Shift register clear before rerunning the VI

Similar Questions

• 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.

• 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.

• 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

• 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

• 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.

• Change the shift register

  I hope someone can direct me on that. I'm stuck.

  The NTC, I want that it start at zero, enter the nested loop

  and when the case statement is equal to one, add 3000,

  so I have a lag on "undesirable" elements in the 1 d

  table I'm parsing...

  TIA!

  the nested loop shift register is not initialized, it is best to initialize it with a constant 0

  Then, the index entry Array subset function is connected after or before the function incriment? I can't decide...

  in any case, if it is connected before the incremint function then the nested loop iteration 1 will send a 0 at the entrance to the function of the subset of the Array index.

  EWW! , a lot of entries, words of functions in the above paragraph lol , can u get me here?

  now I can just understand the problem what exactly you're talking about. the sequence of events for your code will be like this:

  After the nested loop is complete, the output will be available (1 d the function add array element table) to the structure of the case where you want to add 3000 to the value of the shift register and start the loop nested with initialized again records with value = 3000? Am I wrong?

  If I'm right, you must reset the shift by using a control register, create a local variable to him and place it in the business structure then her manipulate it as shown below:

  

  Since the default data of "N" type digital command value is 0, then initially the shift registers will be initialized with 0 as the guy above

  Thank you

• 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

• 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!

• 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.

• 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.

• 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.

• replacement for the shift register chart

  Hi, I have a problem with my program. When I want to compare graphic legend with my present graphic it just overwirite with my graph.when this I add new data and then run my program it does not work. I tried to use shift regoster but my program eror. where should I put the register shift in my program? Thank you

  The shift register is on the outer loop. There is no need for inside while loops. The shift register should contain a table of two parcels of xy. Replace one of the accrding of two plots to the key using "replace the subset of the table".

  You don't need to a simple shift register, you have far too much code.

  (Please attach the screws with a suitable name. My Downloads folder already contains dozends of files "x.vi no title")

• Store the Teststand ActiveX reference in the LV shift register

  It is posted here

  I'm trying to store the references TestStand ActiveX in a shift register not initialized a VI.  In my case, the references are passed into the TestStand VI (not created from in VI).  If I call the same VI to the next step (same sequence and execution of the previous step), since the shift register ActiveX references are not valid.

  The VI remains reserved to run during these two stages and is not unloaded from memory, so the shift register data should remain intact (in fact, the numerical values of the references are actually kept).  LabVIEW is still trying to close any ActiveX reference, even if they were not created from the VI?  Is there a way around this problem?  Or I'm just something wrong?

  

  jsiegel-

  In general, when the code is passed a COM reference and code is to keep the reference to a global or shift register, even after his return from the call, the code must add a reference to the object so that the object server knows that the object must be destroyed not. It is also the responsibility of the code that fits on the reference in the world or a register shift to release the reference to the object when it is no longer necessary.  LabVIEW is not different from any other language.

  So, here are more details. TestStand application LabVIEW for run the VI, TestStand after the reference as a parameter to the method of the server to run the LabVIEW VI. COM creates a proxy for reference and give the reference of proxy for the code module. Your VI then stores the value of the proxy reference in the global or the shift register. When your VI ends and returns to TestStand, COM releases the proxy reference, the value in the global or the shift register is no longer valid.

  Basically, you need to add or duplicate the reference to the object passed in LabVIEW by calling the VariantToData function. Pass the existing reference, set the input type to the same type of the reference, and the result will be the reference in doubles. You can assign the double reference to worldwide or register.

  Normally you must well to release the reference later by reading the value of the global or shift register, explicitly calling the function close reference with which to reference, and then assign A Refnum Constant stepped up to the global level or shift register to nullity. In the case of a module of code, I believe that when TestStand unloads the VI, LabVIEW frees the reference correctly. If this isn't the case, you will get a debug message to unpublished during the TestStand stop object if you have this option enabled.

• Anyway to save the data in a loop without using shift register or feedback loop

  Hi all

  I've been thinking, is it possible to save data in a loop to the next iteration without the use of a shift register or a feedback loop?

  I need the possibility to reload the data within a loop from a file of lvm, but I want to use the same data until I have load some new.

  The reason why I don't want use the node registry or feedback shift is due to speed, although I don't know if the registry change

  in fact moves all the data of one register to another, or if it is stopped until a change data occur.

  I want to reuse data medium and large (6 measure of strength, pressure 2, 1 flow channels) of about 10 s data in each file with samplingsrate of 2 kHz...

  In my testing program, I have several CPU demanding computations and 3D graphics, so I want just to minimize the CPU loading as much as I can for each part of the

  software...

  I am enclosing a small VI to explain what I mean.

  I have now, I shouldn't use the express VI and I'll change that as well - it's just a proof of concept!

  Hope you guys can help me better understand this shift register...

  Thank you!

  -Tommy

  If the speed is the name of the game, go with the flow (data) and stick to a shift of registers or feedback node.  No data is moved, their job is to do pretty much exactly what you describe.  Any other solution, control/locals/globals will imply a copy of the data, and then you will have problems with speed.