2

怎么判断一个事件的第n个事件的时间间隔不超过一定时间呢？例如，一个事件最多可以发生 ‍5 ti‍mes every 10 minutes.

在 STL 中我们可以使用这个

VAR
    counter:CTU;
    timer:TON;
    Event:BOOL;
    bMaxEventHappend:BOOL;
    tElapsedTime:TIME;
END_VAR


counter(CU:=Event);
IF counter.CV=1 THEN
    timer(IN:=TRUE);
END_IF
IF counter.CV=5 THEN
    bMaxEventHappend:=TRUE;
    counter(Reset:=TRUE);
END_IF

//resetProcess
IF counter.CV=1 AND timer.et>=T#10M THEN
    timer(IN:=FALSE);
    counter(Reset:=TRUE);
ELSIF counter.CV=2 THEN
    tElapsedTime:=timer.et;
    IF timer.ET-tElapsedTime >=T#10M THEN
       timer(IN:=FALSE);
       counter(Reset:=TRUE);
    END_IF
ELSIF counter.CV=3 THEN
    tElapsedTime:=tElapsedTime+timer.et;
    IF timer.ET-tElapsedTime >=T#10M THEN
       timer(IN:=FALSE);
       counter(Reset:=TRUE);
    END_IF
ELSIF counter.CV=4 THEN
    tElapsedTime:=tElapsedTime+timer.et;
    IF timer.ET-tElapsedTime >=T#10M THEN
       timer(IN:=FALSE);
       counter(Reset:=TRUE);
    END_IF
ELSIF counter.CV=5 THEN
    tElapsedTime:=tElapsedTime+timer.et;
    IF timer.ET-tElapsedTime >=T#10M THEN
       timer(IN:=FALSE);
       counter(Reset:=TRUE);
    END_IF
END_IF

这种方法似乎不是实现期望的最佳方法。还有另一种最佳方法吗？

任何帮助，将不胜感激。

4

2 回答 2

1

不幸的是，我无法准确理解您要完成的工作。如果我的回答没有给出您的想法，只需更新您的问题并解释您想要做什么。

TYPE MY_EVENT : STRUCT
        TimeStart : TIME; (* Time when event was started *)
        TimeEnd : TIME; (* Time when event was ended *)
        TimeWorked: TIME; (* Time that event was working *)
        Index: USINT; (* Index in array *)
    END_STRUCT
END_TYPE

PROGRAM PLC_PRG
    VAR
        arEvents: ARARY[1..5] OF MY_EVENTS; (* Last 5 events *)
        xEventStart: BOOL; (* Event started *)
        xEventStartM: BOOL; (* Memmory of event started for edge detection *)
        stEvent: MY_EVENT; (* Current event *)
        usiCount: USINT := 1; (* Current index *)
        usiFor: USINT := 1; (* FOR iterator *)
        stSearchEvent: MY_EVENT; (* Searched event *)
    END_VAR

    (* Raising edge of event. Event starts *)
    IF xEventStart AND NOT xEventStartM THEN
        stEvent.TimeStart := TIME();
    END_IF;
    
    (* Falling edge of event. Event ends *)
    IF NOT xEventStart AND xEventStartM THEN
        (* Finalize event *)
        stEvent.TimeEnd := TIME();
        stEvent.Index := usiCount;
        stEvent.TimeWorked := (stEvent.TimeEnd - stEvent.TimeStart);
        
        (* Save event in array *)
        arEvents[usiCount] := stEvent;

        IF usiCount = 5 THEN
            usiCount := 1;
        ELSE
            usiCount := usiCount + 1;
        END_IF;

        stEvent.TimeEnd    := T#0S;
        stEvent.TimeStart  := T#0S;
        stEvent.TimeWorked := T#0S;
    END_IF;
    xEventStartM := xEventStart;

    (* Current event is longer than 1 minute *)
    IF (TIME() - stEvent.TimeStart) > T#1M THEN
        // Do something
    END_IF;

    (* Get longest event out of last 5 events *)
    stSearchEvent.TimeWorked := T#0S;
    FOR usiFor TO 5 DO
        IF (stSearchEvent.TimeWorked < arEvents[usiFor].TimeWorked) THEN
            stSearchEvent := arEvents[usiFor];
        END_IF;
    END_FOR;
END_PROGRAM

这是一个示例，如何将最后 5 个事件存储在数组中，然后如何知道当前事件太长以及如何在数组中查找最长事件。

于 2021-02-18T04:39:30.157 回答
0

最后，我找到了解决这个问题的方法：

PROGRAM MAIN
VAR
    trigger:r_Trig();
    event: BOOL;
    tDuration:TIME:=T#10M;
    tTimeInit: TIME:=TIME()-tDuration;
    aTime:ARRAY[1..5] OF TIME:=[tTimeInit,tTimeInit,tTimeInit,tTimeInit,tTimeInit];
    alarm:BOOL;
    i:INT;
END_VAR


trigger(clk:=event);
IF trigger.Q THEN
    IF (TIME()-aTime[i+1]<tDuration) THEN
        alarm:=TRUE;
    END_IF
    aTime[i+1]:=TIME();
    i:=(i+1) MOD 5;
END_IF

并且似乎有效（已经过测试 tDuration:TIME:=T#10s;）

于 2021-02-18T21:43:42.053 回答