30 1 fundamentals of oscillators

... Frequency fT 44 1. 10.4 MOSFETs y Parameters 44 1. 10.5 Dual Gate MOSFETs 45 1. 11 Diode Detectors 48 1. 11. 1 52 Minimum Detectable Signal Level – Tangential Sensitivity 1. 12 Varactor Diodes 52 1. 13 Passive ... (1 + 20 ) + 11 .6 = 32.6pF 21 (1. 64) As: f 3dB = rb 'e + rbb ' + Rs 2π [CT ](rbb ' + Rs )rb 'e (1. 65) 250 + 10 + 50 = 10 1MHz 2π 32.6 10 12 (10 + 50)250 (1. 66) then: f 3dB = [ ] The value of S 21 ... the BFG25A, are shown in Figure 1. 12 10 S 21 0 .1 10 10 0 10 10 Frequency, MHz Figure 1. 12 S 21 for a typical bipolar transistor operating at 1mA 26 Fundamentals of RF Circuit Design It has therefore...

Ngày tải lên: 08/04/2013, 10:50

... 1- 1 1- 2 1- 3 1- 4 1- 5 1- 6 1- 7 1- 8 1- 9 1- 10 30 CHAPTER I What are the three principal forms of business organization? What are the advantages and disadvantages of each? Would the “normal” rate of ... 20 01 Accounts payable $ 60 2000 $ 30 Accounts receivable 375 315 Notes payable 11 0 60 Inventories 615 415 Accruals 14 0 13 0 Total current assets $1, 000 $ 810 $ 310 $ 220 Net plant and equipment 1, 000 ... taxes (EBIT, or operating income) $ 283.8 $ 263.0 88.0 60.0 $ 19 5.8 $ 203.0 78.3 81. 2 $ 11 7.5 $ 12 1.8 4.0 4.0 Net income $ 11 3.5 $ 11 7.8 Common dividends $ 57.5 $ 53.0 Addition to retained earnings...

Ngày tải lên: 25/01/2014, 23:20

... link (soft link) from t2 file to t1 file Which of the following commands is right? a) ln –s t1 t2 b) ln /a/b/t1 /a/b/t2 c) ln –s /a/b/t1 /a/b/t2 Page of (exam #1) 39) Your umask is set to 012 If ... the middle 15 lines of a 45 line text file named test.txt Which of the following commands will you use? a) head –n 16 -30 test.txt b) tail +16 test.txt > head -15 test.txt c) head –n 30 test.txt ... Which one of the following commands copies files with the dat extension from /tm1 into /tm2? a) mv /tm1/*.dat /tm2 b) cp /tm1/*.dat /tm2 c) cp -p /tm2 < /tm1/*.dat END Page of (exam #1) ...

Ngày tải lên: 18/02/2014, 21:20

... furniture Office Equipment Total capital assets being depreciated Less accumulated depreciation for: Office furniture Office Equipment Decreases 23,772 273, 713 17 ,762 315 ,247 23 ,19 4 243,5 81 116 23, 310 ... www.adultpdf.com 35 REGIONAL OFFICE OF EDUCATION #1 NOTES TO FINANCIAL STATEMENTS JUNE 30, 2009 Note 11 - DUE FROM WCR At June 30, 2009, the Regional Office of Education #1 was due $12 ,490 from the West ... Date 12 - 31- 08 12 - 31- 07 12 - 31- 06 Three-Year Trend Information for the Regular Plan Percentage Annual Pension ofAPC Net Pension Cost (APC) Contributed Obligation 22 ,14 5 10 0% $0 22,933 10 0% o 17 ,768...

Ngày tải lên: 19/06/2014, 19:20

... Interest income Total revenues 90, 411 24, 518 11 4,929 13 5,056 20 ,12 7 79,534 8,6 61 20,244 6,490 84,068 7,864 19 ,2 61 5,934 (4,534) 797 983 556 4,378 16 2 12 1,667 (4,378) 11 4,929 Expenditures: Salaries ... (9,986) $ 3, 312 Fund balance - June 30, 2009 52 (15 ,557) 4 ,10 6 Fund balance - July 1, 2008 $ 5,5 71 52 (794) 51, 192 51, 192 _ 5 _1, 194 $ Net changes in fund balances 35,454 288,5 91 99,897 24 ,16 4 7,078 ... Fund balance - June 30, 2009 (16 4) (66) 13 ,389 Net changes in fund balances 20,000 80 ,18 4 4,378 16 2 12 1,667 307 4, 218 4,702 80 ,11 8 71, 580 2,577 4,277 402 70,702 234 17 3,398 16 4 $ Student Assistance...

Ngày tải lên: 19/06/2014, 19:20

... Variance Positive (Negative) 13 0, 10 2 15 (15 ) 13 0, 10 1 13 0, 10 4 (3) 13 0, 10 1 13 0, 11 9 (18 ) (17 ) ,;$:: ==, :( ~17 : ) $ 2,269 Fund balance, July 1, 2008 $ Fund balance, June 30, 2009 2,252 This is trial ... June 30, 2009 $ $ 2,000 2,000 2,000 I 2,0 01 Supervisory Fund $ $ 14 ,609 14 ,268 3 41 1,680 1, 0 61 619 852 13 2,0 21 1 ,15 6 Bus Driver Fees See accompanying Independent Auditor's Report 51 7, 519 724 ... ENDED JUNE 30, 2009 DISTRIBUTIVE FUND Balance July 1, 2008 Additions Reductions 1, 274 $ 1, 219 , 516 $ 1, 220,646 1, 274 1, 219 , 516 Balance June 30, 2009 1, 220,646 ASSETS $ Cash Total assets $ 14 4 14 4 LIABILITIES...

Ngày tải lên: 19/06/2014, 19:20

... other 18 1 Plural forms of other: other(s) vs the other(s) 18 3 Summary of forms of other 18 6 Chapter MODAL AUXILIARIES 7 -1 7-2 7-3 7-4 7-5 7-6 7-7 7-8 7-9 7 -10 7 -11 7 -12 7 -13 7 -14 ~ ... 15 2 Chapter NOUNS AND PRONOUNS il:~ 6 -1 6-2 6-3 6-4 6-5 6-6 6-7 6-8 6-9 6 -10 6 -1 6- 12 6 -13 6 -14 6 -15 6 -16 c ? Vlll CONTENTS Pronunciation of final -s/-es 15 7 Plural forms of nouns ... TlME RSN 1- 1 1- 2 1- 3 1- 4 1- 5 1- 6 1- 7 Chapter The simple present and the present progressive Forms of the simple present and the present progressive Frequencyadverbs Final-8 12 Spelling...

Ngày tải lên: 01/07/2014, 14:20

... more than six times? draw well-for example, draw a picture of me? 10 cook? 11 walk on your hands? 12 play tennis? 13 program a computer? 14 write legibly with both your right hand and your left ... (three) stories* and is made of (brick) has four legs and is found on a farm is green and we can see it out that window 10 is sweet and you can eat it 13 EXERCISE 11 Expressing posslblllty: COULD ... could also My f i e n d ( Of course, I may But 1' 11 I may this weekend, but I'm not sure HelShe might But helshe'll probably One hundred years from now, will probably 17 - may could POLITE QUESTIONS:...

Ngày tải lên: 01/07/2014, 14:20

... Basic Science 1 6 8 9 11 13 13 14 14 14 15 15 17 19 21 21 22 22 24 24 25 27 27 28 29 29 30 31 33 34 Biomechanics of the Spine Stephen Ferguson Core Messages ... Disorders Fundamentals of Diagnosis and Treatment Norbert Boos · Max Aebi (Editors) Spinal Disorders Fundamentals of Diagnosis and Treatment With 274 Figures in 12 90 Parts and 19 0 Tables Prof Dr ... mean difficult choices in the allocation of treatment modalities Therefore, a basic knowledge of the state of the art of the diagnosis and treatment of spinal disorders is required, not only...

Ngày tải lên: 02/07/2014, 06:20

... therapeutic efficacy of facet joint blocks in relieving pain attributed to facet joints [ 21] Carette et al [ 21] selected 11 0 out of 19 0 patients who experienced pain relief of more than 50 % after ... Sagittal view of CT/discogram showing contrast medium extension to the margin of the disc b Corresponding MRI of the disc The diagnostic value of discography remains a matter of debate of % and a ... needle placement Of 807 injected cervical discs, Grubb et al [47] had a rate of discitis of 0.37 % corresponding to 1. 7 % patients with discitis treated In Zeidmann’s [13 6] review of 400 diagnostic...

Ngày tải lên: 02/07/2014, 06:20

... Promotion Limitations Conclusions 71 74 76 77 77 85 88 91 92 93 94 95 97 97 10 2 10 4 10 5 10 5 10 7 11 1 11 2 11 5 12 2 12 4 12 7 13 2 13 2 13 3 13 6 13 9 14 0 14 3 14 4 14 4 14 5 14 9 15 0 A Commentary on Business Marketing: ... 210 211 212 212 217 218 219 220 220 222 225 229 229 230 235 237 243 243 247 248 249 250 Reply to the Commentaries: Business Marketing Comes of Age 253 David A Reid Richard E Plank Reply to Professor ... Wilson 19 68; Hill, Alexander, and Cross 19 75; Webster 19 84; Hutt and Speh 19 81, 19 96; Anderson and Narus 19 99) Industrial marketing has been around for more than 16 0 years (Frederick 19 34) and...

Ngày tải lên: 06/07/2014, 07:20

... 11 -3 The /etc/profile Script 11 -5 Login Sequence 11 -6 The dtprofile 11 -10 Shells .11 -11 The ENV Environment Variable 11 -14 Command Format .11 -14 ... Variable 11 -14 Exercise: Setting Initialization Files 11 -16 Tasks .11 -16 Exercise Summary .11 -17 Exercise Solutions 11 -18 Check Your Progress 11 -19 Think ... Solutions 10 -19 Check Your Progress 10 - 21 Think Beyond 10 -22 Initialization Files 11 -1 Relevance 11 -2 Features of Initialization Files 11 -3 Overview...

Ngày tải lên: 24/07/2014, 02:20

... Chapter 12 12 -1 12-2 12 -3 12 -4 12 -5 12 -6 12 -7 Chapter 13 13 -1 13-2 13 -3 13 -4 13 -5 13 -6 13 -7 13 -8 13 -9 13 -10 Chapter 14 14 -1 14-2 14 -3 14 -4 14 -5 14 -6 14 -7 14 -8 14 -9 14 -10 14 -11 Appendix A1 -1 A1-2 A1-3 ... 14 0 Using be used/accustomed to and get used/accustomed to 14 0 Contents v fm_ph/prs_AZAR_396 01 10 -11 10 -12 Chapter 11 11 -1 11- 2 11 -3 11 -4 11 -5 11 -6 11 -7 11 -8 11 -9 11 -10 Chapter ... 11 1 Chapter COMPARISONS 11 4 9 -1 9-2 9-3 9-4 9-5 9-6 9-7 9-8 9-9 9 -10 9 -11 9 -12 Chapter 10 10 -1 10-2 10 -3 10 -4 10 -5 10 -6 10 -7 10 -8 10 -9 10 -10 ...

Ngày tải lên: 24/07/2014, 07:21

... from each of the three nodes NODE FBD Py2 (Fx2 )1+ (Fx2)2 Px2 (Fy2 )1+ (Fy2)2 (Fy2 )1 (Fy2)2 2 (Fy1 )1 (Fy3)2 (Fx1 )1 (Fx3)2 3 Py1 (Fx1 )1+ (Fx1)3 (Fx2)2 (Fx2 )1 Px1 (Fy1 )1+ (Fy1)3 (Fy1)3 (Fx1)3 Py3 (Fx3)2+ ... and Frames Problem 93 93 94 10 1 10 2 10 9 11 8 Beam and Frame Analysis: Force Method, Part II Deflection of Beam and Frames Problem 12 1 12 1 13 2 iii Problem Problem 14 3 15 2 Beam and Frame Analysis: ... form as ⎧ Fx1 ⎫ ⎪F ⎪ ⎪ y1 ⎪ ⎨ ⎬= ⎪ Fx ⎪ ⎪ Fy ⎪ ⎩ ⎭ ⎡ k 11 ⎢k ⎢ 21 ⎢ k 31 ⎢ ⎣k 41 k12 k13 k 22 k 32 k 42 k 23 k 33 k 43 k14 ⎤ k 24 ⎥ ⎥ k 34 ⎥ ⎥ k 44 ⎦ ⎧u1 ⎫ ⎪v ⎪ ⎪ 1 ⎨ ⎬ ⎪u ⎪ ⎪v ⎪ ⎩ ⎭ (12 ) Truss...

Ngày tải lên: 05/08/2014, 09:20

... from each of the three nodes NODE FBD Py2 (Fx2 )1+ (Fx2)2 Px2 (Fy2 )1+ (Fy2)2 (Fy2 )1 (Fy2)2 2 (Fy1 )1 (Fy3)2 (Fx1 )1 (Fx3)2 3 Py1 (Fx1 )1+ (Fx1)3 (Fx2)2 (Fx2 )1 Px1 (Fy1 )1+ (Fy1)3 (Fy1)3 (Fx1)3 Py3 (Fx3)2+ ... and Frames Problem 93 93 94 10 1 10 2 10 9 11 8 Beam and Frame Analysis: Force Method, Part II Deflection of Beam and Frames Problem 12 1 12 1 13 2 iii Problem Problem 14 3 15 2 Beam and Frame Analysis: ... form as ⎧ Fx1 ⎫ ⎪F ⎪ ⎪ y1 ⎪ ⎨ ⎬= ⎪ Fx ⎪ ⎪ Fy ⎪ ⎩ ⎭ ⎡ k 11 ⎢k ⎢ 21 ⎢ k 31 ⎢ ⎣k 41 k12 k13 k 22 k 32 k 42 k 23 k 33 k 43 k14 ⎤ k 24 ⎥ ⎥ k 34 ⎥ ⎥ k 44 ⎦ ⎧u1 ⎫ ⎪v ⎪ ⎪ 1 ⎨ ⎬ ⎪u ⎪ ⎪v ⎪ ⎩ ⎭ (12 ) Truss...

Ngày tải lên: 05/08/2014, 09:20

... 0 0 ⎧ Fx1 ⎫ ⎡ k 11 ⎪F ⎪ ⎢ ⎪ y1 ⎪ ⎢k 21 ⎪ ⎪ ⎪ ⎪ ⎢0 ⎨ ⎬ = ⎢ ⎪ ⎪ ⎢0 ⎪ Fx ⎪ ⎢k 31 ⎪ ⎪ ⎢ ⎪ Fy ⎪ ⎢k 41 ⎩ ⎭ ⎣ 0 0 0 k 11 k 21 k 31 k 41 k12 k 22 k 32 k 42 k13 k 23 k 33 k 43 0 0 0 k13 k 23 0 k12 k 22 ... x-direction y-direction 0.033 0.029 0. 011 -0.0 21 -0.007 -0.50 0.60 0 0.33 -1. 00 0.67 Member Elongation (m) Force (MN) -0.0 21 -0.004 -0.008 0. 011 0. 030 0. 012 -0.52 -0 .14 -0 .19 0.36 -0.60 0.23 Note that ... Node E Node EA(MN) ∆x ∆y L C S EA/L 10 0 4 0.0 1. 0 25.00 2 10 0 3 1. 0 0.0 33.33 3 10 0 -4 0.0 -1. 0 25.00 4 10 0 3 1. 0 0.0 33.33 10 0 -4 0.6 -0.8 20.00 0.6 0.8 20.00 10 0 * S Node and E Node represent...

Ngày tải lên: 05/08/2014, 09:20

... 0.83 kN Joint Σ Fy = 0, F1(4/5)+0 .17 =0, F1= –0. 21 kN F1 0.5 kN 0.62 kN 0 .17 kN Note that only one equation from the FBD of joint is needed to find the remaining unknown of F1 The second equilibrium ... direction The pair of forces in the x-direction must be equal and opposite because they are co-linear Joint F4 = 0, F4 F5 F1 F5 = F1= F F2 Joint F10 F7 F7 = 0, F 11 40 F10= F 11 Truss Analysis: ... (3-b) 1. 2 m kN 4kN 5kN kN 1. 6 m (4-a) 0.7 m 0.9 m 0.9 m 0.7 m 0.9 m 0.9 m 0.9 m 2m (3-c) 1. 2 m 1. 2 m 3m 3m (4-b) (4-c) kN kN 2m 2m kN 1m 1m 1m 1m 2m 2m Problem 43 1m 1m 2m kN 2m Truss Analysis:...

Ngày tải lên: 05/08/2014, 09:20

... 0.83 kN Joint Σ Fy = 0, F1(4/5)+0 .17 =0, F1= –0. 21 kN F1 0.5 kN 0.62 kN 0 .17 kN Note that only one equation from the FBD of joint is needed to find the remaining unknown of F1 The second equilibrium ... direction The pair of forces in the x-direction must be equal and opposite because they are co-linear Joint F4 = 0, F4 F5 F1 F5 = F1= F F2 Joint F10 F7 F7 = 0, F 11 40 F10= F 11 Truss Analysis: ... (3-b) 1. 2 m kN 4kN 5kN kN 1. 6 m (4-a) 0.7 m 0.9 m 0.9 m 0.7 m 0.9 m 0.9 m 0.9 m 2m (3-c) 1. 2 m 1. 2 m 3m 3m (4-b) (4-c) kN kN 2m 2m kN 1m 1m 1m 1m 2m 2m Problem 43 1m 1m 2m kN 2m Truss Analysis:...

Ngày tải lên: 05/08/2014, 09:20

... Denoting the elongation of memebr as V3 We have (a) V3= α (∆T)L= 1. 2 (10 -5)/oC (14 oC) (6,000 mm)= 1mm (b) V3=1mm (c) V3= 16 kN (6 m)/[200 (10 6) kN/m2 (500) (10 -6)m2]=0.0 01 m= 1mm Next we need to find ... 1. 0 −0.0 Contribution of Reaction Forces Reaction Number Force Number Equation Number and Value of Entry 2i -1 Coeff 2i Coeff 1 -1. 0 0.0 0.0 2 0.0 -1. 0 -1. 0 Contribution of Externally Applied ... ⎢ ⎣ − 1. 0 − 1. 0 0.0 0.0 0.0 − 1. 0 − 0.6 0 0.8 0 0.6 1. 0 0 − 0.8 0.0 0 0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ 0.0⎥ ⎥ − 1. 0⎥ ⎦ ⎧ F1 ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ F2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ F3 ⎪ ⎪ 0.5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬=⎨ ⎪ F4 ⎪ ⎪− 1. 0⎪ ⎪...

Ngày tải lên: 05/08/2014, 09:20

... Denoting the elongation of memebr as V3 We have (a) V3= α (∆T)L= 1. 2 (10 -5)/oC (14 oC) (6,000 mm)= 1mm (b) V3=1mm (c) V3= 16 kN (6 m)/[200 (10 6) kN/m2 (500) (10 -6)m2]=0.0 01 m= 1mm Next we need to find ... 1. 0 −0.0 Contribution of Reaction Forces Reaction Number Force Number Equation Number and Value of Entry 2i -1 Coeff 2i Coeff 1 -1. 0 0.0 0.0 2 0.0 -1. 0 -1. 0 Contribution of Externally Applied ... ⎢ ⎣ − 1. 0 − 1. 0 0.0 0.0 0.0 − 1. 0 − 0.6 0 0.8 0 0.6 1. 0 0 − 0.8 0.0 0 0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ 0.0⎥ ⎥ − 1. 0⎥ ⎦ ⎧ F1 ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ F2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ F3 ⎪ ⎪ 0.5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬=⎨ ⎪ F4 ⎪ ⎪− 1. 0⎪ ⎪...

Ngày tải lên: 05/08/2014, 09:20