... 9.66 22 .2 1.54 4. 02 6.90 22 .2 25.1 40.3 25 .1 ** 6.63 4 .20 * 37.1 25 .1 8.8 21 .3 24 .97 21 .3 * 44.3 3.95 *** 16.4 21 .3 11.6 21 .0 27 .2 21.0 * 43.1 3.98 *** 14.4 21 .0 51.4 22 .7 *** 2. 58 3.96 59.4 22 .7 ... < 0.001 5.6 11 .2 21 .2 21.4 35.9 13.0 21 .2 21.4 ** 32. 9 38.6 3.90 3.93 *** *** 17.3 16.4 21 .2 21.4 8 .2 20.9 22 .9 20 .9 * 55.7 3.91 *** 16.8 20 .9 13.3 22 .6 23 .3 22 .6 23 .9 4.0 19.4 22 .6 Müller et ... 22 .9 *** 17.9 22 .9 16.5 4.01 *** 22 .0 22 .9 *** 81.4 69.0 22 .7 20 .8 *** *** 6.87 8.43 22 .7 20 .8 70.3 53.1 4.05 3.94 *** *** 10.6 14.3 22 .7 20 .8 4.05 42. 5 25 .0 ** 8. 42 25.0 14.6 4.05 *** 11.8 25 .0...
Ngày tải lên: 11/08/2014, 15:22
... the head and neck N Engl J Med 20 01, 344:1 125 -1131 Syrjanen S: Human papillomavirus infections and oral tumours Med Microbiol Immunol 20 03, 1 92: 123 - 128 Gillison ML: Human papillomavirus and prognosis ... papillomavirus and a subset of head and neck cancers J Natl Cancer Inst 20 00, 92: 709- 720 12 Gillison ML, Lowy DR: A causal role for human papillomavirus in head and neck cancer Lancet 20 04, 363:1488-1489 ... Natl Cancer Inst 20 09, 101:4 12- 423 25 Wang SS, Hildesheim A: Viral and host factors in human papillomavirus persistence and progression J Natl Cancer Inst Monogr 20 03, 31:35-40 26 Weinberger PM,...
Ngày tải lên: 11/08/2014, 20:20
Advanced Verification Flow part 2
... popular tools that employ formal and semi-formal verification methods 15.3 .2 Equivalence Checking After logic synthesis and place and route tools create gate level netlist and physical implementations ... both the RTL and gate level representations of the design and mathematically prove that they are functionally equivalent Thus, functional verification can focus entirely on RTL and there is little ... possible behavior and will wrongly report the design as "proven." Figure 15-8 shows the verification flow with a formal verification tool In the best case, the tool either proves a particular assertion...
Ngày tải lên: 20/10/2013, 16:15
Tài liệu Longman preparation series for the toeic test advanced part 2 ppt
Ngày tải lên: 24/12/2013, 11:17
Tài liệu Mastering skills for the toefl ibt advanced part 2 doc
Ngày tải lên: 22/01/2014, 02:20
CIS 185 Advanced Routing ProtocolsEIGRP Part 2 pot
... address 1 92. 168.1.101 25 5 .25 5 .25 5.0 no ip split-horizon eigrp 110 frame-relay map ip 1 92. 168.1.1 02 1 02 broadcast frame-relay map ip 1 92. 168.1.103 103 broadcast router eigrp 110 network 1 92. 168.1.0 ... address 1 92. 168.1.101 25 5 .25 5 .25 5.0 no ip split-horizon eigrp 110 frame-relay map ip 1 92. 168.1.1 02 1 02 broadcast frame-relay map ip 1 92. 168.1.103 103 broadcast router eigrp 110 network 1 92. 168.1.0 ... frame-relay ip address 1 92. 168.1.103 25 5 .25 5 .25 5.0 frame-relay map ip 1 92. 168.1.101 130 broadcast frame-relay map ip 1 92. 168.1.1 02 130 broadcast router eigrp 110 network 1 92. 168.1.0 The adjacencies...
Ngày tải lên: 05/03/2014, 17:20
Hydrodynamics Advanced Topics Part 2 pdf
... moduli λ and μ and coefficients of shear and volume viscosities are expressed as μ=μ− 2 b2 b ( a1γ − aγ ) b2 b2 , λ = λ + a − a b1 − b2 ( aγ − a γ ) , η = γ a , 2 a2 2 (39) ... equations of items 2. 9.1 and 2. 9 .2 confirm that the normalized turbulent fluxes are expressed as functions of n and only, while the covariances may be expressed as functions of n, , and 2 2. 10 Transforming ... 2 20 Hydrodynamics – Advanced Topics 2 D f n n n n 2 f 2 n z2 1 1 1 (1 n) f n f 2 ...
Ngày tải lên: 18/06/2014, 22:20
Hydrodynamics Advanced Topics Part 2 pot
... moduli λ and μ and coefficients of shear and volume viscosities are expressed as μ=μ− 2 b2 b ( a1γ − aγ ) b2 b2 , λ = λ + a − a b1 − b2 ( aγ − a γ ) , η = γ a , 2 a2 2 (39) ... equations of items 2. 9.1 and 2. 9 .2 confirm that the normalized turbulent fluxes are expressed as functions of n and only, while the covariances may be expressed as functions of n, , and 2 2. 10 Transforming ... 2 20 Hydrodynamics – Advanced Topics 2 D f n n n n 2 f 2 n z2 1 1 1 (1 n) f n f 2 ...
Ngày tải lên: 19/06/2014, 10:20
Advanced Biomedical Engineering Part 2 docx
... by Age Adjusted SI by Age 22 20 20 18 18 SI for Standard Pulse Rate 22 16 SI 14 12 10 16 14 12 10 2 20 40 60 80 100 20 Age Age vs SI Plot Regr 40 60 80 100 Age Age vs Standard SI Plot Regr Fig ... Audiology, Vol 20 , pp 181 -21 1 Ebata, M (20 03) Spatial unmasking and attention related to the cocktail party problem Acoust Sci and Tech , Vol 24 , pp 20 8 -21 9 Egan, J., Carterette, E., and Thwing, ... Normalization shift (dB) [ -20 -15 -10 -5 5] -3 [ -20 -15 -10 -5 5] -5 0. 12 m [ -20 -15 -10 -5 5] -8 1m [ -20 -15 -10 -5 5] 0 .25 m [ -20 -15 -10 -5 5] +3 0. 12 m Lateral target 1m 0 .25 m [ -20 -15 -10 -5 5] +6...
Ngày tải lên: 19/06/2014, 12:20
Advanced Model Predictive Control Part 2 pptx
... Control Systems Technology 12( 2): 23 5 -24 9 Vandenberghe, L (20 10) The CVXOPT linear and quadratic cone program solvers working manuscript Vandenberghe, L & Boyd S (20 04) Convex Optimization, Cambridge ... 599- 623 (20 00) Abonyi, J and Babuska.R and Abotto M and Szeifert F and Nagy L Identification and control of nonlinear systems using fuzzy Hammerstein models, Ind.Eng.Chem.Res, 39 43 02- 4314. (20 00) ... Vehicles 27 25 Ling, K.V.; Wu, B.F & Maciejowski, J.M (20 08) Embedded model predictive control (MPC) using a FPGA, Proc 17th IFAC World Congress, pp 1 525 0-1 525 5 Maciejowski, J.M (20 02) Predictive...
Ngày tải lên: 19/06/2014, 19:20
Fundamental and Advanced Topics in Wind Power Part 2 potx
... V1 V2 ] S(V 12 V 22 2V1V2 2V2 ) S(V 12 3V2 2VV2 ) 0 (27 ) Solving the resulting equation by factoring it yield Eqn 28 (V 12 3V 22 2V1V2 ) (V1 V2 )(V1 3V2 ) Equation ... speed V2, applying the chain rule of differentiation and equating the result to zero as: dP dV2 V1 d [V1 V2 V1 V2 ] S dV2 d [ V 12 V2 V1 V2 ] S dV2 S[ V 12 V2 2V2 V1 ... SV V 12 V 22 SV V1 V2 The last expression implies that: (V1 V 22 ) (V1 V2 )(V1 V2 ) 2 V (V1 V2 ), V , S , or: V (V1 V2 ), (V1 V2 ) or V1 V2 (9) This...
Ngày tải lên: 19/06/2014, 21:20
Advanced Radio Frequency Identification Design and Applications Part 2 potx
... s11 s 12 V1 V1 (51) − = s + s 22 V2 V2 21 − − According to the above matrix, the V1 and V2 can be written into Equation 52 − + + V1 = s11 V1 + s 12 V2 (52a) − V2 (52b) = + s21 V1 + + s 22 V2 − + ... 44 − |V+ + V0 |2 Rchip |V+ |2 |1 + s L |2 Rchip V |I0 |2 Rchip = | |2 Rchip = = 2 Zchip 2| Zchip |2 2| Zchip |2 (44) As mentioned before, the transmission line between the chip and the tag antenna ... Equation 45 and Equation 49 into Equation 52, solving for V1 /V1 and − /V+ gives V2 − V1 s 12 s21 s L (53) + = Γrant = s11 − s s − V1 22 L 21 Operating Range Evaluation of RFID Systems − V2 + V1 Hence,...
Ngày tải lên: 19/06/2014, 23:20
Advanced Topics in Mass Transfer Part 2 pot
... Mixing and Transport Comb-Like and Random Jet Jet Array Stirring Systems Controlled Mixing and Transport in in Comb-Like and Random Array Stirring Systems S (mm) σCu 10 13. 32 14 8.87 49 24 3.73 ... Comb-Like and Random Jet Jet Array Stirring Systems Controlled Mixing and Transport in in Comb-Like and Random Array Stirring Systems 17 59 Datta, M & Landolt, D (20 00) Fundamental aspects and applications ... Controlled Mixing and Transport in Comb-Like and Random Jet Array Stirring Systems S Delbos1 , E Chassaing1 , P P Grand2 , V Weitbrecht3 and T Bleninger4 Institute for Research and Development...
Ngày tải lên: 19/06/2014, 23:20
Advanced Trends in Wireless Communications Part 2 potx
... 1) 2 , Lμ, Lμ; 2Lμ + 1; , 2 + At2 + Bt2 (18) whereas (17) is reduced to: uvMγ ( u +v ) u2 + Au2 ( 3) F , , 1, 1, Lμ, Lμ; 2Lμ + ; 2 u + v2 + Au2 + Av2 2 (u2 + v2 )(4Lμ + 1) D Kiid (u, v) = 2 + ... L u2 + A L u2 u2 , ,··· , ,··· , u + v2 + A u + A v2 + A L u2 + A L v2 + B1 u2 + B1 v2 + B L u + B L v2 (17) For i.i.d branches (16) reduces to: Γ (2Lμ + 1 /2) Jiid (t) = √ Mγ (t2 /2) F1 πΓ (2Lμ ... obtained, as: K asym (u, v) = 2d 2 Cuv π ( u + v2 ) d + 1 (1 − x )d−1 /2 − u2 x u + v2 − d −1 dx 2d−1 Cuv u2 = F1 1, d + 1; d + ; 2 u + v2 π (u2 + v2 )d+1 (2d + 1) (26 ) Fig Average Symbol Error...
Ngày tải lên: 19/06/2014, 23:20
Advanced Transmission Techniques in WiMAX Part 2 ppt
... 1570 29 60 49.4, 1600 26 50 51 .2, 1570 26 50 13 .2, 520 0–5935 10.0, 5305–5865 27 .9, 5060–6705 Fig 15(c) 26 26 26 2 20 20 20 9.5 9.5 9.5 7.0 9.0 11 58.7,1610 29 50 57.8, 1610 29 20 37.4, 1610 23 50 9.3, ... Antennas win WS Lin L2 L3 Fig 15(a) 30 30 21 - 7.5 6.0 11.0 - - 61.0, 1600–3000 58.5, 1 620 29 60 58 .2, 1630 29 70 7.5, 4880– 526 0 5.8, 5180–5490 16.1, 5040–5 920 Fig 15(b) 26 26 26 2 20 20 20 - 6.0 8.0 10 ... bandwidths not just cover the WiFi bands of 2. 4 GHz (2. 4 -2. 484 GHz) and 5 .2 GHz (5.15-5 .25 GHz), but also the licensed WiMAX bands of 2. 5 GHz (2. 5 -2. 69 GHz) and 3.5 GHz (3.4 -3.69 GHz) Fig 20 ...
Ngày tải lên: 20/06/2014, 23:20
Advanced Microwave Circuits and Systems Part 2 ppt
... of the transfer matrix T defined by a1 b1 T= T11 T21 T11 T21 =T b2 a2 = T 12 T 22 = − S211 − S11 S211 T 12 T 22 b2 a2 , − −S211 S 22 − S 12 − S11 S211 S 22 (70) , (71) A thru-only de-embedding method for ... b2 = S11 S21 S 12 S 22 a1 a2 = Sa (64) 26 Advanced Microwave Circuits and Systems S′, S′̃ S, T, S̃ 1′ a1 2 n +2 4′ n+3 6′ 7′ 5′ n+1 3′ b1 n+4 8′ a2 b2 2n-port (2n−3)′ n−1 (2n−1)′ 2n−1 n 2n (2n 2) ′ ... similarity transformation: − − − W1 S211 S 22 S 121 S11 W1 = Λ1 , (67) − − − W2 S 22 S 121 S11 S211 W2 = 2 , (68) where Λ1 and 2 are diagonal matrices W1 and W2 can be computed by eigenvalue decom˜...
Ngày tải lên: 21/06/2014, 06:20