linear interpolation of function handle
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Hi everybody, I need to calculate the value in the range pi <= wave_angle < pi/2 of the a function handle @(freq_wave,wave_angle) which is defined as:
((log(B/T))^-1).*(0.87/CB).^((1+Fn).*cos(wave_angle)).*(1/3-(2/3).*cos(wave_angle))
in the range ((wave_angle >=0) & (wave_angle <= pi/2))
((log(B/T))^-1).*(0.87/CB).^(1+Fn) in the range (wave_angle=pi)&(vel_nave>V_G(freq_wave)./2)&(Fn_rel(freq_wave)>=0.12)
((log(B/T))^-1).*(0.87/CB) in the range (wave_angle=pi)&(vel_nave<V_G(freq_wave)./2)+ ((log(B/T))^-1).*(0.87/CB) .*(wave_angle=pi)&(Fn_rel(freq_wave)<0.12);
but I don't konw which function is the best to use. Thanks in advance for your help.
7 Commenti
Matt J
il 29 Nov 2023
Since you have a continuous formula for the function, there is nothing to interpolate. Just evaluate it at the locations you want.
Gianluca Angelini
il 29 Nov 2023
Torsten
il 29 Nov 2023
Please supply a complete code where all the variables and functions are defined.
In your definition of a1 from above, B, T, CB, Fn, vel_nave, function V_G, function Fn_rel are missing.
Gianluca Angelini
il 29 Nov 2023
As you can see, none of the four conditions is satisfied for your inputs. Thus the result is logical 0.
(Note that I added additional brackets around your conditions).
Extrapolating values for a1 over the limits where a1 is defined is completely arbitrary and thus nonsense.
B=22.5; T=7; CB=0.59; Fn=0.2; vel_nave=10.28; g=1;Lpp=1;
V_G=@(freq_wave) g/(2*freq_wave);
Fn_rel=@(freq_wave) (vel_nave-V_G(freq_wave)/2)/(g*Lpp)^0.5;
a1=@(freq_wave,wave_angle) ((log(B/T))^-1).*(0.87/CB).^((1+Fn).*cos(wave_angle)).*(1/3-(2/3).*cos(wave_angle)).*((wave_angle >=0) & (wave_angle <= pi/2)) +...
((log(B/T))^-1).*(0.87/CB).^(1+Fn) .*((wave_angle==pi)&(vel_nave>V_G(freq_wave)./2)&(Fn_rel(freq_wave)>=0.12)) +...
((log(B/T))^-1).*(0.87/CB) .*((wave_angle==pi)&(vel_nave<V_G(freq_wave)./2))+...
((log(B/T))^-1).*(0.87/CB) .*((wave_angle==pi)&(Fn_rel(freq_wave)<0.12));
freq_wave = 5;
wave_angle = 2.09;
((wave_angle >=0) & (wave_angle <= pi/2))
((wave_angle==pi)&(vel_nave>V_G(freq_wave)./2)&(Fn_rel(freq_wave)>=0.12))
((wave_angle==pi)&(vel_nave<V_G(freq_wave)./2))
((wave_angle==pi)&(Fn_rel(freq_wave)<0.12))
a1(5,2.09)
Gianluca Angelini
il 29 Nov 2023
Risposte (1)
Nipun
il 21 Dic 2023
Hi Gianluca,
I understand that you are trying to evaluate a semantic relation, given the variable values. In the illustrated case, the equation to be used changes according to the range of the input.
It seems like you have a piecewise function defined over different ranges of wave_angle. In MATLAB, you can use the if, elseif, and else statements to define a piecewise function. The if statements can be used to check the conditions for each range and evaluate the corresponding expression.
Here is a code snippet to help you with the same:
function result = your_function(freq_wave, wave_angle, B, T, CB, Fn, vel_nave)
if wave_angle >= 0 && wave_angle < pi/2
result = ((log(B/T))^-1) .* (0.87/CB).^((1+Fn).*cos(wave_angle)) .* (1/3 - (2/3) .* cos(wave_angle));
elseif wave_angle == pi && vel_nave > V_G(freq_wave)/2 && Fn_rel(freq_wave) >= 0.12
result = ((log(B/T))^-1) .* (0.87/CB).^(1+Fn);
elseif wave_angle == pi && vel_nave < V_G(freq_wave)/2
result = ((log(B/T))^-1) .* (0.87/CB);
elseif wave_angle == pi && Fn_rel(freq_wave) < 0.12
result = ((log(B/T))^-1) .* (0.87/CB) .* (wave_angle == pi) .* (Fn_rel(freq_wave) < 0.12);
else
% Define default behavior if none of the conditions are met
result = NaN; % or any default value you prefer
end
end
This function checks the conditions for each range of wave_angle and evaluates the corresponding expression. You can adjust the conditions and expressions based on your specific requirements. The result variable holds the final value of the function for a given set of input parameters.
Since the signature and description of functions "Fn_rel" and others were not provided, I assume that they work as expected.
Hope this helps.
Regards,
Nipun
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