... JOIN SPJ JOIN ( J RENAME CITY AS JCITY ) ) { SCITY, JCITY } 7.24 ( J JOIN SPJ JOIN S ) { P# } 7.25 ( ( ( J RENAME CITY AS JCITY ) JOIN SPJ JOIN ( S RENAME CITY AS SCITY ) ) WHERE JCITY ... constraints, which is why Chapters 7 and 8 precede Chapter 9. Copyright (c) 2003 C. J. Date page 7 .15 AND JCITY =/ SCITY ) { S#, P#, J# } 7. 21 P SEMIJOIN ( SPJ SEMIJOIN ( S...
Ngày tải lên: 06/08/2014, 01:21
... the introductory remark: (Begin quote) The "views" mentioned in the titles of references [22 . 3 -2 2. 5], [22 .10], [22 . 12] , [22 . 16] , [22 .25 ], [22 .28 ], [22 .30], and [22 .35] are not ... Copyright (c) 20 03 C. J. Date page 21 .7 21 .3 See Section 21 .2. 21 .4 See Section 21 .4. 21 .5 See Section 21 .4. 21 .6 No answer provided. 21 .7...
Ngày tải lên: 06/08/2014, 01:21
Echocardiography A Practical Guide to Reporting - part 6 pot
... surgery. Echocardiography: A Practical Guide for Reporting 76 Figure 7.1 Aortic abscess. Parasternal short-axis view showing cavities between the PA and aorta and in the anterior aorta. The aortic valve ... if the absolute values are normal. ã Typical dilatation in Marfan syndrome affects predominantly annulus and sinuses, causing a ‘pear-shaped’ aorta. Arteriosclerotic dilatatio...
Ngày tải lên: 11/08/2014, 00:20
Introduction to Elasticity Part 5 pot
... referred to axes rotated by θ = 45 ◦ from the x-y axes can be computed by matrix multiplication as: A = c 2 s 2 2sc s 2 c 2 −2sc −sc sc c 2 − s 2 = 0 .50 .51 .0 0 .50 .5 1.0 −0 .50 .50 .0 Then = ... s 2 = 0 .50 .51 .0 0 .50 .5 1.0 −0 .50 .50 .0 Then = RAR −1 = 1.00.00.0 0.01.00.0 0.00.02.0 0 .50 .51 .0 0 .50 .5 1.0 −0 .50 .50 .0 ...
Ngày tải lên: 11/08/2014, 09:21
Introduction to Elasticity Part 6 pps
... 0.0 567 0.0441 0.0 567 −0.0189 The total strain is then ij = 1 3 kk δ ij + e ij = 0.00 960 0.0378 0.0441 0.0378 0.0285 0.0 567 0.0441 0.0 567 −0.00930 If we evaluate the total strain using ... said to be orthotropic. 6 Constitutive Equations David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 October 4,...
Ngày tải lên: 11/08/2014, 09:21
Introduction to Elasticity Part 15 pot
... vector can therefore be written equivalently as u 1 ,u 2 ,u 3 . A common and useful shorthand is simply to write the displacement vector as u i ,wherethe isubscript is an index that is assumed to ... grain size E modulus of elasticity, electric field E ∗ activation energy E viscoelastic stiffness operator e electronic charge e ij deviatoric strain F force f s form factor for shear G shear...
Ngày tải lên: 11/08/2014, 09:21
Introduction to Electronics - Part 1 pot
... Circuit 11 4 Example 11 4 For b = 10 0 11 4 For b = 300 11 4 Biasing BJTs - The Four-Resistor Bias Circuit 11 5 Introduction 11 5 Circuit Analysis 11 6 Bias Stability 11 7 To maximize bias stability 11 7 Example ... /2 v od v o1 v o2 + + + ++ - - - v X R C R C β i b2 β i b1 r π r π i b2 i b1 R EB ( β +1) i b2 ( β +1) i b1 v id /2 v id /2 v od v o1...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 2 docx
... current : Introduction to Electronics 13 Amplifier Cascades + - v i1 A voc1 v i1 + - R i1 R o1 i i1 + - v o1 = v i2 A voc2 v i2 + - R i2 R o2 i i2 i o2 v o2 + - Amplifier 1 Amplifier 2 Fig. 25 . ... as fast as we need it to be (i.e., infinitely fast). Introduction to Electronics 27 Differential Amplifiers + - + - + - + - v I1 v I2 v...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 3 doc
... small V r : Introduction to Electronics 63 The Zener Diode Voltage Regulator + + - - v D v OUT i D + - V SS R S R L Fig. 87. Zener regulator with load. + + - - v D v OUT i D + - V SS R S R L Fig. ... well-suited to circuits with nonlinear elements. Introduction to Electronics 41 Op Amp Circuits - Integrators and Differentiators + - v O v i RC i...
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Introduction to Electronics - Part 4 pot
... in particular, where, the latter equation is the approximation for a forward-biased EBJ. Introduction to Electronics 80 BJT Common-Emitter Characteristics v BE + - v CE + - i B i C + + - - Fig. ... cause a large change in collector current - this is how we get this device to amplify!!! Introduction to Electronics 93 Metal-Oxide-Semiconductor FETs (MOSFETs)...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 5 pot
... it takes a circuit-analysis “trick” to see that: Introduction to Electronics 116 Biasing BJTs - The Four-Resistor Bias Circuit R B R C R E V CC V BB + + - - Fig. 1 75. Four-resistor bias circuit ... approximation: Introduction to Electronics 109 Basic FET Amplifier Structure Fig. 166. 2N3819 output characteristics, with curves for gate-source voltages of (from bott...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 6 pot
... models !!! 3. Diode small-signal resistance r d varies with Q -point. 4. The diode small-signal model is simply a resistor !!! Introduction to Electronics 1 36 Bipolar IC Bias Circuits V BE1 I O ... I DSS / V P 2 we can write: Introduction to Electronics 1 46 The Common-Emitter Amplifier R S R 1 R 2 R C R E R L C in C out v s v o + + - - V CC Q 1 v in + -...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 7 pdf
... Introduction to Electronics 176 Low- & Mid-Frequency Performance of CE Amplifiers R S R 1 R 2 R C R EB R L C in C out v s v o + + - - V CC Q 1 v in + - C E R EF Fig. 252. Generic single-supply ... examples is shown at left. Introduction to Electronics 177 Low- & Mid-Frequency Performance of CE Amplifiers v s R S R B r R L ’ β i b v o + + - - B E C i...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 8 docx
... . parameter unit 4000 74C 74HC 74HCT AC ACT t PD ns 80 90 9 10 5 5 P static < 1 à W for all versions I OH mA -1 .0 -0 .36 -4 .0 -4 .0 -2 4 -2 4 I OL mA 2.4 0.36 4.0 4.0 24 24 I IH à A 1.0 1.0 1.0 1.0 1.0 1.0 I IL mA -1 .0 -1 .0 -1 .0 -1 .0 -1 .0 -1 .0 V OH V ... V/ à s. Introduction to Electronics 205 Instrumentation Amplifier R R R R v 2 v 1...
Ngày tải lên: 14/08/2014, 02:21
Introduction to Electronics - Part 9 pdf
... input. R C R C I BIAS V CC -V EE v OD v O1 v O2 Q 1 Q 2 i C1 i C2 + + + - v ID /2 = -1 V v ID /2 = -1 V + + - - -1 V +1 V 0.7 V + - 0.3 V -1 .3 V + - Fig. 324. Differential amplifier with -2 ... output is low. The solution to both of these problems is to replace the pull-up resistor with an active pull-up . Output Voltage, V O...
Ngày tải lên: 14/08/2014, 02:21