p06.dvi

Size: px
Start display at page:

Download "p06.dvi"

Transcription

1 I 6 : 1

2 (1) u(t) y(t) : n m a n i y (i) = b m i u (i) i=0 i=0 t, y (i) y i (u )., a 0 0, b 0 0. : 2

3 (2), Laplace, (a 0 s n +a 1 s n 1 + +a n )Y(s) = (b 0 s m + b 1 s m 1 + +b m )U(s),, Y(s) U(s) = b 0s m +b 1 s m 1 + +b m a 0 s n +a 1 s n 1 + +a n.,. : 3

4 (3), 1 1, Y(s)/U(s) (U(s) Y(s) u(t) y(t) Laplace ) s,. : 4

5 (4). G(s) = N(s) D(s) = b 0s m +b 1 s m 1 + +b m a 0 s n +a 1 s n 1 + +a n N(s),D(s),b 0 0, a 0 0., m, n. : 5

6 (5) denominator, numerator,, D(s), N(s) ( ). G(s), H(s). 1. : 6

7 (6), G(s) = b 0 a 0 s m + b 1 a 0 s m bm a 0 s n + a 1 s n a n a 0 a }{{ 0},. : 7

8 (7) N(s) D(s) ( )1. 1,,.,,. : 8

9 (8) n, m. n m,., n > m,.,. : 9

10 (9) (s β 1 ) (s β m ) G(s) G(s) = g 0 (s α 1 ) (s α n ) {α 1,...,α n }. {β 1,...,β m }. g 0 = β 0 /α 0. : 10

11 (10) G(s) = n i=0 b is n i n i=0 a is n i, (a 0 0, b 0 0). ) n,g(s) = b 0 i=1( b i b 0 a 0 a i s n i + a n 0 i=0 a is n i.,. : 11

12 (11) G(s), = G 0 + G 1 (s), G 0, G 1 (s),., U(s),G 0 U(s)+ G 1 (s)u(s). Laplace. : 12

13 (12) G 0 U(s) Laplace G 0 u(t).,. G 1 (s)u(s) Laplace,G 1 (s) Laplace ( g 1 (t) ) u(t). g 1 (t),. : 13

14 (13) G(s) = g 0 (s β 1 ) (s β m ) (s α 1 ) (s α n ). {α( 1,...,α n }, G(s) = A1 g A ) n. s α 1 s α n. : 14

15 (14) A i (i = 1,...,n), ( n i=1 A i j i j)) (s α ( 1 ). {α 1,...,α n },. : 15

16 (15) (s β 1 ) (s β m ) G(s) = g 0 (s α 1 ) r 1 (s αk ) r k. ( ri )) A ij (s α i ) j ( k, G s = g 0 i=1 j=1.. :

17 (16) A ij. L 1 [ Laplace ], t 0 L 1 1 = tk 1 (s α) k (k 1)! exp[αt], [ ] 1 L 1 = tk 1 s k (k 1)!,. : 17

18 (1) Laplace ( ),, sx(s) = L[x (t)]+x(0) ( 2 ) [ ] t k 1 1 L (k 1)! eαt = (s α) (k, α k 2 ) 3 (L[ ] Laplace ). : 18

19 (2), ( ). x t, k. dk x x (k) dt k ] L [x (k) ( s) = 0 0 x (k) e st dt = [ x (k 1) e st ] 0 ] x (k 1) e st dt = x (k 1) (0)+sL [x (k 1) : 19

20 (3) [ f(t) ] 0 = f( ) f(0)., s, x (k). ] ], L [x (k) = sl [x (k 1) x (k 1) (0) = s ( sl[x (k 2) ] x k 2 (0) ) x (k 1) (0) = ], L [x (k) = s k X(s) s k 1 x(0) x (k 1) (0) :

21 (4). cosβt = eiβt +e iβt, sinβt = eiβt e iβt 2 2i [ ] L cosβt = 1 ( 1 2 s iβ + 1 ) s+iβ [ ] L sinβt = 1 ( 1 2i s iβ 1 ) s+iβ : 21

22 Laplace. ẏ +y = sint, y(0) = 1 : (sy(s) y(0))+y(s) ( 1 : 1 ) 2i s i 1 s+i : ( (s+1)y(s) = ) 1 2i s i s+i Y(s) = 1 + ( 1 1 ) 1 s+1 2i(s+1) s i s+i 2 3 Laplace. : 22

23 1 = A1 + B1 (s+1)(s i) s+1 s i A 1,B 1, A 1 = 1(1 i), B 2 1 = 1(1 i) 2 1 = A2 + B2 (s+1)(s+i) s+1 s+i A 2,B 2, A 2 = 1 (1+i), B 2 2 = 1 (1+i) 2, y(t) = e i ( 12 (1 i)+ 12 (1+i) ) + 1 2i e t ( 1 2 (1 i)eit 1 ) 2 (1+i)e it : 23

24 , 1 i = 2e iπ 4, 1 + i = 2e iπ 4, y(t) = 3 2 e i ( e i(t π 4 ) e i(t π)) 4 2 = 3 2 e sin(t π 4 ) : 24

25 (1),..,. : 25

26 (2),,,,. : 26

27 U(s) Y(s) U(s) 1 Y(s) K s 2 +s+1 1 Y(s) = KU(s) Y(s) = s 2 +s+1 U(s) : 27

28 (4)., : 28

29 (5),.,. : 29

30 (6) x + + x+y x + - x-y y y x x x. : 30

31 (7)., Σ,.. : 31

32 (8),.,.. : 32

33 (9),.. : 33

34 G 1 (s) G 2 (s) = G 1 (s)g 2 (s) G 1 (s) G 2 (s) + + = G 1 (s)+g 2 (s) : 34

35 R(s) + U(s) Y(s) R(s) G(s) G(s) = 1-G(s)H(s) + H(s) Y(s) Y(s) = G(s)U(s), U(s) = R(s)+H(s)Y(s) Y(s) = G(s)R(s)+G(s)H(s)Y(s) Y(s) = G(s) 1 G(s)H(s) R(s) : 35

36 (12) s ( ) : 36

37 ( ) U(s) + + Y(s) Y(s) = U(s)+Y(s) U(s) = 0, Y(s) : ; : 37

38 ( ) U(s) Y(s) = U(s) 2/3 Y(s) 0.5(U(s)+0.5Y(s)) = Y(s), 0.5U(s) = 0.75Y(s), = 2 3 U(s) : 38

39 (1),,.. (branch) (node) 2., ( ) ( ).,. : 39

40 (2).,.,.,.,. : 40

41 + + G(s) H(s) G(s) H(s) : 41

42 (1) n A i j n 1. ( 1) i+j A (i,j), a ij, ã ij. : 42

43 (2) (i,j) ã ji A, adja. Cramer, A, A 1. A 1 = 1 deta adja : 43

44 (1),. 1 1 : ẋ = Ax+Bu, y = Cx+Du., x R n, u R, y R, A R n n, B R n 1, C R 1 n, D R. u y. : 44

45 (2) x, u, y,, A, B, C. A, B, C,. L[ ] Laplace. : 45

46 (3) L[x(t)] = X(s), L[u(t)] = U(s), L[y(t)] = Y(s). Laplace, sx(s) = AX(s)+BU(s)., (si A)X(s) = BU(s). I n. : 46

47 (4) P(s) = det(si A) s,, (si A),. (si A) 1 = 1 adj(si A) P(s) : 47

48 (5) (si A)X(s) = BU(s), X(s) = (si A) 1 BU(s). Y(s) = CX(s)+DU(s), Y(s) = ( C(sI A) 1 B +D ) U(s).,. : 48

49 (6) (C(sI A) 1 B +D) ( ) 1,., A ( ),. : 49

50 (1)., : ẋ = Ax+Bu, y = Cx+Du., x R n, u R m, y R p, A R n n, B R n m, C R p n, D R p m. : 50

51 (2) Laplace, sx(s) = AX(s) + BU(s). si A 1 1, : X(s) = (si A) 1 BU(s) : Y (s) = ( C(sI A) 1 B +D ) U(s) : 51

52 (3) G(s) = C(sI A) 1 B +D.,G(s) m p, s. G(s). : 52

53 (4) ( ), : 53

54 Markov (1) ( (si A) = s I A ) s. 1 1 r = 1+r +r2 + ( r < 1) ( ) : 54

55 Markov (2) s ( )) 1 (si A) 1 = s = 1 s ( I A s (I + ) As A2 + s + 2 = I s + A s 2 + A2 s 3 + : 55

56 Markov (3) G(s) = C(sI A) 1 B +D (si A) 1 G(s) = D+ CB s + CAB s 2 + CA2 B s 3 + : 56

57 Markov (4) G(s) s D, CB, CAB, CA 2 B, Markov. Markov, ( II ) : 57

58 vs (1), ( ),. : 58

59 vs (2) ( ), ( ).,. : 59

60 vs (3),,,. : 60

61 vs (4), ( ),.,,. : 61

62 vs (5) ( ),.,.,,. : 62

63 vs (6),,,. : 63

64 W. S. Levine (ed.), The Control Handbook, 2/e, CRC Press, 2011, ( ),,, 2015,, 1993, H,, 2000 J. Mikusinsuki (, ), ( ), 15,, 1982 : 64

214 March 31, 214, Rev.2.1 4........................ 4........................ 5............................. 7............................... 7 1 8 1.1............................... 8 1.2.......................

More information

213 March 25, 213, Rev.1.5 4........................ 4........................ 6 1 8 1.1............................... 8 1.2....................... 9 2 14 2.1..................... 14 2.2............................

More information

4.2.................... 20 4.3.................. 21 4.4 ( )............... 22 4.5 ( )...... 24 4.6 ( )........ 25 4.7 ( )..... 26 5 28 5.1 PID........

4.2.................... 20 4.3.................. 21 4.4 ( )............... 22 4.5 ( )...... 24 4.6 ( )........ 25 4.7 ( )..... 26 5 28 5.1 PID........ version 0.01 : 2004/04/16 1 2 1.1................. 2 1.2.......................... 3 1.3................. 5 1.4............... 6 1.5.............. 7 2 9 2.1........................ 9 2.2......................

More information

( ) ( )

( ) ( ) 20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))

More information

2012 September 21, 2012, Rev.2.2

2012 September 21, 2012, Rev.2.2 212 September 21, 212, Rev.2.2 4................. 4 1 6 1.1.................. 6 1.2.................... 7 1.3 s................... 8 1.4....................... 9 1.5..................... 11 2 12 2.1.........................

More information

arma dvi

arma dvi ARMA 007/05/0 Rev.0 007/05/ Rev.0 007/07/7 3. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 3. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 3.3 : : : :

More information

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....

More information

曲面のパラメタ表示と接線ベクトル

曲面のパラメタ表示と接線ベクトル L11(2011-07-06 Wed) :Time-stamp: 2011-07-06 Wed 13:08 JST hig 1,,. 2. http://hig3.net () (L11) 2011-07-06 Wed 1 / 18 ( ) 1 V = (xy2 ) x + (2y) y = y 2 + 2. 2 V = 4y., D V ds = 2 2 ( ) 4 x 2 4y dy dx =

More information

数学演習:微分方程式

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

K E N Z OU

K E N Z OU K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................

More information

6 19,,,

6 19,,, 6 19,,, 15 6 19 4-2 à A si A s n + a n s n 1 + + a 2 s + a 1 à 0 1 0 0 1 0 0 0 1 a 1 a 2 a n 1 a n à ( 1, λ i, λ i 2,, λ i n 1 ) T ( λ i, λ 2 i,, λ n 1 i, a 1 a 2 λ i a n λ ) n 1 T i ( ) λ i 1, λ i,, λ

More information

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x 2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin

More information

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2 No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j

More information

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. 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..............

More information

[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt

[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt 3.4.7 [.] =e j(t+/4), =5e j(t+/3), 3 =3e j(t+/6) ~ = ~ + ~ + ~ 3 = e j(t+φ) =(e 4 j +5e 3 j +3e 6 j )e jt = e jφ e jt cos φ =cos 4 +5cos 3 +3cos 6 =.69 sin φ =sin 4 +5sin 3 +3sin 6 =.9 =.69 +.9 =7.74 [.]

More information

2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................

More information

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....................

More information

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

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

More information

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i 1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [

More information

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),

More information

sikepuri.dvi

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

More information

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2 II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh

More information

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 = #A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/

More information

08-Note2-web

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)

More information

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x + (.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d

More information

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P 6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P

More information

1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................

More information

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

More information

untitled

untitled - k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

More information

AHPを用いた大相撲の新しい番付編成

AHPを用いた大相撲の新しい番付編成 5304050 2008/2/15 1 2008/2/15 2 42 2008/2/15 3 2008/2/15 4 195 2008/2/15 5 2008/2/15 6 i j ij >1 ij ij1/>1 i j i 1 ji 1/ j ij 2008/2/15 7 1 =2.01/=0.5 =1.51/=0.67 2008/2/15 8 1 2008/2/15 9 () u ) i i i

More information

chap9.dvi

chap9.dvi 9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =

More information

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 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

More information

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 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 information

(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 )

More information

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x [ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),

More information

‚åŁÎ“·„´Šš‡ðŠp‡¢‡½‹âfi`fiI…A…‰…S…−…Y…•‡ÌMarkovŸA“½fiI›ð’Í

‚åŁÎ“·„´Šš‡ðŠp‡¢‡½‹âfi`fiI…A…‰…S…−…Y…•‡ÌMarkovŸA“½fiI›ð’Í Markov 2009 10 2 Markov 2009 10 2 1 / 25 1 (GA) 2 GA 3 4 Markov 2009 10 2 2 / 25 (GA) (GA) L ( 1) I := {0, 1} L f : I (0, ) M( 2) S := I M GA (GA) f (i) i I Markov 2009 10 2 3 / 25 (GA) ρ(i, j), i, j I

More information

05Mar2001_tune.dvi

05Mar2001_tune.dvi 2001 3 5 COD 1 1.1 u d2 u + ku =0 (1) dt2 u = a exp(pt) (2) p = ± k (3) k>0k = ω 2 exp(±iωt) (4) k

More information

genron-3

genron-3 " ( K p( pasals! ( kg / m 3 " ( K! v M V! M / V v V / M! 3 ( kg / m v ( v "! v p v # v v pd v ( J / kg p ( $ 3! % S $ ( pv" 3 ( ( 5 pv" pv R" p R!" R " ( K ( 6 ( 7 " pv pv % p % w ' p% S & $ p% v ( J /

More information

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63> MATLAB/Simulink による現代制御入門 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/9241 このサンプルページの内容は, 初版 1 刷発行当時のものです. i MATLAB/Simulink MATLAB/Simulink 1. 1 2. 3. MATLAB/Simulink

More information

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g( 06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,

More information

直交座標系の回転

直交座標系の回転 b T.Koama x l x, Lx i ij j j xi i i i, x L T L L, L ± x L T xax axx, ( a a ) i, j ij i j ij ji λ λ + λ + + λ i i i x L T T T x ( L) L T xax T ( T L T ) A( L) T ( LAL T ) T ( L AL) λ ii L AL Λ λi i axx

More information

1 911 9001030 9:00 A B C D E F G H I J K L M 1A0900 1B0900 1C0900 1D0900 1E0900 1F0900 1G0900 1H0900 1I0900 1J0900 1K0900 1L0900 1M0900 9:15 1A0915 1B0915 1C0915 1D0915 1E0915 1F0915 1G0915 1H0915 1I0915

More information

B ver B

B ver B B ver. 2017.02.24 B Contents 1 11 1.1....................... 11 1.1.1............. 11 1.1.2.......................... 12 1.2............................. 14 1.2.1................ 14 1.2.2.......................

More information

.. p.2/5

.. p.2/5 IV. p./5 .. p.2/5 .. 8 >< >: d dt y = a, y + a,2 y 2 + + a,n y n + f (t) d dt y 2 = a 2, y + a 2,2 y 2 + + a 2,n y n + f 2 (t). d dt y n = a n, y + a n,2 y 2 + + a n,n y n + f n (t) (a i,j ) p.2/5 .. 8

More information

1 y(t)m b k u(t) ẋ = [ 0 1 k m b m x + [ 0 1 m u, x = [ ẏ y (1) y b k m u

1 y(t)m b k u(t) ẋ = [ 0 1 k m b m x + [ 0 1 m u, x = [ ẏ y (1) y b k m u ( ) LPV( ) 1 y(t)m b k u(t) ẋ = [ 0 1 k m b m x + [ 0 1 m u, x = [ ẏ y (1) y b k m u m 1 m m 2, b 1 b b 2, k 1 k k 2 (2) [m b k ( ) k 0 b m ( ) 2 ẋ = Ax, x(0) 0 (3) (x(t) 0) ( ) V (x) V (x) = x T P x >

More information

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y 5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =

More information

all.dvi

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

More information

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ 4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ

More information

II 2 II

II 2 II II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................

More information

II 1 II 2012 II Gauss-Bonnet II

II 1 II 2012 II Gauss-Bonnet II II 1 II 212 II Gauss-Bonnet II 1 1 1.1......................................... 1 1.2............................................ 2 1.3.................................. 3 1.4.............................................

More information

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63> 信号処理の基礎 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/081051 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 AI ii z / 2 3 4 5 6 7 7 z 8 8 iii 2013 3 iv 1 1 1.1... 1 1.2... 2 2 4 2.1...

More information

all.dvi

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

More information

6.1 (P (P (P (P (P (P (, P (, P.

6.1 (P (P (P (P (P (P (, P (, P. (011 30 7 0 ( ( 3 ( 010 1 (P.3 1 1.1 (P.4.................. 1 1. (P.4............... 1 (P.15.1 (P.16................. (P.0............3 (P.18 3.4 (P.3............... 4 3 (P.9 4 3.1 (P.30........... 4 3.

More information

Untitled

Untitled http://www.ike-dyn.ritsumei.ac.jp/ hyoo/dynamics.html 1 (i) (ii) 2 (i) (ii) (*) [1] 2 1 3 1.1................................ 3 1.2..................................... 3 2 4 2.1....................................

More information

CG38.PDF

CG38.PDF ............3...3...6....6....8.....8.....4...9 3....9 3.... 3.3...4 3.4...36...39 4....39 4.....39 4.....4 4....49 4.....5 4.....57...64 5....64 5....66 5.3...68 5.4...7 5.5...77...8 6....8 6.....8 6.....83

More information

2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) =

2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) = 1 1 1.1 I R 1.1.1 c : I R 2 (i) c C (ii) t I c (t) (0, 0) c (t) c(i) c c(t) 1.1.2 (1) (2) (3) (1) r > 0 c : R R 2 : t (r cos t, r sin t) (2) C f : I R c : I R 2 : t (t, f(t)) (3) y = x c : R R 2 : t (t,

More information

入試の軌跡

入試の軌跡 4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf

More information

L P y P y + ɛ, ɛ y P y I P y,, y P y + I P y, 3 ŷ β 0 β y β 0 β y β β 0, β y x x, x,, x, y y, y,, y x x y y x x, y y, x x y y {}}{,,, / / L P / / y, P

L P y P y + ɛ, ɛ y P y I P y,, y P y + I P y, 3 ŷ β 0 β y β 0 β y β β 0, β y x x, x,, x, y y, y,, y x x y y x x, y y, x x y y {}}{,,, / / L P / / y, P 005 5 6 y β + ɛ {x, x,, x p } y, {x, x,, x p }, β, ɛ E ɛ 0 V ɛ σ I 3 rak p 4 ɛ i N 0, σ ɛ ɛ y β y β y y β y + β β, ɛ β y + β 0, β y β y ɛ ɛ β ɛ y β mi L y y ŷ β y β y β β L P y P y + ɛ, ɛ y P y I P y,,

More information

25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3

More information

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,

More information

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H 199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

More information

2 0.1 Introduction NMR 70% 1/2

2 0.1 Introduction NMR 70% 1/2 Y. Kondo 2010 1 22 2 0.1 Introduction NMR 70% 1/2 3 0.1 Introduction......................... 2 1 7 1.1.................... 7 1.2............................ 11 1.3................... 12 1.4..........................

More information

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

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

More information

Chap11.dvi

Chap11.dvi . () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x

More information

meiji_resume_1.PDF

meiji_resume_1.PDF β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E

More information

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,. 24(2012) (1 C106) 4 11 (2 C206) 4 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 (). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5... 6.. 7.,,. 8.,. 1. (75%)

More information

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,. 9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,

More information

: , 2.0, 3.0, 2.0, (%) ( 2.

: , 2.0, 3.0, 2.0, (%) ( 2. 2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................

More information

高校生の就職への数学II

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...............................

More information

6.1 (P (P (P (P (P (P (, P (, P.101

6.1 (P (P (P (P (P (P (, P (, P.101 (008 0 3 7 ( ( ( 00 1 (P.3 1 1.1 (P.3.................. 1 1. (P.4............... 1 (P.15.1 (P.15................. (P.18............3 (P.17......... 3.4 (P................ 4 3 (P.7 4 3.1 ( P.7...........

More information

85 4

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

More information

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) 4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7

More information

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1 ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD

More information

30 (11/04 )

30 (11/04 ) 30 (11/04 ) i, 1,, II I?,,,,,,,,, ( ),,, ϵ δ,,,,, (, ),,,,,, 5 : (1) ( ) () (,, ) (3) ( ) (4) (5) ( ) (1),, (),,, () (3), (),, (4), (1), (3), ( ), (5),,,,,,,, ii,,,,,,,, Richard P. Feynman, The best teaching

More information

,, 2. Matlab Simulink 2018 PC Matlab Scilab 2

,, 2. Matlab Simulink 2018 PC Matlab Scilab 2 (2018 ) ( -1) TA Email : ohki@i.kyoto-u.ac.jp, ske.ta@bode.amp.i.kyoto-u.ac.jp : 411 : 10 308 1 1 2 2 2.1............................................ 2 2.2..................................................

More information

Untitled

Untitled II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j

More information

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,, 01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,

More information

23 7 28 i i 1 1 1.1................................... 2 1.2............................... 3 1.2.1.................................... 3 1.2.2............................... 4 1.2.3 SI..............................

More information

21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

More information

10 1 1 (1) (2) (3) 3 3 1 3 1 3 (4) 2 32 2 (1) 1 1

10 1 1 (1) (2) (3) 3 3 1 3 1 3 (4) 2 32 2 (1) 1 1 10 10 1 1 (1) (2) (3) 3 3 1 3 1 3 (4) 2 32 2 (1) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (2) 1 (3) JI S JI S JI S JI S 25 175 J AS 3 (1) 3 70 (2) (3) 100 4 (1)69 (2) (3) (4) (5) (6) (7) (8)70 (9) (10)2 (11)

More information

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... 23

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... 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

1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O

1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O : 2014 4 10 1 2 2 3 2.1...................................... 3 2.2....................................... 4 2.3....................................... 4 2.4................................ 5 2.5 Free-Body

More information

R R 16 ( 3 )

R R 16   ( 3 ) (017 ) 9 4 7 ( ) ( 3 ) ( 010 ) 1 (P3) 1 11 (P4) 1 1 (P4) 1 (P15) 1 (P16) (P0) 3 (P18) 3 4 (P3) 4 3 4 31 1 5 3 5 4 6 5 9 51 9 5 9 6 9 61 9 6 α β 9 63 û 11 64 R 1 65 13 66 14 7 14 71 15 7 R R 16 http://wwwecoosaka-uacjp/~tazak/class/017

More information

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 = 5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =

More information

untitled

untitled 2 : n =1, 2,, 10000 0.5125 0.51 0.5075 0.505 0.5025 0.5 0.4975 0.495 0 2000 4000 6000 8000 10000 2 weak law of large numbers 1. X 1,X 2,,X n 2. µ = E(X i ),i=1, 2,,n 3. σi 2 = V (X i ) σ 2,i=1, 2,,n ɛ>0

More information

D:/BOOK/MAIN/MAIN.DVI

D:/BOOK/MAIN/MAIN.DVI 8 2 F (s) =L f(t) F (s) =L f(t) := Z 0 f()e ;s d (2.2) s s = + j! f(t) (f(0)=0 f(0) _ = 0 d n; f(0)=dt n; =0) L dn f(t) = s n F (s) (2.3) dt n Z t L 0 f()d = F (s) (2.4) s s =s f(t) L _ f(t) Z Z ;s L f(t)

More information

1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0

1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 : 2016 4 1 1 2 1.1......................................... 2 1.2................................... 2 2 2 2.1........................................ 2 2.2......................................... 3 2.3.........................................

More information

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

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

More information

Gmech08.dvi

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

More information

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 = 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

More information

sec13.dvi

sec13.dvi 13 13.1 O r F R = m d 2 r dt 2 m r m = F = m r M M d2 R dt 2 = m d 2 r dt 2 = F = F (13.1) F O L = r p = m r ṙ dl dt = m ṙ ṙ + m r r = r (m r ) = r F N. (13.2) N N = R F 13.2 O ˆn ω L O r u u = ω r 1 1:

More information

(1) 1 y = 2 = = b (2) 2 y = 2 = 2 = 2 + h B h h h< h 2 h

(1) 1 y = 2 = = b (2) 2 y = 2 = 2 = 2 + h B h h h< h 2 h 6 6.1 6.1.1 O y A y y = f() y = f() b f(b) B y f(b) f() = b f(b) f() f() = = b A f() b AB O b 6.1 2 y = 2 = 1 = 1 + h (1 + h) 2 1 2 (1 + h) 1 2h + h2 = h h(2 + h) = h = 2 + h y (1 + h) 2 1 2 O y = 2 1

More information

V 0 = + r pv (H) + qv (T ) = + r ps (H) + qs (T ) = S 0 X n+ (T ) = n S n+ (T ) + ( + r)(x n n S n ) = ( + r)x n + n (d r)s n = ( + r)v n + V n+(h) V

V 0 = + r pv (H) + qv (T ) = + r ps (H) + qs (T ) = S 0 X n+ (T ) = n S n+ (T ) + ( + r)(x n n S n ) = ( + r)x n + n (d r)s n = ( + r)v n + V n+(h) V I (..2) (0 < d < + r < u) X 0, X X = 0 S + ( + r)(x 0 0 S 0 ) () X 0 = 0, P (X 0) =, P (X > 0) > 0 0 H, T () X 0 = 0, X (H) = 0 us 0 ( + r) 0 S 0 = 0 S 0 (u r) X (T ) = 0 ds 0 ( + r) 0 S 0 = 0 S 0 (d r)

More information

A A = a 41 a 42 a 43 a 44 A (7) 1 (3) A = M 12 = = a 41 (8) a 41 a 43 a 44 (3) n n A, B a i AB = A B ii aa

A A = a 41 a 42 a 43 a 44 A (7) 1 (3) A = M 12 = = a 41 (8) a 41 a 43 a 44 (3) n n A, B a i AB = A B ii aa 1 2 21 2 2 [ ] a 11 a 12 A = a 21 a 22 (1) A = a 11 a 22 a 12 a 21 (2) 3 3 n n A A = n ( 1) i+j a ij M ij i =1 n (3) j=1 M ij A i j (n 1) (n 1) 2-1 3 3 A A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33

More information

29

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

More information

Microsoft Word - 11問題表紙(選択).docx

Microsoft Word - 11問題表紙(選択).docx A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx

More information