dfilterh.dvi

Similar documents
main.dvi

main.dvi

main.dvi

main.dvi

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

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

1 7 ω ω ω 7.1 0, ( ) Q, 7.2 ( Q ) 7.1 ω Z = R +jx Z 1/ Z 7.2 ω 7.2 Abs. admittance (x10-3 S) RLC Series Circuit Y R = 20 Ω L = 100

xy n n n- n n n n n xn n n nn n O n n n n n n n n

arma dvi

2: 3: A, f, φ f(t = A sin(2πft + φ = A sin(ωt + φ ω 2πf 440Hz A ( ( 4 ( 5 f(t = sin(2πf 1t + sin(2πf 2 t = 2 sin(2πt(f 1 + f 2 /2 cos(2πt(f 1 f

さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a


untitled

meiji_resume_1.PDF

[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) (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

6. Euler x

表1_表4

untitled

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

1. 線形シフト不変システムと z 変換 ここで言う システム とは? 入力数列 T[ ] 出力数列 一意変換 ( 演算子 ) 概念的には,, x 2, x 1, x 0, x 1, x 2, を入力すると, y 2, y 1, y 0, y 1, y 2, が出力される. 線形システム : 線形シ

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

2012 A, N, Z, Q, R, C

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.

II R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k

1 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

Microsoft Word - 信号処理3.doc

SOWC04....

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

振動と波動

1

平塚信用金庫の現況 2015

,. 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,,.

chap1.dvi

Z...QXD (Page 1)


関数のグラフを描こう

平成 30 年度 ( 第 40 回 ) 数学入門公開講座テキスト ( 京都大学数理解析研究所, 平成 30 ~8 年月 72 月日開催 30 日 [6] 1 4 A 1 A 2 A 3 l P 3 P 2 P 1 B 1 B 2 B 3 m 1 l 3 A 1, A 2, A 3 m 3 B 1,

資料5:聖ウルスラ学院英智小・中学校 提出資料(1)

2007-Kanai-paper.dvi

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)

スタートアップガイド_応用編

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

LMS NLMS LMS Least Mean Square LMS Normalized LMS NLMS AD 3 1 h(n) y(n) d(n) FIR w(n) n = 0, 1,, N 1 N N =

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


x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)

!#" $ %& '( 2.4 q n k k n qn qn [3] )+* & "+-/." ( ) 3.2 Scilab FIR TMS320C DSP 2 Code Composer Studio 1.23 DSP %& '

( ) : 1997

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

訂正目次.PDF

ISBN ISBN ISBN

ISBN ISBN 5 128p ISBN ISBN 2

ISBN ISBN 0 0 ISBN

民事責任規定・エンフォースメント

496



,2,4

1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.

ii

29

LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University

280PA2629AA-00.0

06佐々木雅哉_4C.indd

sikepuri.dvi


*3 i 9 (1,) i (i,) (1,) 9 (i,) i i 2 1 ( 1, ) (1,) 18 2 i, 2 i i r 3r + 4i 1 i 1 i *4 1 i 9 i 1 1 i i 3 9 +




A µ : A A A µ(x, y) x y (x y) z = x (y z) A x, y, z x y = y x A x, y A e x e = e x = x A x e A e x A xy = yx = e y x x x y y = x A (1)

6

untitled

Sample 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

sp3.dvi

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

PS-M3024

ii

広報さっぽろ 2016年8月号 厚別区

1

, 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

untitled


di-problem.dvi

genron-3

impulse_response.dvi

1 n A a 11 a 1n A =.. a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = 0 ( x 0 ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 1.1 Th

(DFT) 009 DFT: Discrete Fourier Transform 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


newmain.dvi

(time series) ( 225 ) / / p.2/66

技術研究報告第26号

AN8934FA

1部

5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (

A

ohpr.dvi


Transcription:

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 s (1) f =!=(2) =F=F s (2) 12.5.2 2 1 x(0);x(1);x(2);::: 3 2 x(0);x(1);x(2);::: y(n) = 1 (x(n)+x(n 0 1)) 2 3 2 1

4 3 y(n) = 1 (x(n)+x(n 0 1) + x(n 0 2)) 3 3 x(n) [] y(n) () 1: 2: 3: 2 4: 3 5 6 5 7 { ( ) { ( ) {? {? 2

5: 5 6: 7 12.5.3 sin(!n) cos(!n) e i!n = cos(!n)+i sin(!n) i 1 + e cos(!n) = ei!n 0i!n 2 0 e sin(!n) = ei!n 0i!n 2 (3) u(n)( 7(a)) ( 1 n 0 u(n) = 0 n<0 ( )(n)( 7(b)) ( 1 n =0 (n) = 0 n 6= 0 7: 3

8: (n) 12.5.4 ) (n 0 2) n =2 1 ( 8 (a)) 8 (b) x(01) = 01; x(0) = 2; x(2) = 1 0 x(n) x(n) = x(n) = x(01)(n +1) +x(0)(n) +x(1)(n 0 1) + x(2)(n 0 2) x(n) x(n) = k=01 x(k)(n 0 k) (4) 12.5.5 x(n) y(n) y(n) =T[x(n)] ( ) (time-invariant system) x(n) y(n) k y(n 0 k) =T [x(n 0 k)] 9(a)(b) 1 j 4

9: (linear system) x 1 (n);x 2 (n) y 1 (n);y 2 (n) a; b T [ax 1 (n)+bx 2 (n)] = at [x 1 (n)] + bt[x 2 (n)] = ay 1 (n)+by 2 (n) ) 9(c) (linear timeinvariant system) (causal system) n 0 y(n 0 ) n n 0 x(n) ) 5

12.5.6 (impulse response) h(n) h(n) =T [(n)] h(n) x(n) ( ) y(n) = m = n 0 k y(n) = k=01 x(k)h(n 0 k) (5) 0 m=1 x(n 0 m)h(m) = h(k)x(n 0 k) (6) k=01 x(n) h(n) (convolution) y(n) =x(n)?h(n) 3 h(0) = h(1) = h(2) = 1 3 (5) y(n) =T [x(n)] (4) y(n) =T[x(n)] = T [ y(n) = = k=01 k=01 k=01 x(k)(n 0 k)] T [x(k)(n 0 k)] (7) x(k)t [(n 0 k)] (8) P (7) (8) x(k) h(n) =T [(n)] h(n 0 k) =T [(n 0 k)] (8) (5) ) 10 6

10: 12.5.7 (5) ( ) 11 11 y(n) =x(n)+2x(n 0 2) 12.5.8 y(n) =x(n)+by(n 0 1) y 12 n =0 1;b;b 2 ;b 3 ; 111 IIR FIR IIR (innite impulse response) FIR(nite impulse response) FIR IIR IIR (6) ( x(n) y(n)) 7

11: 12: 8

h(k) =b k ; (k = 0; 1; 2;:::) y(n) = IIR k=0 b k x(n 0 k) 12 jbj > 1 12.5.9! ( ) x(n) =cos(!n) y(n) =A(!)cos(!n + (!)) { (!) { (A(!)) ((!)) { (A(!)) ((!)) (!) (frequency characteristics) { ( ) A(!) (amplitude characteristics) { (!) (phase characteristics) e i!n = cos(!n)+isin(!n) h(n) (6) 9

y(n) = k=01 h(k)ei!(n0k) = e i!n k=01 h(k)e0i!k (9) = e i!n H(e i! ) (10) H (9) 6 H(e i! )= k=01 h(k)e0i!k (11) H e i! H H(e i! )=A(!)e i(!) (12) A(!) (!) (10) y(n) = e i!n H(e i! )=A(!)e i(!n+(!)) { { (11) (12) 3 y(n) = 1 (x(n)+x(n 0 1) + x(n 0 2)) 3 13: 3 10

e i! = e i(!+2) 0 (!) < (!) 01 (!) < 1 3 13 {! = A(!)! = F =! F 2 p i s! = F = F s =2 { 12.5.10 z H(e i! ) (11) (12) IIR z H(e i! ) z x(n) z X(z) X(z) = (13) n=01 x(n)z0n z x(n) z X(z) x(n) $ z X(z) Z[x(n)] = X(z) (x(n)) (X(z)) (n) z 14(a) (n) z 1 11

14(b) 2(n 0 2) z c(n 0 k) z cz 0k c ) 14(c) x(n) z x(n) ( (4)) z z 14: z z z z { 2 x 1 (n);x 2 (n) z X 1 (z) =Z[x 1 (n)];x 2 (z) = Z[x 2 (n)] Z[ax 1 (n)+bx 2 (n)] = az[x 1 (n)] + bz[x 2 (n)] = ax 1 (z)+bx 2 (z) a b { x(n) z X(z) =Z[x(n)] Z[x(n 0 k)] = X(z)z 0k ; (k : ) 12

z 2 x 1 (n);x 2 (n) z X 1 (z);x 2 (z) ( (5)) x 1 (n)?x 2 (n) = n=01 x 1(n)x 2 (n 0 k) $ z X 1 (z)x 2 (z) (14) y(n) = k=01 x 1(k)x 2 (n 0 k) z Z[y(n)] = = = n=01 y(n)z0n n=01 ( k=01 x 1(k)( k=01 x 1(k)x 2 (n 0 k))z 0n n=01 x 2(n 0 k)z 0n ) n 0 k = p z 0n = z 0p z 0k = = = k=01 x 1(k)( p=01 x 2(p)z 0p z 0k ) k=01 x 1(k)z 0k ( p=01 x 2(p)z 0p ) k=01 x 1(k)z 0k X 2 (z) = X 1 (z)x 2 (z) IIR h(n) z H(z) H(z) =Z[h(n)] x(n) h(n) y(n) y(n) =x(n)?h(n) = k=01 h(k)x(n 0 k) z z Y (z) =H(z)X(z) Y (z) =Z[y(n)];H(z) =Z[h(n)];X(z) =Z[x(n)] H(z) 13

H(z) z H(z) = Y (z) X(z) 3 y(n) = 1 (x(n)+x(n 0 1) + x(n 0 2)) 3 h(n) = 1 ((n)+(n 0 1) + (n 0 2)) 3 h(n) z H(z) = 1 3 (1 + z01 + z 02 ) 2 z Z[y(n)] = Y (z) Z[x(n)] = X(z) Z[x(n 0 1)] = X(z)z 01 ;Z[x(n 0 2)] = X(z)z 02 Y (z) = 1 3 (X(z)+X(z)z01 + X(z)z 02 ) = 1 3 (1 + z01 + z 02 )X(z) H(z) = Y (z) X(z) = 1 3 (1 + z01 + z 02 ) (6) 2 y(n) = N01 X k=0 h(k)x(n 0 k); (N : ) 3 Y (z) = N01 X k=0 X N01 h(k)x(z)z 0k =( X k=0 H(z) = Y (z) N01 X(z) = h(k)z 0k k=0 h(k)z 0k )X(z) z z (order) N 0 1 z H(z) X(z) H(z)X(z) x(n) 2 k<0 0 0 k<0 h(k) =0 FIR N>0 0 k>n h(k) =0 14

) x(0) = 2; x(1) = 02; x(2) = 2 x(n) =0 2 y(n) = 1 (x(n)+x(n 0 1)) 2 z z y(n) =x(n)+by(n 0 1) z Y (z) =X(z)+bY (z)z 01 Y (z) X(z) H(z) Y (z)(1 0 bz 01 )=X(z) H(z) = Y (z) X(z) = 1 1 0 bz01 H(z) z e i! H(e i! )=H(z)j z=e i! x(n) z ( (13)) X(z) = n=01 x(n)z0n h(n) z H(z) x(n) h(n) H(z) H(z) = z e i! H(e i! )= n=01 h(n)z0n n=01 h(n)e0i!n (11) z e i! (11) Z 15 15

y(n)=t[x(n)] z h(n) y(n) = h(k)x(n-k) x(n) = exp(i n) H(z) = Y[z]/X[z] H(exp(i )) z = exp(i )) 15: z 12.5.11 16

16: LPF (Low Pass Filter), BPF (Band Pass Filter), HPF (High Pass Filter), BRF (Band Reject Filter) 17

// lpf1k.java // // // 8kHz 8 PCM // (C) 1999/9/13 static void convert(string infn, String outfn) { FileInputStream fin; FileOutputStream fout; // // : 60.000 [db] // : 100 // : 8.00000 [khz] // : 1.00000 [khz] double h[] ={ 0.00012980, 0.00012571, 0.00000000,-0.00021187,-0.00037517, -0.00032645, 0.00000000, 0.00047473, 0.00079661, 0.00066250, 0.00000000,-0.00089617,-0.00146002,-0.00118280, 0.00000000, 0.00153030, 0.00244607, 0.00194765, 0.00000000,-0.00244520, -0.00385770,-0.00303525, 0.00000000, 0.00373275, 0.00583714, 0.00455654, 0.00000000,-0.00553114,-0.00860537,-0.00668972, 0.00000000, 0.00807882, 0.01255811, 0.00976658, 0.00000000, -0.01185753,-0.01853088,-0.01452100, 0.00000000, 0.01805053, 0.02870358, 0.02300258, 0.00000000,-0.03057276,-0.05112438, -0.04387509, 0.00000000, 0.07433807, 0.15850612, 0.22485221, 0.25000417, 0.22485221, 0.15850612, 0.07433807, 0.00000000, -0.04387509,-0.05112438,-0.03057276, 0.00000000, 0.02300258, 0.02870358, 0.01805053, 0.00000000,-0.01452100,-0.01853088, -0.01185753, 0.00000000, 0.00976658, 0.01255811, 0.00807882, 0.00000000,-0.00668972,-0.00860537,-0.00553114, 0.00000000, 0.00455654, 0.00583714, 0.00373275, 0.00000000,-0.00303525, -0.00385770,-0.00244520, 0.00000000, 0.00194765, 0.00244607, 0.00153030, 0.00000000,-0.00118280,-0.00146002,-0.00089617, 0.00000000, 0.00066250, 0.00079661, 0.00047473, 0.00000000, -0.00032645,-0.00037517,-0.00021187, 0.00000000, 0.00012571, 0.00012980 }; for (i = 0; i < size; i++) { byte temp; buf[0] = (byte)fin.read(); o = 0.0; for (m = 0; m<=nf; m++) { o += h[m]*((double)buf[m]); } fout.write((byte)o); for (m=nf; m>0; m--) { buf[m] = buf[m-1]; } } 18