find the highest amplitude of a signal

I have to find the value of the highest amplitude of my signal. This should be done automatically without using cursors

What is the data type of your signal (Dynamics, wave form, table, etc.).

For example, try:

Table & Min Max

Waveform Max & Min

Statistics VI express

Tags: NI Software

Similar Questions

Find the time delay between two signals

In this code, I sent you a pulse in the acquisition of data using labview. The pulse triggers a thermistor to another analog input input data.

I want to find the delay between the time of the positive edge of the switch, and the exact time of the DAQ begins to take in the signal of the temperature probe.

I enclose the code below.

Thank you.

Hey,.

Here is a link to some information about the sources of clock on the FPGA.

http://digital.NI.com/public.nsf/allkb/94CE338CC80ACF008625791400516BAA?OpenDocument

Hope this helps

find the highest in equal to the data

Hello
I need to find the top tied at the rank of gvien to the sort order for each grade is also given
could you please help me in the present.

with t1 as 
(select 'AB' t1_cd , 'A+' t1_rk from dual union all
select 'BC','BBB-' from dual union all
select 'CD','B-' from dual union all
select 'DE','AA' from dual union all
select 'EF','BB+' from dual),
 t2 as
 ( select 'AB' t2_cd ,'A+' t2_rk, 1 ordr FROM DUAL union all
   select 'AB' , 'A'  ,1 FROM DUAL UNION ALL
   select 'AE' ,'BBB-',3 from dual union all
   select 'BC' ,'BBB-',4 from dual union all
   select 'CD' ,'B-'  ,5 from dual union all
   select 'DE' ,'AA'  ,2 from dual union all
   select 'EF' ,'BB+' ,6 from dual )
SELECT * FROM T1,T2
WHERE t1.t1_cd= t2.t2_cd   -- I need to find the  code and rank which have rank greater than equal to BBB-

Here, in the above query, I need to find records that have classifications > = 'BBB-' sort order is alos in the ordr T2 table column

Output expected

T1_CD     T1_RK     ORDR                            final_Rk

AB     A+     1  < BBB- i.e;4 so output  L
AB     A+     1  < BBB- i.e;4 so output  L
BC     BBB-     4  = BBB- i.e;4 so output  E
CD     B-     5  > BBB- i.e;4 so output  H
DE     AA     2  < BBB- i.e;4 so output  L
EF     BB+     6  > BBB- i.e;4 so output  H

Thank you

Published by: Smile on May 8, 2012 07:36

with t1 as
(select 'AB' t1_cd , 'A+' t1_rk from dual union all
select 'BC','BBB-' from dual union all
select 'CD','B-' from dual union all
select 'DE','AA' from dual union all
select 'EF','BB+' from dual),
 t2 as
 ( select 'AB' t2_cd ,'A+' t2_rk, 1 ordr FROM DUAL union all
   select 'AB' , 'A'  ,1 FROM DUAL UNION ALL
   select 'AE' ,'BBB-',3 from dual union all
   select 'BC' ,'BBB-',4 from dual union all
   select 'CD' ,'B-'  ,5 from dual union all
   select 'DE' ,'AA'  ,2 from dual union all
   select 'EF' ,'BB+' ,6 from dual )
SELECT t1.*, t2.*,
            case when t2.ordr > t3.ordr then 'H'
                   when t2.ordr = t3.ordr then 'E'
                   when t2.ordr < t3.ordr then 'L'
            end code
FROM T1,T2, (select max(ordr) ordr from t2 where t2_rk = 'BBB-') t3
WHERE t1.t1_cd= t2.t2_cd;

T1_CD T1_RK T2_CD T2_RK       ORDR CODE
----- ----- ----- ----- ---------- ----
AB    A+    AB    A+             1 L
AB    A+    AB    A              1 L
BC    BBB-  BC    BBB-           4 E
CD    B-    CD    B-             5 H
DE    AA    DE    AA             2 L
EF    BB+   EF    BB+            6 H   

6 rows selected.

9215 is 16-bit adc and I can not find the highest resolution of it

Dear Sir

I use the signal 0-5 v dc for a majority use, but I saw that the control is done in 5 steps only (on the other resolution of 1 volt in a way) how can I do

Thanks I realize

The highest sum of the values in any 3 consecutive dates.

Hello
Given
create table test (colamt COMP, coldate date);
Insert test values (10, sysdate + 1);
Insert test values (20, sysdate-1);
Insert test values (5, sysdate-5);
Insert test values (25, sysdate);
Insert test values (35, sysdate-11);
Insert test values (10, sysdate + 1);
Insert test values (1, sysdate-1);
Insert test values (5, sysdate-2);
Insert test values (20, sysdate-3);
Insert test values (10, sysdate-4);
commit;
SQL > select * from test by coldate;
COLAMT COLDATE
---------- ---------
35 19 MARCH 15
5 25 MARCH 15
10 26 MARCH 15
20 MARCH 27, 15
5 MARCH 28, 15
20 MARCH 29, 15
1 MARCH 29, 15
25 MARCH 30, 15
10 MARCH 31, 15
10 MARCH 31, 15

I need to find the highest sum of colamt for the whole period of 3 days.
In this case, it's 50, for the first 3 lines, 19 to March 26.
(In the real world, I need to know the amount the highest recovery archived how many bytes were generated for consecutive periods of 3 size my FRA v$ archived_log)
I'm just a DBA and scratch my head to the analytical functions available for me. : ()

Hello

To display the best total of 3 rows and the date range when the total highest has been reached:

WITH got_total_colamt AS

(

SELECT coldate

SUM (colamt) OVER (ORDER BY coldate

LINES BETWEEN 2 PRIOR

AND CURRENT ROW

) AS total_colamt

, LAG (coldate, 2)

COURSES (ORDER BY coldate

) AS start_coldate

OF the test

)

got_rnk AS

(

SELECT coldate, total_colamt, start_coldate

, EVALUATE () OVER (ORDER BY total_colamt DESC) AS rnk

OF got_total_colamt

)

SELECT total_colamt

start_coldate

coldate AS end_coldate

OF got_rnk

WHERE rnk = 1

;

Output:

TOTAL_COLAMT START_COLDA END_COLDATE

------------ ----------- -----------

50 19 March 2015 March 26, 2015

If you do not need to display the dates, the above query can be shortened a little.

Etbin showed how to get 3 days in response #1, usinng the analytical SUM function totals. The analytic RANK function, estimated above, which of the totals is higher (or, in the event of a tie, lines 2 or more have the largest number). Because you cannot nest analytic functions, the analytical SUM function should be called in a separate subquery before calling the analytic RANK function. Since the analytical functions are calculated after the WHERE clause has been applied, you cannot use in a WHERE clause. That is why RANK is compuuted in another subquery and its results used in the WHERE clause of the main query.

Error to find the frequency of signal to tachometer

I use the vi "Extract only your Information" to find the speed of a motor. The entrance is a monarch Laser Tach AC output. Tach reading is very regular and the AC output is 5 volts peak. There are a few negative pics, but these are at the speed of the engine. My output frequency, who should read the changes back from 1800 1800 at a frequency near 5400, but not exactly 5400 (5321, etc.). Y does it have no signal conditioning that I'm missing before signal to the vi "extract", for example, the window it and then use the frequency domain cutting-edge research?

Thanks in advance,

Ron

Matt,

Just to let you know that I put a bandpass filter in my vi. I took the input of the speed estimated of the user and put in parentheses of the bandpass filter to be 25% below 25% above the target and the input of speed tachometer works very well!

Ron

Using an ATI TV Wonder HD 750 PCIe and the Media Center can't find the analog signals of the conductor not to use a decoder

I just installed a diamond ATI TV Wonder HD 750 PCIe on my 64 bit Windows 7 system (it's actually the second tuner card I tried) and still get the same problem. Media Center recognizes the card and both tuners analog and digital, but fails the analogue listening. I get digital very well. I tried to add a channel manually and still no luck. I use cable TV with no cable box and tested the signal on my TV and receive all channels.

I also downloaded the latest Diamond driver and installed all updates more recent windows.

In Media Center, I tried to automaticlly and manually select my signal and each combination, but available with no luck.

Can you please help

Rob

Because, for an analog tuner, Media Center requires that said tuner is an MPEG2 Encoder (or that the manufacturer creates a software encoder which emulates this.) The Theater HD 750 chip is not this, and it seems (after some looking) that ATI has not created a software encoder (side note: they are unlikely to do so, because they left the tuner market.)

Out of curiosity, what was the 'other tuner"which has been tried before?

Honestly, my recommendation would be to return the tuner (if possible) and get a combo tuner or hybrid that is known to work properly. My first recommendation would be for a Hauppauge HVR-2250 (Council dual combo/hybrid), because it works wonderfully in MC as a tuner NTSC/Clear-QAM of combo/hybrid. MC knows how to treat as what it is, and it works very, very well (I have not one.)

That, or an AverTV Combo G2 (which is not as flexible as the ' 2250, where its slightly lower price.)

HTH,
Chris

MS - MVP (Media Center) [If this post can help solve your problem, please click the 'Mark as answer' or 'Useful' at the top of this message.] [Marking a post as answer, or relatively useful, you help others find the answer more quickly.]

I want to detect and use the video entering via the HDMI port on the computer. I was told the HDMI port was not set, but I can't seem to find the signal or the video.

I want to detect and use the video entering via the HDMI port on the computer. I was told the HDMI port was not set, but I can't seem to find the signal or the video. Any suggestions?

As Tom SC replied, you must contact the company that manufactured your computer. In general, the ports HDMI on most computers is an out-bound port, send video / digital audio output.

You should probably use a video capture card that can capture from HDMI source,

find the nth highest salary

Hi guys, I'll try to find the nth highest salary
using the sub query, is there a case in which the following query may fail
Select
* de (
Select
distinct rownum rn,salary from emp_mgr order by rownum) t where t. rn = 3
;

I do something like that, but he got 3 selects, so I'm a worried buit, trying to optimize it

Select * from)

Select rownum rn, t t.salary (salary select distinct from emp_mgr by salary desc)

) rs

where rs.rn = &n;

To find the max() in a max() or the highest value to 1row-1col w. multiple values
Hello

I'm trying to get a set of results that has following criteria
-the lines using the same column WEL_FK, I would like to than the more recent date
and in the most recent date of the highest weight. (Sorte d'un max() dans un max()) for each WEL_FK).
s ' there is a line that has no other matches WEL_FK. s (such as line 8)
It should still return this line.

PER_PK WEL_FK A_DATE WEIGHT
---------- ---------- ---------- ----------
1 1 4 22 - 10 - 2010
2 1 4 2010-11-23
3 1 6 2010-10-22
4 3 2 2010-10-22
5 3 2 2010-11-23
6 3 2 2010-10-22
7 4 5 22 - 10 - 2010
8 6 5 2010-10-22
9 6 6 2010-10-21
10 6 2010 - 10 - 22 4

10 selected lines.

Expected RESULTS
PER_PK WEL_FK A_DATE WEIGHT
---------- ---------- ---------- ----------
2 1 4 2010-11-23 because there the last date within the same foreign key even if not highest weight
5 3 2010 - 11 - 23 2 because he has the last date within the same foreign key, wts. are equal
7 4 5 22 - 10 - 2010 because there is no other line with this WEL_FK
8 6 2010 - 10 - 22 5 because he has the date the most recent with higher weights.

I am able to get close to what I want using the SQL below gives me the correct lines
but don't tell me does not 'weight' or the 'PER_PK' column
I tried to find a way to choose the highest value of 1 ROW and 1 column
with multiple values, but think that I dug myself in a hole.
When I add in the PER_PK above, it sure gives me again every 10 rows.

SELECT wel_fk,
LTRIM (Max (SYS_CONNECT_BY_PATH (a_date, ','))
DUNGEON (DENSE_RANK LAST ORDER BY curr), ',') as employees
FROM (SELECT wel_fk,
a_date,
ROW_NUMBER() over (PARTITION BY wel_fk ORDER BY a_date) AS curr,
ROW_NUMBER() over (PARTITION BY wel_fk ORDER BY a_date)-1 AS prev
TEST)
GROUP BY wel_fk
CONNECTION BY prev = curr PREREQUISITE AND wel_fk = PRIOR wel_fk
START WITH curr = 1;

WEL_FK EMPLOYEES
---------- --------------------------------------------------
1 2010-10-22,2010-10-22,2010-11-23
3 2010-10-22,2010-10-22,2010-11-23
4 on 10-22-2010
6 2010-10-21,2010-10-22,2010-10-22
==========================================================

I explored analytical, connect by is, subqueries, wm_concat, SYS_CONNECT_BY_PATH, etc. with max () for days and get real close
but somehow it is turned off.

If you are able to provide the actual SQL solution, ideally using what I already have, would be appreciated.

Thank you.

the_sql
Try this:
```
WITH src AS (
SELECT per_pk,
       wel_fk,
       a_date,
       weight,
       MAX( a_date ) OVER (PARTITION BY wel_fk ORDER BY a_date
                  RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) latest_date,
       last_value( weight ) OVER (PARTITION BY wel_fk ORDER BY a_date, weight
                  RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) max_weight
FROM TEST
)
SELECT per_pk,
       wel_fk,
       a_date,
       weight
FROM src
WHERE a_date = latest_date
     AND weight = max_weight
;

PER_PK                 WEL_FK                 A_DATE                    WEIGHT
---------------------- ---------------------- ------------------------- ----------------------
2                      1                      2010/11/23                4
5                      3                      2010/11/23                2
7                      4                      2010/10/22                5
8                      6                      2010/10/22                5        
```

Query to find the nth highest salary...

Hi guys,.

I can't understand this query that we got a Web site. It is used to find the nth highest salary of the employee... Can someone explain to me please each and every part of it...

The Charly is:

Select distinct (a.salary)
EMP a
where 1 = (select count (distinct (salary))
b emp
where a.salary < = b.salary)

The "BOLD" part which is the return value... is my main question about this query...

Help, please

It is called a correlated sub query. The inner query is executed for each iteration of the outer query. For example, if the table emp has 14 rows and then for each salary 14 the inner query will run.
The part

select count(distinct(sal))
from emp b
where a.sal<=b.sal

actually counts the number of treatments that are less than or equal to the treatment of the outer query.
Clearly for the King, who gets the highest salary is greater than or equal to 1 (himself). It returns the highest salary.

SQL> select *
  2  from emp a
  3  where 1 >= (select count(distinct(sal))
  4  from emp b
  5  where a.sal<=b.sal)
  6  order by sal desc;

     EMPNO ENAME                               JOB              MGR HIREDATE         SAL       COMM     DEPTNO
---------- ----------------------------------- --------- ---------- --------- ---------- ---------- ----------
      7839 KING                                PRESIDENT            17-NOV-81       5000                    10

Salary of the first two.

SQL> ed
Wrote file afiedt.buf

  1  select *
  2  from emp a
  3  where 2 >= (select count(distinct(sal))
  4  from emp b
  5  where a.sal<=b.sal)
  6* order by sal desc
SQL> /

     EMPNO ENAME                               JOB              MGR HIREDATE         SAL       COMM     DEPTNO
---------- ----------------------------------- --------- ---------- --------- ---------- ---------- ----------
      7839 KING                                PRESIDENT            17-NOV-81       5000                    10
      7902 FORD                                ANALYST         7566 03-DEC-81       3000                    20
      7788 SCOTT                               ANALYST         7566 09-DEC-82       3000                    20

WiFi connections: Laptop HP cannot find the 5 GHz wireless signal?

Hello, this is my first post I just bought my first laptop and my first wireless router and have been setting them up.

Laptop: HP touch laptop Pavilion 15 15-n088ca with windows (fully updated) 8.1

Router: D-Link Dual Band AC1000-band 2.5 GHz & 5 GHz DIR - 810L

I configured the router and worked out the kinks and updated the software. My laptop detects and connects with the 2.5 GHz Band anywhere in the House. My laptop does not even find or see the 5 GHz Band. However my Galaxy Tablet concludes the two bands and can connect at once, thought the 5 GHz seems weaker.

I noticed that my Galaxy tablet is much more WiFi networks that my laptop (that are in the houses around the street).

I have need help troubleshooting and problem solving. I think that my laptop would find the 5 Ghz band and be able to connect. So I hope that it is a software or driver problem and just a setting that needs to be adjusted. Any help would be appreciated. I don't know much about the WiFi drivers, but I'm a survivor of Vista, so I should be able to solve this problem.

I don't know even what kind of program is running the WiFi software on my laptop? Or where to find it?

Thank you

Hello:

Welcome to the HP consumer's Support Forum.

I'm sorry that I'll have to disappoint you on your first post, but according to the specifications of the product on the link below your laptop comes with a single wireless card band (2.4 GHz) with bluetooth.

http://support.HP.com/us-en/document/c03940371

I can tell this by the description of the WLAN card.

He is listed as a 1 x 1 802.11b/g/n WLAN.

Whenever you see a closed wlan b/g/n card, it's a map of the only band (2.4 GHz).

A dual-band card would be qualified bird a/b/g/n or a/g/n.

When I look at the parts list for your laptop on the link below, the last listed item is the wireless card there.

Atheros AR9485 wireless 802.11b/g/n 1 x 1 adapter

http://PartSurfer.HP.com/search.aspx?SearchText=F0Q62UA

Googling the specifications of the card, I found that will confirm what I wrote above.

http://reviews.CNET.com/adapters-NICs/Atheros-ar9485-802-11/4507-3380_7-35558446.html

So there is nothing wrong with the wireless card or router.

To find the first Summit for data channel

Hey guys,.

I need a little help. I wrote a script to process some data. Basically, what the script does is calculate the linear accelerations peak, peak of the angular speeds, HIC values and IF. But there is a problem with my script that I need help. In my script, I use the ChnPeakFind command to find the pics in my chains. This command works fine to find the peak value, but what I want to do is find the FIRST Summit in the data, without worrying whether or not it is a maximum. I always encountered data that varies in which my first pic (impact analysis) may or may not be the highest point in the channel. As you can see in the Graphs.png that I have attached, the ChnPeakFind command found the 2nd Summit (red circle #2) in my chain of angular velocity (ARS resulting). But what I report is the first peak value (red circle #1). I've also attached a screenshot of the section in my code where I calculate the angular velocity. The 'i' in the calculation is just the channel group number. In this case, it is 1. Anyone know how to change the code or have another suggestion how to find the first Summit in a channel without having to stop and go to the tab analysis to manually find the pic?

Thank you!

Hello

If it please see my littl below escript and let me know if this could work for you
```
Dim     oChnResultant

Call Data.Root.Clear)

Call DataFileLoad ("E:\Customer_Examples\01_Head_Impact\Head Impact Data.tdm", "CT", "Load")

Set oChnResultant = data. Root.ActiveChannelGroup.Channels.Add ("Result", DataTypeChnFloat64)

Call ChnXYZAbsValue ("[1] / ARS X","[1] / ARS Y","[1] / ARS Z", oChnResultant)

"Use the filter twice to get rid of the jig

Call ChnSavitzkyGolayFilter (oChnResultant, oChnResultant, 1, 12)

Call ChnSavitzkyGolayFilter (oChnResultant, oChnResultant, 1, 12)

' Delete few maxima to the beginninf of the signal.

Calculate the call ("R = IIF (R)<>

'Research for first peak.

Call ChnPeakFind ("[1] / Time axis","[1]/Resultant","/PeakX","/PeakY",1,"Max.Peaks","Time")
```

How to find the ratio between the two frequencies in the fft

Hi, I'm Jean,

I need to find the ratio of the frequencies of 1st and 2nd graph fft.

Already, I received information of this site only. Mr. Alten helped me.

Now, I have to continue the same problem and need to find the ratio of the frequencies.

Please someone help me guys...

I will attach my vi file and the text file.

John,

Should you report the frequencies or the report of the amplitudes of the components at these frequencies?

The Information.vi of your multiple for the Signal Processing, extract > range of measures of waveform should do what you want.

The frequency resolution is not very good because you have cycles relatively little waveforms (2 ~ for the lower frequency). I added a few zeros to improve resolution, even if a larger sample of data would be better.

Lynn

algorithm to find the top of the parabola

Use case: mower rejection of common-mode using a programmable potentiometer

Problem: The performance of each circuit varies greatly and you are looking for best value (lowest CMRR) gives erroneous results. The performance seems to be relatively flat on some circuits, straight on others. Launching in the middle range of the potentiometer and moving upwards or downwards, I compare the measured value after the change of the potentiometer to the last measured value. By doing this I can get for the lowest CMR.

Setup: entry wave sign 5V 60 Hz on both positive and negative of a differential ADC. The potentiometer is cut and reduces common-mode noise. Go away and the common-mode noise increases.

Question: I would like to use adjustment curve to calculate the shape of the performance of the CMR. Take several steps (no more than 10 then I hope). Get the x y (x = 0 to 1023 trim adjustment potentiometers, Y = amplitude of the measured signal common-mode). Find the vertex.

Is this possible and can someone point to an example. I have not found one.

A parable is a second order polynomial curve. The Summit must be the peak value. The general polynomial Fit.vi to maintain your data. The Coefficients of the polynomial outputs to calculate the location of the tip with a DAB of analytic geometry.

Edit: Note that you may need to do checks to make sure that your data has not generated any curve as a parable.

Lynn

find the highest amplitude of a signal

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