Similar documents
[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

2

main.dvi

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

Fourier (a) C, (b) C, (c) f 2 (a), (b) (c) (L 2 ) (a) C x : f(x) = a (a n cos nx + b n sin nx). ( N ) a 0 f(x) = lim N 2 + (a n cos nx + b n sin

数値計算:フーリエ変換

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.

画像工学特論


,


1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

,. 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,,.

Korteweg-de Vries

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

main.dvi

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 =

Microsoft Word - 信号処理3.doc

V s d d 2 d n d n 2 n R 2 n V s q n 2 n Output q 2 q Decoder 2 R 2 2R 2R 2R 2R A R R R 2R A A n A n 2R R f R (a) 0 (b) 7.4 D-A (a) (b) FET n H ON p H

°ÌÁê¿ô³ØII

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

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

QMI_09.dvi

QMI_10.dvi

,

p = mv p x > h/4π λ = h p m v Ψ 2 Ψ

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

構造と連続体の力学基礎

main.dvi


2 2 L 5 2. L L L L k.....

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

数値計算:常微分方程式

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

u = u(t, x 1,..., x d ) : R R d C λ i = 1 := x 2 1 x 2 d d Euclid Laplace Schrödinger N := {1, 2, 3,... } Z := {..., 3, 2, 1,, 1, 2, 3

Gmech08.dvi

2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p

1 s(t) ( ) f c : A cos(2πf c t + ϕ) (AM, Amplitude Modulation) (FM, Frequency Modulation) (PM, Phase Modulation) 2


I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

(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 +

基礎数学I

θ (t) ω cos θ(t) = ( : θ, θ. ( ) ( ) ( 5) l () θ (t) = ω sin θ(t). ω := g l.. () θ (t) θ (t)θ (t) + ω θ (t) sin θ(t) =. [ ] d dt θ (t) ω cos θ(t

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

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

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

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

A S hara/lectures/lectures-j.html ϵ-n 1 ϵ-n lim n a n = α n a n α 2 lim a n = 0 1 n a k n n k= ϵ

(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t

) ] [ 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

Part () () Γ Part ,

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

phs.dvi

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +

PDF

201711grade1ouyou.pdf

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

m 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

φ 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

untitled

7-12.dvi

sp3.dvi

A

ohpr.dvi

meiji_resume_1.PDF

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google

Untitled

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 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x

1 4 2 EP) (EP) (EP)

A 99% MS-Free Presentation

1 1.1 H = µc i c i + c i t ijc j + 1 c i c j V ijklc k c l (1) V ijkl = V jikl = V ijlk = V jilk () t ij = t ji, V ijkl = V lkji (3) (1) V 0 H mf = µc

() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.

1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3

all.dvi

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

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


吸収分光.PDF

7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a

Bessel ( 06/11/21) Bessel 1 ( ) 1.1 0, 1,..., n n J 0 (x), J 1 (x),..., J n (x) I 0 (x), I 1 (x),..., I n (x) Miller (Miller algorithm) Bess

1 u t = au (finite difference) u t = au Von Neumann

A

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

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>

30

数学Ⅱ演習(足助・09夏)

2012専門分科会_new_4.pptx

. (.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(

2 (2016 3Q N) c = o (11) Ax = b A x = c A n I n n n 2n (A I n ) (I n X) A A X A n A A A (1) (2) c 0 c (3) c A A i j n 1 ( 1) i+j A (i, j) A (i, j) ã i

impulse_response.dvi

RIMS98R2.dvi

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n

(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

1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N m 1.

80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t = 0 i r 0 t(> 0) j r 0 + r < δ(r 0 x i (0))δ(r 0 + r x j (t)) > (4.2) r r 0 G i j (r, t) dr 0

m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

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) =

基礎から学ぶトラヒック理論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

Transcription:

2014 3 10 5

1 5 1.1..................................... 5 2 6 2.1.................................... 6 2.2 Z........................................ 6 2.3.................................. 6 2.3.1.................. 7 2.4....................................... 8 2.5.................................. 8 2.5.1......................... 10 2.5.2........................... 11 2.5.3................. 11 3 ex6 1 12 3.1 ex6 1..................... 12 3.2 ex6 1.c........................................ 14 3.3.......................... 17 3.4 sinc....................................... 18 4 19 A Fourier 19 A.1 1........................................ 19 A.2 2........................................ 20 A.3 3........................................ 20 B Fourier Fourier 20 B.1 Fourier..................................... 20 B.2 Fourier.................................. 21 C Fourier 23 D 23 1 1.1 C 6

[1 6 ex6 1.c 2 2.1 x : R C T s > 0 x n := x(nt s ) (n Z) {x n } n Z C Z T s Sampling period T s T s. Ω R x(t) = e iωt T s > 0 x n := x(nt s ) x n = e iωnts = e inω. ω := ΩT s {x n } e iω e iω = 1 2.2 Z Z,1 N, N 0 := N {0} {x n } x : R C T s > 0 x n = x(nt ) {x n } n Z C Z X(z) := n Z x n z n X = X(z) {x n } Z (z C, ) 2.3 {x n } n {y n } F : {x n } {y n } 7

2.3.1 F F ({x n } + {y n }) = F ({x n }) + F ({y n }), F (c{x n }) = cf ({x n }) k Z F ({x n k }) = {y n k } n n Z X := C Z = {{x n } n Z ; n Z x n C} F : X X h C Z s.t. x C Z F (x) = x h. h h = F (δ) δ = {δ n } n Z δ n := δ n0 = { 1 (n = 0) 0 (n 0) (Kronecker δ ij δ n = δ n0 ) h F h = F (δ) F [1] h n b n FIR J N s.t. b n = 0 (n < 0 or n > J) b m (0 m J) y n = J x n k b k (n Z) k=0 {x n } n 0, {y n } n 0 y n = min{j,n} m=0 x n m b m 8

2.4 x(t) = e iωt {x n } x n = x(nt s ) = e iωnts = e inω, ω := ΩT s {x n } e iω h = {h n }, y = {y n }, y = x h h z H(z) n Z y n = k Z x n k h k = k Z e i(n k)ω h k = e inω k Z e ikω h k = H(e iω )x n. y = H(e iω )x. h h z H(z) e iω H(e iω ) ω ω H(e iω ) (transfer function) ω [0, 2π] ω [ π, π] (1) G(ω) := H(e iω ), θ(ω) := arg H(e iω ) (arg ) G(ω) (gain),θ(ω) (phase shift) 2.5 x(t) = e iωt T s x n := x(nt s ) = e iωnts = e inω (n Z), ω := ΩT s. x = {x n } y = {y n } {h n } y n = x n k h k = ( ) e i(n k)ω h k h k = e inω = H(e iω )e inω. e ikω k Z k Z H {h n } z : H(z) := k Z X(ω) := H(e iω ) ω e (0, π) { 1 ( ω ωe ) X(ω) = ([ ω e, ω e ] ) = 0 ( ω > ω e ) 9 k Z h k z k.

1.0 0.8 0.6 0.4 0.2 3 2 1 1 2 3 1: X(ω) = k Z h n e inω (ω R) X {h n } Fourier Fourier h n = 1 π X(ω)e inω dω (n R) 2π π 2π Fourier i i Fourier Fourier 2π x : R C c n := 1 2π pi π x(θ)e inθ dθ (n Z) x(θ) = k Z c n e inθ (θ R) {h n } h n = 1 2π π π X(ω)e inω dω = 1 2π ωe ω e e inω dω = ω e π sinc nω e. f e f e = ω e 2π f s ω e = 2π f e f s h n = 2πf e/f s π sinc 2nπ f e f s = 2 f e f s sinc 2nπf e f s. 10

b m = w m h m J/2 w m b m h m J/2 n J/2 y n = x n k h k x n k h k J x n k b k+j/2 = x n J/2 m b m = y[n J/2]. k Z m=0 k J/2 k J/2 2.5.1 h n = ω e π sinc nω e X J (ω) := h n e inω n J/2 Mathematica Clear[n, t, w, x]; w[x_]:=(1-cos[2 Pi x])/2; omega=0.5; h[n_]:=omega/pi Sinc[n omega]; draw[j_]:=plot[sum[h[n]exp[-i n t], {n, -J/2, J/2}],{t, -Pi, Pi}, PlotRange -> All]; f(ω) := 50 n= 50 h n e inω f(ω) H(e iω ) 1.0 0.8 0.6 0.4 0.2 3 2 1 1 2 3 2: (J = 100) Gibbs 11

2.5.2 (hann window), von Hann window,2 (raised cosine window) (hanging wndow) w(x) = 1 cos 2πx 2 (0 x 1). 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 3: hann w(x) = 1 cos 2πx 2 void Hanning window(double w[], int N) N ( n ) w (N ) N w n := ( n + 1/2 ) (n = 0, 1,..., N 1) w (N ) N w = {w n } N 1 n=0 N w w(1/2) = 1 Hanning window() Mathematica Clear[w]; w[x_]:=(1-cos[2 Pi x])/2; hanningwindow[n_]:=if[evenq[n], Table[w[n/N], {n, 0, N-1}], Table[w[(n+1/2)/N], {n, 0, N-1}]] 2.5.3 h n h n := h n w(n/100) f(ω) = 50 n= 50 12 h n e inω

draw2[j_]:=plot[sum[w[n/j-1/2]h[n]exp[-i n t],{n, -J/2, J/2}], {t,-pi,pi}, PlotRange -> All] 1.0 0.8 0.6 0.4 0.2 3 2 1 1 2 3 H(e iω ) 4: (J = 100) 3 ex6 1 [1] ex6 1.c ex6 1 sine 500hz 3500hz.wav WAVE 16 PCM 500Hz, 3500Hz 1000Hz, 1000Hz ex6 1.wav 3.1 ex6 1 ex6 1.c wave.h.wav wave {read write} {8 16}bit {stereo mono}() void 2 1 PCM 2 window function.h 1 cos 2πx ex6 1 hann window w(x) = x 13

sinc.h sinc sin πx sin x sinc(sinc x = ) sinc(sinc x = πx x ) fir filter.h FIR ex6 1 (a) FIR filtering(x,y,l,b,j) y = x b x = {x[n]} L 1 n=0,b = {b[m]} J m=0 ( ) y[n] = minj,n m=0 y = {y[n]} L 1 n=0 b[m]x[n m] (n = 0,..., L). (b) FIR LPF(fe, J, b, w) fe {b m } J m=0 {w m } J m=0 14

3.2 ex6 1.c ex6 1.c #include <stdio.h> #include <stdlib.h> #include <math.h> #include "wave.h" #include "window_function.h" #include "sinc.h" #include "fir_filter.h" int main(void) { MONO_PCM pcm0, pcm1; int n, m, J; double fe, delta, *b, *w; wave_read_16bit_mono(&pcm0, "sine_500hz_3500hz.wav"); pcm1.fs = pcm0.fs; /* */ pcm1.bits = pcm0.bits; /* */ pcm1.length = pcm0.length; /* */ pcm1.s = calloc(pcm1.length, sizeof(double)); /* */ fe = 1000.0 / pcm0.fs; /* */ delta = 1000.0 / pcm0.fs; /* */ J = (int)(3.1 / delta + 0.5) - 1; /* */ if (J % 2 == 1) { J++; /* J+1 */ } b = calloc((j + 1), sizeof(double)); w = calloc((j + 1), sizeof(double)); Hanning_window(w, (J + 1)); /* */ FIR_LPF(fe, J, b, w); /* FIR */ /* */ for (n = 0; n < pcm1.length; n++) { for (m = 0; m <= J; m++) { if (n - m >= 0) { pcm1.s[n] += b[m] * pcm0.s[n - m]; } } } wave_write_16bit_mono(&pcm1, "ex6_1.wav"); free(pcm0.s); free(pcm1.s); free(b); free(w); } return 0; fe = 1000.0 / pcm0.fs; /* */ 1000Hz fe (1/8000s) 15

1/8 f s, f e fe f e /f s fe f e fe= f e /f s delta = 1000.0 / pcm0.fs; /* */ 1000Hz f s, δ delta= δ/f s {b m } J m=0 J + 1 J = (int)(3.1 / delta + 0.5) - 1; /* */ if (J % 2 == 1) { J++; /* J+1 */ } 3.1/delta 1 J = round(3.1 / delta); if(j % 2 == 1) { J--; } 3.1/delta 1 [ 3.1 J = δ f s + 1 ] p (p = 0, 1 J ) 2 J J + 1 1 0 [2] 6.2π N = 6.2π 3.1 2π [2] δ/f s 2π ex6 1 δ = 1000Hz,f s = 8000Hz J = 24 16

Hanning_window(w, (J + 1)); /* */ 3.1/delta 1 1 cos 2πx Hanning window(w, N) hann w(x) := 2 w( n ) (N ) w n := N w( n + 1/2 (n = 0, 1,..., N 1) N ) (N ) sinc hann FIR_LPF(fe, J, b, w); /* FIR */ fe(= f e /f s, ), J( ), w(= {w m } J m=0 hann ) 3 b(= {b m} J m=0, ) ( ) b m := w m 2 f e sinc 2πf e(m J/2) (m = 0, 1,..., J) f s f s fs=8000; fe=1000; delta=1000/fs; J=Round[3.1/delta]; If[OddQ[J], J=J+1]; g=plot[2*fe/fs*sinc[2 Pi fe (m - J/2)/fs], {m, 0, J + 1}] 0.25 0.20 0.15 0.10 0.05 5 10 15 20 25 0.05 5: hann g2 = Plot[w[m/(J + 1)]*2*fe/fs*Sinc[2 Pi fe (m - J/2)/fs], {m, 0, J + 1}] 17

0.25 0.20 0.15 0.10 0.05 5 10 15 20 25 6: hann 3.3 [2] f s f e f e [2] M(ω) M(ω) max M(ω) = 1 2 ω cutoff frequency δ p 1 δ p M(ω) 1 + δ p ω [0, ω p ] [0, ω p ] ω p δ s M(ω) δ s ω [ω s, π] [ω s, π] ω p [ω p, ω s ] [1] [ω p, ω s ] (ω p + ω s )/2 f s ω s ω p δ (?) f e = ω p + ω s, ω s ω p = δ 2 18

3.4 sinc sinc.h sinc sinc() sin x (x R\{0}) sinc(x) := x 1 (x = 0) sinc sin πx (x R\{0}) sinc x := πx 1 (x = 0) sinc [2] Mathematica Sinc[] sinc sinc Mathematica sinc g = Plot[Sinc[x], x, -15, 15, PlotRange -> Full] 1.0 0.8 0.6 0.4 0.2 15 10 5 5 10 15 0.2 7: sinc x = sin x x x = nπ (n Z) y = ± 1 x sinc ω p (0, π) [ ω p, ω p ] χ [ ωp,ωp] Fourier ω p π sinc ω pt 1 χ [ ωp,ω 2π p](ω)e iωt dω = 1 ωp e iωt dω 2π ω p = 1 2π [ e iωt it ] ω=ωp = ω p π sinc ω pt. 19 ω= ω p = 1 2π e iωpt e iωpt it = 1 π sin ω p t t

4 A Fourier i A.1 1 f : R n C (2) g(ξ) := 1 f(x)e ixξ dx (ξ R n ) (2π) n/2 R n g = g(ξ) f Fourier Fg, ˆf : (3) Ff(ξ) = f(ξ) = f (4) f(x) = 1 f(x)e ixξ dx (ξ R n ). (2π) n/2 R n 1 g(ξ)e ixξ dξ (x R n ) (2π) n/2 R n Fourier g : R n C (4) f = f(x) g Fourier F g, F 1 g, g : (5) F g = g(x) = 1 g(ξ)e ixξ dξ (x R n ). (2π) n/2 R n 20

A.2 2 t ω x = x(t) X(ω) := x(t)e iωt dt (ω R) X = X(ω) x Fourier Fx, ˆx : X = X(ω) Fx(ω) = x(ω) = x(t)e iωt dt (ω R). x(t) := 1 X(ω)e iωt dω (t R) 2π infty x = x(t) X Fourier F X, F 1 X, X : F X(t) = X(t) := 1 2π X(ω)e iωt dω (t R). X(ω) x(t) A.3 3 x = x(t) X(f) := x(t)e 2πift dt (f R) X = X(ω) x Fourier Fx, ˆx : Fx(f) = x(f) = X = X(t) x(t) = x(t)e 2πift dt (f R). X(f)e 2πift df (t R) x = x(t) X Fourier F X, X : F X(t) = X(t) = X(f)e 2πift df (t R). R B Fourier Fourier B.1 Fourier T x = x(t) x Fourier : a 0 2 + ( a n cos 2nπt T + b n sin 2nπt ), T n=1 21

a n := 1 T T/2 T/2 x(t) cos 2nπt T dt, b n := 1 T T/2 T/2 x(t) sin 2nπt T dt. [ T/2, T/2] [0, T ] x Fourier : n Z exp 2πnit T, c n := 1 T D N (t) := 1 T T/2 T/2 N n= N ( x(t) exp 2πnit ) dt. T exp 2πnit T (N N, t R) D N Dirichlet T x = x(t), N N x N (t) := N n= N c n exp 1πnit T, c n := 1 T T/2 T/2 ( x(t) exp 2πnit ) dt T x N = x D N, x N (t) = T/2 T/2 x(t τ)d N (τ)dτ (t R) x Fourier N x Dirichlet D N B.2 Fourier x = x(t) T [0, T ] N t j = j T N (j = 0, 1,..., N = 1) x j = x(t j ) Fourier c n = 1 T T 0 ( x(t) exp 2nπit ) dt T 22

C n := 1 T T N 1 N j=0 W 1 N ( x j exp 2nπit ) j = 1 N=1 T N ( W := exp 2πi ) N C n = 1 N N 1 j=0 x j W nj. j=0 W N = 1, W j 1 (j = 1, 2,..., N 1) ( x j exp 2nπij ). N N 1 j=0 W kj = { N (k 0 (mod N)) 0 (k 0 (mod N)) m n (mod N) C n = C m C 0, C 1,..., C N 1 {x j } N 1 j=0 X k = N 1 j=0 x j W kj (k = 0, 1,..., N = 1) {X k } N 1 k=0 {x j} N 1 j=0 Fourier {x j} {X k } Fourier (k, j) W kj W X 0 x 0 X 1. = W x 1.. X N=1 x N=1 Fourier W W W = WW = NI N x j = 1 N W 1 = 1 N W = 1 N (W jk ). N 1 k=0 X k W kj (j = 0, 1,..., N 1) 23

not U := 1 N W U = 1 N W U unitary U U = UU = I N. C Fourier {x n } n Z Fourier (discrete-time Fourier transform,dtft) (6) X(ω) := n Z x n e inω (ω R; 2π ). x n n Z x n 2 < X L 2 (0, 2π) (6) L 2 X 2π : k Z X(ω + 2πk) = X(ω). ω [0, 2π]( ω [ π, π]) x n X ( n)fourier (7) x n = 1 2π π Fourier π X(ω)e inω dω (n Z). D [1] : C, (2013). [2], : MATLAB, (2000). 24