fundamentals of wireless communication tse solution manual

... hugemongous. 46 EE 290S: Fundamentals of Wireless Communication Course Notes U.C. Berkeley Fall 2002 Instructor: David Tse Co-written with: Pramod Viswanath September 18, 2002 fades of the channel (i.e. ... 1000 (30 dB). Stupendous amounts of power would be required for more reliable communication. Before we explore further the root cause of the poor performance of this detector, we note that there ... gain, there is a law of diminishing marginal return: as L increases, the marginal benefit decreases because of the law of large numbers, eventually converging to the performance of the AWGN channel....

Ngày tải lên: 20/10/2013, 07:15

... Tradeoff 2: The Wireless Channel Fundamentals of Wireless Communication, Tse& amp;Viswanath Fundamentals of Wireless Communication David Tse 23 2: The Wireless Channel Fundamentals of Wireless ... evaluation. 25 2: The Wireless Channel Fundamentals of Wireless Communication, Tse& amp;Viswanath 2: The Wireless Channel Fundamentals of Wireless Communication, Tse& amp;Viswanath 2. The Wireless Channel ... delays of 1/W . 2: The Wireless Channel Fundamentals of Wireless Communication, Tse& amp;Viswanath 1. Introduction 10 2: The Wireless Channel Fundamentals of Wireless Communication, Tse& amp;Viswanath Wireless...

Ngày tải lên: 11/07/2014, 21:20

... Hz. ã Primary driver behind wireless communication system design. Fundamentals of Wireless Communication David Tse Dept of EECS U.C. Berkeley Main story ã Communication over a flat fading ... distance of the order of the carrier wavelength. This is 0.3 m for Ghz cellular. ã For vehicular speeds, this translates to channel variation of the order of 100 Hz. ã Primary driver behind wireless ... fading ã Wireless communication typically happens at very high carrier frequency. (eg. f c = 900 MHz or 1.9 GHz for cellular) ã Multipath fading due to constructive and destructive interference of...

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... president of the National Academy of Sciences. The National Academy of Engineering was established in 1964, under the charter of the National Academy of Sciences, as a parallel organization of outstanding ... development of models of several heights for men, women, and chil- dren of various ages for use in the characterization of SAR distributions for exposures characteristic of cell phones, wireless ... development of models of several heights for men, women, and children of various ages for use in the characterization of SAR distributions for exposures characteristic of cell phones, wireless...

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... xxiii About the Authors xxvii Part I Overview of WiMAX 1 Chapter 1 Introduction to Broadband Wireless 3 1.1 Evolution of Broadband Wireless 5 1.1.1 Narrowband Wireless Local-Loop Systems 5 1.1.2 First-Generation ... Professor Robert Heath and Mr. Runhua Chen from the University of Texas at Austin. Both Professor Heath and Mr. Chen possess an incredi- ble degree of intuition and understanding in the area of ... comprehensive treatment of WiMAX. It should be of great interest. —Dr. Reinaldo Valenzuela, Director of Wireless Research, Bell Labs Fundamentals of WiMAX is a comprehensive guide to WiMAX from...

Ngày tải lên: 27/04/2013, 08:58

... solve for one variable in terms of the other(s). 6 In the process of establishing the above analogy, we first establish the concepts of the first and second derivatives of an integral (13) about a ... end-point values of y 0 (x)orη(x)). Form the family of arcs y 0 (x)+η(x) x 1 ≤ x ≤ x 2 (16) x 1 [ x 2 ] y 0 Figure 9: Families of arcs y 0 + η Then for sufficiently small values of  say −δ ≤  ... The third part of the problem, the determination of an arc satisfying the sufficient conditions, is frequently the most difficult of all, and is the part for which fewest methods of a general character...

Ngày tải lên: 31/03/2014, 15:35

... nm Therefore the number of modes in the range 820-to-880 nm is N = 880 - 820 0.25 = 240 modes 4-14. (a) Let N m = n/λ = m 2L be the wave number (reciprocal wavelength) of mode m. The difference ... dBm level). Hence the gain is up 0.375 dB, yielding an output of –4.85 dBm + (7.1 + 0.375) dB = 2.63 dBm which is within 0.37 dB of the normal +3 dBm level. 11-13. First let 2 πν i t + φ i = θ i ... , we then have L SM;ff = -10 log       4 4.0446 = - 0.0482 dB 5-19. Plot of Eq. (5-44). 5-20. Plot of the throughput loss. 16 7-27. Using Eq. (E-10) and the relationship 1 1 + x a ...

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... dark because of the costs of compliance. The costs to comply with Sarbox can be several million dollars, which can be a large percentage of a small firms profits. A major cost of going dark ... $25 Solutions Manual Fundamentals of Corporate Finance 8 th edition Ross, Westerfield, and Jordan Updated 03-05-2007 B-28 SOLUTIONS Market-to-book ratio ... costs of raising funds in the public market. 5. The treasurer’s office and the controller’s office are the two primary organizational groups that report directly to the chief financial officer....

Ngày tải lên: 01/07/2014, 14:22

... RETURN OF A 50-50 PORTFOLIO OF HIGH TECH AND COLLECTIONS? OF HIGH TECH AND U.S. RUBBER? ANSWER: [SHOW S5-45 AND S5-46 HERE.] NOTE THAT THE BETA OF A PORTFOLIO IS SIMPLY THE WEIGHTED AVERAGE OF ... FOLLOWS: ∑ = = n 1i ii kPk ˆ . HERE P i IS THE PROBABILITY OF OCCURRENCE OF THE iTH STATE, k i IS THE ESTIMATED RATE OF RETURN FOR THAT STATE, AND n IS THE NUMBER OF STATES. HERE IS THE CALCULATION FOR HIGH ... “YES” JUST THINK OF THE PURCHASE OF A LIFE OR FIRE INSURANCE POLICY. THE FIRE INSURANCE POLICY HAS A NEGATIVE EXPECTED RETURN BECAUSE OF COMMISSIONS AND INSURANCE COMPANY PROFITS, BUT BUSINESSES...

Ngày tải lên: 01/07/2014, 21:28

... 1.8983 FVIF@7% for 11 yrs = 2.1049 Estimated n = 11 yrs. Answers and Solutions: 6 - 1 Chapter 6 Time Value of Money SOLUTIONS TO END -OF- CHAPTER PROBLEMS 10% 7% 6.5% 6-6 0 1 2 3 4 5 | | | | | | 300 ... $100/0.14 = $714.29. When the interest rate is doubled, the PV of the perpetuity is halved. 6-20 a. Begin with a time line: Answers and Solutions: 6 - 3 i = ? 7% following into your calculator: ... into your calculator: N = 5, I = 3, PV = 0, FV = 508.92, and solve for PMT = $93.07. Answers and Solutions: 6 - 5 6-1 0 1 2 3 4 5 | | | | | | PV = 10,000 FV 5 = ? FV 5 = $10,000(1.10) 5 = $10,000(1.61051)...

Ngày tải lên: 01/07/2014, 21:28

... Chapter 9 uses the rate of return concepts covered in previous chapters, along with the concept of the weighted average cost of capital (WACC), to develop a corporate cost of capital for use in ... will be accepted since their rates of return exceed the WACC. Answers and Solutions: 9 - 10 $3.54 million (60%) of equity capital and $2.36 million (40%) of debt capital. In this scenario, ... projects whose returns exceed their risk-adjusted costs of capital. The appropriate costs of capital are summarized below: Required Rate of Cost of Project Investment Return Capital A $4 million...

Ngày tải lên: 01/07/2014, 21:28

... VALUE OF THE TERMINAL VALUE OF THE INFLOWS, COMPOUNDED AT THE COST OF CAPITAL, TO THE PRESENT VALUE OF THE COSTS. HERE IS THE SETUP FOR CALCULATING PROJECT L’S MODIFIED IRR: 0 1 2 3 | | | | PV OF ... THE SLOPE OF THE NPV PROFILE DEPENDS ENTIRELY ON THE TIMING PATTERN OF THE CASH FLOWS LONG-TERM PROJECTS HAVE STEEPER NPV PROFILES THAN SHORT-TERM ONES. THUS, WE CONCLUDE THAT NPV PROFILES CAN ... LEVEL OF INTEREST RATES. THIS IS CALLED THE PROJECT COST OF CAPITAL IN CAPITAL BUDGETING. 4. FIND (A) THE PV OF THE EXPECTED CASH FLOWS AND/OR (B) THE ASSET’S RATE OF RETURN. 5. IF THE PV OF THE...

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