Derivative of state '1' in block 'X/Y/Integrator' at time 0.55 is not finite.

802 次查看（过去 30 天）
Revathi Meda
Revathi Meda 2015-11-3
回答： Hugo 2024-10-2
An error occurred while running the simulation and the simulation was terminated Caused by: Derivative of state '1' in block 'X/Y/Integrator' at time 0.55 is not finite. The simulation will be stopped. There may be a singularity in the solution. If not, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances). How to fix this error?
  4 个评论
显示 2更早的评论
Sam Chak
Sam Chak 2023-4-14
Hi @MASSAMBA
Please post your Simulink model (a snapshot) on a New Question. It allows to use to visually inspect if there is any block that cause the problem.
By the way, most of the time, the true singularity problem cannot be solved by merely reducing the fixed step size or by tightening the error tolerances. Take note!
Sourabh
Sourabh 2023-5-22
编辑：Walter Roberson 2023-5-22
mdl = 'rl_exam'
obsInfo = rlNumericSpec([2 1]);
obsInfo.Name = 'observations';
obsInfo.Description = 'integrated error and error';
actInfo = rlNumericSpec([1 1]);
actInfo.Name = 'PID output';
env = rlSimulinkEnv(mdl,[mdl '/RL Agent'],obsInfo,actInfo)
agentOpts = rlDDPGAgentOptions(...
'SampleTime',0.01,...
'TargetSmoothFactor',1e-3,...
'DiscountFactor',0.99,...
'ExperienceBufferLength',100000,...
'MiniBatchSize',64,...
'NumStepsToLookAhead',2);
agent = rlDDPGAgent(obsInfo,actInfo,agentOpts)
This is the code i am using and i am getting
  • Algebraic state in algebraic loop containing 'rl_exam2/calculate reward/Sum1' computed at time 0.30000000000000004 is Inf or NaN. There may be a singularity in the solution. If the model is correct, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances) or tweak the initial guess of algebraic loop variable values.
as error . i dont know how to attach the simulink file. i am stuck from last 5 days on this, any help would be great.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（10 个）

Sorelys Sandoval
Sorelys Sandoval 2016-8-21
Hi, I had the same problem and the next thing worked for me:
Go to model configuration parameters/Solver/Solver Options:
Type: Variable-step
Solver: ode15s (stiff/NDF)
and that's all. Hope it works for you :D
  20 个评论
显示 18更早的评论
Nicolas CRETIN
Nicolas CRETIN 2024-4-26
编辑：Nicolas CRETIN 2024-5-21
I don't have exactly the same error, but the following has helped in my case: I reduced the relative tolerance in the solver parameters:
hope it can help someone!

请先登录，再进行评论。

Sebastian Castro
Sebastian Castro 2015-11-4
This usually has to do with really large or infinite signals feeding into an Integrator block. Numerical solvers (expectedly) have trouble integrating such large signals and therefore this leads to errors.
Typical causes can be divide by zero (1/0 = inf) or general unstable dynamics that quickly blow up over time. You should check that in your model. It can help to use Port Value Displays as you step through the model to figure out why this happens.
- Sebastian
Abhiram V. P. Premakumar
编辑：Abhiram V. P. Premakumar 2023-6-30
I solved this issue by eliminating the infinite signal, using a switch block, and setting a clock as it's threshold (or a step block with t=1e-3). Denominator of a DIVIDE block being 0 at the beginning of a simulation, was the issue in my case. Using a clock or a step signal starting at t=1e-3s, along with a switch block, can make the divide block in the output of the switch block use a finite value temporarily, until the actual signal becomes finite.
  9 个评论
显示 7更早的评论
OLUWASEUN ADEKOYA
OLUWASEUN ADEKOYA 2023-6-30
@Abhiram V. P. Premakumar It would be great if you could be more elaborate on what specific values you put into what. For example, the clock, the only option I see here is decimation (which ofcourse only permits positive integer values). Is that the threshold value you are referring to?
Also, MATLAB's Simulink signals an error when a port is empty. Should one worry about that?
Finally. did you mean "integrator block" when you refer to "divide block"?
Thanks in anticipation.
Abhiram V. P. Premakumar
Abhiram V. P. Premakumar 2023-6-30
Denominator of a DIVIDE block being 0 at the beginning of a simulation, was the issue in my case. A clock or a step signal starting at t=1e-3s, can make the divide block use a finite value temporarily until the actual signal becomes finite.

请先登录，再进行评论。

Guy Rouleau
Guy Rouleau 2024-7-25
I describe the most common reasons for this error in this blog post:
Derivative of state ‘1’ in block ‘X/Y/Integrator’ at time 0.55 is not finite » Guy on Simulink - MATLAB & Simulink (mathworks.com)
The most common causes are:
  • A modeling error making the equations diverging
  • Too large step size
  • Division by zero
ahmet akgün
ahmet akgün 2021-5-2
Adaptive Sliding-Mode Control for Boost DC–DC Converters simulation simulation fails;''Derivative of state '1' in block 'power_BoostConverter_Adaptive_SMC_math/TSMC Controller/Integrator' at time 0.0 is not finite. The simulation will be stopped. There may be a singularity in the solution. If not, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances) ''
How to fix this error?
  6 个评论
显示 4更早的评论
Ambe Harrison
Ambe Harrison 2022-8-23
Hello,ahmet akgün. I am presently working on a Back steeping controller. Face thesame problem with the integrator. Please have you solved it.
If possible we may exchange more via email, ambeharrison2000@gmail.com
Zakarya Motea
Zakarya Motea 2023-12-10
If your system has multiple chanels Try to tune them one by one. When tuning one chanel let the input for other chanels as zero.

请先登录，再进行评论。

Pavke
Pavke 2023-3-7
Hello guys
I am doing a project in simulink and I have a problem. It is about braking an asynchronous motor with direct current, which I made (it works correctly), and it is necessary to connect that system to the distribution network (I use IEEE Bus 13). I connected the braking system of the asynchronous motor to the distribution network, and the system starts to work, it goes down from 1800 reversions and comes to 1480, and by the end of the simulation it stands at 1480, the engine will not brake and the simulation does not finish to the end and the following error occurs :
An error occurred while running the simulation and the simulation was terminated Caused by: Derivative of state '1' in block 'distributivnamrezamodel/Transfer Fcn' at time 0.991475799999991 is not finite. The simulation will be stopped. There may be a singularity in the solution. If not, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances).
How to fix this problems?
Thank you
  2 个评论
Daniel Joseph
Daniel Joseph 2023-3-28
Try to figure out why is your signal really large/infinite. Is divide by zero correct in this case? maybe you can substitute a very small value like 1e-6 instead of zero, using the switch example shown above by @Abhiram V. P. Premakumar, such that the solver does not get bogged down at very small /large values.

请先登录，再进行评论。

Anton
Anton 2023-4-23
编辑：Walter Roberson 2023-5-22
Hello, I have the same error. Please help me figure out what's wrong.
K1 = 2000;
K2 = 1;
Kk1 = 0.0012;
T1 = 0.2;
T2 = 0.005;
Tk1 = 0.0001;
T3 = 0.1;
Kt = 0.5;
tp = 0.280;
  4 个评论
显示 2更早的评论
Sam Chak
Sam Chak 2023-4-23
编辑：Sam Chak 2023-4-23
Hi @Saleh Altbawi, it is definitely not the solver. When a system is unstable, it doesn't matter what solver is chose. Here is a simple example of an unstable transfer function, with a positive pole.
Saleh Altbawi
Saleh Altbawi 2023-4-24
Hi. The system is not stable at specific time does not mean division on zero. If that true it should not work from the beginning. That mean solver at specific time could not convergence the transfer function. So try to use different kinds of solver. If does not work then try to update your system's parameters.

请先登录，再进行评论。

Yingxin Zhang
Yingxin Zhang 2023-5-22
hey i have the same problem, can you help me?
Camila
Camila 2024-3-7
Im having the same problem
Hugo
Hugo 2024-10-2
When I get this error, I can fix this by using the folling method :
  1. run your model using the step forward function (run step by step)
  2. Starting from the integrator block that makes the model stop, clik on signals to display values
  3. Track the signal until you find where the value diverges
  4. make the necessary corrections.
It works for me. I hope it will help you.

类别

在 Help Center 和 File Exchange 中查找有关 Discontinuities 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by