Getting blank plot in MATLAB

2020 7 14
1 个回答
更新时间：2020 7 15
3 次查看（30 天）

0 个投票

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

4 个评论
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Satish Jawalageri
Satish Jawalageri 2020-7-14
Okay, thanks. I will check it.
Image Analyst
Image Analyst 2020-7-14
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
Satish Jawalageri 2020-7-15
Thanks for your reply.
I will check this but here Fh is integration of fun.

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回答（1 个）

Cris LaPierre
Cris LaPierre 2020-7-14

0 个投票

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)

13 个评论
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Satish Jawalageri
Satish Jawalageri 2020-7-14
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
Cris LaPierre 2020-7-14
Check your function for errors? With what you have provided, I can't help.
Satish Jawalageri
Satish Jawalageri 2020-7-14
Oky, no worries. Thanks.
madhan ravi
madhan ravi 2020-7-14
编辑：madhan ravi 2020-7-14
Huh? Have you shown what compute_velocity & compute_waveacceleration function is? Are we supposed to read your mind?
Satish Jawalageri
Satish Jawalageri 2020-7-14
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
Cris LaPierre 2020-7-14
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
Satish Jawalageri 2020-7-14
编辑：Satish Jawalageri 2020-7-14
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
Satish Jawalageri 2020-7-14
No, it is the same.
I have attached the plot which i got when run the code.
Cris LaPierre
Cris LaPierre 2020-7-14
编辑：Cris LaPierre 2020-7-14
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
Satish Jawalageri 2020-7-15
Thank you very much. I got it.
Satish Jawalageri
Satish Jawalageri 2020-7-15
编辑：Satish Jawalageri 2020-7-15
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
Cris LaPierre 2020-7-15
These appear to be new questions. I suggest creating a new post for each.
Satish Jawalageri
Satish Jawalageri 2020-7-15
Sure, I will.
Thanks.

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