Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 1 Transmission: Digital vs. Analog Problem 1:

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 1 Transmission: Digital vs. Analog Problem 1: Transmit information “faithfully” (with little or no distortion) to the receiver Reality: signal got distorted over distance because1) impedance (within transmission medium) 2) interference (outside force, e.g. cloud, lightening)  T fixed HH

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 3 Amplitude, Frequency, and Phase of a Signal s(t)=Asin(2  ft+  ) Amplitude=A; Frequency=f; Phase= 

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 5 Pulse Code Modulation (PCM) Continuous analog signal is sampled, i.e., captured the signal’s amplitude at fixed time interval, say 8000 times/sec. The frequency we sampled the signal is called the sampling rate. Fixed number bits (8 or 16 typical) is used to represent each sample. Divide the signal amplitude range by 2 bitspersample (256or 2 16 ) discrete levels. Quantization: The measured amplitudes in real numbers are then rounded off to the closest discrete level. Say 8.2  8 Discrete value is then encoded as bit pattern. 00001000.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 6 Gray contour represent DM encoded signal Red stairs represent PCM encoded signal Assume 5 bits/sample Major source of error How to reduce it? Bold line is original signal

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 7 PCM Encoding Example Example: With Stereo, 16 bits/sample, 44kHz sampling rate, PCM encoding, how many bits of data will be generated by a three minute sound recording? Ans: 3min*60sec/min*44000samples/sec*16bits/sample*2(chan nel)=253.44Mbits=31.68MB. Examle: Telephone network, 8 bits/sample, 8kHz sample rate, PCM encoding, how many bits of data will be generated by a three minute sound recording?

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 8 (a) QCIF Videoconferencing (b) Broadcast TV (c) HDTV @ 30 frames/sec = 760,000 pixels/sec @ 30 frames/sec = 10.4 x 10 6 pixels/sec @ 30 frames/sec = 67 x 10 6 pixels/sec 720 480 1080 1920 144 176 Figure 3.3 Network Requirement for Different Information Type: Resolution

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 9 Playout delay Jitter due to variable delay (b) (c) 123456789 123456 Original sequence (a) 123456789 Figure 3.4 Network Requirement for Different Information Type: Delay and Jitter Delay: time it takes to deliver a message from sender to receiver it is related to propagation delay, transmission delay, queueing Samples in realtime media stream needs to be played back regularly. Jitter: variation of delays among msgs, due to traffic load variation can be solved by playout buffer, which accumulates enough msgs before playing(delivering to end user). What is the penalty?

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 10 Attenuated & distorted signal + noise Equalizer Recovered signal + residual noise Repeater Amp. Figure 3.8 Analog Repeater Equalizer: a device used to eliminate/compensate the distortion. Two causes of distortion: High frequency signal component attenuated more than that of low frequency Delays for different frequency components are different

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 11 Amplifier Equalizer Timing Recovery Decision Circuit. & Signal Regenerator Figure 3.9 A Digital Repeater Equalizer: only need to compensate to a point that can detect positive or negative pulse Timing Recovery Circuit: keeps track the pulse interval. Decision Circuit: sample signal at midpoint to determine the polarity (positive or negative) of the pulse. Digital transmission eliminates accumulation of noise. can operate at lower signal level, greater distance  lower cost handles any type of information that can be digitized. takes advantage of error correction and data encryption techniques.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 12 f 0 W A(f)A(f) (a) Lowpass and idealized lowpass channel (b) Maximum pulse transmission rate is 2W pulses/second, called Nyquist Rate 0W f A(f)A(f) 1 Channel With W bandwidth t t Figure 3.11 Amplitude-response Functions A(f) Bandwidth: the range of frequencies passed by a channel. Can still tell which kind of pulse sent. If rate > 2W, difficult to tell.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 13 Sending Pulse with Multiple Levels With 8 levels per pulse we can send log 2 8= 3bits in just one pulse. Now we can send 3*2W bits per second. Let us increase the signal levels!! Unfortunately, there is noise in most channels  We can not have unlimited number of signal levels encoded in each pulse. 4 signal levels8 signal levels typical noise Still can tell they are all level 1 0 1 2 3 7 6 5 4 3 2 1 0 Difficult To tell if It is 3 or 2

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 14 signal noise signal + noise signal noise signal + noise High SNR Low SNR SNR = Average Signal Power Average Noise Power SNR (dB) = 10 log 10 SNR t t t t t t Figure 3.12

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 15 Shannon Channel Capacity For a low pass channel with W bandwidth, if the Signal Noise Ratio is SNR, then the maximum number of bits can be transferred over this noisy channel is C = W log 2 (1+SNR) For W=3.4kHz, SNR=40dB (S/N=10000), C=44.8kbps? 45.178kbps! For 56kbps modem on upstream from user via telephone company to ISP subject to 3.4kHz channel can achieve about 33.6kbps < 44.8kpbs. On downstream from ISP to user, the signal already digital no need for analog to digital conversion  56kbps achievable.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 16 Channel t 0 t h(t) tdtd Figure 3.17 t s(t) = sin(2  Wt)/ 2  Wt T=1/2W T T T T T T T Signal Pulse with zero intersymbol interference.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 17 +A+A -A-A 0 T 2T2T 3T3T 4T4T5T5T 111100 Transmitter Filter Comm. Channel Receiver Filter Receiver r(t) Received signal t Figure 3.19

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 18 (a) 3 separate pulses for sequence 110 (b) Combined signal for sequence 110 t t TTTTTT TTTTTT Figure 3.20 Sampling time This shows when samples at the right 0, 1T, 2T time, we get correct encoded values.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 19 101 0 11001 Unipolar NRZ NRZ-Inverted (Differential Encoding) Bipolar Encoding Manchester Encoding Differential Manchester Encoding Polar NRZ Figure 3.25 Line codes: represent binary digits using digital signals. NonReturn to Zero: It uses average power A 2 /2=(1/2)(+A) 2 +(1/2)(0) 2 Use less average power A 2 /4=(1/2)(+A/2) 2 +(1/2)(-A/2) 2 then the unipolar NRZ A 2 /2 A A/2 Power=V 2 /R Here V=A, R=1; AveragePower= sum(Prob(bitn)*Power(bitn)) No voltage change  bit 0 Voltage change  bit 1; What data are suited for this? Zero bits do not require sending signal; low power consumption! What’s wrong? Always has voltage transition in the middle of the bit. Use for timing sync.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 20 Ethernet Line Coding Manchester encoding is used in Ethernet Line coding. The presence of mid bit transition makes the timing recovery easy to design. Manchester code is a special case of mBnB code, where m bit information is encoded in n> m encoded bits. Here m=1 and n=2. FDDI (Fiber Distribution Data Interface) LAN uses 4B5B line code.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 21 Modulation Band pass channel passes signal within certain range. Modulation is used to send signal over band pass channel. Information Amplitude shift keying Frequency shift keying Phase shift keying f f 2 f 1 f c 0

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 22 111100 (a) Information (d) 2Y i (t) cos(2  f c t) +2A -2A +A -A (c) Modulated Signal Y i (t) 0 T 2T2T 3T3T 4T4T5T5T 6T6T +A -A (b) Baseband Signal X i (t) 0 2T2T 3T3T 6T6T 0 T 2T2T 3T3T 4T4T5T5T 6T6T T 4T4T5T5T t t t Figure 3.29

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 23 (a) Modulate cos(2  f c t) by multiplying it by A k for (k-1)T < t <kT: AkAk x cos(2  f c t) Y i (t) = A k cos(2  f c t) (b) Demodulate (recover) A k by multiplying by 2cos(2  f c t) and lowpass filtering: x 2cos(2  f c t) 2A k cos 2 (2  f c t) = A k {1 + cos(2  f c t)} Lowpass Filter with cutoff W Hz X i (t) Y i (t) = A k cos(2  f c t) Figure 3.30 Here Yi(t) is the modulated signal with A k be the amplitude of cosine wave sent by sender for kth bit. For bit 0, +A is used. For bit 1, -A is used. Sending Digits Using PSK

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 24 AkAk x cos(2  f c t) Y i (t) = A k cos(2  f c t) BkBk x sin(2  f c t) Y q (t) = B k sin(2  f c t) +Y(t) Modulate cos(2  f c t) and sin (2  f c t) by multiplying them by A k and B k respectively for (k-1)T < t <kT: Figure 3.31 Quadrature Amplitude Modulation (QAM) First the information bit stream are divided into two groups: odd symbols Ak and even symbols Bk

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 25 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 with cutoff W/2 Hz AkAk x 2sin(2  f c t) 2B 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)} Lowpass Filter with cutoff W/2 Hz BkBk Figure 3.32 QAM Demodulator

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 26 AkAk BkBk 16 “levels”/ pulse 4 bits / pulse 4W bits per second AkAk BkBk 4 “levels”/ pulse 2 bits / pulse 2W bits per second 2-D signal Figure 3.33 Signal Constellations  A Signal with amplitude=A, phase=3/4  is sent for 01 bits Receive signal with amplitude=A, phase=5/4   sender sent 00 bits 10 (A, 7/4  ) 11 (A, 1/4  )

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 27 AkAk BkBk 4 “levels”/ pulse 2 bits / pulse 2W bits per second 16 “levels”/ pulse 4 bits / pulse 4W bits per second Figure 3.34 A 2A 0011 0000 0001 0010 0100 0101 0110 0111 1000 1001 1010 1011 1100 1111 1101 1110

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 29 Trellis Modulation Use dense signal sets, say 8, but restrict code sequences with which signals can be used, say only 4 are valid. Goal  Error correction, reduce the error rate. At state 0, receiving input 01, Send signal 2, enter state 1. Note that at state 0, Sender only sends 0, 4, 2, 6 (even signal) At state 1 Sender only Sends 1, 3, 5, 7 (odd signal)

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 30 Trellis Modulation The sender and receiver starts at the same state 0. If the sender receives 01011000, it will send signals 2, 3, 7, 2 and changes its states from 0 to 1, to 3, to 2 and back to 0. The receiver will traverse the same state and deliver 01011000. At state 1, receiver receives signals 7, 5, 3, 2. It will interpret them as correct signals and deliver 11110000 to the upper layer. At state 3, receiver receives signals 7, 5, 3, 2. It will say the first signal 7 is ok but the 2 nd signal 5 is incorrect since at state 2, it can only receive even number signal! At state 0 or 2, receiver receivs signals 7, 5, 3, 2. It will say the first signal is already not incorrect since it can only receive even number signal at those two state. Signal 7 has a phase angle of 7/4 .

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 31 Trellis Modulation: At state 0 Receiver receives 2, 3, 7, 2 signals It delivers 01 01 10 00 to its upper layer. 2 3 7 2

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 32 Trellis Modulation: At state 1, receiver receives signals 7, 5, 3, 2 It delivers 11110000 to its upper layers. It enters state 0 7 5 3 2

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 33 Error Correction in Trellis Modulation At state 1, receiver receives a signal with phase angle = 0.40 . What signal should it interpret? What bits should be delivered?

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 34 10 2 10 4 10 6 10 810 10 12 10 14 10 16 10 18 10 20 10 22 10 24 Frequency (Hz) Wavelength (meters) 10 6 10 4 10 2 1010 -2 10 -4 10 -6 10 -8 10 -10 10 -12 10 -14 power & telephone broadcast radio microwave radio infrared light visible light ultraviolet light x rays gamma rays Figure 3.35 Electromagnetic Spectrum

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 35 Attenuation (dB/mi) f (kHz) 19 gauge 22 gauge 24 gauge 26 gauge 6 12 3 9 15 18 21 24 27 30 110 1001000 Figure 3.37 Attenuation vs. Frequency for Twisted Pair

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 36 Center conductor Dielectric material Braided outer conductor Outer cover Figure 3.39 Coaxial Cable Two conductors  Better immunity to interference and crosstalk. Less attenuation  higher bandwidth (54-500MHz on Cable TV) 446MHz / 6MHzperTVchannel=74 channels.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 37 35 30 10 25 20 5 15 Attenuation (dB/km) 0.010.1 1.0 10100 f (MHz) 2.6/9.5 mm 1.2/4.4 mm 0.7/2.9 mm Figure 3.40 Attenuation vs. Frequency for Coaxial Cable note that typo f(kHz) should be f(MHz).

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 39 Hea d end Upstream fiber Downstream fiber Fiber node Coaxial distribution plant Fiber node Bidirectional Split-Band Amplifier Fiber Figure 3.42 Topology of Hybrid Fiber-Coaxial Systems

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 40 Downstream 54 MHz 500 MHz Upstream Downstream 5 MHz 42 MHz 54 MHz 500 MHz 550 MHz 750 MHz (a) Current allocation (b) Proposed hybrid fiber-coaxial allocation Proposed downstream Figure 3.43 Frequency Allocation in Cable TV Systems Cable Modem 2MHz Channel (500kbps-4Mbps) Shared Time-slot Channels Cable Modem 6MHz channel (36Mbps)

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 41 core cladding jacket light cc (a) Geometry of optical fiber (b) Reflection in optical fiber Figure 3.44

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 42 100 50 10 5 1 0.5 0.1 0.05 0.01 0.81.01.21.41.61.8 Wavelength (  m) Loss (dB/km) Infrared absorption Rayleigh scattering Figure 3.45 Attenuation vs. Wavelength for Optical Fiber 850nm 1300nm 1550nm

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 43 (a) Multimode fiber: multiple rays follow different paths (b) Single mode: narrow core  only direct path propagates in fiber ; higher speed direct path reflected path Figure 3.46 Single-mode vs. Multimode Optical Fiber interference 1300nm has a region with 0.5 dB/km and 25 TeraHz. 1550nm has a region with 0.2 dB/km and 25 THz.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 44 Optical fiber Optical source Modulator Electrical signal Receiver Electrical signal Figure 3.47 Optical Transmission System Wavelength-division multiplexing (WDM): Send multiple wavelength light. Dense WDM (DWDM): provide 160 wavelengths, each operating at 10Gbps for a total of 1600 Gbps. Soliton: Special pulse shape that retain shape over long distance. Experiment results: 80Gbps over 10,000km. Field trials: 10 Gbps over 200km.

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 45 Gigabit Ethernet 1000Base-X standard Two variations: 1000 BASE-SX: use “shortwave” lightsource, nomially at 850 nm, multiple mode fiber. distance limit 550 meters. 1000 BASE-LX: use “longwave” lightsource, nomially at 1300 nm, single mode (5km) or multiple mode fiber (550m).

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 46 10 4 10 6 10 7 10 8 10 910 10 11 10 12 Frequency (Hz) Wavelength (meters) 10 3 10 2 10 1 1 10 -1 10 -2 10 -3 10 5 satellite & terrestrial microwave AM radio FM radio & TV LF MF HF VHF UHF SHF EHF 10 4 Cellular & PCS Wireless cable Figure 3.48 ISM bands: 902-928MHz, 2400-2483.5MHz, 5725-5850MHz  wireless LAN PCS bands: 1.7-2.3GHz AMPS bands: 824-849MHz (832*30kHz Tx channels), 869-894MHz (Rx) Satellite: 4/6, 11/14, 20/30 GHz bands (downlink/uplink)

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 47 Channel Encoder User information Pattern Checking All inputs to channel satisfy pattern/condition Channel output Deliver user information or set error alarm Figure 3.49

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 48 Calculate check bits Channel Recalculate check bits Compare Information bits Received information bits Check bits Information accepted if check bits match Received check bits Figure 350

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 49 x = codewordso = non-codewords x x x x x x x o o o o o o o o o o o o x x x x x x x o o o o o o o o o o o o A code with poor distance properties A code with good distance properties (a) (b) Figure 3.51

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 50 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 1 0 0 1 1 1 Bottom row consists of check bit for each column Last column consists of check bits for each row Figure 3.52

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 51 1 0 0 0 0 0 0 0 1 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 1 0 1 0 0 1 1 1 Two errors One error Three errors Four errors Arrows indicate failed check bits Figure 3.53

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 53 Addition: Multiplication: Division: x 3 + x + 1 ) x 6 + x 5 x 3 + x 2 + x x 6 + x 4 + x 3 x 5 + x 4 + x 3 x 5 + x 3 + x 2 x 4 + x 2 x 4 + x 2 + x x = q(x) quotient = r(x) remainder divisor dividend 35 ) 122 3 105 17 Figure 3.55

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 54 Steps: 1) Multiply i(x) by x n-k (puts zeros in (n-k) low order positions) 2) Divide x n-k i(x) by g(x) 3) Add remainder r(x) to x n-k i(x) (puts check bits in the n-k low order positions): quotient remainder transmitted codeword b(x) = x n-k i(x) + r(x) x n-k i(x) = g(x) q(x) + r(x) Figure 3.56

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 55 Generator polynomial: g(x)= x 3 + x + 1 Information: (1,1,0,0) i(x) = x 3 + x 2 Encoding: x 3 i(x) = x 6 + x 5 1011 ) 1100000 1110 1011 1110 1011 1010 1011 010 x 3 + x + 1 ) x 6 + x 5 x 3 + x 2 + x x 6 + x 4 + x 3 x 5 + x 4 + x 3 x 5 + x 3 + x 2 x 4 + x 2 x 4 + x 2 + x x Transmitted codeword: b(x) = x 6 + x 5 + x b = (1,1,0,0,0,1,0) Figure 3.57

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 56 reg 0reg 1reg 2 ++ Encoder for clockinputreg 0reg 1reg 2 0- 000 11=i 3 100 21=i 2 110 30=i 1 011 40=i 0 111 50101 60100 70010 check bits:r 0 = 0r 1 = 1r 2 = 0 r(x) = x Figure 3.58

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 58 1. Single errors:e(x) = x i 0  i  n-1 If g(x) has more than one term, it cannot divide e(x) 2. Double errors: e(x) = x i + x j 0  i < j  n-1 = x i (1 + x j-i ) If g(x) is primitive, it will not divide (1 + x j-i ) for j-i  2 n-k  1 3. Odd number of errors:e(1) =1 If number of errors is odd. If g(x) has (x+1) as a factor, then g(1) = 0 and all codewords have an even number of 1s. Figure 3.60

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 59 4. Error bursts of length b: 0000110 0001101100 0 e(x) = x i d(x) where deg(d(x)) = L-1 g(x) has degree n-k; g(x) cannot divide d(x) if deg(g(x))> deg(d(x)) L = (n-k) or less: all will be detected L = (n-k+1): deg(d(x)) = deg(g(x)) i.e. d(x) = g(x) is the only undetectable error pattern, fraction of bursts which are undetectable = 1/2 L-2 L > (n-k+1): fraction of bursts which are undetectable = 1/2 n-k L ith position error pattern d(x) Figure 3.61

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 60 b e r + (receiver) (transmitter) error pattern b e r + (receiver) (transmitter) error pattern (a) Single bit input (b) Vector input Figure 3.62

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 61 00100000010000 s = H e = = 101101 single error detected 01001000100100 s = H e = = + = 011011 double error detected 100100 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 11100001110000 s = H e = = + + = 0 110110 triple error not detected 011011 101101 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 111111 Figure 3.63

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 62 s = H r = He s = 0 no errors in transmission undetectable errors correctable errors uncorrectable errors (1-p) 7 7p37p3 1-3p 3p3p 7p7p 7p(1-3p)21p 2 Figure 3.64

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 63 If d min = 2t+1, non-overlapping spheres of radius t can be drawn around each codeword; t=2 in the figure b1b1 b2b2 oooo set of all n-tuples within distance t set of all n-tuples within distance t Figure 3.66

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 64 b1b1 b2b2 b3b3 b4b4 b L- 3 b L- 2 b L- 1 bLbL... L codewords written vertically in array; then transmitted row by row b1b1 b2b2 b3b3 b4b4 b L-3 b L-2 b L-1 bLbL... A long error burst produces errors in two adjacent rows Figure 3.66

Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 65 DTEDCE 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) (a)                          1 13 25 14 Figure 3.67

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Leon-Garcia & Widjaja: Communication Networks Copyright ©2000 The McGraw Hill Companies cs522f200 ch3 page 1 Transmission: Digital vs. Analog Problem 1:

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