A shift register reset

Similar Questions

• Shift register resets

  Hello.

  I have a problem with the shift registers (you will find the attached vi).

  I use three shift registers, each of them carries different things:

  1 number

  2 file (use dialogue)

  3 error in (for the file).

  The while loop has a structure inside the event.

  The problem, I'm facing is that whenever the values of the (2) and (3) changes, the change number (1) register, resets. Why is this? Can I prevent it somehow?

  Thank you very much

  Vasileios.

  You provide the value to the exit tunnel.  And given that the tunnel is configured to use default if Unwired, then the default is to enter into the shift register.  Two suggestions here:

  1. Right-click on the tunnels of the output of the structure of your event and uncheck the use default if Unwired.
  2. Right-click on the tunnels to exit and go to the Tunnel exit linked-> create and wire Unwired case.  Then select the entry tunnel which must match the exit tunnel you use.  This will automatically associate the tunnel entry and exit for each case that you do not have an output value for.  It also automatically wires upward in any new case that you create.

  A few other comments:

  1. You need not wait in the while loop.  The event structure will limit your rate of loop for you.
  2. DO NOT USE THE BUTTON ABANDON!  You must stop your loop somehow.  I would recommend passing on the TRUE value in the Stop Logging event and that wire to the conditional.  This way your VI will stop gracefully rather than act as a car stopping running into a tree.

• 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.

• 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.

• 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.

• 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

• Problem backup a table using a shift register

  In the attached VI I'm build a table 1 d of channels, then try to record using a shift register. For some reason any is not backed up my data and I don't understand why.

  Thank you.

  You must connect the table where all empty, else it will reset to an empty array whenever such a case occurs.

  You also need a registry change on the inner loop.

• Reference stored in shift register

  Hi, I'm having trouble with references. The LabVIEW Wiki reads control references (http://labviewwiki.org/Control_References) are the right way to switch the user interface elements in subVIs. So, this is the path that I took.

  I have this ActionEngine UI (see attachment) I want to handle specific tasks. One task is to add the system messages to a simple text field that is in the user interface. I want to use this same ActionEngine UI in several subVIs to add messages to the same text field. So what I tried to do was having an operation of initialization for the user interface AE that stores a reference to the text field in a shift register. It did not work. Whenever I call UI AE to add system in the subVIs messages that I have to go again in the reference. It defeats the purpose of using a shift register, because I still have over the reference in each unique Subvi using EI, when in reality I want to only do as part of initialization.

  My guess is that the reference is released somehow. How I would solve this good?

  You have the reference wired for change of registry entry.  It gets reset to values each time Wired EI is called.  You should put the reference control in the case of 'initialize' and leave uninitialized entry to the unwired (for a shift register) SR.

• 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

• 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.

• 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