main.dvi
|
|
- ああす ささおか
- 5 years ago
- Views:
Transcription
1 4 DFT DFT Fast Fourier Transform: FFT 4.1 DFT IDFT X(k) = 1 n=0 x(n)e j2πkn (4.1) 1 x(n) = 1 X(k)e j2πkn (4.2) k=0 x(n) X(k) DFT 2 ( 1) (2 1) 2 O( 2 ) 4.2 FFT radix2 FFT 1 (4.1)
2 86 4. X(0) w 0 w 0 w 0 X(1) w 0 w 1 w 1 = X( 1) w 0 w 1 w 1 x(0) x(1). x( 1) (4.3) w = e j2π w 0,w1,,w 1 kn = m + r, 0 r 1 w kn = wr (4.4) (4.3) X() =F ()x() (4.5) X() =[X(0),X(1),,X( 1)] T (4.6) x() = [x(0),x(1),,x( 1)] T (4.7) (4.8) [] T F () DFT 2 Decimation-in-frequency F () X() F () 2k k 2k +1 k ˆX() = ˆF ()x() (4.9) ˆX() = [ ˆX 0 (/2), ˆX 1 (/2)] T (4.10) ˆX 0 (/2) = [X(0),X(2),,X( 2)] T (4.11) ˆX 1 (/2) = [X(1),X(3),,X( 1)] T (4.12) ˆX 0 (k) = X(2k), 0 k 2 1 (4.13)
3 4.2 FFT 87 3 ˆX 1 (k) = X(2k +1), 0 k 2 1 (4.14) x() = [x 0 (/2), x 1 (/2)] T (4.15) x 0 (/2) = [x(0),x(1),,x(/2 1)] T (4.16) x 1 (/2) = [x(/2),x(/2+1),,x( 1)] T (4.17) ˆF () = F 1(/2) F 2 (/2) (4.18) F 3 (/2) F 4 (/2) 1. F 1 (/2) F 2 (/2) /2 DFT k n w kn /2, w /2 = e j2π /2 (4.19) F 1 (/2) k n w 2kn = e j2π(2kn)/ = e j2πkn/(/2) = w kn /2 (4.20) 0 k, n 2 1 F (/2) k n F 2 (/2) k n w 2k(n+/2) = e j2π(2k(n+/2))/ = e j2πkn/(/2) j2πk = e j2πkn/(/2) = w kn /2 (4.21) 0 k, n 2 1 F (/2) k n 2. F 3 (/2)F 4 (/2) F 3 (/2) = F 4 (/2) (4.22)
4 88 4. F 3 (/2) k n w (2k+1)n = e j2π(2k+1)n/, 0 k, n 2 1 (4.23) F 4 (/2) k n w (2k+1)(n+/2) = e j2π(2k+1)(n+/2)/ = e j2π(2k+1)n e j2π(/2)/ = e j2π(2k+1)n, 0 k, n 2 1 (4.24) F 3 (/2) k n 1 3. F 3 (/2) F 3 (/2) = F (/2)D(/2) (4.25) D(/2) = diag(w 0,w 1,,w /2 1 ) (4.26) F 3 (/2) k n w (2k+1)n = e j2π(2k+1)n/ = e j2πkn/(/2) e j2πn/ 0 k, n 2 1 (4.27) F (/2) k n e j2πn/ F (/2) diag (e j2π 0,e j2π 1,e j2π 2 ),,e j2π /2 1 ) (4.28) (4.26) 4 DFT F ()
5 4.2 FFT 89 ˆF () = F (/2) F (/2) F (/2)D(/2) F (/2)D(/2) (4.29) = F (/2) 0 I(/2) 0 0 F (/2) 0 D(/2) I(/2) I(/2) I(/2) I(/2) (4.30) I(/2) /2 4.1 D(/2) /2 F (/2) k n 4.1 (4.30) w kn /2, w /2 = e j2π/(/2), 0 k, n 2 1 (4.31) F () DFT F () 4.1 F (/2) FFT 5 (4.11) (4.12)
6 (4.32) =8 DFT F (8) X(0) X(1) X(2) X(3) X(4) X(5) X(6) X(7) = w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 1 w 2 w 3 w 4 w 5 w 6 w 7 w 0 w 2 w 4 w 6 w 0 w 2 w 4 w 6 w 0 w 3 w 6 w 1 w 4 w 7 w 2 w 5 w 0 w 4 w 0 w 4 w 0 w 4 w 0 w 4 w 0 w 5 w 2 w 7 w 4 w 1 w 6 w 3 w 0 w 6 w 4 w 2 w 0 w 6 w 4 w 2 w 0 w 7 w 6 w 5 w 4 w 3 w 2 w 1 x(0) x(1) x(2) x(3) x(4) x(5) x(6) x(7) (4.33) w = e j2π/8 (4.34)
7 X(0) X(2) X(4) X(6) = X(1) X(3) X(5) X(7) 4.2 FFT 91 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 x(0) w 0 w 2 w 4 w 6 w 0 w 2 w 4 w 6 x(1) w 0 w 4 w 0 w 4 w 0 w 4 w 0 w 4 x(2) w 0 w 6 w 4 w 2 w 0 w 6 w 4 w 2 x(3) (4.35) w 0 w 1 w 2 w 3 w 4 w 5 w 6 w 7 x(4) w 0 w 3 w 6 w 1 w 4 w 7 w 2 w 5 x(5) w 0 w 5 w 2 w 7 w 4 w 1 w 6 w 3 x(6) w 0 w 7 w 6 w 5 w 4 w 3 w 2 w 1 x(7) ˆX 0 (4) = F 1(4) F 2 (4) x 0(4) (4.36) ˆX 1 (4) F 3 (4) F 4 (4) x 1 (4) ˆX 0 (4) = [X(0),X(2),X(4),X(6)] T (4.37) ˆX 1 (4) = [X(1),X(3),X(5),X(7)] T (4.38) x 0 (4) = [x(0),x(1),x(2),x(3)] T (4.39) x 1 (4) = [x(4),x(5),x(6),x(7)] T (4.40) w 0 w 0 w 0 w 0 w F 1 (4) = F 2 (4) = 0 w 2 w 4 w 6 w 0 w 4 w 0 w 4 (4.41) w 0 w 6 w 4 w 2 F 4 (4) = w 4 F 3 (4) = F 3 (4) w 0 w 1 w 2 w 3 w = 0 w 3 w 6 w 1 w 0 w 5 w 2 w 7 w 0 w 7 w 6 w 5 (4.42)
8 92 4. w 0 w 1 w 2 w 3 w F 3 (4) = 0 w 3 w 6 w 1 w 0 w 5 w 2 w 7 w 0 w 7 w 6 w 5 w 0 w 0 w 0 w 0 w w = 0 w 2 w 4 w 6 0 w w 0 w 4 w 0 w w 2 0 w 0 w 6 w 4 w w 3 ˆX 0 (4) ˆX 1 (4) = F 1 (4)D(4) (4.43) F = 1 (4) F 1 (4) x 0(4) (4.44) F 1 (4)D(4) F 1 (4)D(4) x 1 (4) = F 1(4) 0 I(4) 0 0 F 1 (4) 0 D(4) I(4) I(4) x 0(4) (4.45) I(4) I(4) x 1 (4) (4.45) F 1 (4) (4.34) w = e j2π/8 w = e j2π/4
9 X(0) X(2) = X(4) X(6) X(1) X(3) = X(5) X(7) 4.2 FFT 93 w 0 w 0 w 0 w 0 x 1 (0) w 0 w 1 w 2 w 3 x 1 (1) w 0 w 2 w 0 w 2 (4.46) x 1 (2) w 0 w 3 w 2 w 1 x 1 (3) w 0 w 0 w 0 w 0 x 2 (0) w 0 w 1 w 2 w 3 x 2 (1) w 0 w 2 w 0 w 2 (4.47) x 2 (2) w 0 w 3 w 2 w 1 x 2 (3) w = e j2π/4 (4.48) x 1 (0) x(0) x(4) x 1 (1) x(1) x(5) = + (4.49) x 1 (2) x(2) x(6) x 1 (3) x(3) x(7) x 2 (0) w 0 x(0) x(4) x 2 (1) w = 1 x(1) x(5) x 2 (2) w 2 (4.50) x(2) x(6) x 2 (3) w 3 x(3) x(7) F 1 (4) DFT F (4) 4.3 F (4) F (8) X(0) w 0 w 0 w 0 w 0 x 1 (0) X(4) w = 0 w 2 w 0 w 2 x 1 (1) X(2) w 0 w 1 w 2 w 3 (4.51) x 1 (2) X(6) w 0 w 3 w 2 w 1 x 1 (3) ˆX 00 (2) = F 1(2) F 2 (2) x 10(2) (4.52) ˆX 01 (2) F 3 (2) F 4 (2) x 11 (2)
10 (4.46) (4.50) ˆX 00 (2) = [X(0),X(4)] T (4.53) ˆX 01 (2) = [X(2),X(6)] T (4.54) x 10 (2) = [x 1 (0),x 1 (1)] T (4.55) x 11 (2) = [x 1 (2),x 1 (3)] T (4.56) F 1 (2) = F 2 (2) = w0 w 0 = 1 1 (4.57) w 0 w F 4 (2) = w 2 F 3 (2) = F 3 (2) = w0 w 1 (4.58) w 0 w 3 F 3 (2) = w0 w 0 w0 0 w 0 w 2 0 w 1 = F 1 (2)D(2) (4.59) (4.52) (4.45) ˆX 00 (2) = F 1(2) F 1 (2) x 10(2) ˆX 01 (2) F 3 (2) F 3 (2) x 11 (2)
11 4.2 FFT 95 = F 1(2) 0 I(2) 0 0 F 1 (2) 0 D(2) I(2) I(2) x 10(2) (4.60) I(2) I(2) x 11 (2) F 1 (2) F 1 (4) DFT 4.4 x(n) X(k) (4.32) X(0) X(1) X(3) X(7) 4.4 DFT F (8) FFT w = e j2π/ FFT 1
12 FFT 2 D(/2 m ),m =2,,L = log 2 / L = log log 2 FFT 4.1 radix2-fft 2 log 2 2 log 2 log 2 3 log 2 O() DFT O( 2 ) FFT O() O( 2 ) FFT O()
13 4.2 FFT FFT FFT x() DFT DFT (4. F () F (/2) D(/2)F (/2) (4. F (/2) D(/2)F (/2) = I(/2) I(/2) I(/2) 0 F (/2) 0 I(/2) I(/2) 0 D(/2) 0 F (/2) x() DFT X() ˆx 0 (/2) = [x(0),x(2),,x( 2)] T (4.63) ˆx 1 (/2) = [x(1),x(3),,x( 1)] T (4.64) X 0 (/2) = [X(0),X(1),,X(/2)] T (4.65) X 1 (/2) = [X(/2+1),X(/2+1),,X( 1)] T (4.66) 4.6 FFT FFT 4.6 FFT
14 radix4-fft FFT radix4-fft DFT ˆX() = ˆF ()x() (4.67) ˆX() = [ ˆX 0 (/4), ˆX 1 (/4), ˆX 2 (/4), ˆX 3 (/4)] T (4.68) ˆX 0 (/4) = [X(0),X(4),,X( 4)] T (4.69) ˆX 1 (/4) = [X(1),X(5),,X( 3)] T (4.70) ˆX 2 (/4) = [X(2),X(6),,X( 2)] T (4.71) ˆX 3 (/4) = [X(3),X(7),,X( 1)] T (4.72) ˆX 0 (k) = X(4k), 0 k 4 1 ˆX 1 (k) = X(4k +1), 0 k 4 1 ˆX 2 (k) = X(4k +2), 0 k 4 1 ˆX 3 (k) = X(4k +3), 0 k 4 1 x() = [x 0 (/4), x 1 (/4), x 2 (/4), x 3 (/4)] T (4.73) x() /4 DFT F ( /4) F ( /4) F ( /4) F ( /4) F ˆF () = 1 (/4) jf 1 (/4) F 1 (/4) jf 1 (/4) F 2 (/4) F 2 (/4) F 2 (/4) F 2 (/4) F 3 (/4) jf 3 (/4) F 3 (/4) jf 3 (/4) (4.74) ˆf(k, n) = f(4k, n) (4.75) ˆf(k + /4,n) = f(4k +1,n) 0 k 4 1 (4.76)
15 4.2 FFT 99 ˆf(k + /2,n) = f(4k +2,n) 0 n 1 (4.77) ˆf(k +3/4,n)=f(4k +3,n) (4.78) F i (/4) /4 DFT F (/4) D i (/4) F i (/4) = F (/4)D i (/4) (4.79) d i (k, k) = e j2πki/ D i (/4) (k, k) (4.80) ˆX 0 (/4) F (/4)D 0 (/4) ˆX 1 (/4) F (/4)D = 1 (/4) ˆX 2 (/4) F (/4)D 2 (/4) ˆX 3 (/4) F (/4)D 3 (/4) I(/4) I(/4) I(/4) I(/4) x 0 (/4) I(/4) ji(/4) I(/4) ji(/4) x 1 (/4) I(/4) I(/4) I(/4) I(/4) x 2 (/4) I(/4) ji(/4) I(/4) ji(/4) x 3 (/4) (4.81) 4.7 ˆX i (/4) F (/4) F () radix4-fft 1 radix4-fft radix4-fft 3 4 log 4 3 log 4 2 log 4 ( ) log 4
16 radix4-fft radix2-fft radix4-fft 4.3 radix-4-fft =4 L 4.3 radix2-fft radix4-fft radix2-fft radix4-fft FFT (4.1) (4.2) DFT Inverse DFT 1/ IDFT DFT exp( jωt) exp(jωt) 1/ IDFT FFT 1/ 1/2
17 4.3 FFT DFT FFT w 0 w 0 w 0 w 0 w F (4) = 0 w 1 w 2 w 3 w 0 w 2 w 0 w 2, w = e j2π/4 (4.82) w 0 w 3 w 2 w 1 (a) F (4) F 1 (2) F 2 (2) F 3 (2) F 4 (2) ˆF (4) = F 1(2) F 2 (2) (4.83) F 3 (2) F 4 (2) (b) (c) F 1 (2) F 2 (2) F 3 (2) F 4 (2) F 3 (2) F 1 (2) (d) ˆF (4) F 1 (2) (e) FFT F 1 (2) 2. DFT FFT w 0 w 0 w 0 w 0 w F (4) = 0 w 1 w 2 w 3 w 0 w 2 w 0 w 2, w = e j2π/4 (4.84) w 0 w 3 w 2 w 1 (a) F (4) F 1 (2) F 2 (2) F 3 (2) F 4 (2) ˆF (4) = F 1(2) F 2 (2) (4.85) F 3 (2) F 4 (2) (b) F 1 (2) F 3 (2) F 2 (2) F 4 (2)
18 (c) F 2 (2) F 1 (2) (d) ˆF (4) F 1 (2) (e) FFT F 1 (2) 3. DFT FFT x(n) x(n) DFT X(k) DFT =2 L DFT X(k) 1 DFT X(k),k =0 1 FFT ( ) = 128 = 1024 L = log 2 FFT DFT % % (1) (2) 1 (3) /2 (4) log 2 (5) L (6) 2 (7) ( 1) (8) (/2)L =(/2) log 2 (9) L = log 2 (10) 7 (11) 8 (12) 9 (13) 10 (14) 9.4 (15) 5.5 (16) 3.1 (17) 1.8 (18) 1 4.
(5 B m e i 2π T mt m m B m e i 2π T mt m m B m e i 2π T mt B m (m < 0 C m m (6 (7 (5 g(t C 0 + m C m e i 2π T mt (7 C m e i 2π T mt + m m C m e i 2π T
2.6 FFT(Fast Fourier Transform 2.6. T g(t g(t 2 a 0 + { a m b m 2 T T 0 2 T T 0 (a m cos( 2π T mt + b m sin( 2π mt ( T m 2π g(t cos( T mtdt m 0,, 2,... 2π g(t sin( T mtdt m, 2, 3... (2 g(t T 0 < t < T
More informationmain.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
More information1 1.1 Excel Excel Excel log 1, log 2, log 3,, log 10 e = ln 10 log cm 1mm 1 10 =0.1mm = f(x) f(x) = n
1 1.1 Excel Excel Excel log 1, log, log,, log e.7188188 ln log 1. 5cm 1mm 1 0.1mm 0.1 4 4 1 4.1 fx) fx) n0 f n) 0) x n n! n + 1 R n+1 x) fx) f0) + f 0) 1! x + f 0)! x + + f n) 0) x n + R n+1 x) n! 1 .
More information2
16 1050026 1050042 1 2 1 1.1 3 1.2 3 1.3 3 2 2.1 4 2.2 4 2.2.1 5 2.2.2 5 2.3 7 2.3.1 1Basic 7 2.3.2 2 8 2.3.3 3 9 2.3.4 4window size 10 2.3.5 5 11 3 3.1 12 3.2 CCF 1 13 3.3 14 3.4 2 15 3.5 3 17 20 20 20
More informationSOWC04....
99 100 101 2004 284 265 260 257 235 225 222 211 207 205 200 197 192 190 183 183 183 183 180 176 171 169 166 165 156 152 149 143 141 141 138 138 136 126 126 125 123 123 122 118 110 109 108 107 107 105 100
More informationuntitled
http://www.mofa.go.jp/mofaj/toko/visa/index.html http://www.cn.emb-japan.go.jp/jp/01top.htm http://www.shanghai.cn.emb-japan.go.jp/ http://www.guangzhou.cn.emb-japan.go.jp/ http://www.shengyang.cn.emb-japan.go.jp/jp/index.htm
More informationXX 1 01 234214 X X 1 0 1 2 3 4 2 1 4000 784 0007533 X X 1 0 1 2 3 4 2 1 4000 7 2 3 7 2 3 2 3 2 2 1 6 2 XXX-XXXX X[ 01 111 9416 39 XXX-XXXX 18.50 3.00 15.50 15.50 0.05 18.50 3.00 15.50,984 1 5 uaj39uuy
More information数値計算:フーリエ変換
( ) 1 / 72 1 8 2 3 4 ( ) 2 / 72 ( ) 3 / 72 ( ) 4 / 72 ( ) 5 / 72 sample.m Fs = 1000; T = 1/Fs; L = 1000; t = (0:L-1)*T; % Sampling frequency % Sample time % Length of signal % Time vector y=1+0.7*sin(2*pi*50*t)+sin(2*pi*120*t)+2*randn(size(t));
More informationp03.dvi
3 : 1 ( ). (.. ), : 2 (1, 2 ),,, etc... 1, III ( ) ( ). : 3 ,., III. : 4 ,Weierstrass : Rudin, Principles of Mathematical Analysis, 3/e, McGraw-Hil, 1976.. Weierstrass (Stone-Weierstrass, ),,. : 5 2π f
More information³ÎΨÏÀ
2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p
More informationTSP信号を用いた音響系評価の研究
1 TSP 98kc068 2 1. 4 2. TSP 2.1 2.2 TSP 2.2.1 ATSP 2.2.2 OATSP 2.3 N 3. 5 5 6 6 7 8 2.3.1 N 8 2.3.2 TSP 9 2.3.3 2.3.4 m 3.1 3.1.1 3.1.2 3.2 3.3 15 18 20 20 20 21 22 24 3.3.1 24 3.3.2 3.4 3.4.1 3.4.2 4.
More informationpower.tex
Contents ii 1... 1... 1... 7... 7 3 (DFFT).................................... 8 4 (CIFT) DFFT................................ 10 5... 13 6... 16 3... 0 4... 0 5... 0 6... 0 i 1987 SN1987A 0.5 X SN1987A
More informationKorteweg-de Vries
Korteweg-de Vries 2011 03 29 ,.,.,.,, Korteweg-de Vries,. 1 1 3 1.1 K-dV........................ 3 1.2.............................. 4 2 K-dV 5 2.1............................. 5 2.2..............................
More information2010 5 No.20 2 / 32
2010 5 No.20 2 / 32 2010 5 No.20 3 / 32 2010 5 No.20 4 / 32 2010 5 No.20 5 / 32 2010 5 No.20 6 / 32 2010 5 No.20 7 / 32 2010 5 No.20 8 / 32 2010 5 No.20 9 / 32 2010 5 No.20 10 / 32 2010 5 No.20 11 / 32
More informationMicrosoft 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],
More informationuntitled
. x2.0 0.5 0 0.5.0 x 2 t= 0: : x α ij β j O x2 u I = α x j ij i i= 0 y j = + exp( u ) j v J = β y j= 0 j j o = + exp( v ) 0 0 e x p e x p J j I j ij i i o x β α = = = + +.. 2 3 8 x 75 58 28 36 x2 3 3 4
More information春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an
More informationCPU Levels in the memory hierarchy Level 1 Level 2... Increasing distance from the CPU in access time Level n Size of the memory at each level 1: 2.2
FFT 1 Fourier fast Fourier transform FFT FFT FFT 1 FFT FFT 2 Fourier 2.1 Fourier FFT Fourier discrete Fourier transform DFT DFT n 1 y k = j=0 x j ω jk n, 0 k n 1 (1) x j y k ω n = e 2πi/n i = 1 (1) n DFT
More information1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =
1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i
More information. (.8.). t + t m ü(t + t) + c u(t + t) + k u(t + t) = f(t + t) () m ü f. () c u k u t + t u Taylor t 3 u(t + t) = u(t) + t! u(t) + ( t)! = u(t) + t u(
3 8. (.8.)............................................................................................3.............................................4 Nermark β..........................................
More informationCDMA (high-compaciton multicarrier codedivision multiple access: HC/MC-CDMA),., HC/MC-CDMA,., 32.,, 64. HC/MC-CDMA, HC-MCM, i
24 Investigation on HC/MC-CDMA Signals with Non-Uniform Frequency Intervals 1130401 2013 3 1 CDMA (high-compaciton multicarrier codedivision multiple access: HC/MC-CDMA),., HC/MC-CDMA,., 32.,, 64. HC/MC-CDMA,
More informationmain.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
More information10 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-- Blackman-Tukey FFT MEM Blackman-Tukey MEM MEM MEM MEM Singular Spectrum Analysis Multi-Taper Method (Matlab pmtm) 3... y(t) (Fourier transform) t=
--... 3..... 3...... 3...... 3..3....3 3..4....4 3..5....5 3.....6 3......6 3......7 3..3....0 3..4. Matlab... 3.3....3 3.3.....3 3.3.....4 3.3.3....4 3.3.4....5 3.3.5....5 3.4. MEM...8 3.4.. MEM...8 3.4..
More information<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
通信方式第 2 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/072662 このサンプルページの内容は, 第 2 版発行当時のものです. i 2 2 2 2012 5 ii,.,,,,,,.,.,,,,,.,,.,,..,,,,.,,.,.,,.,,.. 1990 5 iii 1 1
More informationuntitled
( œ ) œ 2,000,000 20. 4. 1 25. 3.27 44,886,350 39,933,174 4,953,176 9,393,543 4,953,012 153,012 4,800,000 164 164 4,001,324 2,899,583 254,074 847,667 5,392,219 584,884 7,335 4,800,000 153,012 4,800,000
More information1 3 1. 5 2. 5 1. 7 2. 10 3. 11 4. 12 1 12 2) 12 5. 14 1 14 2 14 6. 16 1 16 2 21 3 22 7. 23 1 23 2 24 3 25 4 26 5 27 8. 28 1 28 9. 29 1 29 2 29 3 30 4 30 1. 1 31 1) 31 2 31 3 31 4 32 2. 32 3. 32 4. 1 33
More informationimpulse_response.dvi
5 Time Time Level Level Frequency Frequency Fig. 5.1: [1] 2004. [2] P. A. Nelson, S. J. Elliott, Active Noise Control, Academic Press, 1992. [3] M. R. Schroeder, Integrated-impulse method measuring sound
More informationn ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
More information[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a
13/7/1 II ( / A: ) (1) 1 [] (, ) ( ) ( ) ( ) etc. etc. 1. 1 [1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin
More informationRelaxation scheme of Besse t t n = n t, u n = u(t n ) (n = 0, 1,,...)., t u(t) = F (u(t)) (1). (1), u n+1 u n t = F (u n ) u n+1 = u n + tf (u n )., t
RIMS 011 5 3 7 relaxation sheme of Besse splitting method Scilab Scilab http://www.scilab.org/ Google Scilab Scilab Mathieu Colin Mathieu Colin 1 Relaxation scheme of Besse t t n = n t, u n = u(t n ) (n
More informationuntitled
40 4 4.3 I (1) I f (x) C [x 0 x 1 ] f (x) f 1 (x) (3.18) f (x) ; f 1 (x) = 1! f 00 ((x))(x ; x 0 )(x ; x 1 ) (x 0 (x) x 1 ) (4.8) (3.18) x (x) x 0 x 1 Z x1 f (x) dx ; Z x1 f 1 (x) dx = Z x1 = 1 1 Z x1
More information(interferometer) 1 N *3 2 ω λ k = ω/c = 2π/λ ( ) r E = A 1 e iφ1(r) e iωt + A 2 e iφ2(r) e iωt (1) φ 1 (r), φ 2 (r) r λ 2π 2 I = E 2 = A A 2 2 +
7 1 (Young) *1 *2 (interference) *1 (1802 1804) *2 2 (2005) (1993) 1 (interferometer) 1 N *3 2 ω λ k = ω/c = 2π/λ ( ) r E = A 1 e iφ1(r) e iωt + A 2 e iφ2(r) e iωt (1) φ 1 (r), φ 2 (r) r λ 2π 2 I = E 2
More information(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n
. 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n
More information2. 2 I,II,III) 2 x expx) = lim + x 3) ) expx) e x 3) x. ) {a } a a 2 a 3...) a b b {a } α : lim a = α b) ) [] 2 ) f x) = + x ) 4) x > 0 {f x)} x > 0,
. 207 02 02 a x x ) a x x a x x a x x ) a x x [] 3 3 sup) if) [3] 3 [4] 5.4 ) e x e x = lim + x ) ) e x e x log x = log e x) a > 0) x a x = e x log a 2) 2. 2 I,II,III) 2 x expx) = lim + x 3) ) expx) e
More informationhttp://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 information1 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 informationφ s i = m j=1 f x j ξ j s i (1)? φ i = φ s i f j = f x j x ji = ξ j s i (1) φ 1 φ 2. φ n = m j=1 f jx j1 m j=1 f jx j2. m
2009 10 6 23 7.5 7.5.1 7.2.5 φ s i m j1 x j ξ j s i (1)? φ i φ s i f j x j x ji ξ j s i (1) φ 1 φ 2. φ n m j1 f jx j1 m j1 f jx j2. m j1 f jx jn x 11 x 21 x m1 x 12 x 22 x m2...... m j1 x j1f j m j1 x
More information統計学のポイント整理
.. September 17, 2012 1 / 55 n! = n (n 1) (n 2) 1 0! = 1 10! = 10 9 8 1 = 3628800 n k np k np k = n! (n k)! (1) 5 3 5 P 3 = 5! = 5 4 3 = 60 (5 3)! n k n C k nc k = npk k! = n! k!(n k)! (2) 5 3 5C 3 = 5!
More informationPart () () 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
More information取扱説明書 基本ガイド
2 A G B C D E F H I J K L M N L B C A D F A B C D E G H I Fn F1 F2 F1 Fn F2 F3 F4 Windows TPM Fn Fn F7 F10 F2 F A X 0120-873029 (06)6905-5067 (06)6905-5079
More informationSFGÇÃÉXÉyÉNÉgÉãå`.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
More informationAHP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 information1. 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.
Section Title Pages Id 1 3 7239 2 4 7239 3 10 7239 4 8 7244 5 13 7276 6 14 7338 7 8 7338 8 7 7445 9 11 7580 10 10 7590 11 8 7580 12 6 7395 13 z 11 7746 14 13 7753 15 7 7859 16 8 7942 17 8 Id URL http://km.int.oyo.co.jp/showdocumentdetailspage.jsp?documentid=
More informationFourier Niching Approach for Multi-modal Optimization 2 Yan Pei Hideyuki Takagi 2 Graduate School of Design, Kyushu University 2 2 Faculty of Design,
九州大学学術情報リポジトリ Kyushu University Institutional Repository 多峰性最適化のためのフーリエ ニッチ法 裴, 岩九州大学大学院芸術工学府 高木, 英行九州大学大学院芸術工学研究院 Pei, Yan Graduate School of Design, Kyushu University Takagi, Hideyuki Faculty of Design,
More information1 -- 9 -- 3 3--1 LMS NLMS 2009 2 LMS Least Mean Square LMS Normalized LMS NLMS 3--1--1 3 1 AD 3 1 h(n) y(n) d(n) FIR w(n) n = 0, 1,, N 1 N N = 2 3--1-
1 -- 9 3 2009 2 LMS NLMS RLS FIR IIR 3-1 3-2 3-3 3-4 c 2011 1/(13) 1 -- 9 -- 3 3--1 LMS NLMS 2009 2 LMS Least Mean Square LMS Normalized LMS NLMS 3--1--1 3 1 AD 3 1 h(n) y(n) d(n) FIR w(n) n = 0, 1,, N
More information. ż ż 57 a v i ż ż v o b a ż v i ż v i ż v o ż v o a b 57. v i ż ż v o v o = Ġ v i (86) = ż ż + ż v i (87) v o v i Ġ = ż ż + ż (88) v i v o?? Ġ 6
D:.BUN 7 8 4 B5 6.................................... 6.. C........................... 6..3 ω s............................. 63..4 Bode Diagram.......................... 64..5................................
More information1 -- 9 -- 6 6--1 (DFT) 009 DFT: Discrete Fourier Transform 6--1--1 N x[n] DFT N 1 X[k] = x[n]wn kn, k = 0, 1,, N 1 (6 ) n=0 1) W N = e j π N W N twidd
1 -- 9 6 009 (DFT) 6-1 DFT 6- DFT FFT 6-3 DFT 6-4 6-5 c 011 1/(0) 1 -- 9 -- 6 6--1 (DFT) 009 DFT: Discrete Fourier Transform 6--1--1 N x[n] DFT N 1 X[k] = x[n]wn kn, k = 0, 1,, N 1 (6 ) n=0 1) W N = e
More information00 3 9 ........................................................................................................................................... 4..3................................. 5.3.......................................
More information2014 3 10 5 1 5 1.1..................................... 5 2 6 2.1.................................... 6 2.2 Z........................................ 6 2.3.................................. 6 2.3.1..................
More informationII 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 informationKroneher Levi-Civita 1 i = j δ i j = i j 1 if i jk is an even permutation of 1,2,3. ε i jk = 1 if i jk is an odd permutation of 1,2,3. otherwise. 3 4
[2642 ] Yuji Chinone 1 1-1 ρ t + j = 1 1-1 V S ds ds Eq.1 ρ t + j dv = ρ t dv = t V V V ρdv = Q t Q V jdv = j ds V ds V I Q t + j ds = ; S S [ Q t ] + I = Eq.1 2 2 Kroneher Levi-Civita 1 i = j δ i j =
More information( ) g 900,000 2,000,000 5,000,000 2,200,000 1,000,000 1,500, ,000 2,500,000 1,000, , , , , , ,000 2,000,000
( ) 73 10,905,238 3,853,235 295,309 1,415,972 5,340,722 2,390,603 890,603 1,500,000 1,000,000 300,000 1,500,000 49 19. 3. 1 17,172,842 3,917,488 13,255,354 10,760,078 (550) 555,000 600,000 600,000 12,100,000
More information2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta
009 IA 5 I, 3, 4, 5, 6, 7 6 3. () Arcsin ( (4) Arccos ) 3 () Arcsin( ) (3) Arccos (5) Arctan (6) Arctan ( 3 ) 3. n () tan x (nπ π/, nπ + π/) f n (x) f n (x) fn (x) Arctan x () sin x [nπ π/, nπ +π/] g n
More information2D-RCWA 1 two dimensional rigorous coupled wave analysis [1, 2] 1 ε(x, y) = 1 ε(x, y) = ϵ mn exp [+j(mk x x + nk y y)] (1) m,n= m,n= ξ mn exp [+j(mk x
2D-RCWA two dimensional rigoros copled wave analsis, 2] εx, εx, ϵ mn exp +jmk x x + nk ] m,n m,n ξ mn exp +jmk x x + nk ] 2 K x K x Λ x Λ ϵ mn ξ mn K x 2π Λ x K 2π Λ ϵ mn ξ mn Λ x Λ x Λ x Λ x Λx Λ Λx Λ
More information, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
More information34号 目 次
1932 35 1939 π 36 37 1937 12 28 1998 2002 1937 20 ª 1937 2004 1937 12 º 1937 38 11 Ω 1937 1943 1941 39 æ 1936 1936 1936 10 1938 25 35 40 2004 4800 40 ø 41 1936 17 1935 1936 1938 1937 15 2003 28 42 1857
More informationX 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
More informationI ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT
I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345
More informationt χ 2 F Q t χ 2 F 1 2 µ, σ 2 N(µ, σ 2 ) f(x µ, σ 2 ) = 1 ( exp (x ) µ)2 2πσ 2 2σ 2 0, N(0, 1) (100 α) z(α) t χ 2 *1 2.1 t (i)x N(µ, σ 2 ) x µ σ N(0, 1
t χ F Q t χ F µ, σ N(µ, σ ) f(x µ, σ ) = ( exp (x ) µ) πσ σ 0, N(0, ) (00 α) z(α) t χ *. t (i)x N(µ, σ ) x µ σ N(0, ) (ii)x,, x N(µ, σ ) x = x+ +x N(µ, σ ) (iii) (i),(ii) z = x µ N(0, ) σ N(0, ) ( 9 97.
More informationuntitled
š ( ) 200,000 100,000 180,000 60,000 100,000 60,000 120,000 100,000 240,000 120,000 120,000 240,000 100,000 120,000 72,000 300,000 72,000 100,000 100,000 60,000 120,000 60,000 100,000 100,000 60,000 200,000
More informationアプリケーションソフトウェアガイド PM2200C用
4013356-01 C01 1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Printed
More informationReport C: : ( )
Report2 045713C: : 18 07 23 ( ) 1 3 SCILAB (lpc.sci) Hamming DTF LPC (Levinson-Durbin ) LPC (1) (2) Levinson-Durbin SCILAB lev() SCILAB (3) =0 roots() 0 5KHz 0 5KHz (4) 100 300 8 20 (5) pre emp 1.0 2 fft_len=512;
More informationuntitled
1 SS 2 2 (DS) 3 2.1 DS................................ 3 2.2 DS................................ 4 2.3.................................. 4 2.4 (channel papacity)............................ 6 2.5........................................
More information) ] [ 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
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
More informationuntitled
1 ( 14 1517 18 20 24 2 3 4 5 1947H18536H23 5,2332,155 12H22-H25.3 209144H22-H25.3 H23.4 KI 24,090H25.5 H24.3 5,233H25.5 2,155H22-H23 1,068H22-H23 1.2%H224.1%H24 1947H18536H23 1.34H171.55H22 1H23 553,059
More informationA 2008 10 (2010 4 ) 1 1 1.1................................. 1 1.2..................................... 1 1.3............................ 3 1.3.1............................. 3 1.3.2..................................
More informationC言語による数値計算プログラミング演習
11. 離散フーリエ変換 時間領域における連続関数 x(t) は, 周波数領域の連続関数 (f) を介して, フーリエ変換 (Fourier transform) : i2π ft ( f) = xte ( ) dt 逆フーリエ変換 (inverse Fourier transform): 2 x() t = ( f) e i π ft df と展開される x(t) は区間 [,T] 以外では とすると,
More information4................................. 4................................. 4 6................................. 6................................. 9.................................................... 3..3..........................
More information29
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 informationSample function Re random process Flutter, Galloping, etc. ensemble (mean value) N 1 µ = lim xk( t1) N k = 1 N autocorrelation function N 1 R( t1, t1
Sample function Re random process Flutter, Galloping, etc. ensemble (mean value) µ = lim xk( k = autocorrelation function R( t, t + τ) = lim ( ) ( + τ) xk t xk t k = V p o o R p o, o V S M R realization
More information5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E
5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 2, S 1 N 1 = S 2 N 2 2 (chemical potential) µ S N
More informationI A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
More information30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
More informationpp d 2 * Hz Hz 3 10 db Wind-induced noise, Noise reduction, Microphone array, Beamforming 1
72 12 2016 pp. 739 748 739 43.60.+d 2 * 1 2 2 3 2 125 Hz 0.3 0.8 2 125 Hz 3 10 db Wind-induced noise, Noise reduction, Microphone array, Beamforming 1. 1.1 PSS [1] [2 4] 2 Wind-induced noise reduction
More information1 Fourier Fourier Fourier Fourier Fourier Fourier Fourier Fourier Fourier analog digital Fourier Fourier Fourier Fourier Fourier Fourier Green Fourier
Fourier Fourier Fourier etc * 1 Fourier Fourier Fourier (DFT Fourier (FFT Heat Equation, Fourier Series, Fourier Transform, Discrete Fourier Transform, etc Yoshifumi TAKEDA 1 Abstract Suppose that u is
More information2 Poisson Image Editing DC DC 2 Poisson Image Editing Agarwala 3 4 Agarwala Poisson Image Editing Poisson Image Editing f(u) u 2 u = (x
1 Poisson Image Editing Poisson Image Editing Stabilization of Poisson Equation for Gradient-Based Image Composing Ryo Kamio Masayuki Tanaka Masatoshi Okutomi Poisson Image Editing is the image composing
More informationm dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d
m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m
More informationA B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P
1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A
More informationOHP.dvi
t 0, X X t x t 0 t u u = x X (1) t t 0 u X x O 1 1 t 0 =0 X X +dx t x(x,t) x(x +dx,t). dx dx = x(x +dx,t) x(x,t) (2) dx, dx = F dx (3). F (deformation gradient tensor) t F t 0 dx dx X x O 2 2 F. (det F
More informationv 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 information5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6 cos π 6.7 MP 4 P P N i i i i N i j F j ii N i i ii F j i i N ii li i F j i ij li i i i
i j ij i j ii,, i j ij ij ij (, P P P P θ N θ P P cosθ N F N P cosθ F Psinθ P P F P P θ N P cos θ cos θ cosθ F P sinθ cosθ sinθ cosθ sinθ 5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 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 informationII 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