sinusoidal elevation wave with time

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MOUNIBA REDAH
MOUNIBA REDAH on 25 Mar 2024
Edited: VBBV on 26 Mar 2024
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])
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Mathieu NOE
Mathieu NOE on 25 Mar 2024
Ran in:
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
VBBV on 26 Mar 2024
Edited: VBBV on 26 Mar 2024
Ran in:
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

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Answers (1)

Fangjun Jiang
Fangjun Jiang on 25 Mar 2024
Ran in:
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);

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