Lecture 2 - RF Fundamentals

... Radio FrequencyFundamentals(1 September 20 06) February 20 05 Copyright 20 05 All Rights Reserved 2 Describe the behavior of RF Explain the properties of an RF signal● Understand the ... Spectrum80085013001500Micro = 1 x 1 0-6 Nano = 1 x 1 0-9 Pico = 1 x 1 0- 12 February 20 05 Copyright 20 05 All Rights Reserved 9● Gain describes an increase in the RF signal's Amplitud...

Ngày tải lên: 13/09/2012, 11:16

... 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 ^ ^...

Ngày tải lên: 07/03/2014, 08:20

... b) Codelength m: m=-log2(r) = -log2(0.0000 125 ) = 19.6 =20 bits c) Bitrate: R =20 bits/10 chars = 2. 0 bits/char• Huffman coder: (1+3 +2+ 1+1+4+4 +2+ 1+1)/10 =2. 2 bits/char Properties ... r=0.5*0.1*0 .2* 0.5*0.50.1*0.1*0 .2* 0.5*0.5=0.00000 125 • The number of bits to represent a value in the interval [L, H)=[L, L+r) of size r: m=-log2r = -log2(0.0000 125 ) = 19.6 =20 bits ∏==...

Ngày tải lên: 26/10/2012, 11:57

Ngày tải lên: 29/10/2013, 06:15

... 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 quant.  Uniform ... (NRZ)  Return-to-zero (RZ) 1 0 1 1 0 0 T 2T 3T 4T 5T +V -V +V 0 +V 0 -V 1 0 1 1 0 0 T 2T 3T 4T 5T +V -V +V -V +V 0 -V NRZ-L Unipolar-RZ Bipolar-RZ Manchester Miller Di...

Ngày tải lên: 08/11/2013, 18:15

... 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 ... serially or sequentially bit-by-bit over a sin...

Ngày tải lên: 10/12/2013, 08:15

... 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  0.5 x 2 1 1  0  0. 125 ... 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...

Ngày tải lên: 13/12/2013, 04:15

... 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 )) ... 161.84 52 Total 65 15360.31818 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 19.1411184 2. 0 621 47087 9 .28 2131 1.82E-13 15. 021 509...

Ngày tải lên: 27/01/2014, 11:20

... 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 ... following table. [X](M) Absorbance Transmittance(%) E(L/mole-cm) L(cm) 30 20 00 1.00 0.5 25 00 1.00 2. 5 x 10 -3 0 .2 1.00 4.0 x 10 -5 50 5000 2. 0 x 10 -4 1...

Ngày tải lên: 08/03/2014, 15:35

... pull-up network – a.k.a. static CMOS pMOS pull-up network output inputs nMOS pull-down network Pull-up OFF Pull-up ON Pull-down OFF Z (float) 1 Pull-down ON 0 X (crowbar) CMOS VLSI Design 4th ... 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  Non...

Ngày tải lên: 19/03/2014, 10:20