如何确定SimulationX中混合时间连续时间离散信号的最大值? [英] How to determine the maximum of a mixed time-continuous time-discrete signal in SimulationX?

问题描述

有问题如何确定用Modelica传递的时间上连续时间离散和时间离散的混合信号x(t)的最大值，即

There are questions how to determine the maximum of a mixed time-continuous and time-discrete signal x(t) over passed time with Modelica, i.e.,

y(t)= max {x(s)，[startTime，t]}中带有s.

y(t) = max{ x(s) with s in [startTime,t] }.

这是Modelica错误跟踪器中的未解决问题(请参见 https://trac.modelica .org/Modelica/ticket/109 ).

This is an open issue in the Modelica bug tracker (see https://trac.modelica.org/Modelica/ticket/109).

我将提供特定于SimultionX的解决方案.

I will give a SimultionX-specific solution.

推荐答案

SimulationX通过last运算符扩展了Modelica，该运算符返回参数的最后一个接受值.在事件时间点，它返回在事件迭代之前积分停止的值. last运算符可用于计算当前x值的最大值和最后一个最大值.请参阅以下工作示例.

SimulationX extends Modelica by the last-operator which returns the last accepted value of the argument. At event-time points it returns the value at which the integration stopped before the event iteration. The last-operator can be used to calculate the maximum of the current x value and the last maximum. See the following working example.

model test "test.ism"
    extends SimModel;
    Real x=2*sin(2*pi*time)+sin(20*pi*time)+(if time < 0.5 then 0 else 3) "some input signal with jump";
    Real y=if noEvent( time > time.start ) then max(x,last(y)) else x  "Calculate the maximum with the help of the last-operator";
    Real z(start=0,fixed=true)=-der(z)+y "Just any dymanics.";
end test;

输入信号x和相应的输出信号y如下图所示.

The input signal x and the corresponding output signal y are depicted in the following Figure.

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