2015-02-06 56 views
3

我正在Wolfram System Modeler中创建Max Per Interval块。Modelica - 增量不符合条件

为了方便我解释,我只是最大价值指数增量设置为10

block HighWaterMarkPerInterval 
    extends Modelica.Blocks.Interfaces.SISO; 
protected 
    Integer index; 
    Real currentMax; 
    Real endTimes[1, 45] = [30812532.2, 32037805, 33265581.8, 34493233.8, 35720861.5, 36948483, 38176307.7, 39426940.6, 40654485.4, 41882212.1, 43109672.7, 44337076, 45564265.7, 46793039.6, 48045130.9, 50749960.3, 52040090.6, 53558507.7, 54814537.3, 56331978.2, 57587753.3, 59105952.9, 60362517.8, 61879307.8, 63136031.5, 64363411.4, 65590464.3, 67738027.40000001, 84725789.8, 87831338.09999999, 89030965.40000001, 90258821.8, 91486663.5, 92714210.3, 93941727.7, 95166770.3, 97283519, 99434222.90000001, 100658067.1, 102807019, 104030032.7, 106179193, 107402090, 109550214.2, 110771545.3]; 
algorithm 
    if endTimes[1, index] < time then 
    index := pre(index) + 1; 
    currentMax := 0; 
    else 
    currentMax := 10; // Constant to until I get logic working 
    end if; 
initial algorithm 
    index := 0; 
equation 
    y = currentMax; 
end HighWaterMarkPerInterval; 

当跑,到无穷远马上蝙蝠。我认为我的逻辑有问题,但我无法想象它。

代码应检查是否仍处于间隔时间内,当我们跨越下一个间隔时间时,它将“currentMax”值设置为零。这将重置我在另一个块中实现的最大值。

任何帮助,将不胜感激。谢谢。

编辑:代码节表格示例。

model HighWaterMarkPerInterval 
    annotation(Diagram(coordinateSystem(extent = {{-148.5, -105}, {148.5, 105}}, preserveAspectRatio = true, initialScale = 0.1, grid = {5, 5}))); 
    extends Modelica.Blocks.Interfaces.SISO; 
    Modelica.Blocks.Math.Max maxblock(u1 = currentMax, u2 = u); 
    Real flybyEnds[1, 45] = [30813151,32038322,33266015, truncated for space saving...]; 
    Integer index; 
    Real currentMax; 
initial equation 
    index = 1; 
    currentMax = 0; 
algorithm 
    // When we are in the interval continually grab max block output and output currentMax 
    when {time>=flybyEnds[1, index-1], time <=flybyEnds[1,index]} then 
    currentMax := pre(maxblock.y); 
    y := currentMax; 
    end when; 
    // When we move to the next interval reset current max and move to the next interval 
    when time > flybyEnds[1, index] then 
    currentMax := 0; 
    index := pre(index) + 1; 
    end when; 
end HighWaterMarkPerInterval; 

回答

4

您需要使用when,不if。您可以在Modelica by Example中找到关于它们之间的差异以及它们之间的差异的讨论。

此问题在SO上都有讨论herehere

下面是一个例子(没有经过充分测试,但它显示的基本想法):

model HighWaterMarkPerInterval 
    extends Modelica.Blocks.Interfaces.SISO; 
    parameter Modelica.SIunits.Time sample_rate=3600; 
    Real flybyEnds[45] = {30813151,32038322,33266015,...}; 
    Integer index; 
    Real currentMax; 
initial algorithm 
    // Specify the first time we are interested in... 
    index := 1; 
algorithm 
    // At the start of the simulation, the initial max for the current 
    // interval [0,30813151] is whatever u is. The initial output value 
    // is also the initial value for u 
    when initial() then 
    currentMax := u 
    y := u; 
    end when; 

    // Check at some sample rate (faster than the flyby interval!) 
    // if u > currentMax... 
    when sample(sample_rate, sample_rate) then 
    // New currentMax is the larger of either currentMax or u 
    // when the sample took place 
    currentMax := max(pre(currentMax), pre(u)); 
    end when; 

    // At the end of the "flyby", record the maximum found since 
    // the last flyby and specify the next flyby index. 
    when time>=flybyEnd[index] then 
    // New output is the value of currentMax from this interval 
    y := pre(currentMax); 
    // Now reset currentMax 
    currentMax := pre(u); 
    // Increment index up to the length of flybyEnd 
    index := min(pre(index)+1, size(flybyEnd,1)); 
    end when; 
end HighWaterMarkPerInterval; 
+0

我已经试过落实'when'过去,但我无法弄清楚如何使用时将其复位功能。如果它理解正确,我将需要'when-else'语句的功能来做我想要做的事情。 – user3791725 2015-02-07 00:19:06

+0

你能解释一下“重置它”是什么意思吗? – 2015-02-07 13:39:34

+0

是的,对不起。 'currentMax'应该在每个'endTime'重置。我试图计算每个时间间隔的'currentMax'值,而不是整个模拟中的高水位标记。 'endTimes'对应于间隔结束时间。 – user3791725 2015-02-07 23:45:17