05 Interpolation and Fitting
Xiaolangdi water and sediment regulation issues
data3.txt
1800 1900 2100 2200 2300 2400 2500 2600 2650 2700 2720 2650 32 60 75 85 90 98 100 102 108 112 115 116 2600 2500 2300 2200 2000 1850 1820 1800 1750 1500 1000 900 118 120 118 105 80 60 50 30 26 20 8 5
Interpolation
One problem% % v: water flow S: sediment concentration; V:? Sand amount % assumed water flow and sediment concentration are continuous, a certain amount of time Sand V = v (t) S ( t) % first known water flow and sediment concentration some time, given the estimated amount of sediment at any time and the total sediment volume % of the total amount of sediment is the amount of sediment to do integration % of the time 8: 00-20: 00 = 28800 T1, T2 = T24 = 1.0224 million the format Compact; CLC, Clear; Load data3.txt Liu DATA3 = ([l, 3], :); Liu = Liu '; Liu = Liu (:);% and made water flow order column vectors into sha = data3 ([2,4], :); sha = sha '; sha = sha (:);% silt content and presented in order into a column vector y = sha * liu.; y = y ';% Sand calculated amount into a row vector I =. 1: 24; T = (I-12 is *. 4) 3600 *; T1 = T (. 1); T2 = T (End); % interpolation pp = csape (t, y); % cubic spline interpolation xsh = pp.coefs% calculated interpolation polynomial coefficient matrix, each row is a section on the coefficients of the polynomial TL = quadl (@ (tt) fnval (pp, tt ), t1, t2)% of the total required amount of integral calculation Sand % visual display interpolation polynomial and the old value t0 = t1: 0.1: t2; y0 = fnval (pp, t0) ; plot(t,y,'+',t0,y0)
xsh =
1.0e + 05 *
-0.0000 -0.0000 0.00008 0.5760
-0.0000 -0.0000 0.00008 1.1400
-0.0000 -0.0000 0.00008 1.5750
0.00008 -0.0000 0.00008 1.8700
-0.0000 0.00008 0.00008 2.0700
0.00008 -0.0000 0.00008 2.3520
0.00008 0.00008 0.00008 2.5000
-0.0000 0.00008 0.00008 2.6520
0.00008 -0.0000 0.00008 2.8620
-0.0000 0.00008 0.00008 3.0240
0.00008 -0.0000 0.00008 3.1280
-0.0000 0.00008 -0.0000 3.0740
-0.0000 -0.0000 0.00008 3.0680
0.00008 -0.0000 -0.0000 3.0000
-0.0000 0.00008 -0.0000 2.7140
0.0000 -0.0000 -0.0000 2.3100
0.0000 0.0000 -0.0000 1.6000
-0.0000 0.0000 -0.0000 1.1100
0.0000 -0.0000 -0.0000 0.9100
-0.0000 0.0000 -0.0000 0.5400
0.0000 -0.0000 -0.0000 0.4550
0.0000 -0.0000 -0.0000 0.3000
0.0000 0.0000 -0.0000 0.0800
TL =
1.8440e+11
Fit
% Second problem: determine the relationship between water flow and the amount of Sediment
the format Compact;
% Sediment volume and draw water flow scattergram
CLC, Clear;
Load data3.txt
Liu DATA3 = ([l, 3], :); liu = liu '; liu = liu (:);% is proposed in order to become the water flow and the column vector
sha = data3 ([2,4], :); sha = sha'; sha = sha (:);% proposed sediment concentration and order into a column vector
y = sha * liu;.% calculated amount desilting, there is a column vector
subplot (1,2,1), plot (liu (1:11), y (1:11 ), '*')
the subplot (1,2,2), Plot (Liu (12:24), Y (12:24), '*')
% of the first stage is substantially linear
% of the first stage and are prepared by a two-stage quadratic curve fit and
the remaining standards% which model to select a small difference which model
the format Long E
% or less of the first stage fitting
for J =. 1: 2
nihe1 polyfit {J} = (Liu (1:11), y (1:11) , j);% fit polynomial coefficients arranged from a low power to higher power
yhat1 {j} = polyval (nihe1 {j}, liu (1:11)) ;% determine a predicted value
cha1 (j) = sum (( y (1:11) -yhat1 {j}) ^ 2.);% seeking squared error
rmse1 (j) = sqrt (cha1 (j) / (10-j));% residual standard deviation request
End
celldisp (nihe1)% all elements of the array cells
rmse1
% or less of the second phase of fitting
for j = 1 : 2
nihe2 polyfit {J} = (Liu (12:24), Y (12:24), (J));% using cell array
yhat2 {j} = polyval (nihe2 {j}, liu (12:24) );% determine a predicted value
cha2 (j) = sum (( y (12:24) -yhat2 {j}) ^ 2);.% seek error squared and
rmse2 (j) = sqrt (cha2 (j) / (11 -j));% residual standard deviation request
End
celldisp (nihe2)% all elements of the array cells
rmse2
the format% recovery default number display format short
% final result
% a: Y = 250.5655v-373,384.4661
% two: y = 0.167 v * 2-180.4668v + 72421.0982
