sinusoidal elevation wave with time

I want a sinusoidal elevation wave which is (yo) with the time function (t) that looks like the image 1 the function is smaller and get bigger but the resultthat i get is a stabilized like in image2

Here the code that I did

x0=0.05;

omega0=3;

beta1=0.3;

g=9.81;

t=0:0.01:20;

z0=(beta1*g*(omega0^(-2))-1)^0.5;

b0=4*(omega0^(2))*x0*(g*pi)^(-1);

h=1;

x=1;

% Calculation of y0

y0=0;

n = 1;

while true

% Calculation of the dependent term

a0=((omega0^(2))*x0*(x-0.5*pi))*g^(-1);

dependent_term =((((2*n+1)^(-2))*cos((2*n+1)*x)*cosh(z0*(2*n+1))*((cosh((2*n+1)*z0*h)-(((2*n+1)*sinh(2*n+1)*g*z0*h)*((pi^(2)*z0)^(-1))))^(-1)))*sin(omega0*t));

y0=a0*sin(omega0*t)+b0*dependent_term;

% Summing the dependent term

y0 = y0 + dependent_term;

% Breaking the loop if the dependent term becomes negligible

if abs(dependent_term) < 1e-6

break;

end

n = n + 1;

end

% Display the result Z

plot(t,y0);

% ylim([-0.05 0.1])

[Mathieu NOE]

hello

I have not the possibilty to double check if there is a mistake in the code , but there are some probability that it's the case ...

first thing that looked strange to me is that "dependent term" appears twice in your code , in very different ways , so can you confirm this is what you really want to do ?

% Calculation of the dependent term 
a0=((omega0^(2))*x0*(x-0.5*pi))*g^(-1);
dependent_term =((((2*n+1)^(-2))*cos((2*n+1)*x)*cosh(z0*(2*n+1))*((cosh((2*n+1)*z0*h)-(((2*n+1)*sinh(2*n+1)*g*z0*h)*((pi^(2)*z0)^(-1))))^(-1)))*sin(omega0*t));
y0=a0*sin(omega0*t)+b0*dependent_term; % first summation HERE
% Summing the dependent term 
y0 = y0 + dependent_term;  % second summation HERE

[Mathieu NOE]

then in both lines when you add dependent_term to another variable, and wanted to know what is the realtive weight of both terms that you add

for example , after this line : y0=a0*sin(omega0*t)+b0*dependent_term;

I define the first ratio as : ratio1 = max(abs(b0*dependent_term))./max(abs(a0*sin(omega0*t)));

same , for this second line : y0 = y0 + dependent_term;

I define the second ratio as : ratio2 = max(abs(dependent_term))./max(abs(y0));

so if these ratios are very low, we know that the second term you add in these two lines is probably wrong or not with the right amplitude.

and if we look at the results (see last two figures) , we can actually see that those two ratio values are very low , so what at the end , your y0 is basically only dominated by the sinusoidal term (in bold) : y0=a0*sin(omega0*t)+b0*dependent_term;

code (with ratios display)

x0=0.05;
omega0=3;
beta1=0.3;
g=9.81;
t=0:0.01:20;
z0=(beta1*g*(omega0^(-2))-1)^0.5;
b0=4*(omega0^(2))*x0*(g*pi)^(-1);
h=1;
x=1;
% Calculation of y0
y0=0;
n = 1; 
ratio1(n) = 0;
ratio2(n) = 0;

while true 
    % Calculation of the dependent term 
    a0=((omega0^(2))*x0*(x-0.5*pi))*g^(-1);
    dependent_term =((((2*n+1)^(-2))*cos((2*n+1)*x)*cosh(z0*(2*n+1))*((cosh((2*n+1)*z0*h)-(((2*n+1)*sinh(2*n+1)*g*z0*h)*((pi^(2)*z0)^(-1))))^(-1)))*sin(omega0*t));
    y0=a0*sin(omega0*t)+b0*dependent_term;
  
    ratio1(n) = max(abs(b0*dependent_term))./max(abs(a0*sin(omega0*t)));
  
 % Summing the dependent term 
    y0 = y0 + dependent_term; 
    
    ratio2(n) = max(abs(dependent_term))./max(abs(y0));

    % Breaking the loop if the dependent term becomes negligible 
    if abs(dependent_term) < 1e-6 
       break; 
    end 
    n = n + 1; 
end 
% Display the result Z
plot(t,y0);

% ylim([-0.05 0.1])

figure,semilogy(ratio1)

figure,semilogy(ratio2)

[VBBV]

Hi @MOUNIBA REDAH, Are these problems similar ? The below question

https://www.mathworks.com/matlabcentral/answers/2053667-how-to-code-a-sum-of-function-in-matlab?s_tid=prof_contriblnk

clear
el= 0.3;
L= 0.57 ;
h= 0.15;
t=0:0.5:50;
x=0.02;
omega1=6.0578;
omega=1.1*omega1;
g=9.81;


for k = 1:length(t) 
 K(k) = ((2*k-1)/L)*pi;   
 omegan(k) = sqrt(g*K(k)*tanh(K(k)*h));
 D(k)=(4*omega*(-1)^k-1)/(L*cosh(K(k)*h)*K(k)^2);
 C(k)=(omega*D(k)*pi)/(omegan(k)^2-(omega)^2);
 A(k)=-C(k) - (D(k)/omega);

 z(k,:)=(1/g)*sin(K(k)*(x-(L/2)))*cosh(K(k)*h)*(-A(k)*omegan(k)*sin(omegan(k)*t)+C(k)*omega*sin(omega*t))-(1/g)*A(k)*omega*(x-L/2)*sin(omega*t) ;
end

plot(t,sum(z));grid minor

Answer 1

[Fangjun Jiang]

Maybe this example gives you an idea?

t=0:0.01:20;
f1=1;
y1=sin(2*pi*f1*t);
f2=0.05;
y2=sin(2*pi*f2*t);
y0=y1.*y2;
figure(1);plot(t,y0);

% figure(2)
% subplot(1,3,1); plot(t,y1);
% subplot(1,3,2); plot(t,y2);
% subplot(1,3,3); plot(t,y0);