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How to integrate a modified cfit funciton?

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Tsuwei Tan
Tsuwei Tan el 24 de Dic. de 2017
Comentada: Tsuwei Tan el 27 de Dic. de 2017
Please find the input variables c (sound speed m/s) and z (depth m) at end. Where sound speed is a function of depth c=c(z) I can get the fitted function f by
figure(1)
f=fit(z,c,'smoothingspline');
plot(f,z,c)
xlabel('Depth z ')
ylabel('Spund Speed C(z)')
I can do the integration from 30m to 50m for instance, which is
integrate(f,30,50);
But how can I modify the fitted function f then integrate like
integrate(sqrt(1./f^2-1/1800^2),30,50);
Is there any means to achieve this? Thank you!!!!!
c= [1532.35812332183;1532.09931142412;1531.55663714730;1531.13879204333;1527.17235792085;1524.06757234096;1520.77750440136;1500.98252643213;1494.68127723941;1490.39364113755;1487.89565341648;1487.62521870344;1495.01065240456;1503.30879683038;1503.53436491052];
z= [2.47551587249922;4.48813040298639;6.50074493347357;7.50705219871715;11.0039567548543;13.3491225832363;15.1491225832363;18.8991225832363;22.6491225832363;26.3991225832363;33.8991225832363;41.3991225832363;56.3991225832363;71.3991225832363;78.5991225832363];
  1 comentario
Matt J
Matt J el 24 de Dic. de 2017
Editada: Matt J el 24 de Dic. de 2017
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David Goodmanson
David Goodmanson el 27 de Dic. de 2017
Hi Tsuwei,
Here are two related methods. The first way just does a smoothing spline on the new function that you have defined:
c0 = 1800;
g = fit(z,(1./c.^2 - 1/c0^2),'smoothingspline');
I1 = integrate(g,50,30)
Because of the wonky syntax for the integrate function, the limits 30 and 50 have been reversed so that you get a positive result.
Since sound speed is the primary data, you may feel that it's better to fit sound speed with a smoothing spline rather than some derived quantity. In that case
I2 = integral(@(x) (1./f(x).^2 - 1/c0^2)', 30, 50)
works (note that a transpose sign ' is necessary). Since the original function c is so good, the two methods agree to within a couple of parts per million.
  1 comentario
Tsuwei Tan
Tsuwei Tan el 27 de Dic. de 2017
Thank you David! I used your second method and writes a function as output then I can integrate. I also use Simpson's method to do the integral with a very accurate result. Thank you!

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