Chapter 5 Analog Transmission 1 Mc GrawHill The

Chapter 5 Analog Transmission 1 Mc. Graw-Hill ©The Mc. Graw-Hill Companies, Inc. , 2004

Analog Transmission n Digital data to Analog signal l n Digital Modulation Analog data to Analog signal l Analog Modulation 2 Mc. Graw-Hill ©The Mc. Graw-Hill Companies, Inc. , 2004

5. 1 Modulation of Digital Data Digital-to-Analog Conversion Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation (QAM) Bit/Baud Comparison Mc. Graw-Hill ©The Mc. Graw-Hill Companies, Inc. , 32004

Figure 5. 1 Digital-to-Analog Modulation Amplitude Shift Keying Frequency Shift Keying Quadrature Amplitude Modulation Mc. Graw-Hill Phase Shift Keying ©The Mc. Graw-Hill Companies, Inc. , 42004

Digital-to-Analog Modulation (Digital Modulation) n Digital Modulation n n Digital Demodulation n Mc. Graw-Hill พาบต ขอมลไปบนชดของสญญาณ อนาลอก (signal unit( สญญาณอนาลอกทใชพาขอมล n สญญาณพาหะ (Carrier signal) n สวนใหญทใชเปน sine wave ทำไดโดย เปลยนแปลงคณสมบต ของสญญา ณพาหะ n Amplitude <Amplitude Shift Keying (ASK( n Frequency -> Frequency Shift Keying (FSK) n Phase -> Phase Shift Keying (PSK) การถอดบตขอมลจากชดสญญาณอนาลอกท ไดรบ ©The Mc. Graw-Hill Companies, Inc. , 52004

Bit Rate vs Baud Rate n Bit rate (bps) n n Baud rate (signal unit/sec: baud / sec: baud) n n Mc. Graw-Hill Number of bits per second Number of signal units per second less than or equal to the bit rate ©The Mc. Graw-Hill Companies, Inc. , 62004

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 Mc. Graw-Hill ©The Mc. Graw-Hill Companies, Inc. , 72004

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 Mc. Graw-Hill ©The Mc. Graw-Hill Companies, Inc. , 82004

Amplitude Shift Keying (ASK) n Bit representation n Changing Amplitude of Carrier Signal n One bit, One signal unit n n n ‘ 0’ -> A 1 ‘ 1’ -> A 2 Benefit n Simple (normally used for fiber optic) n Require Less Bandwidth Disadvantage n Mc. Graw-Hill Ex Easily effected by noise ©The Mc. Graw-Hill Companies, Inc. , 92004

Figure 5. 3 ASK Bit rate = baud rate Mc. Graw-Hill 102004 ©The Mc. Graw-Hill Companies, Inc. ,

Carrier Signal กรองความถเฉพาะทผาน transmission ไดเทานน Low Pass Filter Digital Data X LPF m X ASK Modulation 2 cos. A cos. B = cos (A + B) + cos (A - B) X 11 Mc. Graw-Hill LPF Decision making ASK Demodulation ©The Mc. Graw-Hill Companies, Inc. , 2004

Figure 5. 4 Relationship between baud rate and bandwidth in ASK ขนกบ เทคนคการ modulation & fi = S (baud/s) Mc. Graw-Hill 122004 ©The Mc. Graw-Hill Companies, Inc. ,

Example 3 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. Mc. Graw-Hill 132004 ©The Mc. Graw-Hill Companies, Inc. ,

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. Mc. Graw-Hill 142004 ©The Mc. Graw-Hill Companies, Inc. ,

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 Mc. Graw-Hill 152004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 5 Mc. Graw-Hill Solution to Example 5 162004 ©The Mc. Graw-Hill Companies, Inc. ,

Voltage Controlled Oscillator VCO Bit representation Changing Frequency of Carrier Signal One bit, One signal unit Ex ‘ 0’ -> f 1 ‘ 1’ -> f 2 Benefit Less effected by noise Normally used in high frequency radio transmission or coaxial cable Disadvantage Require Large Bandwidth 17 Mc. Graw-Hill FSK Modulation ©The Mc. Graw-Hill Companies, Inc. , 2004

Figure 5. 6 FSK Bit rate = Baud rate Mc. Graw-Hill 182004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 7 Mc. Graw-Hill Relationship between baud rate and bandwidth in FSK 192004 ©The Mc. Graw-Hill Companies, Inc. ,

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 Bit Rate = Baud Rate For FSK BW = Baud rate + fc 1 - fc 0 BW = Bit rate + fc 1 - fc 0 = 2000 + 3000 = 5000 Hz Mc. Graw-Hill 202004 ©The Mc. Graw-Hill Companies, Inc. ,

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 Bit Rate = Baud Rate 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. Mc. Graw-Hill 212004 ©The Mc. Graw-Hill Companies, Inc. ,

Phase Shift Keying (PSK) n Bit representation n Changing Phase of Carrier Signal n One bit, One signal unit n n -> Φ 1 ‘ 1’ -> Φ 2 Less effected by noise compared to ASK n n Normally used in MODEM (MOdulator/DEModulator) Require Bandwidth less than FSK Disadvantage n Mc. Graw-Hill ‘ 0’ Benefit n n Ex Difficult to detect phase shift in case of phase difference (Φ 1Φ 2) is too small 222004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 8 PSK Bit rate = Baud rate 180 o 0 o Mc. Graw-Hill 232004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 9 Mc. Graw-Hill PSK constellation 242004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 10 The 4 -PSK method 4 -PSK = 2 n-PSK=22 -PSK Bit rate = n x baud rate Mc. Graw-Hill 252004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 11 Mc. Graw-Hill The 4 -PSK characteristics 262004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 12 The 8 -PSK characteristics 4 -PSK = 2 n-PSK=23 -PSK Bit rate = n x Baud rate Bit rate = 3 x Baud rate Mc. Graw-Hill 272004 ©The Mc. Graw-Hill Companies, Inc. ,

x X 0 X Pulse 1 X Data 28 Mc. Graw-Hill [0, 1] [-1, 1] 0 -1 1 1 BPSK Modulation ©The Mc. Graw-Hill Companies, Inc. , 2004

X LPF Decision making 2 cos. A cos. B = cos (A + B) + cos (A - B) 29 Mc. Graw-Hill BPSK Demodulation ©The Mc. Graw-Hill Companies, Inc. , 2004

Figure 5. 13 Mc. Graw-Hill Relationship between baud rate and bandwidth in PSK 302004 ©The Mc. Graw-Hill Companies, Inc. ,

Example 8 Find the bandwidth for a 4 -PSK signal transmitting at 2000 bps. 4 -PSK = 22 -PSK Transmission is in half-duplex mode. Bit rate = 2 x Baud rate 2000 = 2 x Baud rate = 1000 baud/s Solution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 1000. But in 8 -PSK the bit rate is 3 times the baud rate, so the bit rate is 3, 000 bps. Mc. Graw-Hill 312004 ©The Mc. Graw-Hill Companies, Inc. ,

Example 9 Given a bandwidth of 5000 Hz for an 8 -PSK signal, Bandwidth = Baud rate what are the baud rate and bit rate? 5000 = Baud rate 8 -PSK = 23 -PSK Bit rate = 3 x Baud rate Solution = 3 x 5000 = 15, 000 bps 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. Mc. Graw-Hill 322004 ©The Mc. Graw-Hill Companies, Inc. ,

Quadrature Amplitude Modulation (QAM) n Bit representation n Combination of ASK and PSK n Changing Amplitude & Phase of Career Signal n One bit, One signal unit n n n ‘ 0’ -> A 1, Φ 1 ‘ 1’ -> A 2, Φ 2 Benefit n Less effected by noise compared to ASK n Require less bandwidth Disadvantage n Mc. Graw-Hill Ex Complex demodulation technique 332004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 14 Mc. Graw-Hill The 4 -QAM and 8 -QAM constellations 342004 ©The Mc. Graw-Hill Companies, Inc. ,

Time domain for an 8 -QAM Figure 5. 15 signal 8 -QAM = 2 n-QAM=23 -QAM Bit rate = n x Baud rate 90 o 180 o 0 o 270 o Mc. Graw-Hill 352004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 16 16 -QAM constellations ITU-T recommendation OSI recommendation Mc. Graw-Hill 362004 ©The Mc. Graw-Hill Companies, Inc. ,

10 (Pulse -1, 1) X=-1 X Digital Data “ 11 01 00 10 01” Series-to. Parallel Y (Pulse 1, 1) X=1 (Pulse 1, 1) 11 X 00 y=1 (Pulse -1, -1) y=-1 01 QAM signal + X Pulse Digital Data 37 Mc. Graw-Hill QAM Modulation ©The Mc. Graw-Hill Companies, Inc. , 2004

X X 2 cos. A cos. B = cos (A + B) + cos (A - B) LPF Parallel-to. Serial LPF Digital Data “ 11 01 00 10 01” 2 sin. A sin. B = - cos (A + B) + cos (A - B) 2 sin. A cos. B = sin (A + B) + sin (A - B) 38 Mc. Graw-Hill QAM Demodulation© The Mc. Graw-Hill Companies, Inc. , 2004

Figure 5. 17 Mc. Graw-Hill Bit and baud 392004 ©The Mc. Graw-Hill Companies, Inc. ,

Table 5. 1 Bit and baud rate comparison Modulation Units Bits/Baud rate Bit Rate Bit 1 N N 4 -PSK, 4 -QAM Dibit 2 N 2 N 8 -PSK, 8 -QAM Tribit 3 N 3 N 16 -QAM Quadbit 4 N 4 N 32 -QAM Pentabit 5 N 5 N 64 -QAM Hexabit 6 N 6 N 128 -QAM Septabit 7 N 7 N 256 -QAM Octabit 8 N 8 N ASK, FSK, 2 -PSK Mc. Graw-Hill 402004 ©The Mc. Graw-Hill Companies, Inc. ,

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 3 bits/baud 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/s Mc. Graw-Hill 412004 ©The Mc. Graw-Hill Companies, Inc. ,

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 log 216 = log 224 = 4. Thus, (1000)(4) = 4000 bps Mc. Graw-Hill 422004 ©The Mc. Graw-Hill Companies, Inc. ,

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 log 2 64 = log 226 = 6. Thus, 72000 / 6 = 12, 000 baud Mc. Graw-Hill 432004 ©The Mc. Graw-Hill Companies, Inc. ,

5. 2 Telephone Modems Modem Standards Mc. Graw-Hill 442004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 18 Telephone line bandwidth A telephone line has a bandwidth of almost 2400 Hz for data transmission. Mc. Graw-Hill 452004 ©The Mc. Graw-Hill Companies, Inc. ,

Note: Modem stands for MOdulator/ DEModulator. MO DEM Mc. Graw-Hill 462004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 19 Modulation/demodulation True Analog ASK, FSK, PSK, QAM Digital Mc. Graw-Hill Digital 472004 ©The Mc. Graw-Hill Companies, Inc. ,

Bandwidth = Baud Rate Bit Rate = Baud Rate for Half-Duplex ASK BWASK = Baud rate. ASK = 2400 Baud/s Bit rate. ASK = Baud rate. ASK = 2400 bps Mc. Graw-Hill 482004 ©The Mc. Graw-Hill Companies, Inc. ,

Bandwidth = Baud Rate Bit Rate = Baud Rate for Full-Duplex ASK BWASK = Baud rate. ASK = 1200 Baud/s Bit rate. ASK = Baud rate. ASK = 1200 bps Mc. Graw-Hill 492004 ©The Mc. Graw-Hill Companies, Inc. ,

Bandwidth = Baud Rate + (fc 1 -fc 0) Bit Rate = Baud Rate for Half-Duplex FSK BWFSK = Baud rate. FSK + (fc 1 – fc 0) Bit rate. FSK = Baud rate. FSK = BWFSK – (fc 1 – fc 0) Mc. Graw-Hill 502004 ©The Mc. Graw-Hill Companies, Inc. ,

Bandwidth = Baud Rate + (fc 1 -fc 0) Bit Rate = Baud Rate for Full-Duplex FSK Max baud rate = Max bit rate = 1200 - (fc 1 -fc 0) 1200 Hz BWFSK = Baud rate. FSK + (fc 1 – fc 0) Bit rate. FSK = Baud rate. FSK = BWFSK – (fc 1 – fc 0) Mc. Graw-Hill 512004 ©The Mc. Graw-Hill Companies, Inc. ,

Bell Modems Guard Band or Guard Frequency 500 200 455 600 200 600 Bit rate = 2 x Baud rate Mc. Graw-Hill 522004 ©The Mc. Graw-Hill Companies, Inc. ,

Bell Modems Bit rate = 2 x Baud rate Bit rate = 3 x Baud rate Bit rate = 4 x Baud rate Mc. Graw-Hill 532004 ©The Mc. Graw-Hill Companies, Inc. ,

ITU Modems International Telecommunication Union (Four-Phase Differential PSK) Mc. Graw-Hill 542004 ©The Mc. Graw-Hill Companies, Inc. ,

ITU Modems Mc. Graw-Hill 552004 ©The Mc. Graw-Hill Companies, Inc. ,

V. 22 bis 16 -QAM Constellation Mc. Graw-Hill 562004 ©The Mc. Graw-Hill Companies, Inc. ,

V. 32 Constellation Mc. Graw-Hill 572004 ©The Mc. Graw-Hill Companies, Inc. ,

V. 33 Constellation Mc. Graw-Hill 582004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 20 Mc. Graw-Hill The V. 32 constellation and bandwidth 592004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 21 Mc. Graw-Hill The V. 32 bis constellation and bandwidth 602004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 22 Mc. Graw-Hill Traditional modems Analog Digital 612004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 23 56 K modems Digital Analog Digital Mc. Graw-Hill 622004 ©The Mc. Graw-Hill Companies, Inc. ,

The 56 K Digital Modem A 56 K modem is based upon one of two standards: - V. 90 - Upstream speed is maximum 33, 600 bps - V. 92 - Newer standard which allows maximum upstream speed of 48 Kbps (under ideal conditions) and can place a data connection on hold if the telephone service accepts call waiting and a voice telephone call arrives Mc. Graw-Hill 632004 ©The Mc. Graw-Hill Companies, Inc. ,

5. 3 Modulation of Analog Signals Amplitude Modulation (AM) Frequency Modulation (FM) Phase Modulation (PM) Mc. Graw-Hill 642004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 24 Mc. Graw-Hill Analog-to-Analog Modulation 652004 ©The Mc. Graw-Hill Companies, Inc. ,

Note: The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BWt = 2 x BWm. AM bandwidth Mc. Graw-Hill Audio bandwidth (4 KHz) 662004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 26 Mc. Graw-Hill Amplitude Modulation 672004 ©The Mc. Graw-Hill Companies, Inc. ,

Amplitude Modulation Mc. Graw-Hill 682004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 27 Mc. Graw-Hill AM bandwidth 692004 ©The Mc. Graw-Hill Companies, Inc. ,

Peak-to-Peak m=1 : 100% AM Mc. Graw-Hill 702004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 28 535 540 Mc. Graw-Hill AM band allocation 550 712004 ©The Mc. Graw-Hill Companies, Inc. ,

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 (Federal Communications Commission) regulations. Solution BWt = 2 x BWm An AM signal requires twice the bandwidth of the original signal: BW = 2 x 4 KHz = 8 KHz Mc. Graw-Hill 722004 ©The Mc. Graw-Hill Companies, Inc. ,

Note: The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BWt = 10 x BWm. FM bandwidth Stereo Audio bandwidth (15 KHz) Mc. Graw-Hill 732004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 29 Mc. Graw-Hill Frequency modulation 742004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 30 Mc. Graw-Hill FM bandwidth 752004 ©The Mc. Graw-Hill Companies, Inc. ,

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). Mc. Graw-Hill 762004 ©The Mc. Graw-Hill Companies, Inc. ,

Figure 5. 31 88. 5 88. 1 88. 2 Mc. Graw-Hill FM band allocation 88. 4 88. 6 772004 ©The Mc. Graw-Hill Companies, Inc. ,

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 (Federal Communications Commission) regulations. Solution BWt = 10 x BWm An FM signal requires 10 times the bandwidth of the original signal: BW = 10 x 4 MHz = 40 MHz Mc. Graw-Hill 782004 ©The Mc. Graw-Hill Companies, Inc. ,