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How to plot for this code to obtain the figure? Please!

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soe min aung
soe min aung on 16 Jun 2020
Commented: soe min aung on 16 Jun 2020
clc
clear all
close all
t2 = 11.9;
eta = @(X,Y) (2^(1/2)*(- 200*pi*(- (42*pi*2^(1/2)*t2)/5 + (2*pi*sin((21*2^(1/2)*t2*(pi^2/2500)^(1/4)*(-(exp(-2*(pi^2/2500)^(1/2))/2 - exp(2*(pi^2/2500)^(1/2))/2)/(exp(-2*(pi^2/2500)^(1/2))/2 + exp(2*(pi^2/2500)^(1/2))/2))^(1/2))/5)*exp(2*(pi^2/2500)^(1/2))*exp(-(pi*X*1i)/50))/((pi^2/2500)^(1/4)*(exp(4*(pi^2/2500)^(1/2)) + 1)*(-(exp(-2*(pi^2/2500)^(1/2))/2 - exp(2*(pi^2/2500)^(1/2))/2)/(exp(-2*(pi^2/2500)^(1/2))/2 + exp(2*(pi^2/2500)^(1/2))/2))^(1/2)) + (2*pi*sin((21*2^(1/2)*t2*(pi^2/2500)^(1/4)*(-(exp(-2*(pi^2/2500)^(1/2))/2 - exp(2*(pi^2/2500)^(1/2))/2)/(exp(-2*(pi^2/2500)^(1/2))/2 + exp(2*(pi^2/2500)^(1/2))/2))^(1/2))/5)*exp(2*(pi^2/2500)^(1/2))*exp((pi*X*1i)/50))/((pi^2/2500)^(1/4)*(exp(4*(pi^2/2500)^(1/2)) + 1)*(-(exp(-2*(pi^2/2500)^(1/2))/2 - exp(2*(pi^2/2500)^(1/2))/2)/(exp(-2*(pi^2/2500)^(1/2))/2 + exp(2*(pi^2/2500)^(1/2))/2))^(1/2))) + (2*pi^(3/2)*exp(-(pi*X*1i)/50)*exp(-pi*Y*1i)*(pi^(1/2)*tanh(2*pi)^(1/2)*exp(2*((2501*pi^2)/2500)^(1/2))*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) + pi^(1/2)*tanh(2*pi)^(1/2)*exp(4*pi)*exp(2*((2501*pi^2)/2500)^(1/2))*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) - 2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*exp(2*pi)*exp((pi*X*1i)/50)*sin((21*2^(1/2)*t2*pi^(1/2)*tanh(2*pi)^(1/2))/5)*((2501*pi^2)/2500)^(1/4) + pi^(1/2)*tanh(2*pi)^(1/2)*exp(2*((2501*pi^2)/2500)^(1/2))*exp((pi*X*1i)/25)*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) + pi^(1/2)*tanh(2*pi)^(1/2)*exp(4*pi)*exp(2*((2501*pi^2)/2500)^(1/2))*exp((pi*X*1i)/25)*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) - 2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*exp(2*pi)*exp(4*((2501*pi^2)/2500)^(1/2))*exp((pi*X*1i)/50)*sin((21*2^(1/2)*t2*pi^(1/2)*tanh(2*pi)^(1/2))/5)*((2501*pi^2)/2500)^(1/4)))/(tanh(2*pi)^(1/2)*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4)*(exp(4*pi) + 1)*(exp(4*((2501*pi^2)/2500)^(1/2)) + 1)) + (2*pi^(3/2)*exp(-(pi*X*1i)/50)*exp(pi*Y*1i)*(pi^(1/2)*tanh(2*pi)^(1/2)*exp(2*((2501*pi^2)/2500)^(1/2))*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) + pi^(1/2)*tanh(2*pi)^(1/2)*exp(4*pi)*exp(2*((2501*pi^2)/2500)^(1/2))*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) - 2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*exp(2*pi)*exp((pi*X*1i)/50)*sin((21*2^(1/2)*t2*pi^(1/2)*tanh(2*pi)^(1/2))/5)*((2501*pi^2)/2500)^(1/4) + pi^(1/2)*tanh(2*pi)^(1/2)*exp(2*((2501*pi^2)/2500)^(1/2))*exp((pi*X*1i)/25)*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) + pi^(1/2)*tanh(2*pi)^(1/2)*exp(4*pi)*exp(2*((2501*pi^2)/2500)^(1/2))*exp((pi*X*1i)/25)*sin((21*2^(1/2)*t2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4))/5) - 2*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*exp(2*pi)*exp(4*((2501*pi^2)/2500)^(1/2))*exp((pi*X*1i)/50)*sin((21*2^(1/2)*t2*pi^(1/2)*tanh(2*pi)^(1/2))/5)*((2501*pi^2)/2500)^(1/4)))/(tanh(2*pi)^(1/2)*tanh(2*((2501*pi^2)/2500)^(1/2))^(1/2)*((2501*pi^2)/2500)^(1/4)*(exp(4*pi) + 1)*(exp(4*((2501*pi^2)/2500)^(1/2)) + 1))))/(160000*pi^2);
eta2 = @(X,Y)((0<=X) & (X<=100) & (-150<=Y) & (Y<=-50)).*eta(X,Y)
% figure
[X1,Y1]=ndgrid(-200:5:200);
surf(X1,Y1,eta2(X1,Y1))
grid on

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Accepted Answer

KSSV
KSSV on 16 Jun 2020
Edited: KSSV on 16 Jun 2020
Z1 = eta2(X1,Y1) ;
Z1 is a complex matrix. USe either real, imag, abs to use surf.
surf(X,Y1,abs(Z1))

  1 Comment

soe min aung
soe min aung on 16 Jun 2020
Thank you so much sir. Could you please give me you email address, I want to ask you about modeling of tsunami generation and propagation. Please help me sir.

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