現(xiàn)代數(shù)字信號(hào)處理 姚天任 第三章答案

1、第三章答案3.1解: (1:由題設(shè):h (n =(10n h n h y (n=1(-n y n y 則u (n =h (n y (n所以可得最陡下降法解:h(n=1 =h *+(I-2R 2h (0- h * 51得 h (Z=2Z1-e (Z + Z1-h (Zh (Z=1-11(Z 2-ZZ e 所以:e (Z = x (Z-2Z1-e (Z- Z1-1-11(z 2-zz e H (Z = 111(211-+-ZZ 所以零點(diǎn)在單位園上,極點(diǎn)在Z = 1-2園上。 (3:要使H(Z穩(wěn)定,則極點(diǎn)在單位園內(nèi)即: 0121><-且3.3(1性能曲面函數(shù): =-=-=-+=10010

2、2222cos(2(2sin(1(1(1(1(2(W W n N n d n N n x n W n W n W n x n d n x n d E n X n d E P n x E n x n x E n x n x E n x E n X n X E R WP RW W n d E n TTT T -101(2W W n x n d n x n d+-+-+=10202585585855852510W W W W+-1055(212502W W121102055(2152545545(2510w w w w w +-+=(2誤差性能曲面matlab 程序: (31(*(*21(*2(*1

3、(*2(*(*2(*1(*2(*2 210112001-+-=-+-+=n x n d n x E w n x n X E w w n x n d n x n X E w n X E w w w w w T (4=-=-=-=-* 2.1029-0.6498 7553.40 0.4422 0.1367-0.1367- 0.4422 7553.402.5 0.77250.7725 2.5 1(1-(n x 1-xn *x(n1-x(n *n x( ( *11221n x n d n x n d n x pR w(591-10 1029.2698.04.7553- 0-10 *(2min =-=-

4、=*w p n d E T 3.42725.3*2*27275.1*2*20.70717071.0 0.7071- 7071.02725.3 7275.1 2.5 .0.77250.7725 2.5 1-(n x 1-x(n *x(n1-x(n * x(n (1120102111021w2=V V V V n x E R TT42161422( 8722242 8722112 ( 2(15.3101021201010101010101022+-+=+-+=-+=-+=n d E P R n d E n T 解:( (3001131212'21124 4(438423287( 3(323

5、296872112872112 2'1''1'02min 2110min 2*2min *1*03131*1*011*-=-=-=-+=+=+=-=-=-=-=-T TTv v v R E v v v v v v Rv v n d E P n d E P R 、(歡迎訪問機(jī)器人淘寶 (n = Ee 2 (n = E(d (n y (n 2 = Ed 2 (n + T (n E X (n X (n T (n 2 Ed (n X T (n (n (n = 0 1 T R = E X (n X (n T = E(x(n + r(n x(n - 1 + r(n - 1

6、(x(n + r(n x(n - 1 + r(n - 1 T X(n = x(n + r(n x(n - 1 + r(n - 1 T E( x(n + r (n( x(n 1 + r (n 1 E(x(n + r(n 2 = E( x(n 1 + r (n 1 2 E( x(n 1 + r (n 1( x(n + r (n 2 2 2 2 2 2 2 E(x(n + r(n = E (sin N n + r (n = E sin N n + E r (n + 2 E sin N nr (n = E 1-cos 4N n 2 + = 1 + = E( x(n 1 + r (n 1 2 2 E(

7、x(n + r (n( x(n 1 + r (n 1 = E x(n x(n 1 + E x(nr (n 1 + E(n x(n 1 + Er(nr(n - 1 = Ex(nx(n - 1 = Esin 2N n sin 2N (n 1 = E 1 (cos 2N cos( 4N n 2N = 1 cos 2N 2 2 1 cos 2 1 + R = 1 2 2 2 1 N 2 + 2 cos N P T = Ed (n X (n = Ed (n( x(n + r (nd (n( x(n 1 + r (n 1 = E2 cos 2N n sin 2N n = 0 sin 2N Ecos 2N

8、(n 1 sin 2N (n 1 = Esin 4N n Esin 2N Ed 2 (n = E4 cos 2 2 N n = 2 1 cos 2 1 + (n = Ed 2 (n + 0 1 1 2 2 2 1 N 0 20 sin 2N 0 1 2 + 1 2 cos N 2 = 2 + (0.5 + ( 0 + 12 + 0 cos 2N + 2 1 sin 2N 歡迎訪問機(jī)器人淘寶 歡迎訪問機(jī)器人淘寶 ( 2 = = 0 T 1 = (0.5 + 2 0 + 1 cos( 2N = 0 0 2 2 1 = (0.5 + 2 1 + 0 cos( N + 2 sin( N = 0 * 0

9、 = 2 cos( 2N sin( 2N /(1 + 2 2 cos( 2N 2 * 2 2 2 2 1 = 2(1 + 2 sin( N /(1 + 2 cos( N 1 1 + cos 2N 1 1 cos 2N 2 2 2 2 (3 R = 1 E - R = 1 =0 1 cos 2N + 2 cos 2N 1 2 2 2 = 1/2 + = 1 / 2 cos( 2N + t r R = 1 + 2 = 1 + 2 0 < u < t r R 系統(tǒng)收斂的u值范圍為 : 0 < u < 1/(1 + 2 (4Tmse1 = 1 / 4u1 = 5.14 Tmse

10、2 = 1 / 4u 2 = 104 M = ut r R = 0.05 1 3.11 答案： (1 e ( n = x ( n h ( n y ( n (n = E (e 2 (n = min (n = E x 2 (n + h 2 (n E ( y 2 (n 2 E x(n y (nh(n = 0.4h 2 (n 4h(n + 11 d ( n = 0 .8 h ( n 4 = 0 dh(n h( n = 5 （2） = (n = 0 .8 h 4 h 最陡下降法推導(dǎo)加權(quán)系數(shù)迭代計(jì)算公式為： h(n + 1 = h(n + µ (n = h(n + u (4 0.8h(n = (

11、1 3.2 µ h(n + 4 µ （3）求加權(quán)系數(shù)表達(dá)式 歡迎訪問機(jī)器人淘寶 歡迎訪問機(jī)器人淘寶 h(n = h * + ( I 2µR n h(0 h * = 10 + (1 0.8µ n h(0 10 1 0 .4 要求 0 < µ < 1 0 < µ < max 3.12 答案： 即0 < µ < 2.5 (1為保證收斂應(yīng)使µ滿足0 < µ < tr R1 tr R 1 = k = 2 k =0 1 即0 < µ < 1 2 (

12、2幾何比(rmse = rk2 = (1 2µ k 2 由于兩R yy的值相同，故自適應(yīng)濾波器的收斂速度相同。 3.15 解： （1）由題意，是兩系數(shù)自適應(yīng)線性組合器，則 y (n = 0 x0 (n + 1 x1 (n 2 2 y 2 (n = 0 x0 (n + 2 0 1 x0 (n x1 (n + 12 x12 (n 3 3 2 2 y 3 (n = 0 x0 (n + 3 0 1 x0 (n x1 (n + 3 0 12 x0 (n x12 (n + 13 x13 (n 4 4 3 3 2 2 y 4 (n = 0 x0 (n + 4 0 1 x0 (n x1 (n + 6

13、 0 12 x0 (n x12 (n + 4 0 13 x0 (n x13 (n + 14 x14 (n 又 e 4 (n = (d (n y (n 4 = d 4 (n 4d 3 (n y (n + 6d 2 (n y 2 (n 4d (n y 3 (n + y 4 (n e 4 (n = d 4 (n 4 0 d 3 (n x0 (n 41 d 3 (n x1 (n 2 2 + 6 0 d 2 (n x 0 (n + 12 0 1 d 2 (n x 0 (n x1 (n + 6 12 d 2 (n x12 (n 3 3 2 2 4 0 d (n x0 (n 12 0 1 d (n x0 (

14、n x1 (n 12 0 12 d (n x0 (n x12 (n 4 13 d (n x13 (n 4 4 3 2 2 2 3 4 + 0 x 0 (n + 4 0 1 x 3 (nx 1 (n + 6 0 12 x 0 (nx 1 (n + 4 0 13 x 0 (nx 1 (n + 14 x 1 (n 0 (n = Ee 4 (n = Ed 4 (n 4 0 Ed 3 (n x0 (n 4 1 Ed 3 (n x1 (n 2 2 + 6 0 Ed 2 (n x 0 (n + 12 01 Ed 2 (n x0 (n x1 (n + 6 12 Ed 2 (n x12 (n 3 3 2 2 4 0 Ed (n x0 (n 12 0 1 Ed (n