Anyone can help me. Could you please check my code with attach file for fig 8 page 74 by using equation 2.30 page 67.
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syms k1 k2
eta0 = 1; % Maximum displacement(m)
L = 100; % Propagate length (km)
W = 100; % Source width (km)
v = 0.14; % Ruptuer velocity to obtain maximum surface amplitude(km/s)
h = 2; % Water depth (km)
g = 0.0098; % Acceleration due to gravity (km/s^2)
t1 = 50/v; % (s)
t = (t1); % (s)
k = sqrt(k1.^2+k2.^2);
w = sqrt(g.*k.*tanh(k.*h));
% A = ((sin(w*t))./(w.*cosh(k*h))).*((eta0*v)./(2*L));
A = ((cos(w*t))./(cosh(k*h))).*((eta0*v)./(2*L));
B = (1-exp(-1i*100*k1))./(1i*k1);
C = ((exp(-1i*100*k1))./(1-((50*k1)./pi).^2)).*(1i*k1).*((50./pi).^2).*((exp(1i*100*k1))-1);
D = (exp(1i*150*k2)-exp(1i*50*k2))./(1i*k2);
E = (1./(1-(((100*k2)./pi).^2))).*(1i*k2).*((100./pi).^2).*(exp(1i*50*k2)+exp(1i*150*k2));
F = (4.*sin(50*k2))./k2;
G = (exp(-1i*50*k2)-exp(-1i*150*k2))./(1i*k2);
H = (1./(1-(((100*k2)./pi).^2))).*(1i*k2).*((100./pi).^2).*(exp(-1i*150*k2)+exp(-1i*50*k2));
K1 = 0:100;
K2 = -150:150;
for x1 = 1:101;
for x2 = 1:301;
k1 = K1(x1);
k2 = K2(x2);
if (k==0)
eta_(x1,x2)= 200*eta0*v*t;
else
eta_(x1,x2)= (A.*(B-C)).*((D-E)+F+(G+H));
end
end
end
Zsym = subs(eta_);
Z = double(Zsym);
ETA = ifft2(Z)
mesh(abs(ETA))
figure(2)
mesh(real(ETA))
figure(3)
mesh(imag(ETA))
% M = 101;
% N = 301;
% Zsym = subs(eta_);
% Z = double(Zsym);
% ETA = abs(ifft2(Z));
% I = -M/2:1:M/2-1;
% J = -N/2:1:N/2-1;
% mesh(I,J,ETA)
% % X = abs(ifft2(Z,K1,K2));
% % X = ifftshift(X);
% % % F = [-K1/2:K1/2-1]/K1;
% % plot(F,X)
in this journal fig 8
1 Comment
darova
on 17 Mar 2021
That's too difficult. Pictures you attached are look the same. Why do you need someone to check it?
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