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Dear
I'm doing an iterative algorithm in MATLAB/Simulink/Matalb Function. However, when i run the simulation, the errors occurred.(please see the attached picture and code). If possible, please help me. Thank you very much for your help.
function gamma = cg(gamma0,voutx,tao,e1,e2,M)
beta1=gamma0(1);
beta2=gamma0(2);
beta3=gamma0(3);
beta4=gamma0(4);
beta5=gamma0(5);%gamma0 is a vector.
xd=20;%设定位置x
T=0.01;%积分步长
falf=fal(tao,0.5,0.1);
fal1=dfal(e1,0.7,0.1);
fal2=dfal(e2,0.95,0.1);
b0=1;
k=1;
N=100000;%times maximum
epsilon=1e-5;%parameter
gamma=gamma0;
while(k<N)
g=gradient(xd,voutx,fal1,fal2,falf,T,b0,M,gamma);%gradient
beta1=beta1-2*(xd-voutx)*M*(beta4*fal1*T*tao)/(1+(beta4*fal1*T+beta5*fal2)*T*b0);
beta2=beta2-2*(xd-voutx)*M*T*falf*(beta4*fal1*T+beta5*fal2)/(1+(beta4*fal1*T+beta5*fal2)*T*b0);
beta3=beta3-2*(xd-voutx)*M*((beta4*fal1*T+beta5*fal2)*T*T+T/b0)*falf/(1+(beta4*fal1*T+beta5*fal2)*T*b0);
beta4=beta4-2*(xd-voutx)*M*fal1/(1+(beta4*fal1*T+beta5*fal2)*T*b0);
beta5=beta5-2*(xd-voutx)*M*fal2/(1+(beta4*fal1*T+beta5*fal2)*T*b0);%the three parameters that need to be adjusted
gamma=[beta1;beta2;beta3;beta4;beta5];%vector
if(norm(g)<epsilon), break; end
end
19 个评论
Walter Roberson
2022-7-16
Were there other error messages as well?
Joe Jones
2022-7-17
Dear
After I ran the model, it only showed the errors in the first Figure, and there was no other error information. My idea is to optimize the parameters of ADRC and I will show you the complete model now.@Walter Roberson
Fangjun Jiang
2022-7-18
The "complete model" doesn't help to show the error. The error indicates there is a "chart" and points to the place, "ADRC_dingdian.../cgadrc2". Click that link and where does it lead to in the model? Show all block names to help identifying the model blocks.
Joe Jones
2022-7-18
Dear
Sorry. All block names have been shown in the picture. When uploading the picture, I modified the name of the blocks. I re-simulated after unifying the name, but the same error is still displayed.I click the link and it leads to the MATLAB function 'frcg'. Thank you! @Fangjun Jiang@Walter Roberson
Fangjun Jiang
2022-7-18
编辑：Fangjun Jiang
2022-7-19
Open the MATLAB Function block code, click "edit data" to see if there is anything obvious regarding the data types of the parameters and signals.
The code seems to have problems.
- In the while loop, k never increases so it is an infinite loop. You need to add k=k+1
- the usage of gradient() function is puzzling. xd is a constant. Why there are so many input arguments?
Dear
I modify the code according to your comments.I click "edit data" and change the 'DataType' to 'double'. Then the errors are solved. However, a new error appears:
"Output argument 'gamma' is not assigned on some execution paths."
My 'gamma' is the output of the matlab function, but my output is not in the if-else statement. If possible, please help me. Thank you. @Fangjun Jiang
function gamma=frcg(gamma0,voutx,tao,e1,e2,M)
k=0;
n=length(gamma0);
N = 10000;
epsilon = 1e-5;
xd=20;
T=0.01;
b0=1;
falf=fal(tao,0.5,0.1);
fal1=dfal(e1,0.7,0.1);
fal2=dfal(e2,0.95,0.1);
g0=zeros(5,1);
d0=zeros(5,1);
while(k<N)
g=tidu(xd,voutx,fal1,fal2,falf,T,b0,M,gamma0,tao);
itern=k-(n+1)*floor(k/(n+1));
itern=itern+1;
if(itern==1)
d=-g;
else
beta=(g'*g)/(g0'*g0);
d=-g+beta*d0; gd=g'*d;
if(gd>=0.0)
d=-g;
end
end
if(norm(g)<epsilon), break; end
alpha=-(g'*d)/(d'*d);
gamma0=gamma0+alpha*d;
g0=g; d0=d;
k=k+1;
gamma=gamma0;
end
Fangjun Jiang
2022-7-19
There is one possibility. If "if(norm(g)<epsilon), break; end" is satisfied during the first iteration, then gamma is never assigned.
Add "gamma=gamma0" as the first line of the function as the default return value to resolve this error.
Joe Jones
2022-7-20
Fangjun Jiang
2022-7-20
Not like that!
Use your previous version of code
Add "gamma=gamma0" as the first line of the function, which means, right below the "function gamma=frcg(gamma0,voutx,tao,e1,e2,M)" line.
Joe Jones
2022-7-21
Dear
I am sorry for misunderstanding your comments.
I modified the code according to your comments. The previous error was indeed resolved, but there is a new one.(please see the attached figure)
I have an ideal: Is it because it may exponentiate negative numbers that it produces complex value?(e.g. a^b: a<0, b is fraction)
I have put the 'tidu' functionn, 'fal' function and 'dfal' function behind main function. If possible, please help me. Thank you ! @Fangjun Jiang@Walter Roberson
function gamma=frcg(gamma0,voutx,tao,e1,e2,M)
gamma=gamma0;%I have added this statement to the first line.
k=0;
n=length(gamma0);
N = 10000;
epsilon = 1e-5;
xd=20;
T=0.01;
b0=1;
falf=fal(tao,0.5,0.1);
fal1=dfal(e1,0.7,0.1);
fal2=dfal(e2,0.95,0.1);
g0=zeros(5,1);
d0=zeros(5,1);
while (k<N)
g=tidu(xd,voutx,fal1,fal2,falf,T,b0,M,gamma0,tao);
itern=k-(n+1)*floor(k/(n+1));
itern=itern+1;
if(itern==1)
d=-g;
else
beta=(g'*g)/(g0'*g0);
d=-g+beta*d0; gd=g'*d;
if(gd>=0.0)
d=-g;
end
end
if(norm(g)<epsilon), break; end
alpha=-(g'*d)/(d'*d);
gamma0=gamma0+alpha*d;
g0=g; d0=d;
k=k+1;
gamma=gamma0;
end
%tidu function
function g=tidu(xd,voutx,fal1,fal2,falf,T,b0,M,gammak,tao)
g1k=(gammak(4)*fal1*T*tao)/(1+(gammak(4)*fal1*T+gammak(5)*fal2)*T*b0);
g2k=T*falf*(gammak(4)*fal1*T+gammak(5)*fal2)/(1+(gammak(4)*fal1*T+gammak(5)*fal2)*T*b0);
g3k=((gammak(4)*fal1*T+gammak(5)*fal2)*T*T+T/b0)*falf/(1+(gammak(4)*fal1*T+gammak(5)*fal2)*T*b0);
g4k=fal1/(1+(gammak(4)*fal1*T+gammak(5)*fal2)*T*b0);
g5k=fal2/(1+(gammak(4)*fal1*T+gammak(5)*fal2)*T*b0);
g = -2*(xd-voutx)*M*[g1k;g2k;g3k;g4k;g5k];
%dfal functionn
function y=dfal(x,alpha,delta)
if x>delta
y=alpha*x^(alpha-1);
else if x<-delta
y=(-1)^(alpha+1)*alpha*x^(alpha+1);
else
y=delta^(alpha-1);
end
end
end
%fal function
function y=fal(x,alpha,delta)
if abs(x)>delta
y=abs(x)^alpha*sign(x);
else
y=x/(delta^(1-alpha));
end
end
Walter Roberson
2022-7-21
Yes, negative to a fraction needs complex numbers.
Do complex numbers make sense for the physics involved?
Joe Jones
2022-7-21
Fangjun Jiang
2022-7-21
Check the dfal() function. I think the output y is supposed to be symmetric to the input x. Your function is NOT.
Check the literature to make sure it is implemented correctly. Plot out dfal() vs. x to verify.
Dear
I modified the code according to your comments. The previous error was indeed resolved, but there is a new one.(please see the attached figure 1)
After I click 'fix' in figure 1 and run it again, the same error is still reported, but it seems that the error type has swapped, but there is still an error (please see the red box in the error figure 2).
I click the blue link and find that it points to the block 'derivative'.(please see figure 3)
May I ask what is the reasonfor this? If possible, please help me. Thank you! @Walter Roberson@Fangjun Jiang
Fangjun Jiang
2022-7-22
- The direct derivative (du/dt) operation usually would cause problem in digital simulation. Use an approximation of derivative if it is absolutely needed. In your model, two derivatives are multiplied but I didn't see where it is used. If it is for observation only, then delete or comment out all of them.
- In the solver settings, reduce the step size or reduce the relative tolerance, you might be able to get the simulation running without error.
- If there is a true algebraic loop, you need to find it and resolve it.
- ADRC is known to have digital simulation stability issues due to its non-linearity. Check the literature to see if there is recommendation on the settings for Simulink simulation.
Dear
Thank you for your comments!
What I want to do is the derivative of the system output x with respect to the output of the ADRC, which I then convert to the product of the two derivatives with respect to time, as shown in Figure 1.
The product of the two derivatives is used in the subsystem 'cg1', The content of the subsystem can be seen in figure 2.
I will modify and check my model according to your comments. If I still have questions, I will come back to you. I hope that you can answer me by then. Thank you!
Best Regards!
Fangjun Jiang
2022-7-25
This is probably where the help from this community ends. The simulation of the Simulink model runs but it is extremely slow and it stopped due to instability of the simulation. You need to consult the expert in your field to resolve the problem. Most likely the parameters of the ADRC is too aggressive. Avoid directly derivative.
Joe Jones
2022-7-26
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