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### Chap 05 dsn

1. 1. 29/10/2012 Computer Communications Ch # 5 – Analog Signals By: Faiza Tariq 31 October 2012 1
2. 2. 29/10/2012 Modulation of Digital Data Digital-to-Analog Conversion Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation Bit/Baud Comparison Modulation Types  Amplitude Modulation  If bit = 0 x(t) = 0  If bit = 1 x(t) = sin(2πft)  Frequency Modulation (FSK)  If bit = 0 x(t) = sin(2πf1t)  If bit = 1 x(t) = sin(2πf2t)  Phase Modulation  If bit = 0 x(t) = sin(2πft)  If bit = 1 x(t) = sin(2πft-θ) 2
3. 3. 29/10/2012 Figure 5.1 Digital-to-analog modulation Figure 5.2 Types of digital-to-analog modulation 3
4. 4. 29/10/2012 Bit Rate vs. Baud Rate  Baud is like car and bit rate like passenger  If 1000 cars go from one place to another place, carrying only passenger i.e. driver then 1000 passengers could be transported  However if each car carries four passenger, then 4000 passengers are transported  No. of cars not no. of passengers determines traffic, i.e. may require wider high way  No. of bauds determines the required bandwidth, not number of bits Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate. Baud Rate  Baud rate is a measure of the number of discrete signals that can be transmitted (or received) per unit of time  A modem’s baud rate measures the number of signals that it is capable of transmitting (or receiving) per second  Baud rate represents the number of times per second that a modem can modulate (or demodulate) the carrier signal to represent bits  Although baud rate and bit rate are sometimes used interchangeably to refer to modem data transfer speeds, these are only identical when each signal transmitted (or received) represents a signal bit  A modem’s bit rate is typically higher than its baud rate because each signal transmitted or received may represent a combination of two or more bits 4
5. 5. 29/10/2012 Bit Rates and Bandwidth  The bandwidth of an analog channel is the difference between the minimum and maximum frequencies it can carry  A voice-grade dial-up circuit can transmit frequencies between 300 and 3400 Hz and thus has a bandwidth of 3100 Hz  For digital circuits, bandwidth is a measure of the amount of data that can be transmitted per unit. Bits per second (bps) is the most widely used measure for digital circuits  Over time, bit rates (bps) have also become on of the key measures of modem performance (e.g. a 56 Kbps modem)  However, modem bit rates are not necessarily an accurate reflection of their data throughput rates 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 the carrier signal or carrier frequency  The receiver device is tuned to the frequency of the carrier signal  Digital information then modulates the signal by changing one or more properties called modulation or shift keying 5
6. 6. 29/10/2012 Amplitude Shift Keying (ASK)  Two binary values are represented by two different amplitudes of the carrier frequency  Both frequency and phase remains constant while amplitude changes  Which voltage represents 1 and 0 is left on designers  ASK transmission is highly susceptible to noise interference because of unintentional voltages introduced onto a line (heat up) ASK  A popular ASK technique is called on/off keying (OOK), one bit represented by no voltage and one by a voltage 6
7. 7. 29/10/2012 Frequency Shift Keying (FSK)  Binary values are represented by two different frequencies  Resulting signal is:  S(t) = A sin (2Лf1t) binary 1  S(t) = A sin (2Лf2t) binary 0  FSK avoids most of the noise problem as ASK, as no voltage spikes, just frequency change  Limiting factors of FSK are physical capabilities of the carrier. Phase Shift Keying (PSK)  Phase of the signal is shifted to represent binary data  It is’nt susceptible to noise degradation that affects ASK  The main advantage of phase modulation is that only single frequency to send both zeros and ones in both direction (two phases in each total four phases required) 7
8. 8. 29/10/2012 Phase Shift Keying (PSK) Bit 0 1 Phase 0 180 Tribit 000 001 010 011 100 101 111 Phase 0 45 90 135 180 270 315 Bit Dibit 00 01 10 11 Phase 0 90 180 270 010 011 001 000 100 101 111 Baud Rate = n Bit Rate = n 0 0 1 0 1 0 0 0 1 0 1 0 1 1 0 0 Dibit 00 Baud Rate = n 10 10 Tribit 001 Quadbit 0011 00 10 Bit Rate = 2n 10 Baud Rate = n 000 111 Baud Rate = n 0001 1110 11 10 Bit Rate = 3n 101 110 Bit Rate = 4n 101 1101 8
9. 9. 29/10/2012 Figure 5.1 Digital-to-analog conversion 9
10. 10. 29/10/2012 Figure 5.2 Types of digital-to-analog conversion Note Bit rate is the number of bits per second. Baud rate is the number of signal elements per second. In the analog transmission of digital data, the baud rate is less than or equal to the bit rate. 10
11. 11. 29/10/2012 Example 5.1 An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Example 5.2 An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need? Solution In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L. 11
12. 12. 29/10/2012 Figure 5.3 Binary amplitude shift keying Figure 5.4 Implementation of binary ASK 12
13. 13. 29/10/2012 Example 5.3 We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1? Solution The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1). Example 5.4 In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction. 13
14. 14. 29/10/2012 Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4 Figure 5.6 Binary frequency shift keying 14
15. 15. 29/10/2012 Example 5.5 We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1? Solution This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means Figure 5.7 Bandwidth of MFSK used in Example 5.6 15
16. 16. 29/10/2012 Example 5.6 We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth. Solution We can have L = 23 = 8. The baud rate is S = 3 MHz/3 = 1000 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1000 = 8000. Figure 5.8 shows the allocation of frequencies and bandwidth. Figure 5.8 Bandwidth of MFSK used in Example 5.6 16
17. 17. 29/10/2012 Figure 5.9 Figure 5.10 Binary phase shift keying Implementation of BASK 17
18. 18. 29/10/2012 Figure 5.11 QPSK and its implementation Example 5.7 Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0. Solution For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz. 18
19. 19. 29/10/2012 Figure 5.12 Concept of a constellation diagram Example 5.8 Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals. Solution Figure 5.13 shows the three constellation diagrams. 19
20. 20. 29/10/2012 Figure 5.13 Three constellation diagrams Note Quadrature amplitude modulation is a combination of ASK and PSK. 20
21. 21. 29/10/2012 Figure 5.14 Constellation diagrams for some QAMs 5-2 ANALOG AND DIGITAL Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to us. Topics discussed in this section: Amplitude Modulation Frequency Modulation Phase Modulation 21
22. 22. 29/10/2012 Analog to Analog Conversion  A higher frequency may be needed for effective transmission of analog signal  Analog data (Voice) is combined (added) with carrier wave signal (carrier signal is used to convey the audio data)  Principal techniques used  Amplitude Modulation (AM)  Frequency Modulation (FM) Figure 5.15 Types of analog-to-analog modulation 22
23. 23. 29/10/2012 Figure 5.16 Amplitude modulation Note The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B. 23
24. 24. 29/10/2012 Figure 5.17 AM band allocation Note The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B. 24
25. 25. 29/10/2012 Figure 5.18 Frequency modulation Figure 5.19 FM band allocation 25
26. 26. 29/10/2012 Figure 5.20 Phase modulation Note The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + β)B. 26
27. 27. 29/10/2012 Amplitude Modulation (AM) Amplitude Modulation (AM) (as shown in (a) Analog Modulation 27
28. 28. 29/10/2012 Amplitude Modulation (AM)  By changing amplitude of the carrier signal  The frequency and the phase of the carrier remains same  AM stations are allowed carrier frequencies anywhere between 530 and 1700 KHz  However each station’s carrier frequency must be separated from those on either side of it by atleast 10 KHz to avoid interference.  The total bandwidth required can be determined from the bandwidth of the audio (modulated) signal  Bw = 2 X Bw Frequency Modulation (FM)  In FM transmission, the frequency of the carrier is modulated.  The bandwidth of the FM signal is 10 times the bandwidth of the modulated (audio) signal.  Bw = 10 X Bw  The bandwidth of an audio signal (speech and music) broadcast in stereo is 15 KHz.  Each FM station therefore needs a minimum bandwidth of 150 KHz. 28
29. 29. 29/10/2012 DTE – DCE Interface  Data Terminal Equipment (DTE):  Data Terminal Equipment include (DTE) includes any unit that function either as a source of binary data or destination of digital data. E.g. PC, printer or fax.  Data Circuit Transmitting Equipment or Data Communication Equipment (DCE):  any functional unit that transmits or receives data in the form of an analog or digital signal  DCEs are simply instruments used to allow DTEs to communicate.  At physical layer level DCE takes data generated by a DTE, converts them to appropriate signal and transmits them across the network through a communication channel e.g. modems or multiplexers Pixelization and Binary Representation  Used in digital fax, bitmapped graphics 1-bit code: 00000000 00111100 01110110 01111110 01111000 01111110 00111100 00000000 29
30. 30. 29/10/2012 Modems  Modulator/Demodulator ITU-T Bell Baud Rate Bit Rate Modulation Technique v.21 103 300 300 FSK v.22 212 600 1200 PSK v.23 202 1200 1200 FSK v.26 201 1200 2400 PSK v.27 208 1600 4800 PSK Data Compression  In some cases data is compressed prior to transmission  For a call if transmission time is reduced, will automatically reduce the call charges.  Advantage minimum use of telephone line or less internet charges  Some modems, called intelligent modems offer compression feature, which select the algorithm to suit the type of the data being transmitted. 30
31. 31. 29/10/2012  Intelligent modems attempt to send fewer bits by searching for repeated pattern & sending some type of shorthand notation.  Receiving modem translates the shorthand back to the original bits creating the illusion that two modems are operating at higher speed.  ITU-T V.42 bis standard can theoretically compress to data to one – fourth its original size under ideal conditions  Since it is hard to predict if the data transmitted will have repeated pattern, the compression at this ideal level is not achieved Character Suppression  When a frame consists of characters and a character is repeated twice or more times, may be replaced by another  The control device at the transmitter scans the frame contents prior to transmission, if contiguous string of three or more characters is located, replaces these sequence  ¶¶¶¶¶¶¶ TO  7 ¶ TO 31
32. 32. 29/10/2012 Huffman Coding  It exploits that not all the characters in a transmitted frame occur with the same frequency. E.g.  In a frame comprising strings of character, certain characters occur more often than others  Instead of using fixed number of bits per character, we use a different encoding scheme in which the most common characters are encoded using fewer bits than less frequent characters  AAAABBCD  4 X 1 + 2 X 2 + 1 X 3 = 14 bits v.34  Standard speed 28,800 bps  With data compression data rate can be twice the standard speed. 32
33. 33. 29/10/2012 Another Look at Compression _ With 4:1 Compression, a V.34 Modem Can Receive Data at 115.2 kbps from the PC _ However the ~30 kbps limit of the phone system is not exceeded. Still transmit at 33.6 kbps. ~35 kbps Maximum Compression in Modem 115.2 kbps 33.6 kbps 56K Modem  Bit rate = 56000 bps  Downloading max. speed = 56000 bps, as ISP uses digital line at other end.  Uploading max speed = 33,600 bps Digital-Analog Conversion Upstream Analog-Digital Conversion Downstream Upstream Digital-Analog Downstream No analog – digital conversion on downstream link makes download speed fast 33
34. 34. 29/10/2012 56 K Modem 56 K Standards (ITU-T)  V.90  56 K dial-up modem  V.92  Less handshaking time  Supports CLI and Call Waiting, by sending IRQ  The quick connect method reduces the negotiation time from over 20 seconds to about 10 seconds.  The second additional feature is a Modem-on-Hold™ (MOH) feature. This codifies a method for the central site modem to request the client modem to go on hold, or vice versa, and is a mechanism whereby call-waiting tones can be better survived by voice-band dial-up 34
35. 35. 29/10/2012 Modem Speed Standards 28 _ Most data modems are also fax modems _ _ V.14 V.29 14.4 kbps 9,600 bps Modem Training _ When two modems connect 30 _ They exchange information about themselves _ This allows them to settle upon the highest supported standard in each category _ They also do line testing during this period _ This is called handshaking or training _ Training takes 10 to 30 seconds 35
36. 36. 29/10/2012 Modem Intelligence _ Computer Can Send Commands to Modem _ _ 31 Dial a number, including how long to wait, compression, etc. Called intelligent modems _ Hayes Developed the first Command Set _ _ _ Most modems follow the same command set We call them “Hayes compatible” Commands start with “AT” _ Other Standards for Fax Modems _ Class 1 and Class 2: extensions to Hayes Telephone Bandwidth is Limited 36 _ Speed is Limited _ Maximum speed is related to bandwidth (Shannon’s Law) _ Maximum speed for phone lines for transmission is a little over 30 kbps _ So modems can’t get much faster 36
37. 37. 29/10/2012 NULL Modem  To connect 2 PCs (DTE) with each other, modems are not needed  An interface is needed to connect these two DTEs to handle exchange  The solution by EIA standard – NULL Modem provides DTE – DTE without DCE TX RX 2 2 3 3 TX RX Example 1 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 37
38. 38. 29/10/2012 Example 2 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 Figure 5.3 ASK 38
39. 39. 29/10/2012 Figure 5.4 Relationship between baud rate and bandwidth in ASK Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is halfduplex. 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. 39
40. 40. 29/10/2012 Example 4 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. Example 5 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 for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5). fc (forward) = 1000 + 5000/2 = 3500 Hz fc (backward) = 11000 – 5000/2 = 8500 Hz 40
41. 41. 29/10/2012 Figure 5.5 Solution to Example 5 Figure 5.6 FSK 41
42. 42. 29/10/2012 Figure 5.7 Relationship between baud rate and bandwidth in FSK Example 6 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 42
43. 43. 29/10/2012 Example 7 Find the maximum 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. Figure 5.8 PSK 43
44. 44. 29/10/2012 Figure 5.9 Figure 5.10 PSK constellation The 4-PSK method 44
45. 45. 29/10/2012 Figure 5.11 The 4-PSK characteristics Figure 5.12 The 8-PSK characteristics 45
46. 46. 29/10/2012 Figure 5.13 Relationship between baud rate and bandwidth in PSK Example 8 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. Solution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. 46
47. 47. 29/10/2012 Example 9 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, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. Note: Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved. 47
48. 48. 29/10/2012 Figure 5.14 The 4-QAM and 8-QAM constellations QAM 48
49. 49. 29/10/2012 Figure 5.15 Time domain for an 8-QAM signal Figure 5.16 16-QAM constellations 49
50. 50. 29/10/2012 Figure 5.17 Bit and baud Table 5.1 Bit and baud rate comparison Modulation Units Bits/Baud Baud rate Bit Rate Bit 1 N N 4-PSK, 4-QAM Dibit 2 N 2N 8-PSK, 8-QAM Tribit 3 N 3N 16-QAM Quadbit 4 N 4N 32-QAM Pentabit 5 N 5N 64-QAM Hexabit 6 N 6N 128-QAM Septabit 7 N 7N 256-QAM Octabit 8 N 8N ASK, FSK, 2-PSK 50
51. 51. 29/10/2012 Example 10 A constellation diagram consists of eight 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 Example 11 Compute the bit rate for a 1000-baud 16-QAM signal. Solution A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus, (1000)(4) = 4000 bps 51
52. 52. 29/10/2012 Example 12 Compute the baud rate for a 72,000-bps 64-QAM signal. Solution A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000 / 6 = 12,000 baud 5.2 Telephone Modems Modem Standards 52
53. 53. 29/10/2012 Note: A telephone line has a bandwidth of almost 2400 Hz for data transmission. Figure 5.18 Telephone line bandwidth 53
54. 54. 29/10/2012 Note: Modem stands for modulator/demodulator. Figure 5.19 Modulation/demodulation 54
55. 55. 29/10/2012 Figure 5.20 The V.32 constellation and bandwidth Figure 5.21 The V.32bis constellation and bandwidth 55
56. 56. 29/10/2012 Figure 5.22 Traditional modems Figure 5.23 56K modems 56
57. 57. 29/10/2012 5.3 Modulation of Analog Signals Amplitude Modulation (AM) Frequency Modulation (FM) Phase Modulation (PM) Figure 5.24 Analog-to-analog modulation 57
58. 58. 29/10/2012 Figure 5.25 Types of analog-to-analog modulation Note: The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BWt = 2 x BWm. 58
59. 59. 29/10/2012 Figure 5.26 Amplitude modulation Figure 5.27 AM bandwidth 59
60. 60. 29/10/2012 Figure 5.28 AM band allocation Example 13 We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations. Solution An AM signal requires twice the bandwidth of the original signal: BW = 2 x 4 KHz = 8 KHz 60
61. 61. 29/10/2012 Note: The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BWt = 10 x BWm. Figure 5.29 Frequency modulation 61
62. 62. 29/10/2012 Figure 5.30 FM bandwidth Note: The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of 150 KHz. The FCC requires the minimum bandwidth to be at least 200 KHz (0.2 MHz). 62
63. 63. 29/10/2012 Figure 5.31 FM band allocation Example 14 We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Ignore FCC regulations. Solution An FM signal requires 10 times the bandwidth of the original signal: BW = 10 x 4 MHz = 40 MHz 63