determine the difference in phase between a reference signal and measure

Similar Questions

• The best way to draw the difference of phase vs. time of two signals

  I have three channels (time, force, displacement) of approximately 1 000 seconds of a sinusoidal test load test data (sampling rate was 100 Hz).  I would draw the difference in phase between the force and displacement (perhaps using a second window 30) according to time.  I tried using a few different analysis functions, but I get what looks like random noise (phase difference between force and displacement is very small and the difference in amplitude is very fantastic - 4 orders of magnitude).  Any suggestion would be appreciated! -Jim

  Hi Jim,.

  The phase channel resulting, that you get with the function FFT DIAdem is in the frequency domain.  If you select a channel data and time in the FFT dialog box, you will get a frequency channel that results as well as the Phase channel resulting.  If you select only a weather channel of waveform data in the FFT dialog box, you will get a Phase of waveform frequency channel resulting.

  Brad Turpin

  Tiara Product Support Engineer

  National Instruments

• How to measure the difference in time between the passage by zero of a sine wave and the rise time of a pulse on a same graphic?

  I have a sine wave of 50 Hz and a pulse of the signal on the same chart. The difference in phase between the two is between 0-90 degrees.

  Now I need to calculate the time difference between (when the sinusoidal wave passes through zero volts) and (when the pulse increases). The frequency will remain about even for the two signals.

  The request is for a three-phase generator. In simple terms, when the difference in time between the passage to zero of the sine wave and pulse increases increases, it means that the load on the generator has increased.

  I am a novice user of LabView (version 9, 2009), maybe it's a very simple problem but I was pulling on my hair for the past few days and couldn't understand anything. Any help would be greatly appreciated. I use DAQ USB-6008 to measure these tensions and the impulse of the generator and a sensor

  I have attached a jpg file (a graphic that I just did with excel to explain). The time 't' is what I'm trying to measure

  

  See you soon

  Zdzislaw

  Awais.h,

  For problems of this kind I recommend start writing the granular steps you would take to manually fix this problem.  You can't say LabVIEW (or any programming language) If you can't succinctly describe the solution to your problem.

  The I want to address this problem is to:

  1. find all the zero crossing points and edges on the rise
  2. for every rising edge find the difference between the timestamp and previous passage by zero

  Here is an implementation of this algorithm LabVIEW:

  

• How to correct the difference of phase because without delay?

  Hello

  I use the DAQ card 6024E to acquire three sine waves of 50 Hz sampling rate of 20 kHz, sample to read: 2000Samples. I need to view and measure the difference in phase between the signals. However, I understand that there is without delay, leading to the release of phase measurement. Taken from the book of graphical programming LabVIEW, it is advisable to entry for common signal and display on the same graph. use graphic cursors to measure the timing of the track-to-track of tilt. Then apply timebase correction values can be applied to data from setting X 0 (initial time).

  What I tried that I have attached in the printscreen. But I don't know how to include the other channel in the same graph to compare the signal. Can someone enlighten me please how to do this?

  I'm using labVIEW 8.0.

  Your help is greatly appreciated.

  POH, (NTU student)

  Hi, thanks for the advice so far... but I always failed to implement the VI of tbd.

  Here's how I solved it. I tried to change the start time in the book cited, but no change for her, even though the gap is increased, there still no change the displayed waveform. timestamps there were used to check that there is a change in the offset of seizures.

  The need to implement so just align vi available in the library will work with an accuracy of all 0.01degrees of the waveforms.

  Well this problem solved!

  See you soon!

  POH

• How to make the difference in percentage between two matrices?

  Hello everyone, I just want to ask how to make the difference in percentage between two binary matrices. This is to compare two images converted to matrices in LabView. Thank you.

  Well, in the current image interesting pixels are zero in the reference image. Here is a small example that would take all the zero 100% pixel, then look at the pixels of same in the other image to see how are always 0.

  Modify if needed.

  

• What is the difference (if any) between iPhoto and iPhoto library? I managed to delete my pictures in freeing up space, but have a backup available. Do I have to download these two files in the pictures folder is one duplicate of the other?

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• How to find the difference "within weeks" between two date values?

  Hi all

  Jdev version 11.1.1.7.1

  I used two < af:inputDate > & a < af:inputText > < af:panelFormLayout > components. My requirement is, I want to display the difference of weeks between these two day values in the inputText component when the user clicks the shape.

  Any suggestion would be appreciated.

  Kind regards

  
  

  Novel

  You can use this method to get the number of days between the day and date, and then divided by 7 to get the number of weeks

  public static long getDifferenceDaysBetweenTwoDates (d1, d2 oracle.jbo.domain.Date oracle.jbo.domain.Date)

  {

  If (d1! = null & d2! = null)

  {

  return (d1.getValue () .getTime () - d2.getValue () .getTime ()) / (24 * 60 * 60 * 1000);

  }

  return 0;

  }

  Check - http://sameh-nassar.blogspot.in/2014/10/dealing-with-dates-in-java.html

  Ashish

• What is the difference in features between the domain of Adobe Acrobat Standard and Pro controller?

  What is the difference in features between the domain of Adobe Acrobat Standard and Pro controller?

  See: http://www.adobe.com/products/acrobat/product-comparison.html

  Monday, June 1, 2015-16:00, jeanm18631066, [email protected]>

• find the difference in days between two dates

  Hello world
  
  I'm trying to find out the difference in days between two dates and the execution of the query that I'm passing
  
  SELECT TO_char(sysdate, 'dd/mm/yyyy') - TO_char('15/11/2011', 'dd/mm/yyyy') DAYS FROM DUAL
  
  the error I get is
  
  ORA-01722: invalid number
  01722 00000 - "invalid number."
  * Cause:
  * Action:
  
  Could someone please help.
  
  Thanks in advance

  user10636796 wrote:
  Hello world

  Thanks a lot for all the replies. I am trying to apply it in a statement to my table like this

  SELECT trunc (sysdate) - TO_char (date_last_recommended, ' dd/mm/yyyy') DAYS OF recommendation;

  SELECT trunc (sysdate) - TRUNC (date_last_recommended) DAYS OF recommendation;

• Calculate the difference in days between two Dates

  Hello

  I'm trying to understand how to calculate the difference in days between two dates using JavaScript in LiveCycle. (Minimum = JavaScript knowledge)

  Where 'Start_Date' and 'Current_Date' are the names of the two dates in the palette of the hierarchy. (the two Date/time field)

  * Current date is using the object > value > execution property > current Date/time

  I need a text or number field showing the difference in days. (Difference_in_Days)

  I noticed the following code is pretty standard among other responses:

  var

  Start_date = new Date (Start_Date);

  var

  Current_Date = new Date (Current_Date);

  var

  nAgeMilliseconds = Current_Date.getTime) - Start_date.getTime ();

  var

  nMilliSecondsPerYear = 365 * 24 * 60 * 60 * 1000 ;

  I know there is lack of code and code above are may not be not correct.

  Please notify.

  OK, that's because of the way that javascript and works of the calculate event.  The field will be filled with whatever the script resolves at the end of execution. Technically, your script does not have a value because the last thing you do is an assignment to a variable.  Change the last line as follows:

  Math.ABS ((firstDate.getTime)

  ((- secondDate.getTime (()) / (oneDay));

  (eliminate the variable assignment) and get rid of the app.alert.  Your script will "return" (have) regardless of the value of calculation from the East and which will be stored in the field.

• Get the difference in days between 2 dates in Essbase 9.3.1

  How to calculate to get the difference in days between 2 dates in the format day of Essbase 9.3.1.la is 20090101 (yyyymmdd). If 2010907-20100917 = 10 days.
  Start and end dates may extend over several years.

  @DateDiff does not exist in the 9.3.1 I think. With this assumption...

  Could you use the CDF 'Functions of Date'? According to the docs (I've never used it):
  @JdaysOffset - returns the number of days between the firstDate and the secondDate.

  See samplecode.oracle.com for the CDF. Just search for CDF and you will find it on the second page (I think) of the FMCS.

  Or see this thread for another approach that I think the CDF under the covers: Re: Formate Date.

  Of course, you could spend to 11 for DateDiff. :)

  Kind regards

  Cameron Lackpour

• Validate the difference in days between the dates in a table

  Greetings,
  
  I have a tabular form and date on each line. When the user saves the form, I want to sort the rows by chronological order of these dates, then go get the first and last date, calculate the difference between them and validate it.
  
  Does anyone know how to do?
  
  From now on, I came up with this:
  
  1. a PL/SQL Page process "before the calculations and Validations' which creates a collection and inserts lines in my tabular form (p_c004 being the date field):

  begin   
    if apex_collection.collection_exists('UPDATE_DATES') then
      apex_collection.delete_collection('UPDATE_DATES');
    end if;
  
    apex_collection.create_or_truncate_collection('UPDATE_DATES');
  
    for i in 1..apex_application.g_f04.count
    loop
        apex_collection.add_member(
            p_collection_name => 'UPDATE_DATES',
            p_c002            => htmldb_application.g_f02(i),
            p_c003            => htmldb_application.g_f03(i),
            p_c004            => htmldb_application.g_f04(i),
            p_c005            => htmldb_application.g_f05(i),
            p_c006            => htmldb_application.g_f06(i),
            p_c007            => htmldb_application.g_f07(i));
    end loop;
  end;

  and
  
  2. a "PL/SQL Error: Validation

  DECLARE
    l_count NUMBER;
  BEGIN
    apex_collection.sort_members('UPDATE_DATES', 4);
    l_count := apex_collection.collection_member_count('UPDATE_DATES');
  
    SELECT
    to_date(c004(l_count),'DD/MM/YYYY') - to_date(c004(1),'DD/MM/YYYY') Difference 
  
    FROM 
    apex_collections 
  
    WHERE 
    collection_name = 'UPDATE_DATES'
    and to_date(c004(l_count),'DD/MM/YYYY') - to_date(c004(1),'DD/MM/YYYY') < 30;
  
  EXCEPTION
    WHEN NO_DATA_FOUND THEN
      raise_application_error(-20101, 'Error');
  END;

  In my understanding, the validation must sort the collection in chronological order of the dates entered, calculate the difference in days between the nearest and farthest date and return it if it's less than 30 years (the difference in days allowed). If it's more than 30 years, the SELECT statement does not return anything and my block of exceptions raises a custom error that will trigger the validation "process of refusal."
  
  But when I try this, it tells me c004 does not exist and if I drop the parentheses, it works, but of course, it does not give the desired result.
  
  Any ideas?
  
  Best regards
  Mathieu

  Assuming that your domain is c004 then I don't see the need for the sorting of the collection.

  You can write a simple SQL select that sent you the desired result.

  Maybe something like this not tested

  SELECT
    max(to_date(c004,'DD/MM/YYYY')) - min(to_date(c004,'DD/MM/YYYY') ) Difference
    FROM apex_collections
    WHERE collection_name = 'UPDATE_DATES';
  

  The test if the result is > 30 days. If that's what you want to know.

• Unlike COUNTY and the difference in data between 2 sets of data

  Hey guys,.
  I have a question here, probably, it revolves around the UNION, INSERSECT and LESS and maybe something else I'm missing.
  
  I'm two sets of questions, and I have 2 paintings.
  

  select count(student_id) from adhoc_student_table1;
  select count(student_id) from adhoc_student_table2;
  
  select student_id from adhoc_student_table1 order by student_id;
  select student_id from adhoc_student_table2 order by student_id;

  These tables contain data and have been sorted by student_id. I want to find the difference in NUMBER between the first set of SELECT and I want to find a difference in the student ID in the second set to SELECT.
  
  Any ideas? As always, any help is appreciated!

  Hello

  These tables do contain data and have been sorted by student_id. I want to find the difference in COUNT between the first set of SELECT and I want to find a difference in Student IDs in the second set of SELECT.
  

  Hope this helps

  select (select count(student_id) from adhoc_student_table1) -
     (select count(student_id) from adhoc_student_table2) from dual
  
  ---If you want the difference in first table not available in second table.
  
  select student_id from adhoc_student_table1 order by student_id
  minus
  select student_id from adhoc_student_table2 order by student_id
  
  -- or you wanted the difference from both table.
  with tab1 as
  (
     select student_id from
        adhoc_student_table1
  )
  tab2 as
  (
      select student_id from
  adhoc_student_table2
  )
  select * from (select student_id from tab1
                       union select student_id from tab2)
  minus
  select * from (select student_id from tab1
                      intersect select student_id from tab2)
  )
  

  If you could post the input samples and expected, maybe you can get a more valuable solution.

  Concerning
  Anurag Tibrewal.

• Align the two signals and measure the Phase Shift

  Hello

  I do an experiment in which I use the NI USB-6221 DAQ card. The jury is able to make 250 k samples/second. I want to measure two voltages in a circuit and find the phase shift between them at frequencies between 1 and 10000. First I ouputted a wave sinusoidal frequency variable through the Commission and applied to a test circuit. Then I used the Board to measure the two tensions consecutively (thus reducing the maximum sampling frequency at 125 k). I used the signals align VI and measured the two phases and then calculates the phase shift (VI attached in Phase 1). It worked well for the test circuit I built in which the phase shift went way logarithmique.20 degrees ~84.5 degrees and then stabilized. At frequencies above 5 000 Hz phase shift must have remained constant, but it varies more or less 1 degree. When the phase shift is 84.5 degrees, present a degree of variability is not particularly explicit. When I asked my program on the circuit that I really wanted to measure, the phase shift went from-. 5 degrees up to about 1.2 degrees. The change in the values of phase shift at high frequencies (> 3000) was environ.2 degrees. Given the small phase shift, this variation is unacceptable. Now I tried to use a sequence to each blood individually (increase the maximum sampling frequency to 250 k) and then align the two signals and measure the phase of each shift. When I use align it and re - sample Express VI to realign the two signals, I get the message "error 20333 analysis: cannot align two waveforms with dt even if their samples are not clocked in phase." Is it possible to align two signals I describe here? I enclose the new VI as Phase 2

  Matthew,

  I think I have an idea for at least part of the problem.

  I took your program data and deleted stuff DAQ.  I have converted the Signal on the chart control and looked then what was going on with the signal analysis.

  The output of the Waveforms.vi line has two waveforms, like the entry.  However, arrays of Y in the two waveforms are empty!  It does not generate an error. After some head scratching, reading the help files and try things out, that's what I think is happening: the time t0 two input signals are 1,031 seconds apart. Since the wavefoms contains 1,000 seconds of data, there is no overlap and may not align them.

  I changed the t0 on two waveforms are the same, and it lines up.  The number of items in the tables is reduced by one. Then I increased the t0 of 0.1 seconds on the first element. The output had both greater than the entry by dt t0 t0 and the size of the arrays was 224998.  Reversing the t0 two elements shifts the phase in the opposite direction.

  What that tells me, is that you can not reliably align two waveforms which do not overlap.

  I suggest that you go to 2-channel data acquisition and that it accept the reduced sample rate.  You won't get the resolution you want, but you should be able to tell if something important happens.

  You may be able to improve the equivalent resolution by taking multiple steps with a slight phase shift. This is similar to the way that old oscilloscopes of sampling (analog) worked. Take a series of measures with the signal you are currently using.  The make enough average to minimize changes due to noise. Then pass the phase of the signal of excitement to an amount that is smaller than the resolution of phase of sampling rate and repeat the measurements.  Recall that I calculated that for a 5 kHz signal sampled at 125kHz, you get a sample every 14.4 degrees. If shift you the phase of 1 degree (to the point/mathematical simulation), you get a different set of samples for excitement.  They are always separated by 14.4 degrees.  Take another series of measures. Transfer phase another degree and repeat.  As long as your sampling clocks are stable enough so that frequency does not drift significantly (and it shouldn't with your equipment), you should be able to get near resolution of what you need.  The trade-off is that you need to perform more measurements and may need to keep track of the phase shifts between the various measures.

  Lynn

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