The document discusses digital transmission fundamentals, including: - Digital representation of analog signals involves sampling, quantization, and pulse code modulation. - The sampling rate must be at least twice the bandwidth of the signal to allow perfect reconstruction. - Quantization maps samples to discrete levels, introducing quantization error. More levels reduce error but increase transmission bandwidth needs. - Digital transmission enables long distance communication by regeneration of the digital signal rather than analog amplification, overcoming distance limitations of analog systems.

1. Chapter 3 Digital Transmission Fundamentals Contain slides by Leon-Garcia and Widjaja

2. Chapter 3 Digital Transmission Fundamentals Digital Representation of Information Why Digital Communications? Digital Representation of Analog Signals Characterization of Communication Channels Fundamental Limits in Digital Transmission Line Coding Modems and Digital Modulation Properties of Media and Digital Transmission Systems Error Detection and Correction

5. Chapter 3 Digital Transmission Fundamentals Digital Representation of Information

10. Color Image = + + Color image Red component image Green component image Blue component image Total bits = 3  H  W pixels  B bits/pixel = 3HWB bits Example: 8  10 inch picture at 400  400 pixels per inch 2 400  400  8  10 = 12.8 million pixels 8 bits/pixel/color 12.8 megapixels  3 bytes/pixel = 38.4 megabytes H W H W H W H W

11. Examples of Block Information 1-8 Mbytes (5-30) 38.4 Mbytes 8x10 in 2 photo 400 2 pixels/in 2 JPEG Color Image 5-54 kbytes (5-50) 256 kbytes A4 page 200x100 pixels/in 2 CCITT Group 3 Fax (2-6) Kbytes- Mbytes ASCII Zip, compress Text Compressed(Ratio) Original Format Method Type

17. Video Frames Broadcast TV at 30 frames/sec = 10.4 x 10 6 pixels/sec 720 480 HDTV at 30 frames/sec = 67 x 10 6 pixels/sec 1080 1920 QCIF videoconferencing at 30 frames/sec = 760,000 pixels/sec 144 176

18. Digital Video Signals 19-38 Mbps 1.6 Gbps 1920x1080 @30 fr/sec MPEG2 HDTV 2-6 Mbps 249 Mbps 720x480 pix @30 fr/sec MPEG2 Full Motion 64-1544 kbps 2-36 Mbps 176x144 or 352x288 pix @10-30 fr/sec H.261 Video Confer-ence Compressed Original Format Method Type

21. Chapter 3 Communication Networks and Services Why Digital Communications?

27. Bit Rates of Digital Transmission Systems Many wavelengths >1600 Gbps Optical fiber 1 wavelength 2.5-10 Gbps Optical fiber 5 km multipoint radio 1.5-45 Mbps 28 GHz radio IEEE 802.11 wireless LAN 2-11 Mbps 2.4 GHz radio Coexists with analog telephone signal 64-640 kbps in, 1.536-6.144 Mbps out ADSL twisted pair Shared CATV return channel 500 kbps-4 Mbps Cable modem 100 meters of unshielded twisted copper wire pair 10 Mbps, 100 Mbps Ethernet twisted pair 4 kHz telephone channel 33.6-56 kbps Telephone twisted pair Observations Bit Rate System

28. Examples of Channels 40 Gbps / wavelength Many TeraHertz Optical fiber 54 Mbps / channel 300 MHz (11 channels) 5 GHz radio (IEEE 802.11) 30 Mbps/ channel 500 MHz (6 MHz channels) Coaxial cable 1-6 Mbps 1 MHz Copper pair 33 kbps 3 kHz Telephone voice channel Bit Rates Bandwidth Channel

29. Chapter 3 Digital Transmission Fundamentals Digital Representation of Analog Signals

33. Example Fourier Series Only odd harmonics have power T 1 = 1 ms 1 1 1 1 0 0 0 0 . . . . . . t x 2 (t) 1 0 1 0 1 0 1 0 . . . . . . t T 2 =0.25 ms x 1 (t) x 1 ( t ) = 0 + cos(2  4000 t ) + cos(2  3(4000) t ) + cos(2  5(4000) t ) + … 4  4 5  4 3  x 2 ( t ) = 0 + cos(2  1000 t ) + cos(2  3(1000) t ) + cos(2  5(1000) t ) + … 4  4 5  4 3 

36. Sampling Theorem Interpolation filter (a) (b) Nyquist: Perfect reconstruction if sampling rate 1/ T > 2 W s Sampler t x(t) t x(nT) t x(t) t x(nT)

37. Digital Transmission of Analog Information Interpolation filter Display or playout 2 W samples / sec 2 W m bits/sec x(t) Bandwidth W Sampling (A/D) Quantization Analog source 2 W samples / sec m bits / sample Pulse generator y(t) Original Approximation Transmission or storage

39. Quantizer Performance M = 2 m levels, Dynamic range( - V , V ) Δ = 2 V/M Average Noise Power = Mean Square Error: If the number of levels M is large, then the error is approximately uniformly distributed between (- Δ /2, Δ 2) ... error = y(nT)-x(nT)=e(nT) input ...      x(nT) V -V σ e 2 = x 2 dx = Δ 2 12 1 Δ ∫ Δ 2 Δ 2

42. Chapter 3 Digital Transmission Fundamentals Characterization of Communication Channels

57. Chapter 3 Digital Transmission Fundamentals Fundamental Limits in Digital Transmission

67. Channel Noise affects Reliability High SNR Low SNR SNR = Average Signal Power Average Noise Power SNR (dB) = 10 log 10 SNR virtually error-free error-prone signal noise signal + noise signal noise signal + noise

70. Chapter 3 Digital Transmission Fundamentals Line Coding

72. Line coding examples NRZ-inverted (differential encoding) 1 0 1 0 1 1 0 0 1 Unipolar NRZ Bipolar encoding Manchester encoding Differential Manchester encoding Polar NRZ

78. Chapter 3 Digital Transmission Fundamentals Modems and Digital Modulation

80. Amplitude Modulation and Frequency Modulation Information +1 -1 Amplitude Shift Keying +1 -1 Frequency Shift Keying t t Map bits into amplitude of sinusoid: “1” send sinusoid; “0” no sinusoid Demodulator looks for signal vs. no signal Map bits into frequency: “1” send frequency f c +  ; “0” send frequency f c -  Demodulator looks for power around f c +  or f c -  1 1 1 1 0 0 0 T 2 T 3 T 4 T 5 T 6 T 0 T 2 T 3 T 4 T 5 T 6 T

82. Modulator & Demodulator Modulate cos(2  f c t ) by multiplying by A k for T seconds: A k x cos(2  f c t ) Y i ( t ) = A k cos(2  f c t ) Transmitted signal during k th interval Demodulate (recover A k ) by multiplying by 2cos(2  f c t ) for T seconds and lowpass filtering (smoothing): x 2cos(2  f c t ) 2 A k cos 2 (2  f c t ) = A k {1 + cos(2  2 f c t )} Lowpass Filter (Smoother) X i (t) Y i ( t ) = A k cos(2  f c t ) Received signal during k th interval

83. Example of Modulation Information Baseband Signal Modulated Signal x(t) A cos(2  ft) -A cos(2  ft) 1 1 1 1 0 0 +A -A 0 T 2T 3T 4T 5T 6T +A -A 0 T 2T 3T 4T 5T 6T

84. Example of Demodulation Recovered Information Baseband signal discernable after smoothing After multiplication at receiver x(t) cos(2  f c t ) +A -A 0 T 2T 3T 4T 5T 6T A {1 + cos(4  ft )} -A {1 + cos(4  ft )} 1 1 1 1 0 0 +A -A 0 T 2T 3T 4T 5T 6T

87. QAM Demodulation Y(t) x 2cos(2  f c t ) 2cos 2 (2  f c t )+2B k cos(2  f c t )sin(2  f c t ) = A k {1 + cos(4  f c t )}+ B k {0 + sin(4  f c t )} Lowpass filter (smoother) A k 2 B k sin 2 (2  f c t )+2A k cos(2  f c t )sin(2  f c t ) = B k {1 - cos(4  f c t )}+ A k {0 + sin(4  f c t )} x 2sin(2  f c t ) B k Lowpass filter (smoother) smoothed to zero smoothed to zero

91. Chapter 3 Digital Transmission Fundamentals Properties of Media and Digital Transmission Systems

101. Cable Network Topology Head end Upstream fiber Downstream fiber Fiber node Coaxial distribution plant Fiber node = Bidirectional split-band amplifier Fiber Fiber

106. Very Low Attenuation 850 nm Low-cost LEDs LANs 1300 nm Metropolitan Area Networks “ Short Haul” 1550 nm Long Distance Networks “ Long Haul Water Vapor Absorption (removed in new fiber designs) 100 50 10 5 1 0.5 0.1 0.05 0.01 0.8 1.0 1.2 1.4 1.6 1.8 Wavelength (  m) Loss (dB/km) Infrared absorption Rayleigh scattering

114. Radio Spectrum 10 4 10 6 10 7 10 8 10 9 10 10 10 11 10 12 Frequency (Hz) Wavelength (meters) 10 3 10 2 10 1 1 10 -1 10 -2 10 -3 10 5 Satellite and terrestrial microwave AM radio FM radio and TV LF MF HF VHF UHF SHF EHF 10 4 Cellular and PCS Wireless cable Omni-directional applications Point-to-Point applications

116. Chapter 3 Digital Transmission Fundamentals Error Detection and Correction

121. Checkbits & Error Detection Calculate check bits Channel Recalculate check bits Compare Information bits Received information bits Sent check bits Information accepted if check bits match Received check bits k bits n – k bits

127. Error-detecting capability 1, 2, or 3 errors can always be detected; Not all patterns >4 errors can be detected 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 1 Arrows indicate failed check bits Two errors One error Three errors Four errors (undetectable)

139. Division Circuit Clock Input Reg 0 Reg 1 Reg 2 0 - 0 0 0 1 1 = i 3 1 0 0 2 1 = i 2 1 1 0 3 0 = i 1 0 1 1 4 0 = i 0 1 1 1 5 0 1 0 1 6 0 1 0 0 7 0 0 1 0 Check bits: r 0 = 0 r 1 = 1 r 2 = 0 r(x) = x Reg 0 + + Encoder for g(x) = x 3 + x + 1 Reg 1 Reg 2 0,0,0,i 0 ,i 1 ,i 2 ,i 3 g 0 = 1 g 1 = 1 g 3 = 1

146. Hamming (7,4) code 7 1 1 1 1 1 1 1 1 1 1 1 3 1 1 1 0 0 0 0 1 1 1 0 4 1 1 0 1 0 1 0 1 1 0 1 4 1 1 0 0 1 0 1 1 1 0 0 4 1 0 1 1 1 0 0 1 0 1 1 4 1 0 1 0 0 1 1 1 0 1 0 3 1 0 0 1 0 0 1 1 0 0 1 3 1 0 0 0 1 1 0 1 0 0 0 4 0 1 1 1 0 0 1 0 1 1 1 4 0 1 1 0 1 1 0 0 1 1 0 3 0 1 0 1 1 0 0 0 1 0 1 3 0 1 0 0 0 1 1 0 1 0 0 3 0 0 1 1 0 1 0 0 0 1 1 3 0 0 1 0 1 0 1 0 0 1 0 4 0 0 0 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 w ( b ) b 1 b 2 b 3 b 4 b 5 b 6 b 7 b 1 b 2 b 3 b 4 Weight Codeword Information

148. Error Detection with Hamming Code 0 0 1 0 0 0 0 s = H e = = 1 0 1 Single error detected 0 1 0 0 1 0 0 s = H e = = + = 0 1 1 Double error detected 1 0 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 0 0 0 s = H e = = + + = 0 1 1 0 Triple error not detected 0 1 1 1 0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1

153. Chapter 3 Digital Transmission Fundamentals RS-232 Asynchronous Data Transmission

155. Pins in RS-232 connector DTE DCE Protective Ground (PGND) Transmit Data (TXD) Receive Data (RXD) Request to Send (RTS) Clear to Send (CTS) Data Set Ready (DSR) Ground (G) Carrier Detect (CD) Data Terminal Ready (DTR) Ring Indicator (RI) 1 2 3 4 5 6 7 8 20 22 1 2 3 4 5 6 7 8 20 22 (b) 13 (a)                          1 25 14