Lecture 2 ArithCode
Lecture 2 Wireless Environment and Wireless LANs
... (OFDM) ● IEEE 8 02. 11b (Wi-Fi) ■ 11 Mbps (with fallback to 5.5, 2 and 1 Mbps) in the 2. 4 GHz band ■ Uses DSSS ● IEEE 8 02. 11g ■ 20 + Mbps in the 2. 4 GHz band Wireless Environment and Wireless LANs 32 WLANs/WPANs: ... for operation at 1 and 2 Mbps ■ DSSS, FHSS or infrared ■ Extensions (IEEE 8 02. 11b, IEEE 8 02. 11a, etc.) allow for operation at higher data rates and (in the cas...
Ngày tải lên: 29/10/2013, 06:15
Modulation and coding course- lecture 2
... AGC Quantization noise variance: 2 Sat 2 Lin 22 2 )()(})]({[ σσσ +==−= ∫ ∞ ∞− dxxpxexqx q E ll L l l qxp q )( 12 2 12/ 0 2 2 Lin ∑ − = = σ Uniform q. 12 2 2 Lin q = σ Lecture 2 14 Uniform and non-uniform ... (M> ;2) requires less bandwidth than binary PCM. For a given average pulse power, binary PCM is easier to detect than M-ary PAM (M> ;2) . Mk 2 log= Lecture...
Ngày tải lên: 08/11/2013, 18:15
Tài liệu lecture 2: Communication Concepts ppt
... http://www.informationfactory.com/digsec1.pdf lecture 2. doc Page 1 (3) CSN200 Introduction to Telecommunications, Fall 1999 Lecture_ 02 Parallel vs. Serial Transmission: Parallel: In ... CSN200 Introduction to Telecommunications, Winter 20 00 Lecture_ 02 Communication Concepts Data Transmission Modes: There are ... mode of communication used on a PC’s serial p...
Ngày tải lên: 10/12/2013, 08:15
Tài liệu Lecture 2: HỆ ĐẾM-CƠ SỐ ppt
... conversion: 9F2 16 = 9 F 2 1001 1111 0010 = 100111110010 2 110100101 2 = 0001 1010 0101 2 = (1 A 5) 16 11010.11 2 = 0001 1010 . 1100 2 = 1 A . C 16 Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ... Lecture 2: HỆ ĐẾM-CƠ SỐ Biên soạn:Th.S Bùi Quốc Bảo (Base on Floyd, Pearson Ed.) Phần phân Lặp lại phép nhân tích phần nguyên 0. 125 x 2 0 .25 0 0 .25 x 2 0.5 0 0...
Ngày tải lên: 13/12/2013, 04:15
Tài liệu Kinh tế ứng dụng_ Lecture 2: Simple Regression Model ppt
... by: b 0 ± t(α /2; n -2) x SE(b 0 ) = (b 0 – t(α /2; n -2) x SE(b 0 ); b 0 + t(α /2; n -2) x SE(b 0 )) b 1 ± t(α /2; n -2) x SE(b 1 ) = (b 1 – t(α /2; n -2) x SE(b 1 ); b 1 + t(α /2; n -2) x SE(b 1 )) ... 0.570665305 R Square 0. 325 65889 Adjusted R Square 0.315 122 31 Standard Error 12. 721 84 023 Observations 66 ANOVA Df SS MS F Significance F Regression 1 50 02. 224 173...
Ngày tải lên: 27/01/2014, 11:20
Lecture 2 MATLAB fundamentals potx
... 12 18 -3 12 18 -3 B = B = 2 5 2 2 5 2 1 1 2 1 1 2 0 -2 6 0 -2 6 D = D = 12 18 -3 12 18 -3 12 18 -3 12 18 -3 2 5 2 2 5 2 2 5 2 2 5 2 1 1 2 1 1 2 1 1 2 1 ... example: c = 2+ 3 ^2+ 1/ c = 2+ 3 ^2+ 1/ ( ( 1 +2 1 +2 ) ) 1 1 st st c = 2+ 3 ^2+ 1/ c = 2+ 3 ^2+ 1/ 3 3 c = 2+ c = 2+ 3 3 ^ ^ 2 2 +1/(1 +2) +1/(1 +2...
Ngày tải lên: 07/03/2014, 08:20
lecture 2 spectroscopy
... 28 0 Azo -N = N- 26 2 Nitro -N=O 27 0 Thioketone -C =S 330 Nitrite -NO2 23 0 Conjugated Diene -C=C-C=C- 23 3 Conjugated Triene -C=C-C=C-C=C- 26 8 Conjugated Tetraene -C=C-C=C-C=C-C=C- 315 Benzene 26 1 Practice ... the analyte in the unknown 6 Typical Beer’s Law Plot y = 0.02x 0 0 .2 0.4 0.6 0.8 1 1 .2 0.0 20 .0 40.0 60.0 concentration (uM) A R 2 = 0.995 Characteristics of Beer’s...
Ngày tải lên: 08/03/2014, 15:35
CMOS VLSI Design - Lecture 2: Circuits & Layout docx
... How many transistors are needed? 20 10 (too many transistors)Y SD SD= + 4 4 D1 D0 S Y 4 2 2 2 Y 2 D1 D0 S CMOS VLSI Design 4th Ed. 1: Circuits & Layout 27 Transmission Gate Mux Nonrestoring ... Circuits & Layout 29 4:1 Multiplexer 4:1 mux chooses one of 4 inputs using two selects – Two levels of 2: 1 muxes – Or four tristates S0 D0 D1 0 1 0 1 0 1 Y S1 D2 D3 D0 D1 D2...
Ngày tải lên: 19/03/2014, 10:20