EndoPaper.pdf
|
|
- ありさ あくや
- 7 years ago
- Views:
Transcription
1 Research on Nonlinear Oscillation in the Field of Electrical, Electronics, and Communication Engineering Tetsuro ENDO.,.,, (NLP), (1973 ),. (, ),..., 191, 1970, ,,, ,, ,,, ( ), ( ), ( ),,. Tetsuro ENDO, Fellow (Department.of Electronics and Bioinformatics, Kawasaki-shi, Japan). Fundamentals Review Vol. No. pp , 1973,.,,,,,,..,,, 10, 1 3, 4.,,. 198,,,,,.,,,, ,,,. ( ),,.,.., Fundamentals Review Vol. No. 31
2 1 ( ) 191/1/ /03/31 196/04/ /03/ /04/ /03/ /04/ /03/ /04/ /03/ /04/ /03/ /04/ /03/ /04/01-197/03/ /04/ /03/ /04/ /07/31 197/04/ /03/ /07/ /03/ /04/ /03/ /04/01-198/03/ /04/01-198/03/ /04/ /03/31 198/04/ /09/30 198/04/ /03/ /04/01-198/0/ /10/ /09/ /04/ /03/31 198/0/ /0/1 1986/04/ /0/0 1986/10/ /0/ /0/ /0/ /0/1-1989/0/ /0/0-1990/0/ /0/0-1991/0/ /0/0-199/0/ 1990/0/ /0/ /0/18-199/0/ 1991/0/ /0/1 199/0/3-1993/0/1 199/0/3-1994/0/ /0/ /0/ /0/ - 199/0/ /0/14-199/0/ /0/ /0/17 199/0/0-1996/0/17 199/0/0-1997/0/ /0/ /0/ /0/ /0/ 1997/0/ /0/ 1997/0/ /0/1 1998/0/3-1999/0/1 1998/0/3-000/0/1 1999/0/ - 000/0/1 1999/0/ - 001/0/9 000/0/ - 001/0/9 000/0/ - 00/0/7 001/0/30-00/0/7 001/0/30-003/0/7 00/0/8-003/0/7 00/0/8-004/0/8 003/0/8-004/0/8 003/0/8-00/0/7 004/0/9-00/0/7 004/0/9-006/0/7 00/0/8-006/0/6 00/0/8-007/0/4 006/0/7-007/0/4 006/0/7-008/0/4 007/0/ - 008/0/4 007/0/ - 008/0/7-008/0/7 -,,,,.,,,.,,.,.,. 4..,.,., , (a), (b). 1 (a), 1(b) 3 Fundamentals Review Vol. No.
3 (a) ( ) ε = (b)., i = g 1 v + g 3 v 3, g 1,g 3 > 0.,,. π/ π/,.,. 1 (a). ẍ + x = ε Ω (1 x )ẋ + Ω 1 x αy Ω ÿ + y = k ε Ω (1 y )ẏ + Ω k y k αx Ω, x v 1, y v, Ω, α (0 α 1), ε (> 0).,. (1) ε.,,, x = a cos(τ ϕ), ẋ = a sin(τ ϕ) y = b cos(τ θ), ẏ = b sin(τ θ). a, b, ϕ, θ (, ε ), 1, (= )., () x y 1. (1) (= ) ( )., (1),, Ω, k 1 (, ), α (1) (). (1).,. ( ) ȧ = εa 1 a + αb sin(θ ϕ) Ω 4 ϕ = Ω 1 Ω ḃ = k εb Ω ( θ = Ω k Ω αb cos(θ ϕ) a ) k αa sin(θ ϕ) 1 b 4 k αb b cos(θ ϕ), ȧ, ϕ, ḃ, θ =0 (a e, ϕ e, b e, θ e ),,. (3) ( δ =0 ) ( δ =0 ),. α δ., δ (k 1)/k σ (Ω 1)/Ω., 3, 4.,,, (= ),.,. α,.. 1(b). (3) Fundamentals Review Vol. No. 33
4 σ= 1 δ a =b =4 1 1± ε sin(ϕ θ) = σ α 1 σ α δ ( δ) ϕ θ (6) π また, 安定条件は a > 4 となる. 抵抗結合の場合の位相特性, 振 幅特性を図示するとそれぞれ図, 6 のようになる. この場合には 図 3 振幅特性 (コンデンサ結合) ε = 0.0. コンデンサ結合の場合とは異なって同期周波数, 振幅, 位相差及び 参考文献 1 より転載 安定性は直接求められる. 抵抗結合の場合には同期状態はただ一 つだけ存在し, 同期周波数は両発振器の固有周波数の間の値をと る. また同期しているときは両発振器の振幅は常に等しく (> ), かつ両発振器の固有周波数が異なるとき最大となる位相差は同期 範囲内において, π/ から π/ まで連続的に変化し, 両発振器 の固有周波数が等しいとき位相差は0となる. 図 4 位相特性 (コンデンサ結合) ε = 0.0. 図 位相特性 (抵抗結合) 参考文献 1 より転載 x +x= ε Ω 1 α (1 x )x + x + y Ω Ω Ω k ε Ω k k α (1 y )y + x y + y = y+ Ω Ω Ω 参考文献 1 より転載 (4) コンデンサ結合の場合と同様に, 式 (4) の右辺の各項は ε オーダ の微少量であるとする. このとき, 右辺の微少量を0としたとき の解は式 () のようになる. これを基に ε = 0 における解を平均 化法により求めると次の平均化された 1 階の微分方程式が得ら 図 6 振幅特性 (抵抗結合) 参考文献 1 より転載 れる. a = εa Ω 1 a 4 + αb cos(ϕ θ) Ω 4. 発振器の結合系の研究 ϕ = Ω 1 αb sin(ϕ θ) Ω Ωa k εb b = Ω 1 b 4 () + k αb cos(ϕ θ) Ω Ω k k αb θ = sin(ϕ θ) + Ω Ωb 次に図 7 に示す発振器のインダクタンスによる結合系の解析を とは別のやり方で解析してみよう. この系は基本的に 図 1(a) の系において両発振器の固有周波数が等しい場合と同じ である. この系を適当に正規化すると次のような 階の連立微分 方程式が得られる. θ = 0 として漸近安定な平衡点を求めると次 上式より, a, ϕ, b, x 1 ε(1 x1 )x 1 + x1 αx = 0 のような定常解が得られる. x ε(1 x )x + x αx1 = 0 34 (7) Fundamentals Review Vol. No.
5 (7) α ε ε (1). (7). ẍ + Bx = εẋ 1 3 εẋ c [ ] x =[x 1,x ] T, x c =[x 3 1 α α 1,x3 ]T, B = α 1 α (8) : x = Py, P 1. (8) ÿ +(P 1 BP)y = εẏ 1 3 ε(p 1 ẋ c ) (9) B λ 1 = 1 α, λ = 1 + α., λ 1 1 p 1 = [ 1/, 1/ ] T, λ 1 [ p = 1/, 1/ ] T, P =[p1, p ] R, P 1 BP,. [ ] P 1 BP = P T λ 1 0 BP = (10) 0 λ B, P 1 P T. (9). ÿ 1 + ω 1 y 1 = εf 1 (y 1,y, ẏ 1, ẏ ) ÿ + ω y = εf (y 1,y, ẏ 1, ẏ ) ω 1 λ 1, ω λ f 1 (y 1,y, ẏ 1, ẏ ) ẏ εg 1(y 1,y, ẏ 1, ẏ ) f (y 1,y, ẏ 1, ẏ ) ẏ 1 3 εg (y 1,y, ẏ 1, ẏ ). (1) g 1, g. [g 1,g ] T = P 1 ẋ c = d(pt x c ) dt. 1 1 ( 1 P T = 1 1, x c = ( 1 y y y 1 1 y (11) (1) (13) ) 3 ) 3 g 1 = g 3 y 1ẏ1 + 3 ẏ1y +3y 1y ẏ 3 y ẏ + 3 ẏy 1 +3y y 1 ẏ 1, f 1,f. f 1 ẏ 1 1 y 1ẏ1 1 ẏ1y y 1y ẏ = f ẏ 1 y ẏ 1 (14) ẏy1 y y 1 ẏ 1 (11) ε =0 y 1 = ρ 1 sin(ω 1 t + θ 1 ), ẏ 1 = ρ 1 ω 1 cos(ω 1 t + θ 1 ) y = ρ sin(ω t + θ ), ẏ = ρ ω cos(ω t + θ ) (1). ε = 0, (1) ρ i ρ i (t),θ i θ i (t), (11) ρ i sin ψ i + ρ i θi cos ψ i =0 ρ i ω i cos ψ i ρ i ω i θi sin ψ i = f i (y 1,y, ẏ 1, ẏ ) ψ i ω i t + θ i, i =1,,. ρ i = ε ω i f i (y 1,y, ẏ 1, ẏ ) cos(ω i t + θ i ) θ i = ε ω i ρ i f i (y 1,y, ẏ 1, ẏ ) sin(ω i t + θ i ),i=1, (16) y i, ẏ i,i =1, (1) sin, cos t. (16) ε>0, ρ i,θ i. ρ i = θ i = i =1, lim T lim T ε ω i T ε ω i ρ i T T 0 T 0 f i (y 1,y, ẏ 1, ẏ ) cos(ω i t + θ i )dt f i (y 1,y, ẏ 1, ẏ ) sin(ω i t + θ i )dt (17),., ψ i (t) =ω i t+θ i,i=1,, (14), (1) (17). ρ i = ε ω i [ω i ρ i cos ψ i 1 ω iρ 3 i sin ψ i cos ψ i 1 ω iρ i ρ i+1 cos ψ i sin ψ i+1 ] ω i+1 ρ i ρ i+1 sin ψ i sin ψ i+1 cos ψ i cos ψ i+1 θ i = ε ω i [ ω i sin ψ i cos ψ i 1 ω iρ i sin3 ψ i cos ψ i 7 1 ω iρ i ρ i+1 cos ψ i sin ψ i sin ψ i+1 ] ω i+1 ρ i+1 sin ψ i sin ψ i+1 cos ψ i cos ψ i+1 Fundamentals Review Vol. No. 3
6 (18),, p(t) = T lim T 0 p(t)dt (19)., i =1, 3 1. (18)., ω 1,ω (, ). ω 1 /ω = m/n m, n,. cos ψ i = + 1 cos ψ i(t) = 1 sin ψ i cos ψ i = sin ψ i(t) =0 sin ψ i cos ψ = cos 4ψ i(t) = 1 8 sin ψ i+1 cos ψ = cos ψ i(t) cos ψ i+1 (t) cos ψ i (t) cos ψ i+1 (t) = 1 4 sin ψ i sin ψ i+1 cos ψ i cos ψ i+1 = 4 sin ψ i(t) sin ψ i+1 (t) =0 sin ψ i = 1 cos ψ i(t) = 1 sin 3 ψ i cos ψ i = 4 sin ψ i(t) 1 8 sin 4ψ i(t) =0 sin ψ i cos ψ i sin ψ i+1 = 4 sin ψ i(t) 1 4 sin ψ i(t) cos ψ i+1 (t) =0 sin ψ i sin ψ i+1 cos ψ i+1 = 4 sin ψ i+1(t) 1 4 sin ψ i+1(t) cos ψ i (t) =0 (0) (0) i =1, i (18) (0), ρ 1, θ 1, ρ, θ. ρ 1 = 1 16 ερ 1(8 ρ 1 ρ ) (1a) ρ = 1 16 ερ (8 ρ ρ 1) (1b) θ 1 = 0 (1c) θ = 0 (1d) 0, (1) y 1 y ω 1 ω,, θ 1 (t),θ (t)., (1a), (1b),. 0.,. (ρ 1s,ρ s )=(0, 0), (, 0), (0, ), ( 8/3, 8/3) ().. J. [ ] J = 1 16 ε 8 3ρ 1s ρ s 4ρ 1s ρ s (3) 4ρ 1s ρ s 8 ρ 1s 3ρ s, ε>0, (, 0), (0, ),. (, 0) y y 1 = sin(ω 1 t + θ 1 ),y = 0, (0, ) y 1 =0,y = sin(ω t + θ ). x 1 1 x = Py, P = 1 1. (, 0) (= ) : x 1 = x = sin(ω 1 t + θ 1 ) (0, ) (= ) : x 1 = x = sin(ω t + θ ) ω 1 = 1 α, ω = 1+α, θ 1,θ. 0 <ε 1, ,.,, 3 4., I V i = g 1 v g 3 v 3 + g v,g 1,g 3,g > 0, 9. 7 N,. ẍ 1 + ε(1 βx 1 + x4 1 )ẋ 1 + x 1 αx =0 ẍ + ε(1 βx + x4 )ẋ + x αx 1 =0 (4) 0 < ε 1, 36 Fundamentals Review Vol. No.
7 f i (y 1,y, ẏ 1, ẏ ) ẏ i βg i(y 1,y, ẏ 1, ẏ ) 1 h i(y 1,y, ẏ 1, ẏ ) (7b). g 1,g,h 1,h. (a) [g 1,g ] T = P 1 ẋ c = d(pt x c ) dt [h 1,h ] T = P 1 ẋ f = d(pt x f ) dt (8) (b) 8 (3) 0 α 1, β.. ẍ + Bx = εẋ εβẋ c 1 εẋ f (a) x =[x 1,x ] T, x c =[x 3 1,x3 ]T, x f =[x 1,x ]T [ ] 1 α (b) B = α 1. (a) : x = Py, P 1. ÿ +(P 1 BP)y = εẏ εβ(p 1 ẋ c ) 1 ε(p 1 ẋ f ) (6) B λ 1 =1 α, λ =1+α. 4.,.. g 1 = g h 1 = h 3 y 1ẏ1 + 3 ẏ1y +3y 1y ẏ 3 y ẏ + 3 ẏy 1 +3y y 1 ẏ 1 y4 1ẏ1+ 1 y 1ẏ1y + 10 y 3 1 y ẏ y4 ẏ+ 1 yẏy y 3y 1ẏ 1 + ẏ1y y 1 y 3 ẏ ẏy y y 3 1ẏ1 (9) (7a) (1), ρ i ρ i (t),θ i θ i (t) ρ θ (16) (7b) f 1,f., ω 1,ω,. U i ρ i,i=1,. U 1 = εu 1 (1 1 8 βu βu U U U 1U ) U = εu (1 1 8 βu 1 4 βu U U U U 1 ) θ 1 =0 θ =0 ÿ i + ω i y i = εf i (y 1,y, ẏ 1, ẏ ),i=1, ω i λ i (7a) (30) 0, y 1,y.. 9 ( ). (U 1s,U s )=(0, 0), (β ± β 8, 0), (0, β ± β 8), ( ) 3β ± 9β 80, 3β ± 9β 80 (β ± 16 β,β 16 β ) (31) 9 I V,,. Fundamentals Review Vol. No. 37
8 (U 1s,U s )=(0, 0), (β + β 8, 0), (0, β + β 8), ( ) 3β + 9β 80, 3β + 9β 80 D ( ). (3) (0, 0)., y y 1 = β + β 8 sin(ω 1 t + θ 1 ),y = 0 (33) y 1 =0,y = β + β 8 sin(ω t + θ ) (34) 10. 3, x, x 1 = y 1 / +y /, x = y 1 / y /. (β + β 8, 0) (= ): β > 8 x 1 = x = β + β 8 sin(ω 1 t + θ 1 ) (0, β + β 8) (= ): β > 8 x 1 = x = β + β 8 sin(ω t + θ ). ((3β + 9β 80)/, (3β + 9β 80)/) (= ): 4 /3 <β<4 x 1 = 0.3β +0.3 β 80 9 sin(ω 1t + θ 1 ) + 0.3β +0.3 β 80 9 sin(ω t + θ ) x = 0.3β +0.3 β 80 9 sin(ω 1t + θ 1 ) 0.3β +0.3 β 80 9 sin(ω t + θ ) ω 1 = 1 α, ω = 1+α, θ 1,θ. 0 <ε 1, 8, β =3.3 (30), 11 S:, R:, D:, ( ), 4. 1, α 0,., ( ), 4. ( α ), (,, )., 4. 3 ( ),, ( )., 4. 3,,. 38 Fundamentals Review Vol. No.
9 , 1 α, ()., ( ),,,.,,,,,.., ( ),, 0 180, x 1,x.,,,,. α., α, α., ε,,.,. 7 T. Miyano and T. Tsutsui, Data synchronization in a network of coupled phase oscillators, Phys. Rev. Lett., vol.98, 0410, 007. ( ) PLL IEEE Circuits & Systems 006 IEEE.,,.,. 6, 7..,.. 1,,,, vol. 48,no.9, pp.11-17, 196. T. Endo and S. Mori, Mode analysis of a multimode ladder oscillator, IEEE Trans. Circuits Syst., vol.cas3, no., pp , ,,,, T. Endo and T. Ohta, Multimode oscillations in a coupled oscillator system with fifth-power nonlinear characteristics, IEEE Trans. Circuits Syst., vol.cas7, no.4, pp.77-83, Y. Uwate and Y. Nishio, Synchronization phenomena in van der Pol oscillators coupled by a time-varying resistor, Int. J. Bifurcation Chaos, vol.17, no.10, pp , M. Yamauchi, M. Wada, Y. Nishio, and A. Ushida, Wave propagation phenomena of phase states in oscillators coupled by inductors as a ladder, IEICE Trans.Fundamentals, vol.e8-a, no.11, pp.9-98, Nov Fundamentals Review Vol. No. 39
24.15章.微分方程式
m d y dt = F m d y = mg dt V y = dy dt d y dt = d dy dt dt = dv y dt dv y dt = g dv y dt = g dt dt dv y = g dt V y ( t) = gt + C V y ( ) = V y ( ) = C = V y t ( ) = gt V y ( t) = dy dt = gt dy = g t dt
More information0.,,., m Euclid m m. 2.., M., M R 2 ψ. ψ,, R 2 M.,, (x 1 (),, x m ()) R m. 2 M, R f. M (x 1,, x m ), f (x 1,, x m ) f(x 1,, x m ). f ( ). x i : M R.,,
2012 10 13 1,,,.,,.,.,,. 2?.,,. 1,, 1. (θ, φ), θ, φ (0, π),, (0, 2π). 1 0.,,., m Euclid m m. 2.., M., M R 2 ψ. ψ,, R 2 M.,, (x 1 (),, x m ()) R m. 2 M, R f. M (x 1,, x m ), f (x 1,, x m ) f(x 1,, x m ).
More information5 36 5................................................... 36 5................................................... 36 5.3..............................
9 8 3............................................. 3.......................................... 4.3............................................ 4 5 3 6 3..................................................
More information基礎数学I
I & II ii ii........... 22................. 25 12............... 28.................. 28.................... 31............. 32.................. 34 3 1 9.................... 1....................... 1............
More informationuntitled
10 log 10 W W 10 L W = 10 log 10 W 10 12 10 log 10 I I 0 I 0 =10 12 I = P2 ρc = ρcv2 L p = 10 log 10 p 2 p 0 2 = 20 log 10 p p = 20 log p 10 0 2 10 5 L 3 = 10 log 10 10 L 1 /10 +10 L 2 ( /10 ) L 1 =10
More information2 1 17 1.1 1.1.1 1650
1 3 5 1 1 2 0 0 1 2 I II III J. 2 1 17 1.1 1.1.1 1650 1.1 3 3 6 10 3 5 1 3/5 1 2 + 1 10 ( = 6 ) 10 1/10 2000 19 17 60 2 1 1 3 10 25 33221 73 13111 0. 31 11 11 60 11/60 2 111111 3 60 + 3 332221 27 x y xy
More information3
- { } / f ( ) e nπ + f( ) = Cne n= nπ / Eucld r e (= N) j = j e e = δj, δj = 0 j r e ( =, < N) r r r { } ε ε = r r r = Ce = r r r e ε = = C = r C r e + CC e j e j e = = ε = r ( r e ) + r e C C 0 r e =
More information330
330 331 332 333 334 t t P 335 t R t t i R +(P P ) P =i t P = R + P 1+i t 336 uc R=uc P 337 338 339 340 341 342 343 π π β τ τ (1+π ) (1 βτ )(1 τ ) (1+π ) (1 βτ ) (1 τ ) (1+π ) (1 τ ) (1 τ ) 344 (1 βτ )(1
More information5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)............................................
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h f(a + h, b) f(a, b) h........................................................... ( ) f(x, y) (a, b) x A (a, b) x (a, b)
More informationNote5.dvi
12 2011 7 4 2.2.2 Feynman ( ) S M N S M + N S Ai Ao t ij (i Ai, j Ao) N M G = 2e2 t ij 2 (8.28) h i μ 1 μ 2 J 12 J 12 / μ 2 μ 1 (8.28) S S (8.28) (8.28) 2 ( ) (collapse) j 12-1 2.3 2.3.1 Onsager S B S(B)
More informationver.1 / c /(13)
1 -- 11 1 c 2010 1/(13) 1 -- 11 -- 1 1--1 1--1--1 2009 3 t R x R n 1 ẋ = f(t, x) f = ( f 1,, f n ) f x(t) = ϕ(x 0, t) x(0) = x 0 n f f t 1--1--2 2009 3 q = (q 1,..., q m ), p = (p 1,..., p m ) x = (q,
More information204 / CHEMISTRY & CHEMICAL INDUSTRY Vol.69-1 January 2016 047
9 π 046 Vol.69-1 January 2016 204 / CHEMISTRY & CHEMICAL INDUSTRY Vol.69-1 January 2016 047 β γ α / α / 048 Vol.69-1 January 2016 π π π / CHEMISTRY & CHEMICAL INDUSTRY Vol.69-1 January 2016 049 β 050 Vol.69-1
More informationohp_06nov_tohoku.dvi
2006 11 28 1. (1) ẋ = ax = x(t) =Ce at C C>0 a0 x(t) 0(t )!! 1 0.8 0.6 0.4 0.2 2 4 6 8 10-0.2 (1) a =2 C =1 1. (1) τ>0 (2) ẋ(t) = ax(t τ) 4 2 2 4 6 8 10-2 -4 (2) a =2 τ =1!! 1. (2) A. (2)
More informationdvipsj.4131.dvi
7 1 7 : 7.1 3.5 (b) 7 2 7.1 7.2 7.3 7 3 7.2 7.4 7 4 x M = Pw (7.3) ρ M (EI : ) M = EI ρ = w EId2 (7.4) dx 2 ( (7.3) (7.4) ) EI d2 w + Pw =0 (7.5) dx2 P/EI = α 2 (7.5) w = A sin αx + B cos αx 7.5 7.6 :
More information467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 B =(1+R ) B +G τ C C G τ R B C = a R +a W W ρ W =(1+R ) B +(1+R +δ ) (1 ρ) L B L δ B = λ B + μ (W C λ B )
More informationI.2 z x, y i z = x + iy. x, y z (real part), (imaginary part), x = Re(z), y = Im(z). () i. (2) 2 z = x + iy, z 2 = x 2 + iy 2,, z ± z 2 = (x ± x 2 ) +
I..... z 2 x, y z = x + iy (i ). 2 (x, y). 2.,,.,,. (), ( 2 ),,. II ( ).. z, w = f(z). z f(z), w. z = x + iy, f(z) 2 x, y. f(z) u(x, y), v(x, y), w = f(x + iy) = u(x, y) + iv(x, y).,. 2. z z, w w. D, D.
More informationesba.dvi
Ehrenberg-Siday-Bohm-Aharonov 1. Aharonov Bohm 1) 0 A 0 A A = 0 Z ϕ = e A(r) dr C R C e I ϕ 1 ϕ 2 = e A dr = eφ H Φ Φ 1 Aharonov-Bohm Aharonov Bohm 10 Ehrenberg Siday 2) Ehrenberg-Siday-Bohm-Aharonov ESBA(
More information数学演習:微分方程式
( ) 1 / 21 1 2 3 4 ( ) 2 / 21 x(t)? ẋ + 5x = 0 ( ) 3 / 21 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 ( ) 3 / 21 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 x(t) = sin 5t? ẋ
More information最 新 測 量 学 ( 第 3 版 ) 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. このサンプルページの 内 容 は, 第 3 版 1 刷 発 行 時 の
最 新 測 量 学 ( 第 3 版 ) 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/047143 このサンプルページの 内 容 は, 第 3 版 1 刷 発 行 時 のものです. 3 10 GIS 3 1 2 GPS GPS GNSS GNSS 23 3 3 2015
More informationuntitled
Y = Y () x i c C = i + c = ( x ) x π (x) π ( x ) = Y ( ){1 + ( x )}( 1 x ) Y ( )(1 + C ) ( 1 x) x π ( x) = 0 = ( x ) R R R R Y = (Y ) CS () CS ( ) = Y ( ) 0 ( Y ) dy Y ( ) A() * S( π ), S( CS) S( π ) =
More informationx : = : x x
x : = : x x x :1 = 1: x 1 x : = : x x : = : x x : = : x x ( x ) = x = x x = + x x = + + x x = + + + + x = + + + + +L x x :1 = 1: x 1 x ( x 1) = 1 x 2 x =1 x 2 x 1= 0 1± 1+ 4 x = 2 = 1 ± 5 2 x > 1
More information1 1 ( ) ( 1.1 1.1.1 60% mm 100 100 60 60% 1.1.2 A B A B A 1
1 21 10 5 1 E-mail: qliu@res.otaru-uc.ac.jp 1 1 ( ) ( 1.1 1.1.1 60% mm 100 100 60 60% 1.1.2 A B A B A 1 B 1.1.3 boy W ID 1 2 3 DI DII DIII OL OL 1.1.4 2 1.1.5 1.1.6 1.1.7 1.1.8 1.2 1.2.1 1. 2. 3 1.2.2
More informationhttp://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 2 N(ε 1 ) N(ε 2 ) ε 1 ε 2 α ε ε 2 1 n N(ɛ) N ɛ ɛ- (1.1.3) n > N(ɛ) a n α < ɛ n N(ɛ) a n
http://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 1 1 1.1 ɛ-n 1 ɛ-n lim n a n = α n a n α 2 lim a n = 1 n a k n n k=1 1.1.7 ɛ-n 1.1.1 a n α a n n α lim n a n = α ɛ N(ɛ) n > N(ɛ) a n α < ɛ
More informationnote01
γ 5 J, M α J, M α = c JM JM J, M c JM e ipr p / M p = 0 M J(J + 1) / Λ p / M J(J + 1) / Λ ~ 1 / m π m π ~ 138 MeV J P,I = 0,1 π 1, π, π 3 ( ) ( π +, π 0, π ) ( ), π 0 = π 3 π ± = m 1 π1 ± iπ ( ) π ±,
More information日本糖尿病学会誌第58巻第1号
α β β β β β β α α β α β α l l α l μ l β l α β β Wfs1 β β l l l l μ l l μ μ l μ l Δ l μ μ l μ l l ll l l l l l l l l μ l l l l μ μ l l l l μ l l l l l l l l l l μ l l l μ l μ l l l l l l l l l μ l l l l
More information168 13 Maxwell ( H ds = C S rot H = j + D j + D ) ds (13.5) (13.6) Maxwell Ampère-Maxwell (3) Gauss S B 0 B ds = 0 (13.7) S div B = 0 (13.8) (4) Farad
13 Maxwell Maxwell Ampère Maxwell 13.1 Maxwell Maxwell E D H B ε 0 µ 0 (1) Gauss D = ε 0 E (13.1) B = µ 0 H. (13.2) S D = εe S S D ds = ρ(r)dr (13.3) S V div D = ρ (13.4) ρ S V Coulomb (2) Ampère C H =
More informationuntitled
Global Quantitative Research / -2- -3- -4- -5- 35 35 SPC SPC REIT REIT -6- -7- -8- -9- -10- -11- -12- -13- -14- -15- -16- -17- 100m$110-18- Global Quantitative Research -19- -20- -21- -22- -23- -24- -25-
More informationx (0, 6, N x 2 (4 + 2(4 + 3 < 6 2 3, a 2 + a 2+ > 0. x (0, 6 si x > 0. (2 cos [0, 6] (0, 6 (cos si < 0. ( 5.4.6 (2 (3 cos 0, cos 3 < 0. cos 0 cos cos
6 II 3 6. π 3.459... ( /( π 33 π 00 π 34 6.. ( (a cos π 2 0 π (0, 2 3 π (b z C, m, Z ( ( cos z + π 2 (, si z + π 2 (cos z, si z, 4m, ( si z, cos z, 4m +, (cos z, si z, 4m + 2, (si z, cos z, 4m + 3. (6.
More informationチュートリアル:ノンパラメトリックベイズ
{ x,x, L, xn} 2 p( θ, θ, θ, θ, θ, } { 2 3 4 5 θ6 p( p( { x,x, L, N} 2 x { θ, θ2, θ3, θ4, θ5, θ6} K n p( θ θ n N n θ x N + { x,x, L, N} 2 x { θ, θ2, θ3, θ4, θ5, θ6} log p( 6 n logθ F 6 log p( + λ θ F θ
More information1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0
A c 2008 by Kuniaki Nakamitsu 1 1.1 t 2 sin t, cos t t ft t t vt t xt t + t xt + t xt + t xt t vt = xt + t xt t t t vt xt + t xt vt = lim t 0 t lim t 0 t 0 vt = dxt ft dft dft ft + t ft = lim t 0 t 1.1
More informationR¤Çʬ¤«¤ëÎÏ³Ø·Ï - ¡Áʬ´ô¤ÎÍͻҤò²Ä»ë²½¤·¤Æ¤ß¤ë¡Á
.... R 2009 3 1 ( ) R 2009 3 1 1 / 23 : ( )!, @tkf, id:tkf41, (id:artk ) : 4 1 : http://arataka.wordpress.com : Python, C/C++, PHP, Javascript R : / ( ) R 2009 3 1 2 / 23 R? R! ( ) R 2009 3 1 3 / 23 =
More informationi 0 1 0.1 I................................................ 1 0.2.................................................. 2 0.2.1...........................
2008 II 21 1 31 i 0 1 0.1 I................................................ 1 0.2.................................................. 2 0.2.1............................................. 2 0.2.2.............................................
More informationPart. 4. () 4.. () 4.. 3 5. 5 5.. 5 5.. 6 5.3. 7 Part 3. 8 6. 8 6.. 8 6.. 8 7. 8 7.. 8 7.. 3 8. 3 9., 34 9.. 34 9.. 37 9.3. 39. 4.. 4.. 43. 46.. 46..
Cotets 6 6 : 6 6 6 6 6 6 7 7 7 Part. 8. 8.. 8.. 9..... 3. 3 3.. 3 3.. 7 3.3. 8 Part. 4. () 4.. () 4.. 3 5. 5 5.. 5 5.. 6 5.3. 7 Part 3. 8 6. 8 6.. 8 6.. 8 7. 8 7.. 8 7.. 3 8. 3 9., 34 9.. 34 9.. 37 9.3.
More information34 2 2 h = h/2π 3 V (x) E 4 2 1 ψ = sin kxk = 2π/λ λ = h/p p = h/λ = kh/2π = k h 5 2 ψ = e ax2 ガウス 型 関 数 1.2 1 関 数 値 0.8 0.6 0.4 0.2 0 15 10 5 0 5 10
33 2 2.1 2.1.1 x 1 T x T 0 F = ma T ψ) 1 x ψ(x) 2.1.2 1 1 h2 d 2 ψ(x) + V (x)ψ(x) = Eψ(x) (2.1) 2m dx 2 1 34 2 2 h = h/2π 3 V (x) E 4 2 1 ψ = sin kxk = 2π/λ λ = h/p p = h/λ = kh/2π = k h 5 2 ψ = e ax2
More information診療ガイドライン外来編2014(A4)/FUJGG2014‐01(大扉)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
More information,..,,.,,.,.,..,,.,,..,,,. 2
A.A. (1906) (1907). 2008.7.4 1.,.,.,,.,,,.,..,,,.,,.,, R.J.,.,.,,,..,.,. 1 ,..,,.,,.,.,..,,.,,..,,,. 2 1, 2, 2., 1,,,.,, 2, n, n 2 (, n 2 0 ).,,.,, n ( 2, ), 2 n.,,,,.,,,,..,,. 3 x 1, x 2,..., x n,...,,
More information日本糖尿病学会誌第58巻第2号
β γ Δ Δ β β β l l l l μ l l μ l l l l α l l l ω l Δ l l Δ Δ l l l l l l l l l l l l l l α α α α l l l l l l l l l l l μ l l μ l μ l l μ l l μ l l l μ l l l l l l l μ l β l l μ l l l l α l l μ l l
More information1 P2 P P3P4 P5P8 P9P10 P11 P12
1 P2 P14 2 3 4 5 1 P3P4 P5P8 P9P10 P11 P12 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 & 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1! 3 2 3! 4 4 3 5 6 I 7 8 P7 P7I P5 9 P5! 10 4!! 11 5 03-5220-8520
More informationDS II 方程式で小振幅周期ソリトンが関わる共鳴相互作用
1847 2013 157-168 157 $DS$ II (Takahito Arai) Research Institute for Science and Technology Kinki University (Masayoshi Tajiri) Osaka Prefecture University $DS$ II 2 2 1 2 $D$avey-Stewartson $(DS)$ $\{\begin{array}{l}iu_{t}+pu_{xx}+u_{yy}+r
More information2003/3 Vol. J86 D II No.3 2.3. 4. 5. 6. 2. 1 1 Fig. 1 An exterior view of eye scanner. CCD [7] 640 480 1 CCD PC USB PC 2 334 PC USB RS-232C PC 3 2.1 2
Curved Document Imaging with Eye Scanner Toshiyuki AMANO, Tsutomu ABE, Osamu NISHIKAWA, Tetsuo IYODA, and Yukio SATO 1. Shape From Shading SFS [1] [2] 3 2 Department of Electrical and Computer Engineering,
More informationMicrosoft Word - Wordで楽に数式を作る.docx
Ver. 3.1 2015/1/11 門 馬 英 一 郎 Word 1 する必要がある Alt+=の後に Ctrl+i とセットで覚えておく 1.4. 変換が出来ない場合 ごく稀に以下で説明する変換機能が無効になる場合がある その際は Word を再起動するとまた使えるようになる 1.5. 独立数式と文中数式 数式のスタイルは独立数式 文中数式(2 次元)と文中数式(線形)の 3 種類があ り 数式モードの右端の矢印を選ぶとメニューが出てくる
More information1 1 1 1 1 1 2 f z 2 C 1, C 2 f 2 C 1, C 2 f(c 2 ) C 2 f(c 1 ) z C 1 f f(z) xy uv ( u v ) = ( a b c d ) ( x y ) + ( p q ) (p + b, q + d) 1 (p + a, q + c) 1 (p, q) 1 1 (b, d) (a, c) 2 3 2 3 a = d, c = b
More information662/04-直立.indd
l l q= / D s HTqq /L T L T l l ε s ε = D + s 3 K = αγk R 4 3 K αγk + ( α + β ) K 4 = 0 γ L L + K R K αβγ () ㅧ ర ㅧ ర (4) (5) ()ᑼ (6) (8) (9) (0) () () (3) (3) (7) Ƚˎȁ Ȇ ၑა FYDFM වႁ ޙ 䊶䊶 䊶 䊶䊶 䊶 Ƚˏȁζ υ ίυέρθ
More informationi ( ) PDF http://moodle.sci.u-toyama.ac.jp/kyozai/ I +α II II III A: IV B: V C: III V I, II III IV V III IV 8 5 6 krmt@sci.u-toyama.ac.jp
8 5 6 i ( ) PDF http://moodle.sci.u-toyama.ac.jp/kyozai/ I +α II II III A: IV B: V C: III V I, II III IV V III IV 8 5 6 krmt@sci.u-toyama.ac.jp ii I +α 3.....................................................
More information1 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
More informationDGE DGE 2 2 1 1990 1 1 3 (1) ( 1
早 稲 田 大 学 現 代 政 治 経 済 研 究 所 ゼロ 金 利 下 で 量 的 緩 和 政 策 は 有 効 か? -ニューケインジアンDGEモデルによる 信 用 創 造 の 罠 の 分 析 - 井 上 智 洋 品 川 俊 介 都 築 栄 司 上 浦 基 No.J1403 Working Paper Series Institute for Research in Contemporary Political
More informationMicrosoft Word - 触ってみよう、Maximaに2.doc
i i e! ( x +1) 2 3 ( 2x + 3)! ( x + 1) 3 ( a + b) 5 2 2 2 2! 3! 5! 7 2 x! 3x! 1 = 0 ",! " >!!! # 2x + 4y = 30 "! x + y = 12 sin x lim x!0 x x n! # $ & 1 lim 1 + ('% " n 1 1 lim lim x!+0 x x"!0 x log x
More information●70974_100_AC009160_KAPヘ<3099>ーシス自動車約款(11.10).indb
" # $ % & ' ( ) * +, -. / 0 1 2 3 4 5 6 7 8 9 : ; < = >? @ A B C D E F G H I J K L M N O P Q R S T U V W X Y " # $ % & ' ( ) * + , -. / 0 1 2 3 4 5 6 7 8 9 : ; < = > ? @ A B
More information2017
2017 1 2 2 3 3 4 3.1?......................... 4 3.2 [5, 6]........... 5 3.2.1......... 5 3.2.2................... 6 3.2.3....... 9 3.2.4.......... 10 3.2.5....... 11 3.2.6 : phase slip [5, 8]..........
More informationL A TEX ver L A TEX LATEX 1.1 L A TEX L A TEX tex 1.1 1) latex mkdir latex 2) latex sample1 sample2 mkdir latex/sample1 mkdir latex/sampl
L A TEX ver.2004.11.18 1 L A TEX LATEX 1.1 L A TEX L A TEX tex 1.1 1) latex mkdir latex 2) latex sample1 sample2 mkdir latex/sample1 mkdir latex/sample2 3) /staff/kaede work/www/math/takase sample1.tex
More informationAccuracy Improvement by Compound Discriminant Functions for Resembling Character Recognition Takashi NAKAJIMA, Tetsushi WAKABAYASHI, Fumitaka KIMURA,
Journal Article / 学 術 雑 誌 論 文 混 合 識 別 関 数 による 類 似 文 字 認 識 の 高 精 度 化 Accuracy improvement by compoun for resembling character recogn 中 嶋, 孝 ; 若 林, 哲 史 ; 木 村, 文 隆 ; 三 宅, 康 二 Nakajima, Takashi; Wakabayashi,
More informationA. Fresnel) 19 1900 (M. Planck) 1905 (A. Einstein) X (A. Ampère) (M. Faraday) 1864 (C. Maxwell) 1871 (H. R. Hertz) 1888 2.2 1 7 (G. Galilei) 1638 2
1 2012.8 e-mail: tatekawa (at) akane.waseda.jp 1 2005-2006 2 2009 1-2 3 x t x t 2 2.1 17 (I. Newton) C. Huygens) 19 (T. Young) 1 A. Fresnel) 19 1900 (M. Planck) 1905 (A. Einstein) X (A. Ampère) (M. Faraday)
More information1 1 2 GDP 3 1 GDP 2 GDP 3 GDP GDP GDP 4 GDP GDP GDP 1 GDP 2 CPI 2
Macroeconomics/ Olivier Blanchard, 1996 1 2 1........... 2 2............. 2 2 3 3............... 3 4...... 4 5.............. 4 6 IS-LM................. 5 3 6 7. 6 8....... 7 9........... 8 10..... 8 8
More informationサイバネットニュース No.121
2007 Spring No.121 01 02 03 04 05 06 07 08 09 10 12 13 14 18 01 02 03 04 05 06 07 L R L R L R I x C G C G C G x 08 09 σ () t σ () t = Sx() t Q σ=0 P y O S x= y y & T S= 1 1 x& () t = Ax() t + Bu() t +
More information- 1 - - 2 - 320 421 928 1115 12 8 116 124 2 7 4 5 428 515 530 624 921 1115 1-3 - 100 250-4 - - 5 - - 6 - - 7 - - 8 - - 9 - & & - 11 - - 12 - GT GT - 13 - GT - 14 - - 15 - - 16 - - 17 - - 18 - - 19 - -
More information平成18年度弁理士試験本試験問題とその傾向
CBA CBA CBA CBA CBA CBA Vol. No. CBA CBA CBA CBA a b a bm m swkmsms kgm NmPa WWmK σ x σ y τ xy θ σ θ τ θ m b t p A-A' σ τ A-A' θ B-B' σ τ B-B' A-A' B-B' B-B' pσ σ B-B' pτ τ l x x I E Vol. No. w x xl/ 3
More information3 3 i
00D8102021I 2004 3 3 3 i 1 ------------------------------------------------------------------------------------------------1 2 ---------------------------------------------------------------------------------------2
More informationサイバニュース-vol134-CS3.indd
NEWS 2012 WINTER 134 No. F=maF ma m af Contents N, X θ 1,θ 2 θ N 0θ i π/2 X i X 0 Θ i Θ 1 = 2θ 1 Θ 2 = 2(θ 1 θ 2) NX N X 0 Θ N N Θ N = 2{θ 1 θ 2θ 3 θ N } Θ N = 2π A 1A 2B 2B 1 mm 3 α α = π /m A 1A
More informationE B m e ( ) γma = F = e E + v B a m = 0.5MeV γ = E e m =957 E e GeV v β = v SPring-8 γ β γ E e [GeV] [ ] NewSUBARU.0 957 0.999999869 SPring-8 8.0 5656
SPring-8 PF( ) ( ) UVSOR( HiSOR( SPring-8.. 3. 4. 5. 6. 7. E B m e ( ) γma = F = e E + v B a m = 0.5MeV γ = E e m =957 E e GeV v β = v SPring-8 γ β γ E e [GeV] [ ] NewSUBARU.0 957 0.999999869 SPring-8
More information1 1 3 1.1 (Frequecy Tabulatios)................................ 3 1........................................ 8 1.3.....................................
1 1 3 1.1 (Frequecy Tabulatios)................................ 3 1........................................ 8 1.3........................................... 1 17.1................................................
More information1: Pauli 2 Heisenberg [3] 3 r 1, r 2 V (r 1, r 2 )=V (r 2, r 1 ) V (r 1, r 2 ) 5 ϕ(r 1, r 2 ) Schrödinger } { h2 2m ( 1 + 2 )+V (r 1, r 2 ) ϕ(r 1, r 2
Hubbard 2 1 1 Pauli 0 3 Pauli 4 1 Vol. 51, No. 10, 1996, pp. 741 747. 2 http://www.gakushuin.ac.jp/ 881791/ 3 8 4 1 1: Pauli 2 Heisenberg [3] 3 r 1, r 2 V (r 1, r 2 )=V (r 2, r 1 ) V (r 1, r 2 ) 5 ϕ(r
More information有機性産業廃棄物の連続炭化装置の開発
( ) Development of the apparatus conveyer type which carbonizes continuously organic industrial waste (About the form of blade in conveyer) 1055047 1 1-1 1 1-2 1-3 2 2 2-1 2-2 2-3 2-4 7 3 3-1 20 3-2 3-3
More informationII 2014 2 (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1
II 2014 1 1 I 1.1 72 r 2 72 8 72/8 = 9 9 2 a 0 1 a 1 a 1 = a 0 (1+r/100) 2 a 2 a 2 = a 1 (1 + r/100) = a 0 (1 + r/100) 2 n a n = a 0 (1 + r/100) n a n a 0 2 n a 0 (1 + r/100) n = 2a 0 (1 + r/100) n = 2
More information25 11M15133 0.40 0.44 n O(n 2 ) O(n) 0.33 0.52 O(n) 0.36 0.52 O(n) 2 0.48 0.52
26 1 11M15133 25 11M15133 0.40 0.44 n O(n 2 ) O(n) 0.33 0.52 O(n) 0.36 0.52 O(n) 2 0.48 0.52 1 2 2 4 2.1.............................. 4 2.2.................................. 5 2.2.1...........................
More informationi 1 1 1.1... 1 1.1.1... 2 1.1.2... 7 1.2... 9 1.3... 1 1.4... 12 1.4.1 s... 12 1.4.2... 12 1.5... 15 1.5.1... 15 1.5.2... 18 2 21 2.1... 21 2.2... 23
2 III Copyright c 2 Kazunobu Yoshida. All rights reserved. i 1 1 1.1... 1 1.1.1... 2 1.1.2... 7 1.2... 9 1.3... 1 1.4... 12 1.4.1 s... 12 1.4.2... 12 1.5... 15 1.5.1... 15 1.5.2... 18 2 21 2.1... 21 2.2...
More information,255 7, ,355 4,452 3,420 3,736 8,206 4, , ,992 6, ,646 4,
30 8 IT 28 1,260 3 1 11. 1101. 1102. 1103. 1 3 1,368.3 3 1,109.8 p.5,p.7 2 9,646 4,291 14.5% 10,p.11 3 3,521 8 p.13 45-49 40-44 50-54 019 5 3 1 2,891 3 6 1 3 95 1 1101 1102 1103 1101 1102 1103 1 6,255
More information7 9 7..................................... 9 7................................ 3 7.3...................................... 3 A A. ω ν = ω/π E = hω. E
B 8.9.4, : : MIT I,II A.P. E.F.,, 993 I,,, 999, 7 I,II, 95 A A........................... A........................... 3.3 A.............................. 4.4....................................... 5 6..............................
More information0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9
1-1. 1, 2, 3, 4, 5, 6, 7,, 100,, 1000, n, m m m n n 0 n, m m n 1-2. 0 m n m n 0 2 = 1.41421356 π = 3.141516 1-3. 1 0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c),
More informationuntitled
1 1 1. 2. 3. 2 2 1 (5/6) 4 =0.517... 5/6 (5/6) 4 1 (5/6) 4 1 (35/36) 24 =0.491... 0.5 2.7 3 1 n =rand() 0 1 = rand() () rand 6 0,1,2,3,4,5 1 1 6 6 *6 int() integer 1 6 = int(rand()*6)+1 1 4 3 500 260 52%
More informationE F = q b E (2) E q a r q a q b N/C q a (electric flux line) q a E r r r E 4πr 2 E 4πr 2 = k q a r 2 4πr2 = 4πkq a (3) 4πkq a 1835 4πk 1 ɛ 0 ɛ 0 (perm
1 1.1 18 (static electricity) 20 (electric charge) A,B q a, q b r F F = k q aq b r 2 (1) k q b F F q a r?? 18 (Coulomb) 1 N C r 1m 9 10 9 N 1C k 9 10 9 Nm 2 /C 2 1 k q a r 2 (Electric Field) 1 E F = q
More information