afei8233 发表于 2008-4-4 14:38

仿真中断--derivative input

我的模型在仿真时,设置时间为4s,可到了2.8s时多一点,仿真中断,无法进行,截图如下:

Derivative input 1 of 'bsjiabf2/BG/baiC/Integrator' at time 2.815980315877682 is Inf or NaN.
Stopping simulation.
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)
这是他的错误叙述,请问错误源自何处?该怎么解决??希望能得到帮助!!谢谢。QQ93644311;邮箱afei8233@yahoo.com.cn

linlin820 发表于 2008-4-4 16:12

代数环问题,可以参考
http://forum.vibunion.com/forum/viewthread.php?tid=28597
http://forum.vibunion.com/forum/viewthread.php?tid=56939

http://www.mathworks.com/support/solutions/data/1-16V6S.html?solution=1-16V6S
http://www.mathworks.com/support/solutions/data/1-2S1NR3.html?solution=1-2S1NR3

afei8233 发表于 2008-4-4 19:28

这应该不是代数环的问题吧?!我理解代数环是怎么一回事!!!还有如下的警告——Warning: Unable to reduce the step size without violating minimum step size of 1e-010 at time 1.000000000000014e-010.Continuing simulation with the step size restricted to 1e-010 and using an effective relative error tolerance of 0.02570638301884068, which is greater than the specified relative error tolerance of 0.001.
Warning: Unable to reduce the step size without violating minimum step size of 1e-010 at time 2.81598151763284.Continuing simulation with the step size restricted to 1e-010 and using an effective relative error tolerance of 1.176948267777999, which is greater than the specified relative error tolerance of 0.001.
Warning: Unable to reduce the step size without violating minimum step size of 1e-010 at time 2.815981517632888.Continuing simulation with the step size restricted to 1e-010 and using an effective relative error tolerance of 1.176935916686738, which is greater than the specified relative error tolerance of 0.001.

afei8233 发表于 2008-4-6 08:11

请给以帮助,谢谢了!!

ligoden 发表于 2008-4-25 08:33

回复 楼主 的帖子

求解微分方程问题,出现了Inf或Nan,设置变步长或者把tolerances error 设大些看看吧

gagmeng 发表于 2008-4-25 09:16

可以试着减小步长!!!:lol :lol

nuaalq 发表于 2009-3-4 16:34

可以试着减小步长!!!:lol :lol

ttwt1210 发表于 2009-3-4 17:01

可以尝试在反馈上加一个memory模块
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