Microsensors Part 3 pptx

... InP, 211 0. 23 H Ch mV s     ; Magnetic Microsensors 23 MGT2: InP, 211 0. 23 H Ch mV s     ; MGT3: GaAs, 211 0. 43 H Ch mV s     . The geometry influence upon off B is shown ... Fig. 2. 13. The NEMI depending on drain current, for threedevices of different materials. MGT1: Si, 211 0.07 H Ch mV s     MGT2: InP, 211 0. 23 H Ch mV s     ; Magneti...

Ngày tải lên: 19/06/2014, 21:20

... Students 2.9 Connected Sentence Cards 3. Instructions 3. 1 Activity Template (Blank) 3. 2 Sample Lesson Plan 3. 3 Sound Connections Demo 3. 6 Going Further 3. 15 Connected Speech Templates ... Exchanges 6. Instructions 6.1 Activity Template (Blank) 6 .3 Multi-Purpose Texts 7. Instructions 7.1 Reading Race – Lesson Plan 7 .3 Discussion Questions 8. Instruc...

Ngày tải lên: 13/12/2013, 21:15

Ngày tải lên: 13/12/2013, 21:15

Ngày tải lên: 14/12/2013, 10:15

... ScrollingDataBinding() { this.AutoScaleBaseSize = new System.Drawing.Size(5, 13) ; this.ClientSize = new System.Drawing.Size(464, 2 53) ; this.Text = "09_ScrollingDataBinding"; this.retrieveButton ... textQuan.Location = new System.Drawing.Point(4, 55); textQuan.Text = ""; textQuan.TabIndex = 3; textQuan.Anchor = AnchorStyles.Top | AnchorStyles.Left | An...

Ngày tải lên: 15/12/2013, 02:16

... contrast, ψ n+1 j =(1+iH∆t) −1 ψ n j (19.2 .34 ) These are both first-order accurate in time, as can be seen by expanding equation (19.2 .32 ). However, neither operator in (19.2 .33 ) or (19.2 .34 ) is unitary. 19 .3 Initial Value ... the form i ∂ψ ∂t = Hψ (19.2 .30 ) where the operator H is H = − ∂ 2 ∂x 2 + V (x)(19.2 .31 ) The formal solution of equation (19.2 .30 ) is ψ(x, t)=e −iHt ψ(x,...

Ngày tải lên: 15/12/2013, 04:15

... like this: f(x)=b 0 + a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 + a 5 b 5 +··· (5.2.1) Printers prefer to write this as f(x)=b 0 + a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 + a 5 b 5 + ··· (5.2.2) In either ... 5 .3 Polynomials and Rational Functions 1 73 Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521- 431 08-5) Copyright (C) 1988-1992 ... without choos...

Ngày tải lên: 15/12/2013, 04:15

... f(x 1 ),f(x 3 ),f(x 2 ), but to the values f(x 1 ),f(x 3 )e Q ,f(x 2 )e 2Q , yielding a new guess for the root, x 4 .The overall updating formula (incorporating the solution 9.2 .3) is x 4 = x 3 +(x 3 −x 1 ) sign[f(x 1 ) ... gives f(x 1 )− 2f(x 3 )e Q + f(x 2 )e 2Q =0 (9.2.2) This is a quadratic equation in e Q , which can be solved to give e Q = f(x 3 )+sign[f(x 2 )]  f(x 3 )...

Ngày tải lên: 15/12/2013, 04:15