how to assign indexes to a function handle?
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Mohammad SD
el 23 de Sept. de 2021
Comentada: Star Strider
el 24 de Sept. de 2021
I'm trying to approximate the double integral of the functions fun_1 and fun_2 over the given region. In the case of one single f it seems ok. But I want to use a vector f insted of a single number to have a vector of TL values. I don't know how to handle this problem. Note that when M >0 and n is a larg number it takes long time to calculate.
The code is:
gamma = 1.4;
R = 286;
T = 273.15;
rho_1 = 1.229;
c_1 = sqrt(gamma*R*T);
rho_2 = rho_1
c_2 = c_1
h = 0.00163;
rho_s = 2750;
M = 0;
m = rho_s*h;
eta = 0.01;
E = 72e9;
v = 0.30;
D = E*h^3/(12*(1-v^2));
f_c1 = c_1^2/(2*pi)*(m/D)^0.5;
f_c2 = c_2^2/(2*pi)*(m/D)^0.5;
n = 1500;
f = linspace(55,7700,n);
for i=1:numel(f)
omega(i) = 2*pi*f(i);
phi_2 = @(phi_1,beta) acos(c_2/c_1.*cos(phi_1).*(1+M.*cos(beta).*cos(phi_1)).^-1);
tau = @(phi_1,beta) ((0.5*(rho_2*c_2/(rho_1*c_1))^0.5...
+0.5*(rho_1*c_1/(rho_2*c_2))^0.5*sin(phi_2(phi_1,beta))./(sin(phi_1).*(1.0...
+M*cos(beta).*cos(phi_1)))+0.5*eta*m*omega(i)*(rho_1*c_1*rho_2*c_2)...
^-0.5.*(f(i)/f_c2).^2.*sin(phi_1).*cos(phi_2(phi_1,beta)).^4).^2.0...
+(0.5*m*omega*sin(phi_2(phi_1,beta))*(rho_1*c_1*rho_2*c_2)^-0.5...
.*(1-(f/f_c2).^2.*cos(phi_2(phi_1,beta)).^4)).^2).^-1;
fun_1 = @(phi_1,beta) tau(phi_1,beta).*sin(phi_1).*cos(phi_1);
q_1(i) = integral2(fun_1,12*pi/180,90*pi/180,0,2*pi);
fun_2 = @(phi_1,beta) sin(phi_1).*cos(phi_1);
q_2 = integral2(fun_2,12*pi/180,90*pi/180,0,2*pi);
tau_avg = q_1/q_2;
TL = -10*log10(tau_avg);
end
2 comentarios
Respuesta aceptada
Star Strider
el 23 de Sept. de 2021
‘But I want to use a vector "f" insted of a single number to have a vector of "TL" values.’
I cannot run that because there are insufficient data provided, and the online Run feature is currently down for scheduled maintenance.
Otherwise, the integral2 function cannot integrate arrays, although integral can. The way to deal with that problem with respect to a double integral is essentially:
f = randn(1, 25);
fcn = @(x,y) sin(2*pi*f.*x) .* exp(0.1*f.*y);
int2 = integral(@(y) integral(@(x) fcn(x,y), 0, 1, 'ArrayValued',1), -1, 0, 'ArrayValued',1);
This returns a vector the size of ‘f’.
.
9 comentarios
Star Strider
el 24 de Sept. de 2021
As always my pleasure!
That paper sounds interesting, however I have no idea what it refers to. It would be interesting to have the PDF file to read.
.
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