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how to assign indexes to a function handle?

조회 수: 48(최근 30일)
Mohammad SD
Mohammad SD 2021년 9월 23일 14:48
댓글: Star Strider 2021년 9월 24일 10:50
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
  댓글 수: 2
Mohammad SD
Mohammad SD 2021년 9월 23일 17:19
@Star Strider Sorry! n = 1500. any number is OK

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Star Strider
Star Strider 2021년 9월 23일 16:03
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’.
.
  댓글 수: 9
Star Strider
Star Strider 2021년 9월 24일 10:50
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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