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UNIT - 3

UNIT - 3. Analog Transmission & Multiplexing Digital – to – Analog conversion, Analog – to – analog conversion, Multiplexing, Spread Spectrum. Analog Transmission. Modulation of Digital Data. Digital-to-analog modulation. Modulation

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UNIT - 3

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  1. UNIT - 3 www.bookspar.com | Website for students | VTU NOTES

  2. Analog Transmission & Multiplexing Digital – to – Analog conversion, Analog – to – analog conversion, Multiplexing, Spread Spectrum. www.bookspar.com | Website for students | VTU NOTES

  3. AnalogTransmission 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 • 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 & Baud Rate (Signal elements) 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. Baudrate is less than or equal to the bit rate. 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 = 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. 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 2L L = 2r L = 28 L= 256 www.bookspar.com | Website for students | VTU NOTES

  11. 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 fc – N baud /2 + fc +Nbaud /2, fc is the carrier frequency. In ASK, BW = (1+d) x Nbaud, where Nbaud 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 fc0 and fc1. Frequencies are between fc0 – N baud /2 & fc1 +Nbaud /2. B.W = (fc1 +Nbaud /2) – ( fc0 - Nbaud /2 )= fc1- fc0 +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 + fc1- fc0 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 + fc1- fc0 Baud rate = BW - (fc1- fc0 ) = 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. 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 00 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 constantshift 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 Note: Quadrature amplitude modulation is a combination of ASK and PSK. 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. 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 23 = 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, log216 = 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. 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. Note: 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. 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. Modem Standards • Most popular modems available are based on V – series standards defined by International Telecommunication Union- Telecommunication Standardization sector( ITU – T) • 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 fc 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: BWt = 10 x BWm. 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

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