APPLIED NUMERICAL METHODS USING MATLAB phần 10 pot

... Recktenwald, G. W., Numerical Methods with MATLAB, Prentice-Hall, Upper Sad- dle River, NJ, 2000. [S-1] Schilling, R. J., and Harris, S. L., Applied Numerical Methods for Engineers Using MATLAB and C, ... 1(F.9) ∞  n=0 (−1) n 2n + 1 = 1 − 1 3 + 1 5 − 1 7 +···= 1 4 π (F .10) (continued overleaf ) Applied Numerical Methods Using MATLAB  , by Yang, Cao, Chung, and Mor...

Ngày tải lên: 09/08/2014, 12:22

... following MATLAB statements. >>n = 0 :100 ; S = sum(2.^-n) (b) Write a MATLAB statement that performs the following computation.  100 00  n=0 1 (2n + 1) 2  − π 2 8 (c) Write a MATLAB statement ... for 100 values of b over the interval [10 7.4 ,10 8.5 ] generated by the MATLAB command ‘ logspace(7.4,8.5 ,100 )’, PROBLEMS 69 (a) Similarly to the MATLAB routine “CtFT1(x,D...

Ngày tải lên: 09/08/2014, 12:22

... 2.2 (a) f 42 ( x ) = tan (p − x ) − x 10 0 20 10 5 0102 015 (b) f 44 b ( x ) = ( x 2 − 25)( x − 10) − 5 2.4 2.6 0 10 −20 −30 −40 −15 10 −50 5 1 510 x 1 x 0 x 1 x 2 x 3 x 2 x 3 x 0 x 0 x 0 x 3 x 3 x 2 x 2 x 1 x 1 1 125 1 0 −1 2 −2 −5 0 510 (d) ... = [0:N-1]; T = 0.1; %sampling period 0 02 5 16 2730 0 10 20 30 0 5 10 32 54 59 0 10 20 30 0 5 10 32 54 59 0 10 20 30 0...

Ngày tải lên: 09/08/2014, 12:22

... Minimum [1 0 2] 0 [1.29 0.57 2] 2.74 [0 0 0] 0 [000] 0 (b1) [101 010] 0 Maximum (good) [0 0 0] 0 No feasible solution (w) (b2) [101 010] 0 Not a minimum, but the max [0.1 0.1 3] 0 One of many minima ... Weird (warning) [0.2 0.3] 0 [10. 25 0] ∞ (d) [2 5] 0 [5.77 8.17] 25.98 [100 10] 0 Minimum PROBLEMS 357 Table 7.3 The Names of MATLAB Built-In Minimization Routines in MATLAB 5.x/6.x Un...

Ngày tải lên: 09/08/2014, 12:22

... “ nm2e04.m”: %nm2e04 N = 4; %the number of unknown variables/equations kmax = 20; tol = 1e-6; At=[ 0100 ;101 0; 0101 ;0 010] /2; x0 = 0; x5 = 10; %boundary values b = [x0/2 0 0 x5/2]’; %RHS vector %initialize all the values ... the range of collected data points, we call it the interpolation/extrapolation. Applied Numerical Methods Using MATLAB  , by Yang, Cao, Chung, and Morris C...

Ngày tải lên: 09/08/2014, 12:22

... 0.0 0100 00000 0.706753 1100 0.0031936183 −0.00035367121 h 4 = 0.00 0100 0000 0.7070714247 0.0003183147 −0.00003535652 h 5 = 0.000 0100 000 0.70 7103 2456 0.0000318 210 −0.00000353554 h 6 = 0.0000 0100 00 ... 0.70 7106 4277 0.0000031821 −0.00000035344 h 7 = 0.00000 0100 0 0.70 7106 7454 0.0000003176 −0.00000003581 h 8 = 0.000000 0100 ∗ 0.70 7106 7842 0.0000000389 0.00000000305 ∗ h 9 = 0.000...

Ngày tải lên: 09/08/2014, 12:22

... (P5 .10. 1)? 10 10 0 10 −2 P e, b (SNR, b ) 10 −4 234567 109 SNR [dB] b = 1 b = 2 b = 3 b = 4 Figure P5 .10 The BER (bit error rate) curves for multidimensional (orthogonal) signaling. %nm5p10.m: ... solutions 2 0 0 246 810 0246 810 3 × 10 4 × 10 –4 (a1) Numerical solutions without modifiers 1 1.5 1 0.5 0 (b1) Relative errors without modifiers true analytical solution y...

Ngày tải lên: 09/08/2014, 12:22

... ????????????) end (i) (x 10 ,x 20 ) = (4.223 × 10 7 , 0)[m] and (x 30 ,x 40 ) = (v 10 ,v 20 ) = (0, 3071)[m/s]. (ii) (x 10 ,x 20 ) = (4.223 × 10 7 , 0)[m] and (x 30 ,x 40 ) = (v 10 ,v 20 ) = (0, 3500)[m/s]. (iii) ... 10 −13 bvp4c() (P6 .10. 3) with bvp2 shoot() NaN (divergent) N/A y(1) = 4, y(2) = 8 bvp2 fdf() bvp4c() 3.5 10 −6 (P6 .10. 4) with bvp2 shoot() y(1) = 1/3, y(4) = 20...

Ngày tải lên: 09/08/2014, 12:22

... using the routine “ fem_basis_ftn()” and plot one of them for node 1 by using the MATLAB command mesh(), as depicted in Fig. 9.8a. Second, without generating the basis functions, we use the MATLAB ... FEM solution drawn by using trimesh () x10 −3 5 0.5 0.5 −0.5 −0.5 −5 0 0 0 1 1 y −1 −1 x u ( x , y ) u ( x , y ) u ( x , y ) (b) 31-point FEM solution by using mesh () x10 −3 5 0....

Ngày tải lên: 09/08/2014, 12:22