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Telecommunications Engineering Topic 3: Modulation and FDMA

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## Telecommunications Engineering Topic 3: Modulation and FDMA

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**Telecommunications EngineeringTopic 3: Modulation and FDMA**James K Beard, Ph.D. (215) 204-7932 jkbeard@temple.edu http://astro.temple.edu/~jkbeard/ Topic 3**Attendance**Topic 3**Topics**• The Survey • Homework • Problem 3.30, p. 177, Adjacent channel interference • Problem 3.35 p. 178, look at part (a); part (b) was done in class • Problem 3.36 p. 178, an intermediate difficulty problem in bit error rate using MSK • Topics • Why follow sampling with coding? • Shannon’s information theory Topic 3**Survey Thumbnail**• Nine completed surveys • Six incomplete or just looks • Five did not open the survey • Selection biases results? • Only 45% of class gives all results • Other 55% can be best, worst, or more of the same Topic 3**Multiple Choice Questions**• All are replies on 1-5 basis • No answers were 1 or 5 • Questions • My background is appropriate • I understand sampled time/frequency domains • I am comfortable with readings and text Topic 3**Multiple Choice Summary**Topic 3**Study Difficulties Reported**• Problems • 2.5 p. 29 • 2.6 p. 33 • 2.14 p. 67 • 2.20 p. 77 • 2.22 p. 81 • Examples • 2.17 p. 75 • Theme example 1 p. 82 Topic 3**Suggestions**• More examples and homework problems worked through – already in progress • Go over more difficult homework problems before they are assigned • Warn and correct wrong answers – AIP • Discuss WHY as well as HOW Topic 3**Survey Summary**• Everybody is OK • Based on 40% sample • But, nobody answered 5’s • Need • More coverage of how and why • Fill in for prerequisites Topic 3**Homework**• Problem 3.30, p. 177, Adjacent channel interference • Problem 3.35 p. 178, look at part (a); part (b) was done in class • Problem 3.36 p. 178, an intermediate difficulty problem in bit error rate using MSK Topic 3**Problem 3.30 Page 177**• Adjacent channel interference Topic 3**Solution**• Power of signal in correct filter • Power of signal in adjacent channel Topic 3**Problem 3.35 p. 178 (a)**• Formulas in table 3.4 page 159 • Begin BPSK with • Integrate over Rayleigh distribution Topic 3**Evaluate the Integral**• Average BER is • Evaluation of integral is left as ETR Topic 3**Problem 3.36 p. 178**• Use MSK with a BER of 10-4 or better • AWGN • Use Table 3.4 or Figure 3.32, pp. 159-160 • SNR requirement is about 8.3 dB • Rayleigh fading • Use Table 3.4 p. 159 • Solve for SNR of about 34 dB Topic 3**Coding Follows Sampling**• Sampling • Simply converts base signal to elementary modulation form • Formatting for performance is left to coding • Coding • Removal of redundancy == source coding • Channel coding == error detection and correction capability added Topic 3**Shannon’s Information Theory**• First published in BSTJ article in 1948 • Builds on Nyquist sampling theory • Adds BER concepts to find maximum flow of bits through a channel limited by • Bandwith • SNR • Channel capacity maximum is Topic 3**Other Important Results**• Channel-coding theorem • Given • A channel capacity CB/S • Channel bit rate less than channel capacity • Then • There exists a coding scheme that achieves an arbitrarily high BER • Rate distortion theory – sampling and data compression losses exempt from channel-coding theorem Topic 3**Concept of Entropy**• Definition – Average information content per symbol • Importance • Fundamental limit on average number of bits per source symbol • Channel-coding theorem is stated in terms of entropy Topic 3**Equation for Entropy**Topic 3**Study Problems and Reading Assignments**• Reading assignments • Read Section 4.6, Cyclic Redundancy Checks • Read Section 4.7, Error-Control Coding • Study examples • Example 4.1 page 197 • Problem 4.1 page 197 Topic 3**Problem 2.4 p. 28**• 4 GHz microwave link • Towers 100 m and 50 m tall, 3 km apart • Midway between, tower 70 m tall • Radius of Fresnel zone, eq. (2.38) p. 27 • Distance d1 = d2 = 1.5 km • Raise both towers Topic 3**Problem 2.5 p. 29**• Similar to 2.4 but LOS is clearly obstructed • Fresnel-Kirchoff diffraction parameter eq. (2.39) is • Diffraction loss is 24 dB • For 400 MHz, v = 1.096, loss = 16 dB Topic 3**Term Projects**• Areas for coverage • Propagation and noise • Free space • Urban • Modulation & FDMA • Coding • Demodulation and detection • Will deploy over Blackboard this week Topic 3**Term Project Timeline**• First week • Parse and report your understanding • Give estimated parameters including SystemView system clock rate • Second week • Block out SystemView • Signal generator • Modulator • Through mid-April • Flesh out as class topics are presented • Due date TBD Topic 3**EE320 Digital Telecommunications**Quiz 1 Report February 21, 2005 Topic 3**The Curve**Topic 3**The Answers**• See previous lectures/slides • Questions 1, 4 • See Excel spreadsheets • Questions 2, 3, 5 • See Mathcad spreadsheet • Fresnel integrals for diffraction loss Topic 3**Question 2**Topic 3**Question 3**Topic 3**Question 5**Topic 3**The Quiz in the Text (1 of 2)**• Question 1 • Text pp 3-5 • Lectures and slides several times • Question 2 • Antenna gain equations (2.2), (2.3) pp 14 • Also equations (2.9), example 2.2, pp 16-17 • Question 3 • Section 2.3.2 and exa,[;e 2.3 pp 24-29 • Two lectures, example worked in class, practice quiz Topic 3**The Quiz in the Text (2 of 2)**• Question 4 • Problem 3.30 p. 177 • Given in class • Answer was to give equation that was given in the problem statement • Question 5 • Problem 3.36, given in class • Use table 3.4, figure 3.33, pp. 159-161 Topic 3**EE320 Convolutional Codes**James K Beard, Ph.D. Topic 3**Bonus Topic: Gray Codes**• Sometimes called reflected codes • Defining property: only one bit changes between sequential codes • Conversion • Binary codes to Gray • Work from LSB up • XOR of bits j and j+1 to get bit j of Gray code • Bit past MSB of binary code is 0 • Gray to binary • Work from MSB down • XOR bits j+1 of binary code and bit j of Gray code to get bit j of binary code • Bit past MSB of binary code is 0 Topic 3**Polynomials Modulo 2**• Definition • Coefficients are ones and zeros • Values of independent variable are one or zero • Result of computation is taken modulo 2 – a one or zero • The theory • Well developed to support many DSP applications • Mathematical theory includes finite fields and other areas Topic 3**Base Concept – Signal Polynomial**• Pose data as a bit stream • Characterize data as impulse response of a filter with weights 1 and 0 • Characterize as z transform • Substitute D for 1/z in transfer function • Example • Signal 11011001 • Filter is 1 + (1/z) +(1/z)3+(1/z)4+ (1/z)6 • Signal polynomial is 1 + D + D2+D4+D6 • Signal and filter polynomials provide base method for understanding convolutional codes Topic 3**Base Concept –Modulo 2 Convolutions**• Scenario • Bitstream into convolution filter • Filter weights are ones and zeros • Output is taken modulo 2 – i.e. a 1 or 0 • Result: A modulo 2 convolution converts one bit stream into another Topic 3**Benefits of Concept**• Convolution is product of polynomials • Conventional multiplication of polynomials is isomorphic to convolution of the sequence of their coefficients • Taking the resulting coefficients modulo 2 presents us with the output of a bitstream into a convolution filter with output modulo 2 • These special polynomials have a highly developed mathematical basis • Implementation in hardware and software is very simple Topic 3**Error-Control Coding**• Two categories of channel coding • Forward error-correction (EDAC) • Automatic-repeat request (handshake) • CRC Codes • Hash codes of the message • Error detection, but not correction Topic 3**Topics in Convolutional Codes**• The node diagram is a block diagram • Polynomial representations • Represent signals, convolutions and special polynomials and polynomial operations • Give us a simple way to understand and analyze convolutions • Trellis diagrams • Give us a mechanism to represent convolution operations as a finite state machine • Provide a first step in visulaization of the finite state machine • Node diagrams • Provide a simple visualization of the finite state machine • Provide a basis for very simple implementation Topic 3**Convolutional Code Steps**• Reduce the message to a bit stream • Operate using modulo-2 convolutions • Convolution filter with short binary mask • Take result modulo 2 • Implemented with one-bit shift registers with multiplexer (see Figure 4.6 p. 196) Topic 3**Path 2**Input 1/z 1/z Path 1 Haykin & Moher Figure 4.6 p. 196 Example 4.1 Page 197 Output Topic 3**Example 4.1 (1 of 4)**• Response of Path 1 • Response of Path 2 • Mutiplex the outputs bit by bit • One side output, then the other • Produce a longer bit stream Topic 3**Example 4.1 (2 of 4)**• Signal • Message bit stream (10011) • Message as a polynomial • Multiply the message polynomial by the Path 1 and Path 2 filter polynomials • Obtain two bit streams from resulting polynomials • Multiplex (interleave) the results Topic 3**Example 4.1 (3 of 4)**• Polynomial multiplication results • Messages • Path 1 (1011111) • Path 2 (1111001) • Multiplexing them (11, 10, 11, 11, 01, 01, 11) Topic 3**Example 4.1 (4 of 4)**• Length of coded message is • Twice the order of the product polynomials +1 • 2.(length of message + length of shift registers - 1) = 2.(5 + 3 - 1)=2.7=14 • Shift registers have memory • Simplest way to clear is to feed zeros • Number of clocks is number of stages • Zeros between message words are tail zeros Topic 3**Problem 4.1**• Signal and polynomial 1 • Signal and polynomial 2 • Result Topic 3**Modulo 2 Convolution Diagrams**Topic 3**Trellis and State Diagrams**• Trellis diagram Figure 4.7 p. 198, and state table 4.2 p. 199 • Horizontal position of node represents time • Top line represents the input • Each row represents a state of the two-path encoder – a finite state machine • Trace paths produced by input 1’s and 0’s • Paths produced by 0’s are solid • Paths produced by 1’s are dotted Topic 3