Analog Transmission System

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1 Analog Transmission System

2 Q ช องทางการส งส ญญาณแบบใดล ะ ถ อว าเป นช องส งส ญญาณ Analog A ช อง wire หร อ wirelss น า

3 Analog Transmission Media WIRE สายโทรศ พท (ADSL,VDSL) Coaxial Cable (DOCSIS)

4 Analog Transmission Media WIRELESS

5 Q แล วด อย างไรล ะว าเป นช องส ง ส ญญาณ Analog

6 Bandpass Channel Most of Analog Transmission Media is Bandpass Channel Cannot transmit Digital Signal Low frequency data pattern will be all lost Need Signal Conversion

7 Q ถ าอยากส งส ญญาณ Digital ไปบน ช องส ญญาณ Analog ล ะ ทาไงด A Conversion ส

8 Signal Conversion Digital signal with inf bandwidth Analog signal with limited bandwidth

9 Q Bandwidth ของ Analog Channel ต างๆท เราใช งานจร งม น เป นย งไงบ าง

10 A Mobile Channel ก อนม ย ใกล ต วด

11 Mobile Frequency Spectrum

12 Cellular communications GSM900 (AIS) MHz (uplink) MHz (downlink) 124 RF channels (channel numbers 1 to 124) spaced at 200 khz GSM1800/CDMA1800 (DTAC/AIS/TRUE) 1,710 1,785 MHz (uplink) 1,805 1,880 MHz (downlink) 374 channels (channel numbers 512 to 885) spaced at 200 khz CDMA2100 (3G/4G)(DTAC/AIS/TRUE) 1,885-2,025 MHz (uplink) 2,110-2,200 MHz(downlink) Spaced at 5 MHz

13 A TV Channel ล ะ

14 Television Channel

16 A ช อง Satellite บ าง

17 Terrestrial / SATELLITE microwave

18 A ช องน ๆ Bluetooth ไง ใครๆก ใช

19 Bluetooth Operate in noisy radio frequency environments GHz GHz omni-directional Point-to-multipoint frequency-hopping scheme 79 hops (RF channels) 1 MHz apart. bandwidth is reduced in Japan, France and Spain The maximum frequency hopping rate is 1600 hops/s link range Usually 10 centimeters to 10 meters Can be extended to more than 100 meters by increasing the transmit power Small amounts of data 1Mbps over short distances (up to 10 meters).

20 A NFC: near-field communication

21 NFC: near-field communication

22 NFC: near-field communication

23 A RFID น ก เคยได ย น

24 RFID: radio frequency identification

25 A เอาอ นน ด วย ช อง WiFi

26 WiFi Band

27 A ช องว ทย (Radio Station) ล ะ

28 Radio Frequency

29 Frequency Utilization Bandwidth Limitation Transmitting Direct Digital Signal is not possible

30 Figure 5.1 Digital-to-analog conversion 5.30 Analog Selected Channel Frequency

31 Digital Modulation Digital to Analog Conversion Changing Analog Sinewave signal properties to represent digital data What sinewave properties to be changed? Amplitude Frequency Phase

32 Q แล วม นทาไงอ ะ ท ใช Analog Signal property มาเป นต วแทนข อม ล Digital A ก ทา modulation ไง

33 What property has been changed?

34 Q Modulation ม นทาไงล ะ งงอย ด A ค อม นทาได หลายแบบตาม Analog Property

35 Figure 5.2 Types of digital-to-analog conversion 5.35 เปล ยน Amplitude เปล ยน frequency เปล ยน Phase เปล ยน Amplitude & Phase

36 ASK (Amplitude Shift Keying) Changing Amplitude of channel frequency 2-ASK (Binary ASK) 0 -> A0 1 -> A1 4-ASK 00 -> A0 01 -> A1 10 -> A2 11 -> A3

37 5.37 Figure 5.3 Binary amplitude shift keying (Binary ASK, On-Off Keying (OOK) RFID/NFC Transmission

38 Figure 5.4 Implementation of binary ASK 5.38

39 cos cos = ½(cos ( + ) + cos ( - )) m(t) cos(2 f c t) cos(2 f c t) = ½[m(t)(cos(2 f c t+2 f c t)+cos(2 f c t-2 f c t))] = ½(m(t)(cos(4 f c t)+cos(0))) Amplitude Shift Keying

40 Signal rate / Data Rate / Bandwidth with ASK Signal rate = # signal unit / sec (baud) Data Rate = #bits/sec (bps) N = (#bits/ signal unit) x (#signal unit/s) N = r x baud N = r x S 2-ASK (OOK) r = 1 (bit/signal unit)

41 4-ASK r = 2 (bits/signal unit)

42 Signal rate / Data Rate / Bandwidth with ASK Signal rate = # signal unit / sec (baud) ข นก บ Channel Bandwidth BW = (1+d)S 0<= d <= 1 S <= BW <= 2S (BW/2) <= S <= BW

43 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 f c = 250 khz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1). N = r x S 5.43

Example 5.4 5.44 In data communications, we normally use full-duplex links with communication in both directions.

44 Example 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.

45 Summarize: Amplitude Shift Keying Bit representation Changing Amplitude of Carrier Signal Benefit Simple (normally used for fiber optic / RFID) Require Less Bandwidth Disadvantage Easily effected by noise

46 FSK (Frequency Shift Keying) Changing Frequency of channel frequency 2-FSK (Binary FSK) 0 -> f0 1 -> f1 4-FSK 00 -> f0 01 -> f1 10 -> f2 11 -> f3

47 Figure 5.6 Binary frequency shift keying 5.47 S <= Sub-carrier spacing (2Df) <= 2S

48 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 5.48 S <= Sub-channel spacing (2Df) <= 2S 0<= d <= 1

49 Figure 5.7 Bandwidth of MFSK used in Example

50 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 = 2^3 = 8. The baud rate is S = N/r = 3 Mbps/3 (bps / bit-per-signal unit) = 1 Mbaud. S <= sub-channel spacing <= 2S Choose d=0 -> (2Δf = S = 1 MHz) This means that the carrier frequencies must be 1 MHz apart.

Figure 5.8 Bandwidth of MFSK used in Example 5.6 5.

51 Figure 5.8 Bandwidth of MFSK used in Example The bandwidth is B = = Figure 5.8 shows the allocation of frequencies and bandwidth.

52 PSK (Phase Shift Keying) Changing Frequency of channel frequency 2-PSK (Binary PSK) 0 -> zeta_0 1 -> zeta_1 4-PSK 00 -> zeta_0 01 -> zeta_1 10 -> zeta_2 11 -> zeta_3

53 Figure 5.9 Binary phase shift keying 5.53

54 Figure 5.10 Implementation of BPSK 5.54

55 Phase Shift Keying Acos(2 fct ); binary '0' st () Acos(2 fct); binary '1' X cos( 2 f t) cos(2 f t ) C C Binary Data [0, 1] X (Digital Signal) [-1, 1] X cos(2 f t) cos(2 ft ) C C

56 Phase Shift Keying X cos(2 f t)cos(2 f t) X f t f t f t f t C C 1 cos(2 C 2 C ) cos(2 C 2 C ) 2 1 X cos 4 fct X 2

57 Figure 5.11 QPSK and its implementation 5.57 MSB LSB Phase combination of 2 bits

58 Example 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.

59 5.59 Figure 5.12 Concept of a constellation diagram

60 Example Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals. Solution Figure 5.13 shows the three constellation diagrams.

61 Figure 5.13 Three constellation diagrams 5.61

62 DATA COMMUNICATIONS, Department of Computer Engineering, KMITL B. A. Forouzan, Data Communications and Networking, 4th Quadrature Amplitude Modulation AS K FS K PS K QA 62 M Bit representation Combination of ASK and PSK Changing Amplitude & Phase of Career Signal One bit, One signal unit Benefit Less effected by noise compared to ASK Require less bandwidth Disadvantage Complex demodulation technique

63 5.63 Note Quadrature amplitude modulation is a combination of ASK and PSK.

64 Figure 5.14 Constellation diagrams for some QAMs 5.64

01076251 DATA COMMUNICATIONS, Department of Computer

65 DATA COMMUNICATIONS, Department of Computer Engineering, KMITL B. A. Forouzan, Data Communications and Networking, 4th Quadrature Amplitude Modulation AS K FS K PS K QA 65 M Quadrature amplitude modulation is a combination of ASK and PSK.

66 Quadrature Amplitude Modulation

67 Quadrature Amplitude Modulation ITU-T recommendation OSI recommendation

68 Quadrature Amplitude Modulation

Quadrature Amplitude Modulation 11 01 00 10 01 X cos(2 fct) Y sin(2 fct) cos Bcos A sinbsin A cos( A B) X Binary Data [0, 1] + )) X (Digital Signal) [-1, 1] Y Binary Data [0, 1] Y (Digital Signal)

69 Quadrature Amplitude Modulation X cos(2 fct) Y sin(2 fct) cos Bcos A sinbsin A cos( A B) X Binary Data [0, 1] + )) X (Digital Signal) [-1, 1] Y Binary Data [0, 1] Y (Digital Signal) [-1, 1] Xcos + Ysin cos cos +sin sin =cos( - ) cos + -1 sin cos( +135) cos + 1 sin cos( -135) cos + -1 sin cos( +45) cos + 1 sin cos( -45) +45

70 Quadrature Amplitude Modulation (x,y)=(0,0) x=0=>-a -Acos( ) X cos(2 f t) Y sin(2 f t) C C Asin( ) A -Acos( ) Acos( ) y=0=>-a -Asin( ) * (0,0) -Asin( )

71 QAM X cos(2 f t) Ysin(2 f t) cos(2 f t) C C C X cos(2 fct)cos(2 fct) Y sin(2 fct)cos(2 fct) X 1 cos(0) cos(4 f ) 1 Ct Y sin(4 fct) X X cos(4 fct) sin(4 fct) 2 X cos(2 fct) Y sin(2 fct) cos Bcos A sinbsin A cos( A B) X cos(2 f t) Ysin(2 f t) sin(2 f t) C C C X cos(2 fct)sin(2 fct) Y sin(2 fct)sin(2 fct) 1 1 X sin(4 fct) Y cos(0) cos(4 fct) X sin(4 fct) Y Y cos(4 fct) 2

72 Q ถ า Amplitude ของ subcarrier ไม เท าก นล ะ จะเก ดอะไรข น A ลองทาด ส

73 SubCarrier Amplitude difference วาด Constellation Diagram 4-QAM: 4 subcarriers (2 amplitudes / 4 phases A1=1, A2=2 A1=2, A2=1

74 Q ถ า Amplitude ของ subcarrier ไม เท าก นล ะ จะเก ดอะไรข น A ลองทาด ส

75 Data Communications through a telephone line

Figure 9.11 Bandwidth division in ADSL 9.76 o Transmission: twisted-pair (1 pair) o Divides 1.

76 Figure 9.11 Bandwidth division in ADSL 9.76 o Transmission: twisted-pair (1 pair) o Divides MHz bandwidth into three bands (256 channels; KHz per channel) o POT (voice) (channel 0) o Upstream (channel 6-30; 25 channels), o Downstream (channel ; 225 channels)

Figure 9.10 Discrete Multitone Technique (DMT) : modulation technique standard for ADSL 9.

77 Figure 9.10 Discrete Multitone Technique (DMT) : modulation technique standard for ADSL control channel 24 data transfer 1 control channel 224 data transfer ~ 4 KHz / channel 60 Kbps/ channel

78 9.78 ANSI standard for ADSL Upstream ( KHz -> 25 channels) Each FDM sub channel: 4 KHz Discrete Multitone Technique (DMT): 15 bits per baud Data rate: 60 Kbps / channel Upstream data rate (no noise): 25 x 60Kbps = 1.5 Mbps data rate (with noise) : 64 Kbps 1 Mbps Downstream ( KHz -> 200 channels) Downstream data rate: 200 x 60 Kbps = 12 Mbps data rate (with noise): 500 Kbps 8 Mbps

79 Mobile Data Modulation PSK, QAM OFDM

80 5-2 ANALOG AND DIGITAL 5.80 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

81 Figure 5.15 Types of analog-to-analog modulation 5.81

82 Figure 5.16 Amplitude modulation 5.82

83 5.83 Note The total bandwidth required for AM can be determined from the bandwidth of the audio signal: B AM = 2B.

84 Figure 5.17 AM band allocation 5.84

85 5.85 Note The total bandwidth required for FM can be determined from the bandwidth of the audio signal: B FM = 2(1 + β)b.

86 Figure 5.18 Frequency modulation 5.86

87 87 DSBSC (Double Sideband Suppressed Carrier) Audio OSC. 2 f Max = µ X DSBSC Carrier 2 f = =

88 DSBSC (Double Sideband Suppressed Carrier) 88

89 89 v A DATA COMMUNICATIONS, Department of Computer Engineering, KMITL B. A. Forouzan, Data Communications and Networking, 4th rad/sec v A/2 = rad/sec

90 Amplitude Modulation AM G [ m( t)] g c( t) G [cos( t)] g cos( t) DATA COMMUNICATIONS, Department of Computer Engineering, KMITL B. A. Forouzan, Data Communications and Networking, 4th 90 Gcos( t)cos( t) g cos( t) G G cos( t t) cos( t t) g cos( t) ))

91 91 Amplitude Modulation DATA COMMUNICATIONS, Department of Computer Engineering, KMITL B. A. Forouzan, Data Communications and Networking, 4th Peak-to-Peak m=1 : 100% AM

92 92 Amplitude Modulation DATA COMMUNICATIONS, Department of Computer Engineering, KMITL B. A. Forouzan, Data Communications and Networking, 4th

93 Figure 5.19 FM band allocation 5.93

94 Figure 5.20 Phase modulation 5.94

95 5.95 Note The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: B PM = 2(1 + β)b.

Introduction to Communications Part Two: Physical Layer Ch5: Analog Transmission. Goals of This Class. Warm Up. Outline of the Class

Introduction to Communications Part Two: Physical Layer Ch5: Analog Transmission Kuang Chiu Huang TCM NCKU Spring/2008 2009/4/11 KuangChiu Huang 1 Goals of This Class Through the lecture of analog transmission,