现在好了，将第二次时间杨梅倒计时名称理解为CTD倒计时。
All right now understand the second time of time arbutus countdown name as CTD countdown.

现在这非常相似，唯一的区别是它是向下计数的。
Now this is pretty much similar took out the only difference is that it counts in a downward direction.

它从较高值开始变为零。
It goes from the higher value and goes to zero.

如果我们假设该块的值为 5，则递减至 4 3 2 并返回至零，这表示这是递减计数。
If we assume that the piece is a value of 5 a decrement back to 4 3 2 and goes back to zero that says this is counted down.

好的，这是向上计数和向下计数之间的区别，我们将在中看到一个练习示例。
OK this is a difference between count up and going down and we'll see an exercise example in.

假设计数器减 1，输入发射器的值从 0 变为 1。
So let's say is the count down to counter counts down by 1 and the value of input emitter changes from 0 to 1.

因此，如果您看到它是一个框图，则此处的 CTD 表明它已下降，您会发现该板下降后向上计数和继续计数之间的差异。
So if you see it's a block diagram a CTD here which indicates it's come down and you will find the difference between count up and go on by this board which is lowered.

好的，这是计数中的重置。
OK this is the reset in count up.

现在会发生什么。
Now what happens.

最初我们必须加载该值。
Initially we have to load the value.

所以负载计数输入是。
So load count input is.

在您的简历中使用这个或这个 PV 值。
Use this or this value PV in your CV.

好的，因为它不会从零开始，它将从 PV 值预设值开始，因此您必须通过负载计数为 Gounder 通电，以便 spiri 将进入 C-v 情况或 1 CB 有 3 Narberth 它每个输入信号。
OK because it has come down to it will not start from zero it will start from the PV value preset value values so you have to energize the Gounder with load count both so that the spiri will go into C-v case or 1 CB has 3 Narberth it with every input signal.

因此减 2，变为 1，然后返回到 0。
So decremented 2 and goes to 1 and goes back to zero.

现在如果这回到零会发生什么。
Now what will happen if this goes back to zero.

偏移将在 Q As 的条件下发生，就好像输出信号为高电平一样。
The skew will be on the condition of Q As if output signals high.

如果 CV 小于或等于零。
If CV is less than or equal to zero.

因此，如果 CV 等于 0，则输出变高。
So if the CV as equal to zero the output goes high.

嗯，这是我目前的反驳。
Um this is my current counter-argue.

好的，让我们看一下时序图。
OK so let's have a look at the timing diagram.

它表示预设计数是否为 3。
It says if the preset count is 3.

我们以这个例子为例。
Let's take this example.

最初你有很多，但现在没关系。
Initially you have a lot but it's ok now.

只要您可以看到 CV 已加载 3，即预设值。
This long as you can see that the CV is loaded with 3 which was the preset value.

好吧，现在每个 CD 输入 CD 都是这样的。
Ok now that every CD input CD is this.

这是我的输入，因为这是这里的输入信号，但每个输入。
This is my input for this is the input signal here but every input.

这将转到 2。
This goes to 2.

这是递减到 1 和 2 0。
This is decrementing to 1 and 2 0.

当它变为零时，马刺输出为真，因为这里的 C 是 C-v。
The moment it goes to zero the Spurs output goes true because C is C-v here.

这小于等于零或等于零。
This is less than an equal into zero or equal to zero.

所以这等于零。
So this is equal to zero.

这就是为什么它是真的。
That's why it is true.

所以当你付出很多的时候。
So when you give a lot.

但场景再次变为三。
But again the scenery changes to three.

因此，再次加载 PV，它变为 3，并且该语句变为 false。
So this is loaded with PV again again it goes to three and this statement goes false.

所以输出回到零。
So the output goes back to zero.

同一时刻就OK了。
OK at the same instant.

现在请注意，这里有 CD 和负载上的输入。
Now notice here you have both inputs on CD and load.

因此，如果负载为真并且您给出一个输入脉冲并且下降，则该值将保持在 3。
So if load is true and you give an input pulse and going down the value will remain at 3.

这不会减少。
This will not decrement.

好的，这类似于当您复位打开并且计数器输入打开时，计数器被复位，定时器及时复位。
OK this is similar to when you are reset is on and the counter input is on the counter is reset the timer is reset in time.

但在这种情况下，如果负载上的 Lorenzana 和 CD 具有更高的优先级，那么当您关闭负载输入然后执行 CD 输入时，该值将保持为 3。
But in this case if Lorenzana and CDs on load has higher priority so the value will remain 3 when you turn off the load input and then you do a CD input.

那么这就带你去做。
Then this will take you to do.

好的，让我们来了解一下这个 CTD。
OK so let's understand this CTD.

我的一个 SES 右拉蒂格的例子，利用 CD 来计算条数并从 Kule 0.0 门通过。
My An example with SES right Lartigue by utilizing CD to count the number of bars and passing from the gate Kule 0.0.

好的，你有一个输入，它是一种传感器，只需折扣从大门通过的人数。
OK so you have an input which is kind of a sensor with just discounting the number of people passing from the gate.

一旦有 10 人通过大门，电机就应该关闭，大门应该关闭，否则意味着电机应该关闭。
And once you have 10 people passes the gate the motor should be closed the gate should be closed or it means the motor should be off.

我们的电机应该打开，在这种情况下，我们正在分配电机应该打开，以便门将关闭。
Our motor should be on in this case we are in assigning the motor should be on so that the gate will be closed.

好的，一旦通过，水就会打开。
OK so once that passes water will be on.

让我们举个例子。
So let's have an example.

所以我们在这里有 CDC 博客，看到这是由传感器激活的倒计时，该传感器要么不为零。
So we have the CDC blog here and see this is a countdown as activated by a sensor which is either not zero.

所以这实际上是在统计经过的人。
So this is actually counting the people passing.

好吧，如果一个人通过了，简历就会减少。
OK so if one person pass the CV will decrement.

因此，最初操作员将在最初实际的低信号下做什么。
So initially what the operator will do operator with initially actual the low signal.

所以负载将主这个巢穴到 C.v.
So the load will Lord this den to C.v.

好的，所以它可能会按下“开始”按钮，或者可能会将其称为负载。
OK so it may press the Start button or it may mention it as a load.

这取决于他，但它会启动这个输入。
It depends upon him but it will actuate this input.

外部的一个或两个都可以将价值加载到您的简历中。
Either that or two from the outside OK which will load the value into your CV.

现在传感器开始工作，人们通过了。
Now the sensor start working and the people pass.

然后它会递减到 9，然后是 8，然后回到零，然后又回到零。
Then it will decrement to 9 then 8 and then goes back to zero and then it goes to zero.

连接正确，电机启动，关闭门。
The connection will be true and the motor will be on which will close the gate.

好吧，这就是枪不能正常工作的原因。
OK so this is how gun don't work right.

那么让我们让西门子 TIAA 能够理解这个逻辑。
So let's make this logic you know Siemens TIAA to understand.

所以这是同样的逻辑。
So this is the same logic.

好的，我要下载它，这样你就可以看到 Isard 的狗是 1 0 2 0，为了演示，我也这样做了。
OK I'm going to download it so you can see the dogs this is Isard are 1 0 2 0 and for demonstration I'm doing that too.

《Phife 网球》的号码很长，所以毫无疑问它会在我的电脑上出现，这就是它的样子。
Phife tennis's long number so it's down no doubt in my PC and that's how it would look.

注意因此程序已加载，并且它是金钱，让我们想象一下这一点。
Note So the program is loaded and it's money that it and let's visualize that right.

所以这是有道理的。
So this makes sense.

所以我希望你能看到输入和输出。
So I hope you can see the inputs and outputs.

所以最初没有任何东西可以把它穿上或推下来。
So initially nothing puts it on or push it off.

所以我要做的就是降低该值，这样您就可以看到这里的值是当时的值，因为我正在进行一些练习。
So what I'm going to do is I'm going to lower the value so you can see that the value here is then because I was using some exercise.

所以我将使用此输入在我的日历中加载 5。
So I will use this input to load 5 in my calendar.

那么到底是用还是不用呢。
So is it or not to use it or not do it on.

你可以看到这里计数器已经加载了值5。
And you can see that the counter has been loaded the value 5 here.

好的。
OK.

现在我不再需要这个负载了。
Now I don't need this load any anymore.

我将把它关闭，计数器中的值是 5。
I will turn it off the value 5 is in the counter.

现在，如果输入变为五到五。
Now if the input goes five to five.

我的意思是电源继续。
I mean power goes on.

所以这是一次。
So this is one time.

所以你可以看到协议四下的内容。
So you can see that under the agreement four.

然后再次变为 3，然后变为 2，再次变为 1，当它变为 0 时，您会看到调制解调器位于门上，门已关闭。
Then again this goes to three and then goes to two and again goes to one and when it goes to zero you see that the modem is on the gate is closed.

嘿，如果我把它打开，它就会变成负值。
Hey if I even turn it on it goes to negative.

并且您知道 TV 小于 PV 小于零或等于零。
And you know that TV is less than PV less than zero or equal to zero.

这就是为什么英国公开赛现在开盘，以及它会如何关闭的原因。
So that's why the Open is on k now how the it would be off.

您必须再次按下它，以使条件变为假。
You have to press it again so that the condition goes false.

所以我点击加载，苹果就响了，因为条件表明 CV 是 CD 不小于或等于零。
So I click on load and the Apple goes off because the condition says CV is CD is not less than or equal to zero.

阿佐夫真是太尴尬了。
That's very awkward Azoff.

因此它的 CV 与零的关系要么是可弯曲的。
So it's either bendable in the relation of CV with zero.

好的。
OK.

否则，在没有任何历史记录的情况下，CVB 与 PV 之间的关系会看到它，或者因为它是折扣停机时间。
Otherwise in gownd of the relation what CVB it PV with no history of it see it or because it it's a discount downtime gone down under.

好的。
OK.

所以这就是解决方案，如果你想在 FBD 逻辑中看到这个实验，类似地，你已经在这里放弃了一个块，输入传感器就在这里，就像这个负载要么与这个负载不太相似，并且你在这里有一个预设值在。
So this was the solution and if you want to see that in FBD logic this experiment Similarly you have ceded a block here the input sensor is here like this one load is either not too similar as this one and you have a preset value here to stay on.

其他救援人员也将采取与此类似的行动。
And the other rescuers are going to go similar to this one.

我们有一个等号。
We have an equal sign.

这是关于穿着柜台的事情，让我们回顾一下这一课。
So this was about gownd down counter and let's review this lesson.

这节课我们学到了什么。
What did we learn in this lesson.

它用于以向下的方向放入。
It is used to come in put in a downward direction.

那很好。
That's pretty much fine.

但输入脉冲是从零到一。
But the input pulse is from zero to one.

但是反反弹会迪克曼这是什么意思是PV的值Panamint或CV的电流输入值等于或小于零输出作为一我们已经理解，如果基本规律的值从0变化到1的值PV 已加载。
But the counter-rally will Dickerman this what it means is the value of PV Panamint or CV current input value is equal to or less than zero output as one we have understood that if the value of fundamental law changes from 0 to 1 the value of PV is loaded.

我们知道，当您将负载更改为某个值时，我们已经看到了时序图，该值是这些 C.v. 中先前加载的值。
We know that we have seen the timing diagram the moment you change load to one value which is in the previous loaded in those C.v.

所以Obert将会被重置。
So the Obert will be reset.

所以这是关于向下的，在下一个视频中我们将看到向上向下计数，它是两个计数器的组合。
So this is about going down and in the next video we'll see Count up down which is a combination of both counters.

我希望这个练习对 CTD 有意义。
I hope this this exercise makes sense to CTD here.

如果有任何疑问，你可以让我来指挥，然后你可以在每个部分提供的任务中练习你的逻辑。
Any doubt you can put me in command and later you can practice your gonged logic's in the quest provided in every section.

所以我真的很想和你一起去克里斯，克里斯可能有垫子，食物会以某种方式扭曲垫子，所以你需要知道这个单位。
So I would really click with you to go to Chris and Chris may have cushions which food somehow twisted cushions so that unit you need to know.

你需要拿笔和纸写下拉蒂格，看看哪里发生了什么，然后你就可以解决逻辑。
You need to take a pen and paper and write the Lartigue and see what's happening where and you can then you can solve the logic.

所以这会让你在任务中对甘德有更多的练习。
So this will give you more practice about Gander's in the quest.

所以我在下一个视频中看到。
So I see in the next video.

谢谢。
Thank you.