APPLIED NUMERICAL METHODS USING MATLAB phần 8 pot
APPLIED NUMERICAL METHODS USING MATLAB phần 8 pot
... 0.53 1. 08 0.46 —
1
−1.21 −1.21 −1.12 −1. 18 −1.21 −1.21 −1.15 1.26
1
−0. 58 −0. 58 −0.76 −0.64 −0. 58 −0. 58 −0.71 −1.70
c
o
3
−1.76 −1.76 −1.44 −1.65 — −1.76 −1.76 −1.54 −1 .84 —
−0.00 −0.00 −3 .82
1
−0.00 ... −1.33 −1 .84 −0.65
1
−0.00 0.00 0.29 −3 .82 −1.41
−0.00 −0.00 −0.00 −0.00 0.00 −0.00 −0.00 −0.00 −22.1 −16.4
1.21 1.21 1.12 1. 18 — 1.21 1.21 1.15 −1.26 —
x
o
1
0. 58 0. 58...
APPLIED NUMERICAL METHODS USING MATLAB phần 2 potx
... (1. 986 0e-15)
1.5000
0.3143 1. 788 9
A
2
x = b
2
0.6 286
0.9429
A
3
x = b
3
A
4
x = b
4
2.5000 1.2247
0.0000
A
5
x = b
5
A
6
x = b
6
(cf) When the mismatching error ||A
i
x − b
i
||’s obtained from MATLAB ... two
integrals (P1 .8. 2a) and (P1 .8. 2b) by using the in-line/M-file functions.
function xfx = xGaussian_pdf(x,m,sigma,x0)
xfx = (x - x0).*exp(-(x - m).^2/2/sigma^2)/sqrt(2*pi)/sig...
APPLIED NUMERICAL METHODS USING MATLAB phần 4 pot
... polynomials of degree 1, 3, 5, and 7.
x −3 −2 −1 0 1 2 3
y −0.2774 0 .89 58 −1.5651 3.4565 3.0601 4 .85 68 3 .89 82
We make the MATLAB program “do_polyfit.m”, which uses the routine
“
polyfits()” to ... %LS
subplot(220+i)
plot(x,y,’k*’,xi,yi,’b:’)
end
%xy1.dat
-3.0 -0.2774
-2.0 0 .89 58
-1.0 -1.5651
0.0 3.4565
1.0 3.0601
2.0 4 .85 68
3.0 3 .89 82
Example 3.6. Polynomial Curve Fit by LS...
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, ... t
m
u
s
(t)
m!
s
m+1
(8) cos ωt u
s
(t)
s
s
2
+ ω
2
Applied Numerical Methods Using MATLAB
, by Yang, Cao, Chung, and Morris
Copyr ight
2005 John Wiley & S...
APPLIED NUMERICAL METHODS USING MATLAB phần 3 docx
... interpolation/extrapolation.
Applied Numerical Methods Using MATLAB
, by Yang, Cao, Chung, and Morris
Copyr ight
2005 John Wiley & Sons, I nc., ISBN 0-471-6 983 3-4
117
126 INTERPOLATION AND CURVE FITTING
(a) 4 /8/ 10
th
... -0 .85 52
0.49 98 -0.6456 -0.5774 0 0.2403 0 .85 52 0.5 184
0 .80 90 0.1100 0.5774 0 0
>>err = U*S*V’-A %to check if the result is right
err =...
APPLIED NUMERICAL METHODS USING MATLAB phần 5 pptx
... (3.66e-15)
Flops 2 489 30 28
x
0
= [1 0.5] x
(P4.7.3) ||f(x)||
Flops 1476 382 1
x
0
= [1 0.5] x [0.5024 0.1506]
(P4.7.4) ||f(x)|| 8. 88e-16 (1.18e-6)
Flops 1127 1932
x
0
= [1 0.5] x
(P4.7.5) ||f(x)||
Flops 288 4 ... (4.6576e-15)
Flops 18, 273 21,935
x
0
= [1 1 1] x
(P4 .8. 5) ||f(x)||
Flops 681 1 5525
x
0
= [1 1 1] x [2.0000 1.0000 3.0000]
(P4 .8. 6) ||f(x)|| 3.4659e -8 (2.6130e -8)
Flops...
APPLIED NUMERICAL METHODS USING MATLAB phần 6 docx
... 4.6212e-2 2. 982 2e-2 8. 4103e-2
0.0001 9.4278e-3 9.4277e-3
0.00001 2. 185 3e-1 2. 985 8e-3 8. 4937e-2
(P5.12.1,2)
0.001 1.2393e-5 1.3545e-5
0.0001 8. 3626e-3 5.0315e-6 6. 484 9e-6
0.00001 1. 384 6e-9 8. 8255e-7
(P5.13.1) ... 1.3545e-5
0.0001 8. 3626e-3 5.0315e-6 6. 484 9e-6
0.00001 1. 384 6e-9 8. 8255e-7
(P5.13.1) N/A 8. 881 8e-16 0 8. 881 8e-16
5.12 Surface Area of Revolutionary 3-D...
APPLIED NUMERICAL METHODS USING MATLAB phần 7 doc
... 2.9036] [2 .89 66, 2.9036]
(f
o
=−156.66) (f
o
=−156.66) (f
o
=−156.66) (f
o
=−156.66)
[−0.5,−1.0] [2.9035, −2.74 68] [−2.74 68, −2.74 68] [−2.74 68, −2.74 68] [2.9029, 2.90 28]
(f
o
=−1 28. 39) (f
o
=−100.12) ... 0.5 989
fo_steep = 0. 289 9 %not a minimum
co_steep = -1.27 68
-0.5 989
-1.5 386
0.1525 %violating
-0.0001
Warning: Gradient must be provided
Maximum # of function evaluation...
APPLIED NUMERICAL METHODS USING MATLAB phần 9 docx
... 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 ... 5000 (E9 .8. 1)
with the initial conditions and boundary conditions
u(x, y, 0) = 0fort = 0 (E9 .8. 2a)
u(x, y, t) = e
y
cos x − e
x
cos y for x = 0,x= 4,y= 0,y= 4 (E9 .8. 2b)
The procedu...
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