Getting blank plot in MATLAB

Hi, could I know where I might be wrong in the following code as Im getting blank graph. The parameter vel and acc are time dependent which in turn Fh is function of time.

function [Fh] = hydro(t)

Cd = 0.6;
Ca = 0.9699;
R = 4.7;

for t=0:0.1:10
    vel = compute_wavevelocity(t)
    acc = compute_waveacceleration(t)
    fun = sym((0.5*997*Cd*2*R*abs(vel)*vel)+(Ca*997*pi*(R^2)*acc)+(pi*(R^2)*997*acc));
    Fh(t==0:0.1:10) = eval(int(fun,0,-120));
end
plot(t,Fh), xlabel('time(s)'), ylabel('Fh')
title ('Fh vs Time')
end

[madhan ravi]

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval Absolutely not necessary.

[Satish Jawalageri]

Okay, thanks. I will check it.

[Image Analyst]

Do you really even need symbolics? Or could you do it numerically, something like this:

Cd = 0.6;
Ca = 0.9699;
R = 4.7;
all_t=0:0.1:10
for k = 1 : length(all_t)
    t = all_t(k);
    vel = compute_wavevelocity(t)
    acc = compute_waveacceleration(t)
    fun = (0.5*997*Cd*2*R*abs(vel)*vel)+(Ca*997*pi*(R^2)*acc)+(pi*(R^2)*997*acc);
    Fh(k) = fun;
end
plot(all_t,Fh), xlabel('time(s)'), ylabel('Fh')
title ('Fh vs Time')

[Satish Jawalageri]

Thanks for your reply.

I will check this but here Fh is integration of fun.

Answer 1

[Cris LaPierre]

This is because t is a scalar. When you plot, it has the value it was assigned on the last loop of the for loop: 10.

Try changing the x input in your plot command:

plot(0:0.1:10,Fh)

[Satish Jawalageri]

Oh yeah! Perfect.

Thanks.

And one more, I am getting zero value in vel and acc corresponding to few values of t (which is not correct correct in this case). Could I know what might be the issue?

Thanks.

[Cris LaPierre]

Check your function for errors? With what you have provided, I can't help.

[Satish Jawalageri]

Oky, no worries. Thanks.

[madhan ravi]

Huh? Have you shown what compute_velocity & compute_waveacceleration function is? Are we supposed to read your mind?

[Satish Jawalageri]

Sorry! Code for both velocity and acceleration.

Velocity:

function velocity = compute_wavevelocity(t)
       T = 10;
       WN = (9.8*(T^2))/(2*pi());
       k = (2*pi())/WN;
       omega = (2*pi())/T;
       y = [320;310;300;290;280;270;260;250;240;230;220;210;200];
       d = 320;
       H = 4;
       vel = (omega*H/2)* ((cosh(k*y))/(sinh(k*d)))* cos((k*0)-(omega*t));
       
       velocity = mean(vel,1);
       
end

Acceleration:

function acceleration = compute_waveacceleration(t)
        T = 10;
       WN = (9.8*(T^2))/(2*pi());
       k = (2*pi())/WN;
       omega = (2*pi())/T;
       y = [320;310;300;290;280;270;260;250;240;230;220;210;200];
       d = 320;
       H = 4;
        acc = ((omega^2)*H/2)* ((cosh(k*y))/(sinh(k*d)))* sin((k*0)-(omega*t));
        
        acceleration = mean(acc,1);
               
    end

[Cris LaPierre]

It is cyclical, occurring every 2.5 seconds. It appears to be related to your expression for vel and acc. Specifically, cos((k*0)-(omega*t)) and sin((k*0)-(omega*t)). Since k*0 is always zero, and since sin and cos return a value of 0 in increments spaced π apart, I'd say that the result you are observing is to be expected based on the code you have shared.

Check that your equation is correct.

[Satish Jawalageri]

Thanks for your reply.

I got that correct, but when I run hydro(t) I am getting proper values of Fh till half time (ie here till 5sec) and then, few values get to zero, i dont know why. Further, I checked fun values which are correct for given time. So Fh(t==0:0.1:10) = eval(int(fun,0,-120)); Is it correct to mention like this?

[Satish Jawalageri]

No, it is the same.

I have attached the plot which i got when run the code.

[Cris LaPierre]

This appears to be due to how you are assigning your values to Fh in your for loop. It's likely taking too long, so the result is some values get skipped. Since this is an unconventional way to do it anyway, I suggest updating your code to the following.

function Fh = hydro(t)

Cd = 0.6;
Ca = 0.9699;
R = 4.7;
t=0:0.1:10;

for i = 1:length(t)
    vel = compute_wavevelocity(t(i));
    acc = compute_waveacceleration(t(i));
    fun = sym((0.5*997*Cd*2*R*abs(vel)*vel)+(Ca*997*pi*(R^2)*acc)+(pi*(R^2)*997*acc));
    Fh(i,1) = double(int(fun,0,-120));
end
plot(t,Fh), xlabel('time(s)'), ylabel('Fh')
title ('Fh vs Time')

Let me anticipate the next issue.

You have hardcoded your function to run for 10 seconds, meaning that it does not matter what value you call hydro with. You will always get the result for 10 seconds.

[Satish Jawalageri]

Thank you very much. I got it.

[Satish Jawalageri]

Could I know how to run this state space model and ode45 function and Is it correct what I have done. Here, F1 and F2 are fuction of Fh.

State space model: (Output.m)

function [sdot] = Output(t,x)

Ze = zeros(3);
I = eye(3);

M = [8070000,0,-629468070;0,8070000,112980;-629468070,112980,6.800000000000000e+10];
Ad = [8.053095218400001e+06,0,-4.831857131040000e+08;0,2.167940435676214e+05,0;-4.831857131040000e+08,0,3.865485704832000e+10];
C = [0,0,0;0,3.241885080000000e+05,0;0,0,1.301151158327999e+09];
Cm = [4.12e+04,0,-2.82e+06;0,1.19e+04,0;-2.82e6,0,3.11e+08];
MA = M+Ad;
Q = C+Cm;


Fwt = 2.793977550000000e+04;    %Wind force on turbine tower

Fg = -79086000;                 %Gravitational force
Fbuoy = 7.844814740000000e+07;  %Buoyancy force
Fp = 2.712318560000001e+06;     %Heave force
Fmx = -334731.8545;             
Fmz = -3517000;                 

t = 0:0.1:10

for i = 1:length(t)
    
    Fh = hydro(t(i))
    F1(1,i) = Fmx+27939.6+6.5*10^5+Fh(1,i);
    F2(1,i) = Fmz+Fg+Fbuoy+Fp;
    F3(1,i) = -112510430.2+3.44*10^6+266.5*10^5+(Fh(1,i)*18);

 A = [Ze, I; -inv(MA)*Q, Ze];
 B = [Ze; inv(MA)];
 F = [F1;F2;F3];
  
 Svec = [x(1);x(2);x(3);x(4);x(5);x(6)];
% 
 sdot(1,i) = A*Svec + B*F;
end

end 

ODE45:

%Initial Condition 
x(10) = 0; x(20) = 0; x(30) = 0;
x(40) = 0; x(50) = 0; x(60) = 0;

IC = [x(10);x(20);x(30);x(40);x(50);x(60)];

%time span
t0 = 0; tf = 10;
tspan = t0:0.1:tf;

[t, x] = ode45(@Output,tspan,IC);
IC = [IC;x(end)];
%x(1) = x(:,1);
%x(2) = x(:,2);
%x(3) = x(:,3);
%x(4) = x(:,4);
%x(5) = x(:,5);
%x(6) = x(:,6);

%Plot Results 
figure(1), clf 
plot(t,x(:,1)), xlabel('time(s)'), ylabel('surge(m)')
title ('Surge vs Time')

[Cris LaPierre]

These appear to be new questions. I suggest creating a new post for each.

[Satish Jawalageri]

Sure, I will.

Thanks.