Điện tử viễn thông LGW2EChapter3Presentation khotailieu

163 27 0
Điện tử viễn thông LGW2EChapter3Presentation khotailieu

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Thông tin tài liệu

Chapter 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 Digital Networks z Digital transmission enables networks to support many services TV E-mail Telephone Questions of Interest z How long will it take to transmit a message? z z z Can a network/system handle a voice (video) call? z z How many bits/second does voice/video require? At what quality? How long will it take to transmit a message without errors? z z z How many bits are in the message (text, image)? How fast does the network/system transfer information? How are errors introduced? How are errors detected and corrected? What transmission speed is possible over radio, copper cables, fiber, infrared, …? Chapter Digital Transmission Fundamentals Digital Representation of Information Bits, numbers, information z Bit: number with value or z z z z n bits allows enumeration of 2n possibilities z z z z n bits: digital representation for 0, 1, … , 2n Byte or Octet, n = Computer word, n = 16, 32, or 64 n-bit field in a header n-bit representation of a voice sample Message consisting of n bits The number of bits required to represent a message is a measure of its information content z More bits → More content Block vs Stream Information Block z Information that occurs in a single block z z z z z Text message Data file JPEG image MPEG file Size = Bits / block or bytes/block z z z kbyte = 210 bytes Mbyte = 220 bytes Gbyte = 230 bytes Stream z Information that is produced & transmitted continuously z z z Real-time voice Streaming video Bit rate = bits / second z z z kbps = 103 bps Mbps = 106 bps Gbps =109 bps Transmission Delay z z z z z z L R bps L/R tprop d c number of bits in message speed of digital transmission system time to transmit the information time for signal to propagate across medium distance in meters speed of light (3x108 m/s in vacuum) Delay = tprop + L/R = d/c + L/R seconds Use data compression to reduce L Use higher speed modem to increase R Place server closer to reduce d Compression z z Information usually not represented efficiently Data compression algorithms z z Represent the information using fewer bits Noiseless: original information recovered exactly z z Noisy: recover information approximately z z z E.g zip, compress, GIF, fax JPEG Tradeoff: # bits vs quality Compression Ratio #bits (original file) / #bits (compressed file) Color Image W H Color image = H W W W Red component image Green component image Blue component image + H + H Total bits = × H × W pixels × B bits/pixel = 3HWB bits Example: 8×10 inch picture at 400 × 400 pixels per inch2 400 × 400 × × 10 = 12.8 million pixels bits/pixel/color 12.8 megapixels × bytes/pixel = 38.4 megabytes Examples of Block Information Type Method Format Original Compressed (Ratio) Text Zip, compress ASCII KbytesMbytes (2-6) Fax CCITT Group A4 page 200x100 pixels/in2 256 kbytes 5-54 kbytes (5-50) JPEG 8x10 in2 photo 4002 pixels/in2 38.4 Mbytes 1-8 Mbytes (5-30) Color Image m = Hamming Code z z Information bits are b1, b2, b3, b4 Equations for parity checks b5, b6, b7 b5 = b1 + b3 + b4 b6 = b1 + b2 b7 = z z + b4 + b2 + b3 + b4 There are 24 = 16 codewords (0,0,0,0,0,0,0) is a codeword Hamming (7,4) code Information Codeword Weight b1 b2 b3 b4 b5 b6 b7 w(b) b1 b2 b3 b4 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 1 1 1 1 1 0 0 0 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 Parity Check Equations z Rearrange parity check equations: = b5 + b5 = b1 + b3 + b4 + b5 = b6 + b6 = b1 + b2 = b7 + b7 = z In matrix form: + b4 + b6 + b2 + b3 + b4 + b7 b1 b2 = 1011100 b3 = 1101010 b4 = H bt = 0 = 0111001 b5 b6 b7 z z All codewords must satisfy these equations Note: each nonzero 3-tuple appears once as a column in check matrix H Error Detection with Hamming Code 1011100 s=He= 1101010 0111001 1011100 s=He= 1101010 0111001 1011100 s=He= 1101010 0111001 0 0 0 0 Single error detected = 1 = + = 1 1 = + 0 0 + 1 1 = Double error detected Triple error not detected Minimum distance of Hamming Code z z Previous slide shows that undetectable error pattern must have or more bits At least bits must be changed to convert one codeword into another codeword Set of ntuples within distance of b1 z z o o b1 o Distance o o o b2 o o Set of ntuples within distance of b2 Spheres of distance around each codeword not overlap If a single error occurs, the resulting n-tuple will be in a unique sphere around the original codeword General Hamming Codes z For m > 2, the Hamming code is obtained through the check matrix H: z z z Each nonzero m-tuple appears once as a column of H The resulting code corrects all single errors For each value of m, there is a polynomial code with g(x) of degree m that is equivalent to a Hamming code and corrects all single errors z For m = 3, g(x) = x3+x+1 Error-correction using Hamming Codes (Transmitter) b + R (Receiver) e Error pattern z z z The receiver first calculates the syndrome: s = HR = H (b + e) = Hb + He = He If s = 0, then the receiver accepts R as the transmitted codeword If s is nonzero, then an error is detected z z z Hamming decoder assumes a single error has occurred Each single-bit error pattern has a unique syndrome The receiver matches the syndrome to a single-bit error pattern and corrects the appropriate bit Performance of Hamming ErrorCorrecting Code z Assume bit errors occur independent of each other and with probability p s = H R = He 7p s=0 No errors in transmission (1–p)7 Undetectable errors 7p3 s=0 1–3p Correctable errors 7p(1–3p) 3p Uncorrectable errors 21p2 Chapter Digital Transmission Fundamentals RS-232 Asynchronous Data Transmission Recommended Standard (RS) 232 z z z z Serial line interface between computer and modem or similar device Data Terminal Equipment (DTE): computer Data Communications Equipment (DCE): modem Mechanical and Electrical specification Pins in RS-232 connector 13 (a) • • • • • • • • • • • • • • • • • • • • • • • • • 14 (b) DTE 25 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) 20 Data Terminal Ready (DTR) 20 22 Ring Indicator (RI) 22 DCE Synchronization z Synchronization of clocks in transmitters and receivers z z clock drift causes a loss of synchronization z 1 0 0 0 Data T Example: assume ‘1’ and ‘0’ are represented by V volts and volts respectively z Correct reception Data Incorrect reception due T to incorrect clock (slower clock) S Clock S’ 1 Clock Synchronization (cont’d) z z Incorrect reception (faster clock) How to avoid a loss of synchronization? z z Asynchronous transmission Synchronous transmission 1 1 Data T S’ Clock 0 Asynchronous Transmission z z Avoids synchronization loss by specifying a short maximum length for the bit sequences and resetting the clock in the beginning of each bit sequence Accuracy of the clock? Data bits Line idle Start bit 3T/2 T T T T Receiver samples the bits T T T Stop bit Synchronous Transmission z Sequence contains data + clock information (line coding) z z z z i.e Manchester encoding, self-synchronizing codes, is used R transition for R bits per second transmission R transition contains a sine wave with R Hz R Hz sine wave is used to synch receiver clock to the transmitter’s clock using PLL (phase-lock loop) 0 1 Voltage time

Ngày đăng: 12/11/2019, 20:04

Từ khóa liên quan

Tài liệu cùng người dùng

  • Đang cập nhật ...

Tài liệu liên quan