Solving Trigonometric Function Equations with Time varying Parameters by MATLAB
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Hello, my goal is to solve the equations about cos (x) and sin (x), plot the solution results into a curve, and the Lac in the equation will change with time.
I tried to use fsolve to solve the equation, but the result is only a fixed value, not a value that changes with time in theory. Here is my code. I look forward to receiving your guidance
%The function I created is as follows
function q=myfun(p)
x=p;
t=0:0.001:5;
Lac=0.91+0.01*sin(t);
q=2 * ( 0.0822*cos(x) - 0.2838*sin(x)) + Lac.^2 - 0.9209;
end
%Here is the command that I call the function
[x]=fsolve(@myfun,[0])
采纳的回答
You were almost there
t = (0:0.001:5)';
x = fsolve(@(x)myfun(x,t),zeros(size(t)));
plot(t,x)
xlabel('t')
ylabel('x')
function q = myfun(x,t)
Lac = 0.91+0.01*sin(t);
q = 2*(0.0822*cos(x)-0.2838*sin(x))+Lac.^2-0.9209;
end
5 个评论
LIULIYAN
2022-10-10
Thank you very much for your reply. Your answer is very helpful to me. May I ask you one more question.
Lac is not 0.91+0.01 * sin (t), but a variable solved by SIMULINK model (see figure). My ultimate goal is to calculate radian x in real time in SIMULINK according to the real-time changes of Lac.
I tried to use the code you provided to establish the S-function function ("Positivesolution" in the figure). The code is as follows. I found that the running speed was very slow. I ran SIMULINK for 30S (Fixed step size=0.001), but it took several decades to complete the execution, which obviously did not meet the real-time requirements.
Please ask whether the following code can be optimized to improve the real-time performance of SIMULINK？
Or, according to the method in the initial question, how to use the data in simout1.data to assign values to Lac to calculate radian x offline？
Look forward to your guidance, thank you again
function[sys,x0,str,ts] = Positivesolution(t,x,u,flag)
switch flag,
case 0,
[sys,x0,str,ts]=mdlInitializeSizes;
case 1,
sys=mdlDerivatives(t,x,u);
case 3,
sys=mdlOutputs(t,x,u);
case {2,4,9}
sys=[];
otherwise
error(['Unhandled flag=',num2str(flag)]);
end
function[sys,x0,str,ts]=mdlInitializeSizes
global cij bj c
sizes = simsizes;
sizes.NumContStates=1;
sizes.NumDiscStates=0;
sizes.NumOutputs=1;
sizes.NumInputs=1;
sizes.DirFeedthrough=1;
sizes.NumSampleTimes=0;
sys=simsizes(sizes);
x0 = [0.873];
str = [];
ts = [];
function sys=mdlDerivatives(t,x,u)
sys(1)=x(1);
function sys=mdlOutputs(t,x,u)
Lac=u(1);
%q=2 * ( 0.0822*cos(x) - 0.2838*sin(x)) + Lac.^2 - 0.9209;
x = fsolve(@(x)2 * ( 0.0822*cos(x) - 0.2838*sin(x)) + Lac.^2 - 0.9209,zeros);
sys(1)=x(1);
Davide Masiello
2022-10-10
编辑：Davide Masiello
2022-10-10
If I understand correctly, you want to change the code so that Lac is not a defined function but an array of values calculated in some other code block before. Am I right?
LIULIYAN
2022-10-10
Well then, instead of passing t to the function and defining Lac(t), you just pass it the array of values like below
Lac = 0.91+0.01*sin((0:0.001:5)') % Lac is now an array of doubles
x = fsolve(@(x)myfun(x,Lac),zeros(size(Lac)));
plot((0:0.001:5)',x)
xlabel('t')
ylabel('x')
function q = myfun(x,Lac)
q = 2*(0.0822*cos(x)-0.2838*sin(x))+Lac.^2-0.9209;
end
LIULIYAN
2022-10-11
更多回答（1 个）
Way better than using fsolve is to compute the analytical solution. That is, if t is the time varying parameter, then we have
syms t x
Lac=0.91+0.01*sin(t);
q = 2 * ( 0.0822*cos(x) - 0.2838*sin(x)) + Lac.^2 - 0.9209;
Now, we wish to solve q==0, for x, but as a function of t.
xsol = solve(q == 0,x,'returnconditions',true)
So there are infinitely many solutions, valu=id for any integer value of k.
xsol.x
That may look a bit messy, but we can first take only the primary solution, where k == 0. Agaoin, add any integer multiple of 2*pi to these solutions.
syms k
xprimary = subs(xsol.x,k,0)
Note there appear to be complex terms in this. We might want to plot the solutions, looking to see if and where imaginary terms enter in. When we do so, we will see the imaginary terms are essentially always zero. So they ended up canceling out always.
fplot(imag(xprimary(1)),[0,5],'b')
fplot(real(xprimary(1)),[0,5],'b')
hold on
fplot(real(xprimary(2)),[0,5],'r')
ylim([-3,1])
grid on
xlabel t
ylabel x(t)
These solutions are in terms of radians. If you want degrees, multiply by 180/pi. but since you used sin and cos in your equaritnis, I can only assume you want to see it in terms of radians.
Finally, IF you want a functional form you can call, then just use matlabFunction. Thus...
x1_t = matlabFunction(real(xprimary(1)));
x2_t = matlabFunction(real(xprimary(2)));
The positive solution was the second one. As you see, we can evaluate it directly for ANY value of t.
x2_t(3)
Thus 0.1286 radians.
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