Shift register do not add

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• 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.

• 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

• 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).

• 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.

• Problem initializing shift register

  Hello

  I'm using labview 7.1 to control a spectrometer. In my program, I have a certain function that runs 32 times and adds up, and all works again n times. I use two 'loop' for s in the other. In the inner loop, I have a shift register to send information from one iteration to another, so that it adds. But I guess there is a problem with the shift register because I get just a blank chart after the loop runs. I used an indicator to check the output of the function, and it certainly works. I guess that the problem happens when I add up the results of two iterations. Sometimes when I would forget initialize the shift register that I use to have this problem, but this time I can't initialize the shift register because the option create a constant is disabled for some reason any. can someone help me please! Thank you!

  labview_dummy wrote:

  I'll try using a case structure and see if it helps.

  Seems to work very well here:

  

  labview_dummy wrote:

  I used the flat sequence structure because I donot want the spectrometer close vi to run before the loop that are parallel and not in series (they are all two cables to the first envelope and not to each other. Thank you!

  Can do "in series" by plugging the entrance in the loop FOR, for example.

• 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.

• 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

• 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!

• 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.