main.dvi
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- あまめ ごみぶち
- 5 years ago
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1 2 e jωt, 0 ωt < 2π x(n) X(z) = x(n)z n (2.1) z z z = e σ+jωt = re jωt, r = e σ (2.2) ωt 0 2π σ e σ+jωt = re jωt r = e σ <r<1(σ <0) 1 <r(0 <σ) r =1(σ =0) 2.1.2
2 X(z) = x(n)z n = x(n)r n e jωnt (2.3) x(n)r n S = x(n)r n < (2.4) x(n) x(n) r z z 1, n 0 x(n) =u(n) = 0, n < 0 (2.5) x(n) S = x(n)r n 1 r N = lim N 1 r 1 (2.6) r 1 < 1 S z e jωnt =1 (2.7) z 1 < 1 z > 1 1
3 X(z) = x(n)z n 1 =, z > 1 (2.8) 1 z X(z) = z X(z) =0 z x(n) = a n u(n) (2.9) X(z) = x(n)z n = a n z n = (az 1 ) n (2.10) az 1 < 1 X(z) = 1 1 az 1 = z z a (2.11) z > a a z =0 z = a 2.3
4 x(n) X(z) = n 2 n=n 1 x(n)z n (2.12) n 1 < 0 z p, p > 0 z = n 2 > 0 z p, p < 0 z =0 2 X(z) = x(n)z n, n 1 < (2.13) n=n 1 z = z 1 S = x(n)z1 n < (2.14) n=n 1 n 1 0 z > z 1 z S = x(n)z n < x(n)z1 n < (2.15) n=n 1 n=n 1
5 32 2. n 1 < 0 (2.13) 1 X(z) = x(n)z n + x(n)z n, n=n 1 n 1 < (2.16) z = z > z 1 a x(n) X(z) z 2. x(n) x(n) < 3. x(n) X(z) z = e jωt x(n) x(n) 4. e jωt
6 h(n) S = h(n) < (2.17) h(n) h(n) S < a < 1 z > a X(z) a h(n)
7 34 2. h(n) h(n) =a n,n 0 az 1 < 1 h(n) 2.5 a > x(n) Z[x(n)]
8 z p z > max{ z p } X(z) =Z[x(n)], R x < z (2.18) Y (z) = Z[y(n)], R y < z (2.19) a b Z[ax(n) +by(n)] = (ax(n)+by(n))z n = a x(n)z n + b y(n)z n = ax(z)+by (z), max{r x,r y } < z (2.20)
9 X(z) =Z[x(n)] x(n + n 0 ) Z[x(n + n 0 )] = x(n + n 0 )z n = x(m)z (m n0) = z n0 [ m= m= x(m)z m ](2.21) Z[x(n + n 0 )] = z n0 X(z) (2.22) n 0 > 0 x(n + n 0 ) z =0, z = Z[a n x(n)] = X(a 1 z), a R x < z (2.23) X(z) z p = r p e jωp z z = r z e jωz (2.24) (2.25) X(a 1 z) a = a e jφ (2.26) z pa = a r p e j(ωp+φ) (2.27) z za = a r z e jωz+φ (2.28) 2 X(z) x(n) z
10 dx(z) dz = = z 1 x(n) dz n dz = nx(n)z n 1 nx(n)z n (2.29) zdx(z)/dz nx(n) 3 x(n) x (n) x (n) ( Z[x (n)] = = x (n)z n ) x(n)(z ) n = X(z ) (2.30) Z[x (n)] = X (z ) (2.31) y(n) = h(k)x(n k) (2.32) k= Y (z) =H(z)X(z) (2.33) 2 y(n) =h(n)x(n) (2.34) Y (z) = h(n)x(n)z n (2.35)
11 38 2. = = 1 2πj [ 1 h(n) 2πj [ c c c ] X(v)v n 1 dv z n (2.36) h(n)( z v ) n ] v 1 X(v)dv = 1 H( z 2πj c v )X(v)v 1 dv (2.37) = 1 H(v)X( z 2πj v )v 1 dv (2.38) (2.36) x(n) 2.4 (2.38) x(n) h(n) z = re jφ v = ρe jθ (2.39) (2.40) v θ v c θ [ π, π] dv dθ = jρejθ = jv (2.41) dv jv = dθ (2.42) π Y (re jφ )= 1 H(ρe jθ )X( r 2π π ρ ej(φ θ) )dθ (2.43) Y φ H X θ (2.35) (2.38) z =1
12 h(n)x (n) = 1 2πj H(v)X ( 1 v )v 1 dv (2.44) c h(n) x(n) v = e jωt (2.45) dv dωt = jejωt = jv (2.46) dv = dωt (2.47) jv h(n)x (n) = 1 π H(e jω )X (e jω )dωt (2.48) 2π π h(n) =x(n) (2.49) x(n) 2 = 1 π X(e jω ) 2 dωt (2.50) 2π π 2.4 X(z) x(n) X(z) z x(n)
13 40 2. X(z) = N 1 i=0 a iz i M 1 i=0 b iz = i N 1 i=0 a i z i = M 1 i=0 b i z i x(n)z n, b 0 =1 (2.51) x(n)z n (2.52) z z n a 0 = x(0) a 1 = b 1 x(0) + x(1). (2.53) 0 = b M 1 x(n M +1)+b M 2 x(n M)+ + x(n) x(0) x(1) x(2) x(n) z N 1 i=0 X(z) = a iz i M 1 i=0 b iz i L M 1 = α i z i β j + 1 p i=0 j=1 j z 1 (2.54) L = N M 0 M 1 X(z) b i z i =0 p j i=0 β j 1 p j z 1 = β j (p j z 1 ) n = β j p n j z n (2.55)
14 M 1 x(n) =α n + β j p n j, 0 n L (2.56) x(n) = M 1 j=1 j=1 β j p n j, L < n (2.57) N = M =2 X(z) = a 0 + a 1 z 1 + a 2 z 2 1+b 1 z 1 + b 2 z 2 = α β p 1 z 1 + β1 1 p 1 z 1 = α 0 + β 1 (p 1 z 1 ) n + β1 (p 1z 1 ) n (2.58) x(0) = α 0 +2R[β 1 ] x(n) =β 1 p n 1 + β 1 p n 1 =2R[β 1 p n 1 ], n 1 (2.59) X(z) = x(n)z n (2.60) z k 1 C 1 X(z)z k 1 x(n) dz = z n+k 1 dz (2.61) 2πj 2πj c 1 z k 1 1, k =0 dz = 2πj c 0, k 0 c (2.62) (2.61) n = k 1 n k (2.61) x(k)
15 42 2. x(k) = 1 X(z)z k 1 dz (2.63) 2πj c C 2.7 Z Im 1 C Re C F (z)dz C (2.64) 2πj c F (z) F (z) K ψ k (z) F (z) = (z z k ) r (2.65) k k=1 ψ k (z) =0 z k z k r k F (z) K [ 1 d r k 1 ] ψ k (z) Res[F (z)] = (r k 1)! dz r (2.66) k 1 z=z k k=1 (2.63) F (z) =X(z)z k 1 (2.67) n 0
16 X(z) =, z > a (2.68) 1 az 1 x(n) = 1 z n 1 dz (2.69) 2πj c 1 az 1 = 1 z n dz (2.70) 2πj c z a n 0 C z = a 1 F (z) = zn z a (2.66) [ 1 d 0 z n ] x(n) = (1 1)! dz 0 z=a (2.71) = a n, n 0 (2.72) H(z) h(n) (2.33) x(n) y(n) Y (z)/x(z) H(z) = h(n)z n (2.73) H(z) = Y (z) X(z) M 1 j=0 b j y(n j) = N 1 i=0 (2.74) a i x(n i) (2.75) x(n) y(n) z 2.3.2
17 44 2. M N b j z j Y (z) = a i z i X(z) (2.76) j=0 i=0 H(z) = Y (z) X(z) = N 1 i=0 a iz i M 1 j=0 b jz j (2.77) IIR (1) y(n) =a 0 x(n)+a 1 x(n 1) + a 2 x(n 2) b 1 y(n 1) b 2 y(n 2) (2.78) z H(z) =Y (z)/x(z) Y (z) = (a 0 + a 1 z 1 + a 2 z 2 )X(z) (b 1 z 1 + b 2 z 2 )Y (z) (2.79) H(z) = Y (z) X(z) = a 0 + a 1 z 1 + a 2 z 2 1+b 1 z 1 + b 2 z 2 (2.80) z 1,z a 0,a 1,a 2,b 1,b w(n)
18 IIR (2) w(n) =x(n) b 1 w(n 1) b 2 w(n 2) (2.81) y(n) = a 0 w(n)+a 1 w(n 1) + a 2 w(n 2) (2.82) z W (z) (2.83) W (z) =X(z) b 1 z 1 W (z) b 2 z 2 W (z) (2.84) Y (z) = a 0 w(z)+a 1 z 1 W (z)+a 2 z 2 W (z) (2.85) H(z) = Y (z) X(z) = a 0 + a 1 z 1 + a 2 z 2 1+b 1 z 1 + b 2 z 2 (2.86) H(z) =0 H(z) = z h(n) =0,n<0 n 0 z =0 z = z (2.77) z
19 46 2. N i=1 H(z) =a (1 z iz 1 ) M j=1 (1 p jz 1 ) (2.87) z i p j h(n) h(n) z = e jωt z H(z) H(z) = 1 az 1 1 bz 1 = z a z b (2.88) a b b < 1 H(z) H(e jω ) = ejωt a e jωt b (2.89) z z 2.10 ω 1 T 1 < H(e jω ) ω 2 T H(e jω ) < ( ) 0.8 ±π/4 ±π/2 π
20 Z Im e 2 jω T e 1 jω T a 0 b 1 Re 2.10 z = e jωt H(z) = (1 ejπ z 1 )(1 e jπ/2 z 1 )(1 e jπ/2 z 1 ) (1 0.8e jπ/4 z 1 )(1 0.8e jπ/4 z 1 ) = (2.90) (1 + z 1 )(1 + z 2 ) cos(π/4)z z 2 (2.91) 2.11( ) ωt = π/4, 7π/4,f = f s /8, 7f s /8 ωt = π/2,π,3π/2,f = f s /4,f s /2, 3f s /4 ωt =2π f = f s ωt =0 π f =0 f s / h(n) =0, n < 0
21 48 2. Z Im 0 π/4 π/4 1 Re j H( e ω T ) 0 π/ 2 π 3π / 2 0 fs /4 f S /2 f / 4 3 S 2π f S ωt ωt f 2.11 h(0) = a 0 (2.92) h(n) =a 0 ( b) n + a 1 ( b) n 1, n 1 H(z) h(n) H(z) = h(n)z n = a 0 + [a 0 ( b) n + a 1 ( b) n 1 ]z n n=1 = a 0 + a 0bz 1 1+bz 1 + a 1z 1 1+bz 1, bz 1 < 1
22 n =- h (n ) < [8] [5] z = e [6] [1] [4] jωt [7] [2] [3] [9] 2.12 H(z) =a 0 = a 0 + a 1 z 1 1+bz 1 (2.93) = ( b) n z n + a 1 n=1 ( b) n z n a 0 1+bz 1 + a 1z 1 1+bz 1 = a 0 + a 1 z 1 1+bz 1 (2.94) H(z) = a 0 + a 1 z 1 1+b 1 z 1 = a 0 1+b 1 z 1 + a 1z 1 1+b 1 z 1 (2.95) 1 2 a 0 1+b 1 z 1 = a 0[1+( bz 1 )+(b 1 z 1 ) 2 + ] (2.96) a 1 z 1 1+b 1 z 1 = a 1z 1 [1+( bz 1 )+(b 1 z 1 ) 2 + ] (2.97) z 1 H(z) =a 0 +(a 0 ( b 1 )+a 1 )z 1
23 (a 0 ( b 1 ) 2 + a 1 ( b 1 ))z 2 + (2.98) = h(0) + h(1)z 1 + h(2)z 2 + (2.99) (2.92) h(n) 2 H(z) H(z) = Y (z) X(z) = a 0 + a 1 z 1 1+bz 1 (2.100) (1 + bz 1 )Y (z) =(a 0 + a 1 z 1 )X(z) (2.101) y(n)+by(n 1) = a 0 x(n)+a 1 x(n 1) (2.102) y(n) =a 0 x(n)+a 1 x(n 1) by(n 1) (2.103) (a) w(n) a 0 x (n) w (n) y(n) + + b T a 1 (a) x(n) a 0 + y(n) T T a 1 b x ( n 1) y( n 1) (b) 2.13 x(n) y(n) w(n) w(n) =x(n) bw(n 1) (2.104) y(n) = a 0 w(n)+a 1 w(n 1) (2.105)
24 x(n) w(n) w(n 1) = b x(n) w(n) y(n) = a 0 w(n)+a b (2.106) (2.107) w(n) a1 y(n) b = a 0 a1 b (2.108) w(n) (2.105) (2.103) y(n) y(n) a1 b y(n) =a y(n 1) a1 b x(n 1) 0 + a a 0 a1 1 b a 0 a1 b (2.109) = a 0 x(n)+a 1 x(n 1) by(n 1) (2.110) (2.104) (2.105) z W (z) X(z) Y (z) z (2.104) (2.105) 2.13(b) x(n) y(n) y(n) =a 0 x(n)+a 1 x(n 1) by(n 1) (2.111) (2.104) (2.105) w(n) w(n 1) W (z) Y (z)/x(z) W (z) =X(z) bz 1 W (z) (2.112) Y (z) = a 0 W (z)+a 1 z 1 W (z) (2.113) W (z) = X(z) 1+bz 1 (2.114) Y (z) = (a 0 + a 1 z 1 )W (z) = a 0 + a 1 z 1 X(z) (2.115) 1+bz 1
25 52 2. H(z) = Y (z) X(z) = a 0 + a 1 z 1 1+bz 1 (2.116) (2.110) w(n) w(n 1) W (z) 4 h(n) y(n) = n h(m)x(n m) (2.117) m=0 (2.104) (2.105) n =0, 1, 2, x(n) w(n 1) w(n) y(n) 5 H(e jω )= h(n)e jωnt (2.118) h(n) = 1 2π π π H(e jω )e jωnt dωt (2.119) 6 z = e jωt 7 (2.93) 2.13
26 h(n) h(n) < (2.120) z p z p < 1 z p =1 z p > (a) (b) (c) (d)
27 54 2. (e) (f) z =( ) 2. h(n) z H(z) h(n) 2r n cos(ω 0 nt ), 0 n h(n) = (2.121) 0, n < 0 (a) h(n) z H(z) z 1 (b) (c) (d) (e) H(z) r =0.9 ω 0 T = π/2 H(z) ωt =0,π/4,π/2,π/4,π x(n) y(n) x(n) =ae j(ωnt+φ), a =0.7,ωT = π/4,φ=0.3[rad](2.122) 3. (a) H(z) = 1 0.3z 1 ( z 1 )(1 0.8z 1 ) H(z) h(n) (2.123) (b) (c) h(n) H(z) H(z) h(n) H(z)
28 (d) (e) (f) (g) (h) (i) (j) h(n) e jωt H(z) e jωt H(z) x(n) y(n) H(z) w(n) x(n) y(n) 1, n =0, 1, 2 x(n) = 0, n 0, 1, 2 (2.124) h(n) x(n) y(n) (a) (b) (c) (d) (e) (f) y(n) x(n) w(n) H(z) =Y (z)/x(z) H(z) H(z) h(n) 2.15 x(n) x(n) =a cos(ωnt + φ) a ωt φ x(n) y(n)() () b =0.5 a 0 =1 a 1 =1y(n) θ tan θ =
29 56 2. x (n) w(n) y(n) + + a 0 b T a IIR 1 0 x(n) ~ n x(n)
30 z k 1 dz (2.125) 2πj c C C e jθ z = e jθ dz dθ = jejθ dz = je jθ dθ (2.126) (2.127) (2.128) C θ 0 2π (2.125) 1 2πj 2π 0 2π e jθ(k 1) je jθ dθ = 1 e jθk dθ 2π 0 1, k =0 = 0, k 0 (2.129) 1 1 dz (2.130) 2πj c (z z p ) k 1 (z z p ) k 1 dz (2.131) 2πj c C z p
main.dvi
6 FIR FIR FIR FIR 6.1 FIR 6.1.1 H(e jω ) H(e jω )= H(e jω ) e jθ(ω) = H(e jω ) (cos θ(ω)+jsin θ(ω)) (6.1) H(e jω ) θ(ω) θ(ω) = KωT, K > 0 (6.2) 6.1.2 6.1 6.1 FIR 123 6.1 H(e jω 1, ω
main.dvi
5 IIR IIR z 5.1 5.1.1 1. 2. IIR(Infinite Impulse Response) FIR(Finite Impulse Response) 3. 4. 5. 5.1.2 IIR FIR 5.1 5.1 5.2 104 5. IIR 5.1 IIR FIR IIR FIR H(z) = a 0 +a 1 z 1 +a 2 z 2 1+b 1 z 1 +b 2 z 2
x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
main.dvi
3 Discrete Fourie Transform: DFT DFT 3.1 3.1.1 x(n) X(e jω ) X(e jω )= x(n)e jωnt (3.1) n= X(e jω ) N X(k) ωt f 2π f s N X(k) =X(e j2πk/n )= x(n)e j2πnk/n, k N 1 (3.2) n= X(k) δ X(e jω )= X(k)δ(ωT 2πk
dfilterh.dvi
12.5 3 ISBN4-7856-1194-4 C3055 12.5.1 x(t) =A sin(t) rad/sec T s =1=F s t = nt s n x(nt s )=Asin(nT s ) F s! =T s x(nt s )! x(n) x(n) =A sin(nt s )=Asin(!n)! rad { F (Hz) =2F { f! =2f f {! f F! = T s ==F
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](/thumbs/101/149159646.jpg "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
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
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
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 = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1](/thumbs/94/121802164.jpg "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
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
18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb
![18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb](/thumbs/93/114079295.jpg "18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb")
r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)
,. 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,
X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 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
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
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液晶の物理１：連続体理論（弾性，粘性）
The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers
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n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
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
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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)) ( )
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W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
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simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =
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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
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zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
『共形場理論』
T (z) SL(2, C) T (z) SU(2) S 1 /Z 2 SU(2) (ŜU(2) k ŜU(2) 1)/ŜU(2) k+1 ŜU(2)/Û(1) G H N =1 N =1 N =1 N =1 N =2 N =2 N =2 N =2 ĉ>1 N =2 N =2 N =4 N =4 1 2 2 z=x 1 +ix 2 z f(z) f(z) 1 1 4 4 N =4 1 = = 1.3
(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z
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) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
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1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
( ) 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) (
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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
1. 1 BASIC PC BASIC BASIC BASIC Fortran WS PC (1.3) 1 + x 1 x = x = (1.1) 1 + x = (1.2) 1 + x 1 = (1.
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2.5 (Gauss) (flux) v(r)( ) S n S v n v n (1) v n S = v n S = v S, n S S. n n S v S v Minoru TANAKA (Osaka Univ.) I(2012), Sec p. 1/30
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Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0
Tips KENZOU 28 7 6 P no problem 2 w = f(z) z w w z w = (z z ) b b w = log (z z ) z = z 2π 2 z = z w = z /2 z = re iθ θ (z = ) 2π 4π 2 θ θ 2π 4π z r re iθ re i2π = r re i4π = r w r re iθ/2 re iπ = r re
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29
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4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t
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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
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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
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(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
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Chap11.dvi
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Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
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II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11
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[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
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) 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
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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
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y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
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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
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2005 Limit Distribution of Quantum Walks and Weyl Equation 2006 3 2 1 2 2 4 2.1...................... 4 2.2......................... 5 2.3..................... 6 3 8 3.1........... 8 3.2..........................
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7 f(z) H f(z)dz = f(z), f (z)dz = 9 3 4 7 7.. 7. 7.. 3 f(z) D D z 7. f(z) = f () d (7.) z f(z) - =( z) = z = z Res(z) =lim!z ( z) z = f(z): (7.) 7. (7.) f(z) 7. 5 7.. f(z) z = a jz aj = r z = a r Ref(a)
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +
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Radiation from moving charges#1 Liénard-Wiechert potential Yuji Chinone 1 Maxwell Maxwell MKS E (x, t) + B (x, t) t = 0 (1) B (x, t) = 0 (2) B (x, t)
Radiation from moving harges# Liénard-Wiehert potential Yuji Chinone Maxwell Maxwell MKS E x, t + B x, t = B x, t = B x, t E x, t = µ j x, t 3 E x, t = ε ρ x, t 4 ε µ ε µ = E B ρ j A x, t φ x, t A x, t
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D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y
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= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k
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prime number theorem
For Tutor MeBio ζ Eite by kamei MeBio 7.8.3 : Bernoulli Bernoulli 4 Bernoulli....................................................................................... 4 Bernoulli............................................................................
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6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..