untitled

AnechoWeb_JA_01

EMCRF EMC Contents Update : AUGUST 24 / 5 2 / 5 emcrf_ ja /254 Deviation (db) Deviation (db) Center Front 2m Back 2m Right 2m Left 2m Center Front 2m Back 2m Right 2m Left 2m Tx =m Tx =m 2 2 - - 2-2 Tx

Gmech08.dvi

51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

1 16 10 5 1 2 2.1 a a a 1 1 1 2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5

(1/18) OA ETC EMI TDK TDK 10m IP ICM IP-100BX IS-060 IP-130BX IS-SM050 ISIS-SM ETC IP-BX ISFE27F5 IR material ETC ETCISFE27F5 TDK IR

EMC / IP-BL/IP/IP-BX/ID/ICM/IB/IS/IS-SM/IK/ICT IR/ISF/ITB Issue date: September 2013 RoHS RoHShttp://www.tdk.co.jp/rohs/ (1/18) OA ETC EMI TDK TDK 10m IP ICM IP-100BX IS-060 IP-130BX IS-SM050 ISIS-SM ETC

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r

LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)

338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x

Gmech08.dvi

145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

all.dvi

38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t

http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................

030801調査結果速報版.PDF

15 8 1 15 7 26 1. 2. 15 7 27 15 7 28 1 2 7:13 16:56 0:13 3km 45 346 108 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3. 3.1 26 7 10 1 20cm 2 1 2 45 1/15 3 4 5,6 3 4 3 5 6 ( ) 7,8 8 7 8 2 55 9 10 9 10

untitled

( ) (mm) (GHz)=300( ) 30 300GHz=1 10mm ( 2GHz2Mbps) Gbps= Mbps ( m),? S G=P/Pi30dB=1000 Gm=4πS/λ 2, S= 80λ 2 Gm=30dB η=g/gm, S= 80λ 2,G=27dB η=50% (GHz) 80 70 60 50 40 30 20 10 16 19 22 25 28 31 34 37

untitled

0 ( L CONTENTS 0 . sin(-x-sinx, (-x(x, sin(90-xx,(90-xsinx sin(80-xsinx,(80-x-x ( sin{90-(ωφ}(ωφ. :n :m.0 m.0 n tn. 0 n.0 tn ω m :n.0n tn n.0 tn.0 m c ω sinω c ω c tnω ecω sin ω ω sin c ω c ω tn c tn ω

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2

- 1-150 khz18 GHz CATV MATV IEC 60728-2 A B (ITE) 2 (3) 4.1 (1) 3 (CISPR) 1 (CISPR 16-1-1 2.1 2006) (CISPR 16-1-2 1 2003 12004) (CISPR 16-1-3 2.0 2004) (CISPR 16-1-4 2.0 2007) 30 MHz 1000 MHz (CISPR 16-1-5

N cos s s cos ψ e e e e 3 3 e e 3 e 3 e

3 3 5 5 5 3 3 7 5 33 5 33 9 5 8 > e > f U f U u u > u ue u e u ue u ue u e u e u u e u u e u N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 > A A > A E A f A A f A [ ] f A A e > > A e[ ] > f A E A < < f ; >

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b

( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

n n (n) (n) (n) (n) n n ( n) n n n n n en1, en ( n) nen1 + nen nen1, nen ( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( ) ( n) Τ n n n ( n) n + n ( n) (n) n + n n n n n n n n

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k

c 2009 i

I 2009 c 2009 i 0 1 0.0................................... 1 0.1.............................. 3 0.2.............................. 5 1 7 1.1................................. 7 1.2..............................

CISPR 32の概要 (ed. 2)

CISPR 32 (ed. 2) e 2017 7 25 1 1 2 2 2.1.................. 2 2.2............... 2 2.2.1.......... 3 3 3 3.1.............. 4 3.2 EUT............. 4 3.2.1 EUT.............. 4 3.2.2.............. 4 3.3...............

LLG-R8.Nisus.pdf

d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =

学習内容と日常生活との関連性の研究－第２部－第４章－１

69 V A V + A V A 2A 2 http://www.jba-hp.jp/ http://www.kbn3.com/ http://www.usba.org/ 70 (1) (1996)35 7 pp.28-33 (2) (1994) 71 () 3 1 1 99 8 1 10 1 11.3 2.5 1 100 11.4 30.9 1 72 (1) http://www.stat.go.jp/data/zensho/1999/zuhyou/a906-6.xls

29 1 6 1 1 1.1 1.1 1.1( ) 1.1( ) 1.1: 2 1.2 1.2( ) 4 4 1 2,3,4 1 2 1 2 1.2: 1,2,3,4 a 1 2a 6 2 2,3,4 1,2,3,4 1.2( ) 4 1.2( ) 3 1.2( ) 1.3 1.3 1.3: 4 1.4 1.4 1.4: 1.5 1.5 1 2 1 a a R = l a l 5 R = l a +

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =

A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B

9 7 A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B x x B } B C y C y + x B y C x C C x C y B = A

重力方向に基づくコントローラの向き決定方法

( ) 2/Sep 09 1 ( ) ( ) 3 2 X w, Y w, Z w +X w = +Y w = +Z w = 1 X c, Y c, Z c X c, Y c, Z c X w, Y w, Z w Y c Z c X c 1: X c, Y c, Z c Kentaro Yamaguchi@bandainamcogames.co.jp 1 M M v 0, v 1, v 2 v 0 v

4‐E )　キュリー温度を利用した消磁：熱消磁

( ) () x C x = T T c T T c 4D ) ) Fe Ni Fe Fe Ni (Fe Fe Fe Fe Fe 462 Fe76 Ni36 4E ) ) (Fe) 463 4F ) ) ( ) Fe HeNe 17 Fe Fe Fe HeNe 464 Ni Ni Ni HeNe 465 466 (2) Al PtO 2 (liq) 467 4G ) Al 468 Al ( 468

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C602E646F63>

スピントロニクスの基礎 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/077461 このサンプルページの内容は, 初版 1 刷発行時のものです. i 1 2 ii 3 5 4 AMR (anisotropic magnetoresistance effect) GMR (giant magnetoresistance

35

D: 0.BUN 7 8 4 B5 6 36 6....................................... 36 6.................................... 37 6.3................................... 38 6.3....................................... 38 6.4..........................................

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A

untitled

- 1 - - 2 - - 3 - - 4 - - 5 - - 6 - - 7 - - 8 - - 9 - - 10 - - 11 - - 12 - - 13 - - 14 - - 15 - - 16 - - 17 - - 18 - - 19 - - 20 - - 21 - - 22 - - 23 - - 24 - - 25 - - 26 - - 27 - - 28 - - 29 - - 30 -

chap1.dvi

1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f

untitled

(a) (b) (c) (d) Wunderlich 2.5.1 = = =90 2 1 (hkl) {hkl} [hkl] L tan 2θ = r L nλ = 2dsinθ dhkl ( ) = 1 2 2 2 h k l + + a b c c l=2 l=1 l=0 Polanyi nλ = I sinφ I: B A a 110 B c 110 b b 110 µ a 110

Microsoft Word - 章末問題

1906 R n m 1 = =1 1 R R= 8h ICP s p s HeNeArXe 1 ns 1 1 1 1 1 17 NaCl 1.3 nm 10nm 3s CuAuAg NaCl CaF - - HeNeAr 1.7(b) 2 2 2d = a + a = 2a d = 2a 2 1 1 N = 8 + 6 = 4 8 2 4 4 2a 3 4 π N πr 3 3 4 ρ = = =

untitled

+ From Tradeoffs of Receive and Transmit Equalization Architectures, ICC006,Bryan Casper, Intel Labs Transmitter Receiver 0 magnitude (db) 0 0 30 40 50 60 0 4 frequency (GHz). Receiver Transmitter FFE

TOP URL 1

TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +

Part () () Γ Part ,

Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

untitled

MRR Physical Basis( 1.8.4) METEK MRR 1 MRR 1.1 MRR 24GHz FM-CW(frequency module continuous wave) 30 r+ r f+ f 1.2 1 4 MRR 24GHz 1.3 50mW 1 rf- (waveguide) (horn) 60cm ( monostatic radar) (continuous wave)

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier

[Ver. 0.2] 1 2 3 4 5 6 7 1 1.1 1.2 1.3 1.4 1.5 1 1.1 1 1.2 1. (elasticity) 2. (plasticity) 3. (strength) 4. 5. (toughness) 6. 1 1.2 1. (elasticity) } 1 1.2 2. (plasticity), 1 1.2 3. (strength) a < b F

x y x-y σ x + τ xy + X σ y B = + τ xy + Y B = S x = σ x l + τ xy m S y = σ y m + τ xy l σ x σ y τ xy X B Y B S x S y l m δu δv [ ( σx δu + τ )

1 8 6 No-tension 1. 1 1.1................................ 1 1............................................ 5.1 - [B].................................. 5................................. 6.3..........................................

高等学校学習指導要領

高等学校学習指導要領

untitled

9118 154 B-1 B-3 B- 5cm 3cm 5cm 3m18m5.4m.5m.66m1.3m 1.13m 1.134m 1.35m.665m 5 , 4 13 7 56 M 1586.1.18 7.77.9 599.5.8 7 1596.9.5 7.57.75 684.11.9 8.5 165..3 7.9 87.8.11 6.57. 166.6.16 7.57.6 856 6.6.5

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx

橡実験IIINMR.PDF

(NMR) 0 (NMR) 2µH hω ω 1 h 2 1 1-1 NMR NMR h I µ = γµ N 1-2 1 H 19 F Ne µ = Neh 2mc ( 1) N 2 ( ) I =1/2 I =3/2 I z =+1/2 I z = 1/2 γh H>0 2µH H=0 µh I z =+3/2 I z =+1/2 I z = 1/2 I z = 3/2 γh H>0 2µH H=0

grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )

2 9 2 5 2.2.3 grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = g () g () (3) grad φ(p ) p grad φ φ (P, φ(p )) y (, y) = (ξ(t), η(t)) ( ) ξ (t) (t) := η (t) grad f(ξ(t), η(t)) (t) g(t) := f(ξ(t), η(t))

Gmech08.dvi

63 6 6.1 6.1.1 v = v 0 =v 0x,v 0y, 0) t =0 x 0,y 0, 0) t x x 0 + v 0x t v x v 0x = y = y 0 + v 0y t, v = v y = v 0y 6.1) z 0 0 v z yv z zv y zv x xv z xv y yv x = 0 0 x 0 v 0y y 0 v 0x 6.) 6.) 6.1) 6.)

r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t

1 1 2 2 2r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t) V (x, t) I(x, t) V in x t 3 4 1 L R 2 C G L 0 R 0

IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................

橡博論表紙.PDF

Study on Retaining Wall Design For Circular Deep Shaft Undergoing Lateral Pressure During Construction 2003 3 Study on Retaining Wall Design For Circular Deep Shaft Undergoing Lateral Pressure During Construction

Keysight Technologies 誘電体測定の基礎

Keysight Technologies Application Note LCR1.1 THz (MUT) 1. PCB PCB SAR RAM IC 3 Keysight - Application Note... 2... 4... 4... 7... 8... 1... 11... 11... 12 Debye... 12 Cole-Cole... 13... 13... 14... 15...

AC Modeling and Control of AC Motors Seiji Kondo, Member 1. q q (1) PM (a) N d q Dept. of E&E, Nagaoka Unive

AC Moeling an Control of AC Motors Seiji Kono, Member 1. (1) PM 33 54 64. 1 11 1(a) N 94 188 163 1 Dept. of E&E, Nagaoka University of Technology 163 1, Kamitomioka-cho, Nagaoka, Niigata 94 188 (a) 巻数

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ

Classic HD:ŠŸŠp”Ò:Discovery:‘‚ŠÞ:‚²„¤:Ÿ_Ł¶:Simulation_for_HRO.dvi

HRO 52 2002 7 19 HRO (Ham Radio Observation) HRO HRO 1 2 100km Microsoft Excel HRO 1999 12 12 16 HRO HRO 1 2 1.1........................................... 2 1.2 HRO............................ 2 1.3 HRO...................................

untitled

8- My + Cy + Ky = f () t 8. C f () t ( t) = Ψq( t) () t = Ψq () t () t = Ψq () t = ( q q ) ; = [ ] y y y q Ψ φ φ φ = ( ϕ, ϕ, ϕ,3 ) 8. ψ Ψ MΨq + Ψ CΨq + Ψ KΨq = Ψ f ( t) Ψ MΨ = I; Ψ CΨ = C; Ψ KΨ = Λ; q

(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {

7 4.., ], ], ydy, ], 3], y + y dy 3, ], ], + y + ydy 4, ], ], y ydy ydy y y ] 3 3 ] 3 y + y dy y + 3 y3 5 + 9 3 ] 3 + y + ydy 5 6 3 + 9 ] 3 73 6 y + y + y ] 3 + 3 + 3 3 + 3 + 3 ] 4 y y dy y ] 3 y3 83 3

吸収分光.PDF

3 Rb 1 1 4 1.1 4 1. 4 5.1 5. 5 3 8 3.1 8 4 1 4.1 External Cavity Laser Diode: ECLD 1 4. 1 4.3 Polarization Beam Splitter: PBS 13 4.4 Photo Diode: PD 13 4.5 13 4.6 13 5 Rb 14 6 15 6.1 ECLD 15 6. 15 6.3

08-Note2-web

r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)

85 4

85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V

(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )

all.dvi

5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0

[ ] [ ] [ ] [ ] [ ] [ ] ADC

[ ] [ ] [ ] [ ] [ ] [ ] ADC BS1 m1 PMT m2 BS2 PMT1 PMT ADC PMT2 α PMT α α = n ω n n Pn TMath::Poisson(x,[0]) 0.35 0.3 0.25 0.2 0.15 λ 1.5 ω n 2 = ( α 2 ) n n! e α 2 α 2 = λ = λn n! e λ Poisson Pn 0.1

1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x

. P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r

c y /2 ddy = = 2π sin θ /2 dθd /2 [ ] 2π cos θ d = log 2 + a 2 d = log 2 + a 2 = log 2 + a a 2 d d + 2 = l

c 28. 2, y 2, θ = cos θ y = sin θ 2 3, y, 3, θ, ϕ = sin θ cos ϕ 3 y = sin θ sin ϕ 4 = cos θ 5.2 2 e, e y 2 e, e θ e = cos θ e sin θ e θ 6 e y = sin θ e + cos θ e θ 7.3 sgn sgn = = { = + > 2 < 8.4 a b 2

1.500 m X Y m m m m m m m m m m m m N/ N/ ( ) qa N/ N/ 2 2

1.500 m X Y 0.200 m 0.200 m 0.200 m 0.200 m 0.200 m 0.000 m 1.200 m m 0.150 m 0.150 m m m 2 24.5 N/ 3 18.0 N/ 3 30.0 0.60 ( ) qa 50.79 N/ 2 0.0 N/ 2 20.000 20.000 15.000 15.000 X(m) Y(m) (kn/m 2 ) 10.000

2005 1

2005 1 1 1 2 2 2.1....................................... 2 2.2................................... 5 2.3 VSWR................................. 6 2.4 VSWR 2............................ 7 2.5.......................................

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

( ) s n (n = 0, 1,...) n n = δ nn n n = I n=0 ψ = n C n n (1) C n = n ψ α = e 1 2 α 2 n=0 α, β α n n! n (2) β α = e 1 2 α 2 1

(3.5 3.8) 03032s 2006.7.0 n (n = 0,,...) n n = δ nn n n = I n=0 ψ = n C n n () C n = n ψ α = e 2 α 2 n=0 α, β α n n (2) β α = e 2 α 2 2 β 2 n=0 =0 = e 2 α 2 β n α 2 β 2 n=0 = e 2 α 2 2 β 2 +β α β n α!

SO(2)

TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6

4 5.............................................. 5............................................ 6.............................................. 7......................................... 8.3.................................................4.........................................4..............................................4................................................4.3...............................................

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z

f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f

22 A 3,4 No.3 () (2) (3) (4), (5) (6) (7) (8) () n x = (x,, x n ), = (,, n ), x = ( (x i i ) 2 ) /2 f(x) R n f(x) = f() + i α i (x ) i + o( x ) α,, α n g(x) = o( x )) lim x g(x) x = y = f() + i α i(x )

untitled

1 17 () BAC9ABC6ACB3 1 tan 6 = 3, cos 6 = AB=1 BC=2, AC= 3 2 A BC D 2 BDBD=BA 1 2 ABD BADBDA ABC6 BAD = (18 6 ) / 2 = 6 θ = 18 BAD = 12 () AD AD=BADCAD9 ABD ACD A 1 1 1 1 dsinαsinα = d 3 sin β 3 sin β

K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................

= hυ = h c λ υ λ (ev) = 1240 λ W=NE = Nhc λ W= N 2 10-16 λ / / Φe = dqe dt J/s Φ = km Φe(λ)v(λ)dλ THBV3_0101JA Qe = Φedt (W s) Q = Φdt lm s Ee = dφe ds E = dφ ds Φ Φ THBV3_0102JA Me = dφe ds M = dφ ds

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

SFG 1 SFG SFG I SFG (ω) χ SFG (ω). SFG χ χ SFG (ω) = χ NR e iϕ +. ω ω + iγ SFG φ = ±π/, χ φ = ±π 3 χ SFG χ SFG = χ NR + χ (ω ω ) + Γ + χ NR χ (ω ω ) (ω ω ) + Γ cosϕ χ NR χ Γ (ω ω ) + Γ sinϕ. 3 (θ) 180

講義ノート 物性研究 電子版 Vol.3 No.1, (2013 年 T c µ T c Kammerlingh Onnes 77K ρ 5.8µΩcm 4.2K ρ 10 4 µωcm σ 77K ρ 4.2K σ σ = ne 2 τ/m τ 77K

2 2 T c µ T c 1 1.1 1911 Kammerlingh Onnes 77K ρ 5.8µΩcm 4.2K ρ 1 4 µωcm σ 77K ρ 4.2K σ σ = ne 2 τ/m τ 77K τ 4.2K σ 58 213 email:takada@issp.u-tokyo.ac.jp 1933 Meissner Ochsenfeld λ = 1 5 cm B = χ B =

DVIOUT

A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)

Microsoft Word - 信号処理3.doc

Junji OHTSUBO 2012 FFT FFT SN sin cos x v ψ(x,t) = f (x vt) (1.1) t=0 (1.1) ψ(x,t) = A 0 cos{k(x vt) + φ} = A 0 cos(kx ωt + φ) (1.2) A 0 v=ω/k φ ω k 1.3 (1.2) (1.2) (1.2) (1.1) 1.1 c c = a + ib, a = Re[c],

keisoku01.dvi

2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.

gk9c02a.dvi

y y y Field Analysis in a Lossy Dielectric Sandwiched between Flanged Rectangular Waveguide and Conducting Plane Makoto HIRANO y, Masaharu TAKAHASHI y, and Minoru ABE y. ( ) [] [],[3] y Department of Electronics

I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co

16 I ( ) (1) I-1 I-2 I-3 (2) I-1 ( ) (100 ) 2l x x = 0 y t y(x, t) y(±l, t) = 0 m T g y(x, t) l y(x, t) c = 2 y(x, t) c 2 2 y(x, t) = g (A) t 2 x 2 T/m (1) y 0 (x) y 0 (x) = g c 2 (l2 x 2 ) (B) (2) (1)

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

6,,3,4,, 3 4 8 6 6................................. 6.................................. , 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p,

211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,

sikepuri.dvi

2009 2 2 2. 2.. F(s) G(s) H(s) G(s) F(s) H(s) F(s),G(s) H(s) : V (s) Z(s)I(s) I(s) Y (s)v (s) Z(s): Y (s): 2: ( ( V V 2 I I 2 ) ( ) ( Z Z 2 Z 2 Z 22 ) ( ) ( Y Y 2 Y 2 Y 22 ( ) ( ) Z Z 2 Y Y 2 : : Z 2 Z

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP

高校生の就職への数学II

II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................

29

9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n

総研大恒星進化概要.dvi

The Structure and Evolution of Stars I. Basic Equations. M r r =4πr2 ρ () P r = GM rρ. r 2 (2) r: M r : P and ρ: G: M r Lagrange r = M r 4πr 2 rho ( ) P = GM r M r 4πr. 4 (2 ) s(ρ, P ) s(ρ, P ) r L r T

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

- 1 -

- 1 - - 2 - - 3 - - 4 - - 5 - - 6 - - 7 - - 8 - - 9 - - 10 - - 11 - - 12 - - 13 - - 14 - - 15 - - 16 - - 17 - - 18 - - 19 - - 20 - - 21 - - 22 - - 23 - - 24 - - 25 - - 26 - EC NEC - 27 - NEC - 28 - R NEC

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds

127 3 II 3.1 3.1.1 Φ(t) ϕ em = dφ dt (3.1) B( r) Φ = { B( r) n( r)}ds (3.2) S S n( r) Φ 128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds