0 and f c1. Frequencies are between fc0 – N baud /2 & f c1 +Nbaud /2. B.W = (f c1 +Nbaud /2) – ( f c0 - Nbaud /2 )= f c1- f c0 +Nbaud The bandwidth required for FSK transmission = the frequency shift+ baud rate of the signal www.bookspar.com | Website for students | VTU NOTES

20. Example Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half- duplex mode, and the carriers are separated by 3000 Hz. Solution For FSK BW = baud rate + f c1  f c0 BW = bit rate + fc1  fc0 = 2000 + 3000 = 5000 Hz www.bookspar.com | Website for students | VTU NOTES

21. Example Find the bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode. Solution Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = Baud rate + fc 1  fc 0 Baud rate = BW  (fc 1  fc 0 ) = 6000  2000 = 4000 But because the baud rate is the same as the bit rate, the bit rate is 4000 bps. www.bookspar.com | Website for students | VTU NOTES

22. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example Jan / Feb 2005 Derive the expression for the bandwidth of a Frequency Shift Keying ( FSK ) signal And hence determine the maximum bit rate of transmission, if the bandwidth of the medium is 12,000 Hz and the difference between the two carrier signal is 2000 Hz. Assume that the transmission is full-duplex mode. www.bookspar.com | Website for students | VTU NOTES

23. Phase Shift Keying ( PSK/BPSK ) In PSK, the Amplitude & frequency of the carrier signal remain constant and phase changes to represent the bit 1 or 0.The phase of the signal during each bit duration is same. In the figure a phase of 0 0 represents bit 0 and a phase of 180 0 represents bit 1. Bandwidth & Baud rate are same. www.bookspar.com | Website for students | VTU NOTES

24. PSK constellation OR Phase – state diagram The PSK is also called as 2-PSK, or BPSK, because two different phases ( 0 & 180 degrees) are used. PSK Constellation or Phase-State diagram Relation between phases & bits. www.bookspar.com | Website for students | VTU NOTES

25. The 4-PSK ( Quadrature PSK) method OR QPSK QPSK use four variations ( 0,90,180 & 270). Each phase shift represent 2 bits. www.bookspar.com | Website for students | VTU NOTES

26. The 4-PSK characteristics Constellation Diagram www.bookspar.com | Website for students | VTU NOTES

27. The 8-PSK characteristics 8-PSK has eight variations with a constant shift of 45 0. Each shift represent 3 bits of data. www.bookspar.com | Website for students | VTU NOTES

28. Example Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Solution For PSK the baud rate is the same as the bandwidth. There fore, baud rate is 5000. In 8-PSK the bit rate is 3 times the baud rate. So, the Bit rate = 5000 x 3 = 15,000 bps. www.bookspar.com | Website for students | VTU NOTES

29. Quadrature amplitude modulation is a combination of ASK and PSK. Note: Quadrature Amplitude Modulation www.bookspar.com | Website for students | VTU NOTES

30. The 4-QAM and 8-QAM constellations Phase change along x-axis, Amplitude change along y-axis. In QAM the number of phase shift is greater than the number of amplitude shift. Time - Domain Plots www.bookspar.com | Website for students | VTU NOTES

31. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 16-QAM constellations 4 3 www.bookspar.com | Website for students | VTU NOTES

32. Example A constellation diagram consists of 8 equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? Solution The constellation indicates 8-PSK with the points 45 degrees apart. Since 2 3 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is = 4800 / 3 = 1600 baud www.bookspar.com | Website for students | VTU NOTES

33. Example Compute the bit rate for a 1000-baud 16-QAM signal. Solution A 16-QAM signal has 4 bits per signal unit. Because, log 2 16 = 4. Thus, (1000)(4) = 4000 bps Example Compute the baud rate for a 72,000-bps 64-QAM signal.Solution A 64-QAM signal has 6 bits per signal unit. Because, log 2 64 = 6. Thus, 72000 / 6 = 12,000 baud www.bookspar.com | Website for students | VTU NOTES

34. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example January/February 2005 Calculate the number of Levels (Signal ) required to transmit the maximum bit rate. What is the BAUD rate. Example July/August 2005 Explain Quadrature Amplitude Modulation. What is the advantage? www.bookspar.com | Website for students | VTU NOTES

35. Telephone Modems Modem Standards Telephone lines can carry frequencies between 300 and 3300 Hz, giving a bandwidth of 3000 Hz. The effective bandwidth of a telephone line being used for data transmission is 2400 Hz, covering the range 600 – 3000 Hz. This bandwidth is called as base bandwidth. During data transmission, any bandwidth is to be modulated to base bandwidth. The device used for the purpose is called MODEM. www.bookspar.com | Website for students | VTU NOTES

36. Telephone line bandwidth The signal Bandwidth must be smaller than the cable bandwidth. www.bookspar.com | Website for students | VTU NOTES

37. We need to modulate the voice signal to use data bandwidth. Devices used to do so are called MODEMS Modem stands for modulator/demodulator. Modulator creates a band-pass analog signal from binary data. Demodulator recovers the binary data from the modulated signal. Note: www.bookspar.com | Website for students | VTU NOTES

38. Modulation/demodulation A modulator converts a digital signal into an analog signal using ASK, FSK, PSK or QAM. A demodulator converts an analog signal into a digital signal. www.bookspar.com | Website for students | VTU NOTES

39. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Modem Standards Most popular modems available are based on V – series standards defined by International Telecommunication Union- Telecommunication Standardization sector( ITU – T) 1)V.32 - This uses both modulation and encoding technique. The data stream is divided into 4-bit sections with a baud rate of 2400. The resulting speed is 4 x 2400 = 9600 bps. 2) V.32bis - It supports 14,400 bps transmission. 3) V.34bis – It supports a bit rate of 28,800 4) V.90 ( 56 k ) – It has a transmission capacity of 56,000 bps. Used for internet communication. www.bookspar.com | Website for students | VTU NOTES

40. Modulation of Analog Signals www.bookspar.com | Website for students | VTU NOTES

41. Analog-to-analog modulation Why modulation is needed? In Radio transmission, Govt assigns a narrow bandwidth to each radio station. The analog signal produced by each station is a low-pass signal. To be able to listen, the low-pass signal need to be modulated before transmission. www.bookspar.com | Website for students | VTU NOTES

42. Types of analog-to-analog modulation AM – Amplitude Modulation FM – Frequency Modulation PM – Phase Modulation www.bookspar.com | Website for students | VTU NOTES

43. Amplitude Modulation In amplitude modulation, the carrier signal is modulated so that its amplitude varies with the changing amplitude of modulating signal. The frequency and phase of the carrier signal remain the same. www.bookspar.com | Website for students | VTU NOTES

44. Amplitude modulation www.bookspar.com | Website for students | VTU NOTES

45. AM Bandwidth www.bookspar.com | Website for students | VTU NOTES

46. AM Band allocation AM stations are allowed the carrier frequencies between 530 kHz to 1700 kHz. Each stations f c must be separated from other at least by 10 kHz to avoid B/W over lapping. www.bookspar.com | Website for students | VTU NOTES

47. Example We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Solution An AM signal requires twice the bandwidth of the original signal: BW = 2 x 4 KHz = 8 KHz www.bookspar.com | Website for students | VTU NOTES

48. Frequency Modulation (FM) In Frequency Modulation, the Frequency of the Carrier Signal is modulated to that of the modulating signal. The Amplitude and Phase of the Carrier signal remains constant. FM Bandwidth The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BW t = 10 x BW m. www.bookspar.com | Website for students | VTU NOTES

49. Frequency modulation www.bookspar.com | Website for students | VTU NOTES

50. FM bandwidth www.bookspar.com | Website for students | VTU NOTES

51. FM stations are allowed Carrier frequencies between 88 MHz & 108 MHz. Stations must be separated by at least 200 kHz to keep B/W overlapping. www.bookspar.com | Website for students | VTU NOTES

52. FM band allocation www.bookspar.com | Website for students | VTU NOTES

53. Example An audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Solution An FM signal requires 10 times the bandwidth of the original signal: BW = 10 x 4 MHz = 40 MHz www.bookspar.com | Website for students | VTU NOTES

54. Multiplexing www.bookspar.com | Website for students | VTU NOTES

55. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Three users at a location communicate with three other users using three separate lines. This arrangement completely dedicates network resources. This becomes inefficient and expensive as the number of users increases. www.bookspar.com | Website for students | VTU NOTES

56. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Multiplexing is introduced when a transmission link has sufficient bandwidth to carry several connection lines. In the fig, a multiplexer combines the signals from three users in a single transmission link. www.bookspar.com | Website for students | VTU NOTES

57. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Multiplexing It is a technique that allows simultaneous transmission of multiple signals across a single data link. In multiplexed system n - lines share the bandwidth of one link. www.bookspar.com | Website for students | VTU NOTES

58. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Link, Channel & Line Link – means physical path. Channel – means the portion of the link that carries a transmission. Line-means portion of the channel. www.bookspar.com | Website for students | VTU NOTES

59. Dividing a link into channels In the Fig, 04 lines direct their transmission stream to a Multiplexer. MUX combines them into a single stream and transmitted. At the receiving, the stream is fed into a Demultiplexer. The DEMUX separates the stream back into its components and directs them into corresponding lines. www.bookspar.com | Website for students | VTU NOTES

60. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 There are different approaches of multiplexing the information from multiple connections to a single transmission line. www.bookspar.com | Website for students | VTU NOTES

61. Categories of multiplexing There are 3 types of Multiplexing. 1)Frequency Division Multiplexing ( FDM ) 2) Wave Division Multiplexing (WDM) 3) Time Division Multiplexing ( TDM ) www.bookspar.com | Website for students | VTU NOTES

62. 1) Frequency Division Multiplexing ( FDM) In FDM, Signals generated by each sending device modulate different carrier frequencies. These modulated signals are then combined into a single composite signal that can be transported by the link. Guard Band – is the strip of unused Bandwidth to prevent signals from overlapping. In the figure, the transmission path is divided into three channels. www.bookspar.com | Website for students | VTU NOTES

63. FDM process Each telephone generates a signal of similar frequency. Inside the multiplexer, these similar signals are modulated onto different carrier frequencies f1, f2, f3. The resulting modulated signals are then combined in to a single composite signal that is sent out over a media. www.bookspar.com | Website for students | VTU NOTES

64. FDM demultiplexing example The demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent component signals. The individual signals are then passed to a demodulator that separates them from their carriers and passes them to the receivers. www.bookspar.com | Website for students | VTU NOTES

65. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example VTU 2006 Define Multiplexing. Explain FDM. www.bookspar.com | Website for students | VTU NOTES

66. Example-1 Assume that a voice channel occupies a bandwidth of 4 KHz. We need to combine three voice channels into a link with a bandwidth of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain with out the use of guard bands. Solution Modulate each of the three voice channels to a different bandwidth. The diagram is as shown in Figure www.bookspar.com | Website for students | VTU NOTES

67. Figure www.bookspar.com | Website for students | VTU NOTES

68. Example-2 Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference? Solution For five channels, need at least four guard bands. There fore, minimum required bandwidth = 5 x 100 + 4 x 10 = 540 KHz. The diagram is as shown in Figure www.bookspar.com | Website for students | VTU NOTES

69. Figure www.bookspar.com | Website for students | VTU NOTES

70. Example-3 Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM Solution The satellite channel is analog. Divide 1MHz into four channels, each channel having a 250-KHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. One solution is 16-QAM modulation. Figure shows one possible configuration. www.bookspar.com | Website for students | VTU NOTES

71. Example www.bookspar.com | Website for students | VTU NOTES

72. The FDM hierarchy To maximize the efficiency, Telephone connections use hierarchy of multiplexed signals from low-band width lines to high bandwidth lines. The hierarchy is made up of Groups, Super groups, Master groups, and Jumbo group. www.bookspar.com | Website for students | VTU NOTES

73. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 FDM was introduced in the year 1930s. The basic analog multiplexer combines 12 voice channels in one line. Each voice signal occupies about 4 kHz. The channels are assigned with 4 kHz of bandwidth to provide guard bands between channels. The multiplexer modulates each voice signal so that it occupies a 4 kHz slot in the band between 60 & 108 kHz. The combined signal is called a group. A collection of 5 groups is called Super group. A In a super group 5 groups, each of bandwidth 48 kHz, into the frequency band from 312 to 552 kHz. 10 super groups be multiplexed to master group of 600 voice signals. This occupies a band 564 to 3084 kHz. Examples of FDM Broadcast radio- FM, AM, & television with 10 kHz, 200 kHz, & 6 MHz Cellular Phones- A pool of frequency slots of 30 kHz each are shared by the users with in a geographic cell. www.bookspar.com | Website for students | VTU NOTES

74. WDM Wave Division Multiplexing www.bookspar.com | Website for students | VTU NOTES

75. Wave Division Multiplexing WDM is used in the transmission of OFC networks. OFC has high data rate capacity. Each telephone generate signals of similar frequency. Inside the Wave division multiplexer, these signals are modulated onto different narrow band carrier frequencies f1, f2, f3. The resulting modulated signals are then combined into a single Optical signal of wider band of light and transported over OFC channel. www.bookspar.com | Website for students | VTU NOTES

76. Prisms in WDM multiplexing and demultiplexing A prism has the property of Combining and splitting of light sources. A prism bends a beam of light based on the angle of incidence and frequency. Using this technique, a multiplexer can be made to combine several input beams into one output beam of a wider band of frequencies. A demultiplexer can also be made to reverse the process. www.bookspar.com | Website for students | VTU NOTES

77. TDM TDM Time – Division Multiplex www.bookspar.com | Website for students | VTU NOTES

78. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 TDM is a digital multiplexing technique. Each connection occupies a portion of time in the link. If the source signal is an analog, it can be changed to digital data and then multiplexed using TDM. In synchronous TDM, the data flow of each input connection is divided into units. Each unit occupies one input time slot. Combination of units is called a frame. www.bookspar.com | Website for students | VTU NOTES

79. Time – Division Multiplex (TDM) In the figure, connections 1,2,3, & 4 occupy a portion of the link. www.bookspar.com | Website for students | VTU NOTES

80. TDM frames - For n- input connections, & T- input time, each connection gets a time slot of T/n. During that time a connection gets a chance of transmitting data for T/n of time, called one unit. The link combines these units into a frame. www.bookspar.com | Website for students | VTU NOTES

81. Example Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (i) the duration of 1 bit before multiplexing, (ii) the transmission rate of the link, (iii) the duration of a time slot, and (iv) the duration of a frame? Solution i) The duration of 1 bit = 1/ (1 Kbps) = 0.001 s = 1 ms ii) The rate of the link= 4 times the rate of connection= 4 Kbps. iii) The duration of each time slot= 1/4 the bit duration = 1/4 ms = 250  s. iv) The duration of a frame= ¼ +1/4 + 1/4 +1/4 = 1 ms. www.bookspar.com | Website for students | VTU NOTES

82. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example VTU Jan 2006 In a TDM system, four channels of 1 kbps each are multiplexed together. Each time slot, called as a unit, carries 1 bit of information. Data from each time slot is transmitted as a frame. Find: i) Transmission rate of the output link of the multiplexer. Ii) The duration of a frame. www.bookspar.com | Website for students | VTU NOTES

83. Interleaving in TDM TDM can be visualized as two fast rotating switches, one on the multiplexing side and the other on the demultiplexing side. The switches are synchronized and rotate at the same speed but in the opposite directions. On the multiplexing side, as the switch opens in front of a connection, that connection has the opportunity to send a unit onto the path. This process is called interleaving. On the demultiplexing side, as the switch opens in front of a connection, that connection has the opportunity to receive a unit from the path. www.bookspar.com | Website for students | VTU NOTES

84. Example Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate of the link. Solution (Next slide) www.bookspar.com | Website for students | VTU NOTES

85. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 1)Each Frame carries 1 byte from each channel. So, the size of each frame = 4 bytes = 32 bits. 2)Duration of a Frame = 1/ 100 bytes per sec = 0.01 S 3) Frame rate = 4 x100/4 = 100 frames. 4) Bit rate = Number of frames transmitted per sec = 100 frames = 100 x 32 = 3200 bps. www.bookspar.com | Website for students | VTU NOTES

86. Example: A multiplexer combines four 100- Kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate, the frame duration, bit rate &the bit duration? Solution (Next slide) www.bookspar.com | Website for students | VTU NOTES

87. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Each Channel send 100 kbps = 100,000 bits / s The link carries = 100,000/ 2= 50,000 frames per Sec. Frame rate = 50,000 Frame duration = 1/ 50,000 = 20 micro Sec Bit rate = 50000 x 8 (frame size ) = 400,000 bits = 400 Kbps Bit duration = 1/ 400,000 = 2.5 micro Sec www.bookspar.com | Website for students | VTU NOTES

88. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example VTU July 2007 Define Multiplexing. How does TDM multiplex & Demultiplex the signals. Exam: VTU July 2007 Four sources each create 250 characters per second. If the interleaved unit is a character and one synchronization bit is added to each frame, find i) the data rate of each source ii) Duration of each character in each source iii) The frame rate iv) Duration of each frame v) Number of bits in each frame vi) The data rate of the link. www.bookspar.com | Website for students | VTU NOTES

89. Framing bits & Synchronization If the multiplexer and demultiplexer are not synchronized, a bit belongs to one channel may be received by the wrong channel. For this reason, one or more synchronization bits are added to the beginning of each frame. These bits are called framing bits. These framing bits follow a pattern ( Alternating 0 & 1) from frame to frame. This helps the demultiplexer to synchronize with the incoming stream, & it can separate time slots accurately. www.bookspar.com | Website for students | VTU NOTES

90. Example We have four sources, each creating 250 characters per second. If the interleaved unit is a character and one synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. Solution See next slide. www.bookspar.com | Website for students | VTU NOTES

91. Solution (continued) We can answer the questions as follows: 1. The data rate of each source is 2000 bps = 2 Kbps. 2. The duration of a character = 1/250 s = 4 ms. 3. The frame rate = 250 frames per second. 4. The duration of each frame = 1/250 s, or 4 ms. 5. No. of bits in each frame = 4 x 8 + 1 = 33 bits. 6. The data rate of the link = 250 x 33= 8250 bps. www.bookspar.com | Website for students | VTU NOTES

92. McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 BIT PADDING Suppose device A could use one time slot, while the faster device B could use two. Since the time slot is fixed, the different data rate must be integer multiples of each other. When the speeds are not integer multiples of each other, they can be made to behave as if they were, by a technique called Padding. For example, One device with a bit rate of 2.55 times that of the other devices. By adding enough bits, make three times the other devices. The extra bits are then discarded by the demultiplexer. www.bookspar.com | Website for students | VTU NOTES

93. Digital Signal hierarchy Lower order multiplexed signals are called Tributaries. www.bookspar.com | Website for students | VTU NOTES