Tachome的最佳方法是什么？

这将取决于输入信号的频率。如果你有一个非常广泛的范围，你可能需要使用这两种方法。



      以下为原文

    That would depend on the frequency of the input signal. If you have a very wide range you may need to use both methods.

不是开关，软件逻辑。在锁定的转子或非常缓慢的速度，你不能计数脉冲，因为不会有任何或它们可能是非常缓慢的。高频脉冲宽度较短。你得到的分辨率将取决于CPU时钟。你需要看看PIC它的外设。和时钟。做一些数学来决定你得到什么，是否适合你的需要。您将需要编写某种类型的自动测距。



      以下为原文

    Not a switch, software logic.
At locked rotor or very slow speeds you can not count pulses since ther will not be any or they may be very slow.
Pulse width at high frequency will be short. The resolution you get would depend on the CPU clock.
You need to look at the pic it's peripherials. And clock. Do some math to determine what you get and if it suits your needs.
You will need to code some type of auto ranging.

MITYELTU，只是在脉冲方法之间使用时间一路。但对于高转速，累加“脉冲之间的时间”求和计数脉冲。在每个更新周期中，通过脉冲计数除以脉冲之间的汇总时间（如果脉冲计数是至少一个）。通过这种方式，你将得到混合值，这将有效地在预定义的（更新速率）周期内为高RPM和低RPM的脉冲之间的脉冲计数。所有的方法都会得到最精确的值。即使在高转速下，分辨率也会很粗糙，计数的数量会产生很好的平均值。不要错过整数累加器变量的大小。在某些情况下，标准的16位整数可能是不够的。此外，使用宽变量知道在处理中断时非原子访问。也使脉冲捕获和求和之间的时间尽可能快地错过在高RPMS尽可能少的脉冲。如果将有规则的脉冲丢失，结果可能包含混叠，一些系统误差。



      以下为原文

    MitYeltu, just use time between pulses method all the way. But for high RPM accumulate "time between pulses" sum and count the pulses. On every update period divide summary time between pulses by pulses count (if pulses count is at least one). This way you'll got hybrid value which effective will be pulse count over predefined (update rate) period for high RPM and time between pulses for low RPM. All the way you'll got most precise value. Even if at high RPM the resolution will be coarse, the number of counts will produce fine average.
Do not miss the size of integer accumulator variable. In some cases standard 16-bit integer may be not sufficient. Also using wide variable be aware of non-atomic access while working with interruptions. Also make the time-between-pulse capturing and summing as fast as possible to miss as few pulses as possible at high RPMs. If there will be regular pulse missing the result may contain an aliasing, some systematic error.

脉冲之间的时间将需要一个时间来处理零rpm。一些PIC定时器/IC外围设备将时间为1到4个脉冲。



      以下为原文

    Time between pulses would require a time out to deal with zero rpm.
Some PIC timer /IC peripherials will time 1 to 4 pulses.

好啊。因此，如果我在脉冲之间进行计数和计时，将会有很多浮点数学。这是很多时钟周期。如果我假设最大值为20krpmax，则在脉冲之间有50U。如果我使用一个10MHz晶体4xPLL（18F4620），我得到一个10MHz的指令时钟（100NS）。这意味着最大500个指令周期。我在C编程与MPLABX和XC8在免费模式，这实际上是使用阿德优化器的代码生成。对于浮点AF视觉的浮点数学和7段显示刷新率要求，我是否有足够的时间积累、计算和显示而不丢失脉冲？



      以下为原文

    OK. So if I do both counting and timing between pulses, there will be alot of floating point math. That's alot of clock cycles. If I assume 20krpm max, I have 50uS between pulses. If I use a 10MHz crystal with 4xPLL (18F4620) I get a 10MHz instruction clock (100nS). That means a max 500 instruction cycles. I am programming in C with MPLABX and XC8 in free mode which actually uses a DE-optimizer for code generation. With the floating point math and 7-segment display refresh rate requirement for persistence af vision, am I going to have enough time to accumulate, calculate and display without missing pulses?

为什么在浮点上呢？你需要分转速吗？多少小数？数学应该用整数来完成。



      以下为原文

    Why do it in floating point?  Do you need fractional RPM? how many decimals?
The Math should be done with integers.

嗨，20K rpm不是问题…而是你的浮法数学。如上所述，您应该恢复到整数MATH.BTW，您可以轻松地从PIC18F4620迁移到PIC18F46K22，它是PIN到PIN兼容的，并且在64 MHz（+60%速度增加）下运行。K22有更多的外设，但如果需要的话，很容易从4620迁移到46k22。



      以下为原文

    Hi, 
 
20k rpm is not the issue...but your floating math surely. As mentioned above you should revert to integer maths.
BTW, you can easily get some performance bump by migrating from PIC18F4620 to PIC18F46K22 which is  pin to pin compatible and runs at 64MHz (+60% speed increase). K22 has some more peripherals but it is easy to migrate from 4620 to 46K22 if needed.
 
Regards

我建议执行整数数据捕获和累加，即整数计数器和整数脉冲计数。这两个值都是整数。因此，你可以把每一个脉冲作为整数来计算和计数。在捕获中断内完成所有整数数学运算。在中断执行常规RPM值更新之后，应该有足够的空闲TCys，例如在背景中的浮点处每秒2…4次。



      以下为原文

    I'd recommend to perform integer data capture and accumulation, i.e. integer counter sum with integer pulse count. Both values are integer by their's nature. So you are able to sum and count each pulse as integers for a fraction of 500Tcys. Do all integer math inside of capture interrupts. There should be a plenty of spare Tcys after an interrupts to perform regular RPM value update, say 2..4 times per second in floating point in background.

因此，如果我理解正确，如果我去掉浮点数学，只使用整数，我怎么才能得到一个RPM的分数。或者我假设马达总是以整数旋转运转？这对我来说似乎并不特别准确。我希望能有0.1转每分钟的分辨率。计时器当然可以处理它。也许我应该转到一个更快的控制器。虽然我不喜欢3.3V控制器，但由于项目的大部分需要重新设计，我可能需要权衡。我将按照上面的建议看46K22。



      以下为原文

    So if I understand you correctly, if I do away with the floating point math and use only integers, how can I get the fractions of an RPM. Or am I assuming that the motor will always be running at whole number revolutions? That does not seem particularly accurate to me. I was hoping for perhaps 0.1 rpm resolution. The timers can certainly handle it. perhaps I should move to a faster controller.
 
While I do not like the 3.3V controllers, as considerable portions of the project will require redesign, I may need the tradeoff. I will look into the 46K22 as suggested above.

如果整数1234表示123.4，则可以在打印期间添加小数。



      以下为原文

    If  the integer 1234 represents 123.4
you can add the decimal during printing.

20000转/秒是333转每秒，所以你有3毫秒之间的脉冲。你可以使用100的ESEC定时器在20000转/分下得到30个计数，这比0.1转/分的分辨率要好得多。在1 rpm，脉冲之间会有60秒，并且（10 kHz）定时器将达到600000计数。你不可能在低于10转/分的转速下运行电机，这将是60000个计数，而无符号整数将是好的。即使在这样的速度下，更新也会在6秒内缓慢地进行。对于低速测量，你可能需要使用多瓣致动器来获得每转10个脉冲。但是你需要选择一个能处理最高转速的传感器。有磁场效应的霍尔效应传感器，依赖于导电材料的存在（不一定是磁性）的接近传感器，和光学传感器。我用接近检测器和PIC制作了一个空气动力摇摆机（机器车间项目）的转速表。蓝牙模块读取附近的笔记本电脑的RPM和占空比：HTTPS://www. Youtub.com /手表？V= 8NaHE12F4IE



      以下为原文

    20,000 RPM is 333 revolutions per second, so you have 3 mSec between pulses. You could use a 100 uSec timer to get 30 counts at 20,000 RPM, which is plenty to get better than 0.1 RPM resolution. At 1 RPM, there will be 60 seconds between pulses, and the (10 kHz) timer would reach 600,000 counts. It is unlikely that you will run the motor at less than 10 RPM, which would be 60,000 counts, and an unsigned integer would be good. Even at that speed, the updates would be annoyingly slow at 6 seconds.
 
For low speed measurement, you may need to use a multi-lobed actuator to get perhaps 10 pulses per revolution. But you need to choose a sensor that will handle the highest RPM. There are Hall effect sensors which operate on magnetic fields, proximity sensors which rely on the presence of conductive material (not necessarily magnetic), and optical sensors.
 
I made a tachometer for an air-powered wobbler engine (machine shop project), using a proximity detector and a PIC with a Bluetooth module to read RPM and duty cycle on a nearby laptop:
 
https://www.youtube.com/watch?v=8naEh12f4IE

关于我上面的帖子，你不知道。我建议收集和总结rpm捕获数据作为整数，因为它们实际上是整数。但同时具有整数脉冲和脉冲间计时器的总和，你可以执行浮点账户。关键的特点是FAST频繁整数数学（只有两个SunSPER一次革命）和缓慢但很少浮点结果计算。通常你不需要每秒更新10000次RPM值。对于人类阅读来说，每秒不超过4次就足够了。因此，收集硬件数据作为整数和（非常快），然后计算浮点结果每秒几次（慢，但你有足够的周期）。这样，你就可以得到精确的浮点RMP结果，而不用通过每次浮点计算来重载CPU。



      以下为原文

   
 
Saying about my above post no, you don't. I've recommended to collect and sum the RPM capture data as integers due to they are actually an integers. But having both integer number of pulses and the sum of timer ticks between pulses you may perform floating point account. The key feature is fast frequent integer math (only two sums per one revolution) and slow but seldom floating point result calculation. Usually you don't need to update RPM value 10000 times per second. For human reading not more than 4 times per second is enough. So, collect the hardware data as integer sums (very fast) then compute floating point result few times per second (slow but you have a plenty of cycles for that). This way you'll got precise floating point RMP result while not overloading the CPU by floating point calculations each revolution.

嗯，我已经设法完成了部分工作。然而，我认为这是真正有用的唯一方法是在脉冲之间调整时间。计数脉冲一段时间必然导致RPM值，这是我使用的周期的倍数。换句话说，如果我在较高频率下使用0.25秒的计数周期，并得到125的脉冲计数，则RPM＝125×4＝500。我所能做的最好的是使用这个分辨率4 RPM。所以，回到平均时间的脉冲之间的时间，如Nilolay建议。现在-如何实现。我怀疑唯一的办法是使用CCP模块。正确的？为了使它容易，让我们假设每转1个脉冲和实际RPM值足够高，这样TMR1不会溢出（只是你知道，我改变了我的内部OSC的外部20MHz晶体-我有困难的PLL，所以我放弃了它）。如果没有CCP或TMR1的预分频，则得到76.3的最小RPM。正确的？所以，为了论证，我们说100转每分钟。这是每脉冲10毫秒的间隔。因此，对于每个时间段，我的CCP中断并且TMR1的值被轮询。那个值应该是50000，对吧？我从来没有使用CCP模块来捕获，当我较早尝试它时，它并没有给我任何有意义的东西。价值观在不断变化。那么，我该如何设置呢？我不能在这一刻发布代码，因为它在不同的PC上，但是我做的是：启动CCP模块，等待第一个脉冲（CCP中断），启动TMR1，等待第二个脉冲（CCP中断），停止TMR1，得到TMR1值，清除TMR1，重复。（实际上我没有重新启动CCP模块。所以我重复等待第一个脉冲语句。我怀疑我的方法有点古怪，这就是我从TMR1获得的不断变化的价值的原因。那么，我如何设置这个东西，使它返回一个可靠的值？数据表并没有给出很多关于这个的信息。



      以下为原文

    Well, I have managed to get part of this working. However, the only way I see this as having any real usefulness is by montioring the time between pulses. Counting pulses for some period inevitably leads to an rpm value that is a multiple of whatever period I am using. In other words, if I use a counting period of 0.25 seconds at higher frequencies and get a pulse count of 125, then rpm = 125*4 = 500. The best I could ever do is a resolution of 4 rpm using this.
 
So, back to averaging the time between pulses over some period of time as Nilolay suggested.
 
Now - how to implement. I suspect the only way to do this is to use the ccp module. Right? To make it easy, lets assume 1 pulse per revolution and an actual rpm value high enough so that tmr1 does not overflow (just so you know, I changed my internal osc for external 20MHz crystal - I was having difficultines with the PLL so I dropped it). I get a min rpm of 76.3 assuming no prescale for ccp or tmr1. Right? so let's just say 100 rpm for the sake of argument. That's a 10mS interval per pulse. So, for each time period, my ccp interrupts and the value of tmr1 is polled. That value should be 50000, right?
 
I've never used the ccp module to do capture, and when I tried it earlier, it did not give me anything that made any sense. The values were constantly changing. So, how do I set this up? I can't post the code at this moment as it's on a different pc, but what I did was this:
 
start ccp module, wait for first pulse (ccp interrupt), start tmr1, wait for second pulse (ccp interrupt), stop tmr1, get tmr1 value, clear tmr1, repeat. (I don't actually restart the ccp module. so I repeat to the wait for first pulse statement).
 
I suspect my method is a bit nutty and that's the reason for the ever changing values I was getting from tmr1. So, how do I set this thing up so that it returns a reliable value? The data sheet does not give alot of info on this.

你会想使用计时器门输入，它可以给你一个计数，一个中断，当它结束。



      以下为原文

    You would want to use the Timer gate input that may give you a count an an interrupt when it it end.

正如我之前提到的，60转/秒是每秒一转，假设你有一个检测每转一个脉冲的传感器，你需要一秒钟来读取。如果你有一个计时器运行每秒600个计数，你可以得到0.1 RPM的分辨率与每秒一读。但是，如果你想读取20000转/秒或333转/秒（3毫秒/Rev），则需要在10千赫下运行定时器，得到30个计数，这是20000/30或667 rpm的分辨率。因此，你可能想要在每个传感器信号上执行一个ISR，并执行两个函数。ULD将在1秒间隔内计数脉冲，这将使分辨率为20000/333或60 rpm（0.3%），但在2000 rpm时，只计算33个脉冲，分辨率为2000/33或60 rpm（3%）。革命。如果它是一个16位定时器，它将溢出65536个计数，即6.5秒。这对应于0.152 r/s或9.150 rpm。因此，这将给你一个分辨率为950/655 36＝0.00014转/分。在更高的速度，你会得到更少的分辨率，所以在20000 rpm，你将读取30计数的分辨率为667转/分钟，在低于9.15转的速度，16位定时器将溢出，但你可以很容易地使用一个无符号长和加法65536它每次它。但是速度较低的问题是更新时间，这将大于6.5秒。如果你想测量低速，并有合理的更新时间，你将需要在轴上使用多片致动器。更多信息：http://wwwopop22.com／doc…uESa TeaTraceNo.Pdfhttp://e。N.WikIOG.Org/Wik/转速表



      以下为原文

    As I mentioned before, 60 RPM is one revolution per second, and assuming you have a sensor that detects one pulse per revolution, you would need one second to get a reading. If you have a timer running 600 counts per second, you could get 0.1 RPM resolution with one reading per second. But if you want to read 20,000 RPM, or 333 revolutions per second (3 mSec/Rev), you would need to run the timer at 10 kHz to get 30 counts, which is a resolution of 20,000/30 or 667 RPM.
 
So, you may want to have an ISR that is executed on each sensor signal, and perform two functions.
 
One function would be to count pulses over a 1 second interval, which will give a resolution of 20,000/333 or 60 RPM (0.3%), but at 2000 RPM you would count only 33 pulses and your resolution would be 2000/33 or 60 RPM (3%).
 
You would also have a timer running at 10 kHz that would be read, and set to zero, on each revolution. If it is a 16 bit timer it will overflow at 65536 counts which is 6.5 Sec. That corresponds to 0.152 R/S or 9.150 RPM. So this will give you a resolution of 9.150/65536 = 0.00014 RPM. At higher speeds you will get less resolution, so at 20,000 RPM you will read 30 counts with a resolution of 667 RPM.
 
At speeds less than 9.15 RPM, the 16 bit timer will overflow, but you can easily use an unsigned long and add 65536 to it each time it does. But the problem with lower speeds is the update time, which will be greater than 6.5 seconds.
 
If you want to measure low speed, and have a reasonable update time, you will need to use a multi-lobed actuator on the shaft.
 
For more information:
 
http://www.opto22.com/doc...ues_Technical_Note.pdf
 
https://en.wikipedia.org/wiki/Tachometer
 
 

我建议：0）尽可能快地运行捕获计数器；1）编写自由运行的捕获中断，它简单地增加脉冲计数器和求和捕获的快速时钟；2）写捕获定时器溢出中断，在低RPM的情况下将定时器翻转数添加到滴答和。这将允许完美地测量任何RPM。只需使用足够大的整数变量。如果多个并发中断是可能的，请不要忘记原子访问；3）使用不同的定时器和另一个中断来定期更新值。ISR可以像一个位标志设置一样简单。不要在中断中处理RPM；4）在主循环中检查来自（3）的标志。如果设置了标志，则阻止原子访问中断、复制计数器和求和、零计数器和求和，使中断恢复。希望复制速度足够快，不要错过任何中断。现在CPU在主循环中有一个时间来计算浮点结果；5）如果RPM的速度比没有检测到的脉冲慢，则计算没有拖动的间隔数。首先计算脉冲RPM值，除以无脉冲间隔计数。这样你就可以测量一个比更新周期慢的RPM。极限是快速计数器和可变大小和良好的意义。



      以下为原文

    I'd recommend:
0) run capture counter as fast as possible;
1) write free-running capture interrupt which is as simple as incrementing pulse counter and summing captured fast timer ticks;
2) write capture timer overflow interrupt which adds timer rollover number to ticks sum in case of low RPM. This will allow flawlessly meter any RPM. Just use large enough integer variable. Do not forget about atomic access if multiply concurrent interrupts are possible;
3) Use different timer and another interrupt to update the value regularly.The ISR may be as simple as one bit flag setting. Don't process the RPM inside an interrupt;
4) In main loop check the flag from (3). If flag is set then block the interrupts for atomic access, copy counter and sum, zero counter and sum, enable interrupts back. Hope the copying will be fast enough to not miss any interruption. Now the CPU have a time in the main loop to compute floating point result;
5) If RPM is so slow than none pulses detected then count number of intervals without of pilulses. At first pulse compute RPM value dividing it by no-pulse interval count. This way you will be able to.measure an RPM which is slower than update period. The limits are the fast counter ticks sum variable size and good sense.

读取轴上的低RPM（和方向）的另一种方法是具有可变宽度带，使得传感器看到增加（或减少）的线性模拟信号（锯齿）。这可以用一个连续的旋转电位器来完成，但是它会很容易磨损。光学或电子装置会更好。然后RPM的下限取决于A/D的分辨率和采样率，也可以使用同步器。它通过三相信号的相对振幅提供绝对轴位置。一个两相的解决方案也是可能的，其中信号将是正弦位置的正弦和余弦。



      以下为原文

    Another way to read low RPM (and direction) on a shaft is to have a variable width band such that the sensor sees an increasing (or decreasing) linear analog signal (sawtooth). This could be accomplished with a continuous rotation potentiometer, but it would be susceptible to wear. An optical or electronic means would be better. Then the lower limit to RPM would depend on the resolution of the A/D and its sampling rate.
 
A synchro can be used as well. It provides absolute shaft position by means of relative amplitudes of a three phase signal. A two-phase solution is also possible, where the signals will be the sine and cosine of the angular position.
https://en.wikipedia.org/wiki/Synchro

HI，许多PIC器件用于电机控制应用，具有正交编码器接口的硬件接口。QEI是一个2相位数字同步器，显然与PSTechPaul在前一次消息中提到的有关。QEI可能有4个位置/旋转的任何分辨率，高达几百个。您可以在轴锍黑色的圆周周围画一个色带一半。R一半白色或银色，然后在它旁边的另一条带状物，相同，但是交错90度。还有2个光学传感器，或者仅仅是2个传感器交错90度。还有其他传感器可能使用开槽轮或带磁性传感器的齿环。它将通过增加或减小位置寄存器来提供关于运动方向或旋转方向的信息。RPM是：（NeXi位置）OLDYPACTION（/No.ByToStudio *TimeType）与普通计数器方法一样，计算每个中断的和或差并不十分必要。事件：从计数器到32位寄存器只复制（16位）位置计数器。当（硬件）计数器从高到低值滚动时，增加高16位。类似于定时器。可以保留多组旧值，用于快速响应低分辨率的更新和高分辨率。具有较低响应的离子显示。正交编码器接口也可在软件中实现，用于低频信号。迈西尔



      以下为原文

    Hi,
Many PIC devices intended for Motor Control applications, have hardware interface for Quadrature Encoder Interface. (mostly among 16-bit devices.)
QEI is a 2 phase Digital Synchro, clearly related to what PSTechPaul mentioned in the previous message.
 
QEI may have any resolution from 4 positions / revolution, up to several hundred.
You may paint a ribbon half around the circumference of a shaft matte Black, the other half White or silver.
Then another ribbon beside it, the same, but staggered 90 degrees. Then 2 optical sensors.
Or just 2 sensors staggered  90 degrees.
There are other sensors possible, using slotted wheel, or toothed ring with magnetic sensors.
 
It will give information about direction of movement or rotation, by incrementing or decrementing the position register.
RPM is:  (New_position - Old_position) /(Number_of_counts * Time_passed)
 
Same as with plain counter methods, it is not really nessesary to calculate sum or difference for each interrupt. event.
Just copy (16 bit ) position (revs) counter from counter to a 32 bit register.
When (hardware) counter roll over from high to low value, increment the High 16 bits.
Similar for Timer.
 
It is possible to keep several sets of Old values,
for fast response update with low resolution, and high resolution display with slower response.
 
Quadrature encoder interface may also be implemented in software, for low frequency signals.
 
   Mysil