How suspend you a shift register?

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• How to program the shift register to play only when a new user is detected user?

  Hello

  I'm currently developing a program of position control in labview. The program is quite simple, in which case the user will enter the distance on which he wants the table in the labview program and labview will send the signal to move a motor that will turn a ball screw to move a table horizontally to the targeted position. The criterion is that the profile of the engine depends on the distance to move, if a biphase (acceleration and deceleration) or three phase (acceleration, steady speed, deceleration) to reach the position of the target.

  The problem occurs when the user wants to enter a new entry second position) for the table, as the input by the user is the position that the table should be, but the necessary input to determine what profile the engine follows depends on the distance that the table moves to the target position. Therefore, I need a function to save the entry by the user temporarily and reminds that when a new user input is detected. Hereby, I would be able to use the difference of the input (input [n + 1] [n] input) and animal feed to determine what profile the engine follows and the entry by the user can be kept in the position he wants to the table to get (to compare with encoder).

  I thought to use for shift registers do, but I am not able to perform the deduction ([n + 1] - [n]) only when it detects a new entry. When I try to use registry to offset, it moves to the target location, and we only reached it will go to the original position. For example, when a user entry 90, this means that the table must be moved to the point 90. The shift register is initialized to 0, it will move to the point 90 (90-0 = 90), but arriving at 90, the shift register sends a signal of 90 (90-90 = 0) and the table back to its original position.

  Is it possible that I can delay the reading of the shift register only when a new entry is detected or there at - it another way for me to achieve what I want?

  I tried searching the forum site and neither discussion but could not find similar problems. Thank you for your help in advance.

  As I understand it, the use of shift registers with a structure of the event (to detect a user event when the user enters a new value) should solve the problem. Do not forget to post your request (or a version of it that isolates the issue) when you arrive at the lab, if we can get a clear visual of the issue you are facing.

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• initialize the shift register

  Hello

  How to initialize the shift register (inside the second loop for) so that it starts from 0 whenever the program runs. I tried to attach a constant 0 in left shift register, but which resets the registry whenever it passes through the inner loop.

  Thank you

  If you only want to reset once at the beginning and not for each iteration of the loop for external, you must add another register to lag on the outer for loop and wire lag 0 to that registry.

• 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.

• 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

• I have my serial number and that you wish to register online, but the message he has already been saved. How can I solve this? We just installed Adobe Acrobat Professional on my PC

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• 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

• How do you replace the values in a control of the ring?

  Hello

  I have a control of the ring (text values) in Labview 2009 and it is populated by programming with 20 points. The end user selects from the list items or fills with a new item based on a checkbox to add a new item. What I want to do, that's when section 20 is entered and the user selects the check box next to the point 21 to add I want around rap on the first item in the list and crush him. Is this possible and a sample VI would be great demonstrate how this is done. Any help would be greatly appreciated!

  Thanks in advance

  Charlie

  Replace the subset of the table.  You need to keep track of the element to replace in a shift register.  I would also like to keep the list in a shift register to make your life a little easier.

• 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

• 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).

• Increment of the counter without using the shift register

  Is there a way to count = count + 1, with using a node registry or feedback shift?  I have a structure of the event within a while loop (see table).  When the button is touched (kickoff event), a test is carried out.  Based on the test in the case of a string is passed outside of the event to a case statement of success/failure.  The user can run one of these 'events' as many times as he or she wishes.  And in an order any... Test1 (fails)... test 1 (fails)... again... test1 (go) again... test 2 (pass) then test 1 (pass).  When the stop button is reached, I would like to know how many times each pass/fail is produced for each test.  Try to stay away from shift registers, as I have 18 w key so a possible failure for each, so I would need 36 shift registers.

  In my example, the printed final statement would be:

  TEST PASS/fail # x test was performed

  test1 failure 1

  2 in case of failure test1

  Test1 pass 3

  Test2 pass 1

  4 passes test1

  etc.

  Do not see an attachment.

  You can use a shift register to store a table.

  Each button would be mapped to an array index. For each button pushed you the value of the index table, + 1, replace the array element and pass out back for the shift register.