这确实是 Gounder 操作 CD udy 现在这是计数和射击计时器的组合。
And this really is the Gounder operations CD udy Now this is a combination of both count up and gunned down timer.

所以你可以看到，你知道我们在 Gunders 上有相似之处，从上到下，每个内侧 DCB 都会受到影响。
So you can see that you know we have a similarity in Gunders and ground up and going down that with every inboard DCB is affected.

但这里CV是递增的，这里是递减的。
But here CV is incremented and here is decremented.

好的。
OK.

因此，当您组合这两个计数器时，您可能有两个输入。
So when you combine these two counters you may have two inputs.

见你，见船。
See you and see the boat.

因此，一个将增加它，另一个增加迪克曼，你将有两个输出。
So one will increment that and another Dickerman that and you're going to have two outputs.

好吧，它基本上是两套公寓的组合。
OK so it's basically a combination of two condos.

假设这是一个框图并了解它是坏发射器，所以让我们继续。
So let's say it's a block diagram and understand it's bad emitters so let's proceed.

因此，它保存了 0 到 1 转换时的向上和向下计数。
So it saves the count up and gone down gone dirt's count up and down on 0 to 1 transition.

所以这非常相似。
So this is pretty much similar.

对于向上计数和向下计数计数器，您都有 0 到 1 的转换。
You have 0 to 1 transition for both count up and come down counters.

好的。
OK.

因此，让我们注意到这里最初有一个数据块和该块名称。
So let's notice here initially you have a data block and that block name.

所以这是一个计数。
So this is a count up.

这是来自 C.D，你计算输入信号，这将增加我的值。
This is from C.D you count up input signal that this will this will increment my value.

如果为零就可以了。
Okay if it is zero.

如果这里有一个信号，该信号将增加到 1。
And if you have a signal here that will increment to 1.

好的。
Okay.

所以如果我就在这里，它就会变成 0 1 和 2。
So if I just right here it goes 0 1 and 2.

好吧，下一张 CD 已经下来了。
Okay the next CD which is a come down.

那么如果是2.
So if it is at 2.

因此，这将使其变成二加一，您可以看到这一切正在结束，并且 CD 正在递减该值。
So this will make it two and one and you can see that this is ending and CD is decrementing the value.

这就是为什么它被称为CD，这就是你和你有一个负载上帝。
That's why it's called CD and that's the you and you have a load God.

这与我们在枪杀中看到的类似，并且至少有一种情况，我们要加载的负载将复制 B.V 的该值。
This is similar to what we saw in gunned down and you have at least one case of what load we'll do load will copy this value of B.V.

如果您想加载它，您可以使用较低的值将支架复制到 CV。
in your CV in case if you want to load it you can use a lower value to copy the stand to CV.

这是一个重置，一个重置将重置 CV 值而不是 PV 值，它将将该理论值重置为零。
And this is a reset one reset will reset this value of the CV not the PV it will reset this value of theory to zero.

这是一个重置函数，加载函数是将这个值 10 加载到您的 CV 中。
This is a function of reset and function of load is to load this value of 10 in your CV.

所以这是很多人做的。
So this is done by a lot.

来。
Come.

好的。
OK.

现在你有两个输出。
Now you have two outputs.

Q 你 QAD NCVS 的计数器值如此。
Q You QAD NCVS the counter value so.

问：你说每秒如果 CV 大于或等于 PV，这与我们在渐新世计数中看到的非常相似。
Q You say you are per second that this one if CV is greater than or equal to PV is this pretty much similar to what we saw in count up Oligocene.

这作为当前值大于或等于之一。
This as one of the current value is greater than or equal into.

否则为零。
Otherwise this is zero.

奥特伍德的金额为零。
The amount is this Outwood zero.

在第二季度中，我们看到下降的情况是，如果 CV 小于或等于 0，我们在这一季度中会说。
And in Q2 what we saw in going down are we're saying in this one if CV is less than or equal to zero.

好的。
Okay.

因此，如果有价值，那么只有当 CV 小于或等于 0 或我们得出的值时，证券才是 1。
So if there's value the security is one only if the CV is less than or equal to zero or what we come down to.

这基本上是两者的结合。
This is basically a combination of both.

你有重置负载的内侧组合，你必须输出，然后你有一个现值，它将显示你的电流。
You have combination of inboards combination of reset load and you have to output and then you have a present value which will show you the current going to you.

好的。
Okay.

因此，在下一张幻灯片中，我们所说的是使用后面的 Efrati 进行向上计数操作，采用 PV S10 并采用另一个输出，该输出将在 0 到 10 之间的海滩上。
So in the next slide what we have it say Stets the operation of count up down using the later for Efrati take PV S10 and take another output which would be on the beach between zero to 10.

所以我们几天来都对这个逻辑持谨慎态度。
So we have a little bit of caution here for several days this logic.

这就是我要在 TIAA 中提出的逻辑。
So this is the logic which I'm going to make in TIAA.

好的。
Okay.

所以我们已经有了一个按日历排列的数据库。
So we have a database already by calendar.

我将把启动博客倒数重命名为倒数。
I will rename the start up blog count up down.

好的。
Okay.

我将更换基民盟来向你们汇报。
And I'll change the CDU to see to you the came.

现在您已经看到所有参数都在这里。
Now you've seen that all the parameters are here.

所以计数我可以将计数更改为大约。
So count up I can change the count up to around.

确实是在绳索上。
Indeed on the ropes.

我确实没有看到你。
I indeed don't see you.

因此，将其重命名为“up”。
So rename this down to come up.

好的。
Okay.

现在，正如您所知，对于可用于我的倒计时的一个，我会指向一个并将其重新定位下来。
Now as you know to one that can be used for my count down I would point to one and retarget it come down.

好的重置。
Okay reset.

我可以使用 0.2 重命名的重置负载来完成此操作，我可以使用下一个输入来完成此操作，该输入可以是一个或三个绿色。
I can do it using 0.2 renamed reset load I can do with the next input which could be either in or three green in it.

好的。
OK.

过程中会给你一个预设值，可以取五个。
Process will give you a preset value can take five.

没关系。
That's fine.

现在 QAD 我可以在这里给出一个巨大的输出，你知道没有一个。
Now QAD I can give an output here huge you know not one.

这可能还有更多的事情要做。
This could be more to do.

并且它被存储为零。
And this is stored at zero.

好的。
OK.

所以我希望这个逻辑没问题。
So I hope this logic is fine.

那么让我们测试一下这个逻辑。
So let's test this logic.

我们来了解一下它的运作。
Let's understand its operation.

所以我们需要下载并且需要可视化它。
So we need to download and we need to visualize it.

好的。
All right.

那么让我们开始吧。
So let's do it.

如此忙碌的强子不会在这里播放音乐。
So busy Hadron won't the music here.

完美的。
Perfect.

现在你会注意到我的暗示，你知道这是真的。
Now you'll notice that my cues you know that one is true.

为什么这是真的。
Why this is true.

这是正确的，因为此输出提示如果 CV 基本上为零，则其大小为真。
This is true because this output cue the size of it would be true if CV was essentially zero.

它小于或等于零。
It's less than or equal to zero.

所以这是零。
So this is zero.

这就是为什么输出为 true，因为这个条件为 true。
That's why the output is true because this condition is true.

条件是 QE 等于 CV 小于等于 0。
It's conditions as if QE is CV is less than equal to zero.

他们也应该归零，这是真的。
They should be too for to zero this is true.

因此，让我们做一件事，让我们增加它。
So let's do one thing let's increment that.

所以我要打开 0.0 看看会发生什么。
So I'm going to turn on 0.0 and see what will happen.

所以我已经完成了一个输入。
So I had done on input by one.

所以这是因为这不再是零，这不是真的，因为这不是等熵。
So this goes for is because this is not zero anymore and this is not true because this is not equal entropy.

所以我的值在零到五之间。
So my value is between zero and five.

这是一个，好的，让我们将其增加二三四五。
This is one OK let's increment it two three four and five.

现在该值是五。
Now the value is five.

所以我的输出是开启的。
So my output is on.

这就是计数操作。
This is the count operation.

因此，即使我超过了六个，它也会保持打开状态。
So if I even go beyond six it will remain on.

好的。
OK.

所以你只是错误地打开了。
So you turn on just by mistake.

现在让我们使用下降将他固定在这个值上。
Now let's stick him in this value using going down.

所以我必须打开第二点。
So I have to turn on the second point.

这是适合您的尺寸，您可以看到您在归零时归零，因为这不等于正常。
This is size for the moment it goes for you you can see that you zero in on zeros off because this is not equal to be OK.

三二一零。
Three two one zero.

其次是我们的设计，因为 Q2 的条件为真。
Second our design because the condition of Q2 is true.

所以我希望你能理解。
So I hope you can understand.

到目前为止我只使用了这两个输入。
So far I have used just these two inputs.

所以这是正确的，因为值为零，并且这会经历并且值 5 和 B 现在处于该点。
So this is true because the value is zero and this goes through and the values 5 and B are now at this point.

如果我什至打开我的CD，它也会是负数。
If I even go even turn on my turn on my CD it will be negative.

但另一个将保持打开状态，因为它小于或等于零。
But the other will remain on because this is less than or equal and zero.

好的，因为数量小于零。
OK because there's less than zero.

这个值小于零。
This is less than zero.

我的近亲繁殖仍然存在。
My outbreed remains on.

好的。
OK.

现在我们可以使用这个信号来快速重置和加载值。
Now we can use this signal to quickly reset and load the value.

假设我立即加载值 5，超出了一点。
Suppose if I just load the value five immediately my little beyond.

所以这有点强迫我。
So this is kind of a forcing me on.

我会打开其中一个或三个。
I would turn on either that or three.

当它消失的那一刻，五个安装仍然打开。
The moment it goes five mount remains on.

如果我想开机，我可以使用重置。
And if I want to be on I can use the reset.

所以我不使用重置。
So I don't use reset.

我的第二点是。
My second point is on.

他们可以使用这两个输入立即打开两个输出，您可以使用“向上计数”信号来打开基于卡片的输出。
They can use these two inputs to turn on two out immediately and you can use the Count up on down signal to turn on the output based on cards.

所以这是两者的结合并且具有相当不错的功能。
So this is a combination of two and with a quite decent function.

好的。
OK.

现在这是关于基本操作的。
Now this was about the basic operation.

我们在逻辑中有一条线将输出保存为另一个输出。让我们采取一些线索，如果 Gounder 的值在 0 到 5 之间，则应该打开，我们如何找到它。
We have a line in the logic which saves the output another output Let's take cues that are not do say that should be on if the value of Gounder is between zero and five how we can find that.

好的，我们可以为此制定一个非常简单的逻辑。
OK so that we can make a very simple logic for that.

这方面的顶级特工是。
The top operatives on this are.

这很奇怪。
And that's odd.

但截至今天，如果这两个产品有百分比折扣，那么托德输出应该会开启。
But as of today if these two are percent off then Todd output should be on.

这两个输出均关闭且值在 0 到 5 之间。
And these two outputs are off and the values between 0 and 5.

所以这两者的“与”运算会使函数取反。
So the and operation of these two would negate function.

它们是在输出上完成的。
They are done on over to an output.

好的。
OK.

所以你会在这里拿一个分支否定它并退出。
So you will take a branch here negate it and out.

好的。
OK.

所以我们将在这里使用该标签。
So we will use the tag here.

一个或一个，这里是两个。
One or one and here are two.

在这里看看可能是什么。
And here see down to which could be.

例如，我将 0 到 5 之间的值中的三个以上的 10 设为开启或关闭。
For example I have put more than three 10 of the values between 0 and 5 up or shut off.

在这种情况下，一切都必须进行。
In that case all to be on.

因此，让我们想象一下逻辑或逻辑，当输入它时，你知道这只是你知道那很好。
So let's visualize that down on the logic or the logic and when entering it either is going to well you know it's just you know that's fine.

所以最初您可以看到 Gounder 中的值为 5。
So initially you can see that the value in the Gounder is five.

所以让我将值设置为零。
So let me set the value to zero.

这是另一个值 0，您可以看到当该值是 5 时，水源已打开。 Moeder 1 已打开。
It's another value zero and you can see that the water do is on when the value is five Moeder one is on.

所以让我递减你知道你可以认为这个值是现在的。
So let me decremented on you know you can think that the value is for now.

所以这里是为了。
So here it is for.

并且此输出已打开。
And this output is on.

这就是我们想要的，因为在 4 点时，我们对伊朗及其尼加德函数的态度都不真实，这会在 3 点时改变我的情绪。
This is what we want because at 4 none of these do our approach on Iran and its Nygard function that true which turns on my mood at three.

因此，即使我转到三二加一，这也意味着当我转到零经文歌时，马拉蒂仍然在墙上，他会关闭，因为我的电机运行条件是它很早就归零了，这适用于它。
So even if I go to three two and one the it means on Malati remains on the wall when I go to zero motets he goes off because my motor goes on condition is this goes to zero this early on and this goes for it.

这是开放的。
This is open.

就是这样。
That's what it is.

因此，这就是您始终可以使用这个简单示例扩展 Gounder 运营以扩大客户规模的逻辑。
So that's the logic that you can always extend your Gounder operations to make your customers larger using this simple example.

好的。
OK.

那么让我们回到演示。
So let's go back to the presentation.

是的。
Yeah.

所以这是理想的解决方案，这也是我在这里提到的。
So this was the ideal solution and this is what I mentioned here as well.

并查看我们的书籍并检查可以称为上限的总数。
And see off to our books and goes through the total of what can be named as upper limit.

不要住在米德兰。
Don't live in Midland.

好的。
Okay.

如果他是 FBD 解决方案，那几乎是一样的。
And if he is FBD solution it's pretty much the same.

博客上说我会看到你有 0.0，因为这是一个哦，所以这里没有人。
The blogs say I'd see you you have 0.0 because it's an oh so there's nobody here.

CD 你要么没有人死。
CD You have either no one has died.

复位为零。
Reset has zero.

不。
No.

2同样的美元0.3 PV是10，你有两个输出下限是scudi 0.1。
2 same dollars 0.3 PV is 10 and you have two outputs down limit is scudi 0.1.

这是现值和零。
This is present value and the zero.

这是。
And this is.

问：您已连接到运动使用虹膜学，这是在这里。
Q You connected to motoric use iridology and this is here.

这个逻辑是0.0和0.1的负函数。
And this logic is negate function of 0.0 and 0.1.

所以这里有一个问题。
So this is with a problem here.

你知道解药是一归零而不是二。
You know antidotal is a one goes to zero not two.

因此很容易从中得出解决方案。
So that's pretty easy to make the solution from that.

或者你可以比较一下。
Or you can compare that.

这是关于 Gounder 的倒数。
So this is about count up down Gounder.

因此，让我们回顾一下，如果 C 的值小于等于 0，则 CV 的谷值等于或大于 PV 计数输出 Q2 的值是一种折扣操作。
So let's review the valley of CV is equal or greater than the value of PV count output Q2 is one discount operation if the value of C is less than equal to zero.

Q 正在倒计时操作。
Q Is on the countdown operations.

这是将PV值中的lot值从0到1的组合操作作为新的CV加载到计数器中。
This is a combo operation of the value of lot from 0 to 1 in the value of PV is loaded to the counter as a new CV.

这又是基础操作，因此我无法倒计时操作，因为我们已在此处加载，或者如果重置艺术家的值从零到一旦设置为零，则会出现声音操作。
This is again ground up operation so I can't count down operations because we have load here or if the value of reset artist from zero to once it is set to zero theres a sound operation.

所以这是我之前说过的向上计数和向下计数的组合，非常有用。
So this is a combination of both count up and gone down I stated before and pretty much useful.

如果你想要上限和下限三个值。
If you want three values at upper limit limit and down.

好的。
Okay.

感谢您观看本课。
Thank you for watching this lesson.

如果您有任何疑问，可以对下一个视频发表评论，您还可以在课程中找到此演示文稿的副本。
If you have any doubt it can put any comment on the next video and you can also find a copy of this presentation in the course.