Shift register Alternative

Similar Questions

• Functional Global Variables: an indicator can be used instead of a shift register?

  It is a simple question, but I can't find an answer to it. The model agreed to a functional Global Variable is to use an uninitialized as in this example shift register:

  

  ('Référence IN' and ' Reference to "is actually a pile of references.) There is also a "Se Refnum" case, which comes the straight through the tunnels shift register.

  My question is, why can't we do store the indicator data? It is much simpler to use a shift register (IMHO a non obvious way to store global data!):

  

  The case "Se Refnum" does absolutely nothing. Other functions such as erasure of data can be implemented just as easily. The advantage of the FGV to help avoid race conditions is maintained because you always use the VI to access the data.

  JonP says:

  
  

  Not so much, the Inidicator can happily live outside the case structure, together and Clear would be just assign different values.

  If you have only a case structure, the indicator could not live outside of it.  In order to maintain the indicator data, your design requires that it is not written in for a case of Get.  If you have an exterior structure deal that decides on 'Get' or 'set or clear' and (in the case of "Set or clear") contains the terminal of the indicator and a classiquee case that decides on 'Set' or 'clear '.  However, I would consider this a design below using the standard template of the FGV.

  The difficulty with retrieving the value if you want to do a read operation / writing change. But LV provides many ways to retrieve data from an indicator (one you don't mention is the 'Value' property), do you mean that's all "incorrect"?

  Yes (I mean that they are all incorrect).

  You could hack your way around the design to work with a single structure of matter and the terminal of the indicator being outside using a method to read the value of the indicator and through a tunnel to the structure of the case through wiring for the tunnel of writing indicator in the case of 'Get '.  However, who will require a local Variable or value of property node.  As I said, these (I only mentioned the local Variable originally) are not good choices for performance and scalability.  If you are not aware of the functional differences between the terminals, local Variables and nodes of property value, refer to this article (obviously advantages/disadvantages such as redraw objects on the front panel are not relevant here).

  I guess you could say that indicators should only be written, but it is difficult to be pure!

  No, it's not, just use the classic design of the FGV!

• problem with the shift register

  for purposes of simplicity, I remove code without problems

  Description:
  1. my state machines, for some reason, a lot of cases.
  2. the order of each case may not change.
  3. the main prupose is to get the difference as on the front panel
  4. prior to entry difference, I first is measured in the case where 2 and TRY using shift register to pass data, ideally at least 5
  5. However, the "previous" value is updated too soon, unlike 'get' is always ZERO.

  for example
  You can see the shift register on the left side has two components, ideally, the 'get' difference should display 2-0 = 2.
  However, given that the second element of the registry team updates too early, my objective cannot reach and meet up with ALWAYS 0

  I think it's my misuse of the shift register for the computer to several cases.

  I pasted this problem for 4 hours... kind of stupid, but could not understand

  in a mulit-case state machine, how to properly passes data to the case (5 in this example) INSTRUCTION to ensure that I get the correct calculation

  GOLD: because I am only looking to the current value and the previous value, are there other ways to get this problem is?

  Thank you

  It is a case where execution highlighting can be helpful. Turn on execution highlighting by clicking the light bulb on the block diagram toolbar. Then, run the VI.  You will see the left side of the team to register the change as the state machine goes through States 3 and 4.  At the time where what happens to 5, all data in the change record is identical.

  One of them might be to use two registers to offset, one for the current value and the other for the previous value.

  Lynn

• Shift register clear before rerunning the VI

  I have this VI very simple read the voltage of an acquisition of data until I stop the loop, then it saves the data to a .wav file.

  My problem is: whenever I stop the VI, I expect the registry to shift to clear. But if I restart the VI it saves the old data in the shift register and the new data in the .wav file. I want to just the new data when I run the .VI, without having to close the program each time.

  Help, please.

  Right-click on the control to the table, and choose "clear picture" (or any option os tht called) and try again. It's a cluster of an array of types of waveform data. (Later), you can right-click and choose "display as icon".

  Ben

• Table of shift register

  I need to create a program that takes an array of six elements and the use of shift registers if he pulled out a table that maintains the first element of the same thing, and then adds the second and third elements together, and then adds the second to the 6 ° element.  So if the input array is 2, 12, 10, 5, 20 and 25, the output array is 2, 22 and 72.  I think that using shift registers and a loop would work, but I can't make it work.

  Thanks for any help

  Ok.  Given that it is for a class, I will do more than any code.

  Comments.

  1. a shift register is a local memory form. So, ask yourself what this problem needs to be recalled?

  2. entry and exit tables are different sizes. No sense to wire the table via the shift register because the shift register will 'remember' how the input array is large.

  3. Reflection on comment 1, how you will get the information in the table to put it in the register shift?

  4. what happens if apply you the same rules but allowed the input array having a size greater than or equal to 6?  That's what is meant by an evolutionary approach to a problem. The code I posted works for length 6.  For any other length program must be changed. A scalable program work without modification for any input of size.

  It is often very useful to begin with a series of conditions, which you said we now clear. Then plan how you would the problem with pencil and paper. Then implement this plan in the software.

  Lynn

• Replace the subset of table w/Shift Register too slow for my Application

  This is similar to other posts, but I have not found one that addresses the limitations of an approach of shift register.

  I have a part of an application that I'm turning to 500-1, 000Hz.  The process extracts a block of data from the ADC, and I need to store this data, then collects more of the ADC.

  I've set up what seems to be the bottleneck of the present in the attached .zip file.  When I run the attached code with profiling (see. PNG file), it tells me that it takes on average 3.8ms for the Subvi to run.  At this rate, I can only run around 260 Hz.  I have a Subvi similar to this in my code and the code can run at 1000 Hz without this Subvi, but slows down to about 160 Hz when it is activated.

  Is it possible that I may collect data about 1.7 KB tables and run at the speed I need?  Any input would be appreciated.

  For purposes of reference. Debugging should be disabled on the sub - vi.

  Looking at the code you provided - don't just disable debugging -! you have some other options of "exécution" to reset.

  

  This should speed up the Sub - vi SR will be your friend again!

• Initial value shift register

  I wrote a simple program with a while loop that contains a shift register that record the value of the previous iteration.  When I stop the program and restart the program even once, I thougth the shift register value will reset to 0 (for a type of data I32)?  Apparenetly, the shift register keeps the value of the previous run until I stop the program.  Why is this?  Why change doesn't register reset?

  
  

• How to make shift register init happens only once, so that the data can persist across multiple tracks of a loop?

  Here's the situation:

  We are repeatedly followed eight real-world signals and comparing them to a threshold value.  We do this via a loop For inside a While loop.  The loop For runs eight times per pass.  We have implemented a binary table 1 d and the use of the index of the loop For as the array index, by putting a Boolean result in the table using the function replace table subset.  We want to keep the data in the table to be 'sticky', in the sense that any True value is locked, so even if a fake comes later, this array element true.  However, since we initialize the array in order for the replacement to the work table, we see that whenever the loop For again, it resets the table and destroys the history.

  I have attached a simple VI to illustrate the concept, using a random number generator as a stand-in for the real world signals.  How we change this VI do and entered real lock through multiples for loop runs, indefinitely?

  In case it is not obvious, I am a relative beginner, so please keep count in your response.

  Thank you

  B

  scottbbb wrote:

  For B, although I love the simplicity of it, I have a question: it solves the problem of the re-initialization?  What the shift of the While loop register get initialized - only once during its launch?

  Yep, the shift register Initializes only at the beginning.  You could say that every time the while loop is called (not each iteration) the shift register is reset with the wrong table.

  And, Yes, GOLD will always keep a REAL when it is TRUE.

  Usually, the simplest solution is the best.

• Shift register do not add

  Hi all

  I am trying to use the shift register adding all the waveform detected. The additional result will be divided the current number of waveform to the average score. However, it seems that the shift register does not at all. See the image below, if the waveform number is 1000, the intensity of the signal averaging is of only 1/1000 of the signal detected. Theoreically, if no noise present and the shift register has added each detection, the average signal should be identical to the detected signal. Can someone help me solve this problem?

  Thank you very much

  

  First of all you should be dividing by the end of the inner loop, I not + 1 of the outer loop.

  If you want an average on the outer loop as well, you need a registry change in this loop.

  Lynn

• Table/Shift register losing values

  Hello

  I have attached the VI.

  The vi has a table with values in them. Simply, it looks a number when this number of games, it moves on a column and add everything in that cell in a table. He continues to do this until he sees 'END' at which point she stops.

  The problem I have is that everything seems fine for the first two "values" after which no matter what additional value he acquired he loses everything and keeps only the last known value.

  Your current code takes the unique element of your table and makes a picture out of it.  You then insert the name Test.  So everything that precedes the family name went at that time.

  Here is the code updated the minimum which keeps the TABLE in the shift register and simply adds new test name in the table.

• Cancellation of registration user stored in the shift register event generates the error 1 if Subvi runs intermittently

  Hello

  I'm trying to understand the behavior of the attached excerpt from a larger overall vi functional.

  In a State, I'm generating a user event and in a State later unsubscribe from the event and destroy it.

  Now, if I went through the VI together in one step (i.e., step through events? set to FALSE), the VI runs without generating an error. However, if I run the VI by intermittent and output after each execution of the loop, the vi generates error 1. Why is it so? Please notify. Thank you.

  Peter

  Why are you registration and deregistration of events user, but you have no event structure in your VI who use them?

  When you run events, probaby the event you registered disappears when your VI stops running.  If you were able to keep your VI in the foreground running, then the life of the event would persist.

  The event number still exists in the uninitialized shift register, but it does not say more once your high level VI stops and you get the error 1.

  If it was really a global functional VI, you would terminals connected to the connector table in your VI, you would call this as a Subvi as part of a main VI and life event would persist and you wouldn't mistake 1.

• How to clear a table used in a shift register quickly?

  Hi all

  I'm sure it's a quick, but I'm missing the concept.

  I have a table of boolean in a shift register. I have a case where I basically have to 'reset' the table rather than adding on... What is the best way to do this?

  Thank you

  Cayenne

  Use a constant empty Boolean matrix to clear your table.

  Steve

• Shift register

  Hi all

  

  

  This is my block diagram and front looks like.the given, I received are spectrophotometer UV, using RS232. FYI, there is no error with the spectrophotometer design.readings cannot be read by LabVIEW.the problem now is that reading spectrophotometer read only read to a time.here is the principle of operation of the spectrophotometer.note IM using spectrophotometer to measure the absorbance of the chemical solution.

   

  
  

   

  as u go push support for the sample on the position 1 to 4, the data will appear on labview.the 1-4 sample solution is different.so, I want to show all the data at once for comparison. such as display them on the Microsoft graph chart and data will be saved on im always new excel.since with labview, I have no idea how am I supposed to do.

   

  or maybe you have an idea or suggestion?

   

  as an attachment, I also tried shift register, but the result will display the same value on all the indicators.how should I do to read the data and display them one by mean of one.which in position 1, after the displayed result stop us the process, then switch to position 2 and visualize the data.and it continue until position 4 at the end show us all the data in a graph or chart.

  your help is greatly appreciated

  If you use the manual sample positioning you can take one of the solutions, we have provided and add the following.

  Place the code we have provided inside (flat or stacked) sequence and add a frame before it.  In this new context, add a while loop that has a button connected to the termination of the Terminal.  Essentially, the code will sit and wait until you press the button to take the measurement. (You can add a timer wait small to reduce the use of CPU, let's say 50 or 100 ms).

  Note that this is a very raw solution as long as the treatment is inside this loop until the button is pressed.  This prevents any other code to run.  The event structures are more robust.

  If you want to automate this process, a motorized linear actuator would work well (although probably not worth it for 4 measures).

• Shift register do not reset

  Can someone take a look at this .vi and tell me why the shift register is not set to zero. It's a program that I am tempted to write to find areas of leading-edge information in a psd.

  Thank you

  Student researcher

  NVM. found the problem. my time was incorrrect loop condition.

• Reset the shift register according to value

  Hey there ' All, I've been thrown in Labview to work, so I made a few minis and other programs on my own. The last one I wrote is a system of notification by e-mail, because it's something I plan use a lot. The idea is, I have a random number generator that collects a number every 100ms using a timed loop. When a number is greater than a certain threshold, say 0.5, an email is sent and a flag is hoisted. From there on, I don't want to send any other emails until the number is BELOW a certain level, say, 0.3. With what I wrote, I get emails to send no problem and I can't stop them sending (using a combination of Boolean indicator and a shift register), but I can't understand how to to reset the indicator to allow emails as soon as I "roll" a number under my lower threshold.

  Is there a way I can achieve this? I would be deeply obliged for any help. I have attached the VI I did below, apologies, if it is messy and so on.

  Use a Select statement after the structure of the case.  If the reset condition is True, then thread a real constant through the register shift.  If the reset condition is False, then just wire the existing through the register shift.