Problem initializing shift register

Similar Questions

• 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

• 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?

  
  

• 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.

• 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.

• 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.

• 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

• 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

• 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.

• 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.