How to generate new result points based on known model curve?

Hi All,

i do have an existing curve built based on known x and y.

I want to use the model curve to generate new data points. considering:

x=[10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100]

and y with lower and upper limits of:

y=[109.601, ... , 10.96007]

On model curve point 570 & 26.5 would correspond to 100 & 109.601 while point 138 & 385.1 to 10 & 10.96007

Existing curve based on:

x=[570, 565.2, 560.4, 555.6, 550.8, 546, 541.2, 536.4, 531.6, 526.8, 522, 517.2, 512.4, 507.6, 502.8, 498, 493.2, 488.4, 483.6, 478.8, 474, 469.2, 464.4, 459.6, 454.8, 450, 445.2, 440.4, 435.6, 430.8, 426, 421.2, 416.4, 411.6, 406.8, 402, 397.2, 392.4, 387.6, 382.8, 378, 373.2, 368.4, 363.6, 358.8, 354, 349.2, 344.4, 339.6, 334.8, 330, 325.2, 320.4, 315.6, 310.8, 306, 301.2, 296.4, 291.6, 286.8, 282, 277.2, 272.4, 267.6, 262.8, 258, 253.2, 248.4, 243.6, 238.8, 234, 229.2, 224.4, 219.6, 214.8, 210, 205.2, 200.4, 195.6, 190.8, 186, 181.2, 176.4, 171.6, 166.8, 162, 157.2, 152.4, 147.6, 142.8, 138]
y=[26.5, 36.5, 46.3, 55.9, 65.2, 74.3, 83.2, 91.8, 100.2, 108.4, 116.4, 124.2, 131.8, 139.2, 146.4, 153.5, 160.3, 167, 173.5, 179.9, 186.1, 192.1, 198, 203.7, 209.3, 214.8, 220.1, 225.3, 230.3, 235.2, 240, 244.7, 249.2, 253.7, 258, 262.2, 266.3, 270.3, 274.2, 278.1, 281.8, 285.4, 288.9, 292.4, 295.7, 299, 302.2, 305.3, 308.3, 311.2, 314.1, 316.9, 319.6, 322.3, 324.9, 327.4, 329.9, 332.3, 334.6, 336.9, 339.1, 341.3, 343.4, 345.5, 347.5, 349.4, 351.4, 353.2, 355, 356.8, 358.5, 360.2, 361.8, 363.4, 365, 366.5, 368, 369.4, 370.8, 372.2, 373.5, 374.8, 376.1, 377.3, 378.5, 379.7, 380.9, 382, 383.1, 384.1, 385.1]

Thank you.

Answer 1

[dpb]

x=[570, 565.2, 560.4, 555.6, 550.8, 546, 541.2, 536.4, 531.6, 526.8, 522, 517.2, 512.4, 507.6, 502.8, 498, 493.2, 488.4, 483.6, 478.8, 474, 469.2, 464.4, 459.6, 454.8, 450, 445.2, 440.4, 435.6, 430.8, 426, 421.2, 416.4, 411.6, 406.8, 402, 397.2, 392.4, 387.6, 382.8, 378, 373.2, 368.4, 363.6, 358.8, 354, 349.2, 344.4, 339.6, 334.8, 330, 325.2, 320.4, 315.6, 310.8, 306, 301.2, 296.4, 291.6, 286.8, 282, 277.2, 272.4, 267.6, 262.8, 258, 253.2, 248.4, 243.6, 238.8, 234, 229.2, 224.4, 219.6, 214.8, 210, 205.2, 200.4, 195.6, 190.8, 186, 181.2, 176.4, 171.6, 166.8, 162, 157.2, 152.4, 147.6, 142.8, 138];
y=[26.5, 36.5, 46.3, 55.9, 65.2, 74.3, 83.2, 91.8, 100.2, 108.4, 116.4, 124.2, 131.8, 139.2, 146.4, 153.5, 160.3, 167, 173.5, 179.9, 186.1, 192.1, 198, 203.7, 209.3, 214.8, 220.1, 225.3, 230.3, 235.2, 240, 244.7, 249.2, 253.7, 258, 262.2, 266.3, 270.3, 274.2, 278.1, 281.8, 285.4, 288.9, 292.4, 295.7, 299, 302.2, 305.3, 308.3, 311.2, 314.1, 316.9, 319.6, 322.3, 324.9, 327.4, 329.9, 332.3, 334.6, 336.9, 339.1, 341.3, 343.4, 345.5, 347.5, 349.4, 351.4, 353.2, 355, 356.8, 358.5, 360.2, 361.8, 363.4, 365, 366.5, 368, 369.4, 370.8, 372.2, 373.5, 374.8, 376.1, 377.3, 378.5, 379.7, 380.9, 382, 383.1, 384.1, 385.1];
hAx=subplot(2,1,1);
plot(x,y,'x')
xlabel('X'),ylabel('Y')
ix=find(x==138);            % locate the two target points; find the second
[x(ix) y(ix)]

ans = 1×2

138.0000 385.1000

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hold on
ix=[1;ix];                  % the first is by inspection; same idea as above if not
scatter(x(ix),y(ix),'r','filled')  % show them on plot
%  570 & 26.5 would correspond to 100 & 109.601 while point 138 & 385.1 to 10 & 10.96007
X=[100;10];                 % desired scaled X, Y end points
Y=[109.601;10.96007];
bx=polyfit(x(ix),X,1);      % the linear mapping of x-->X range
by=polyfit(y(ix),Y,1);      % and ditto for y-->Y

xs=polyval(bx,x);           % new x at each previous
ys=polyval(by,y);           % ditto y
hAx(2)=subplot(2,1,2);
plot(xs,ys,'*')             % plot the rescaled results
xlabel('Xscaled'),ylabel('Yscaled')
xlim(polyval(bx,xlim(hAx(1)))), ylim([0 120]), grid on % set xlim to match unscaled to compare

You can fit the transformed data to some interpolating function or just interpolate with the new data arrays over the scaled range; your choice.

The "trick" is to just do a linear transformation from one scale to the other; with the two points one can calculate the two slopes and interecepts by hand, but in MATLB it's simpler to just use the builtin poly function twins...

[John D'Errico]

This is how I would interpret the request. But it is just a wild guess.