convolution vi

Similar Questions

• Multiplication of cross-checking (circular convolution)

  Hello

  To make the circular convolution of vectors, I would like to make:

  FFT-> multiplication cross-check (W = u * V in MATLAB)-> IFFT

  for example [1 1 1 1 0 0 0 0] with himself-> [1 2 3 4 3 2 1 0]

  Is there an easy way to do this multiplication on the 50g, or a cross-check order direct cconv?

  Thanks in advance

  Hi, yYou can use the same commands with the 50G

  FFT

  IFFT

  For cross-checking multiplication on the 50g, try HADAMARD / / shift MATRICES OPER NXT HADAM

  RPL mode

  [1 2 3] [4-5-6] HADAMARD

  [4-10-18]

  In rpl:

  [ ... ]

  [ ... ]

  FFT

  HADAMARD

  IFFT

• effects of sampling frequency on the convolution of waveforms (need feedback from a GURU of signal processing)

  I have included my code as version 8.5 for those who have not yet upgraded to 8.6.  I have also included some screenshots so that you can replicate the results I got.  I hope that some signal processing guru can shed light on what I mention it further.

  This VI convolves the signal of impulse response of a simulated servomotor which is essentially a damped sine the input pulse which is a step function.  The signal resulting convolved should be IDENTICAL to that of the step response of the engine which is RED on the display 1.  As you can see the convolution that results in table 2 shows the same structure of frequency, but its magnitude is INCORRECT.  As you can see in the catches of 2 screen sizes differ by a factor of 2 & done the sampling frequency of the wave.  Why the sampling frequency, impact on the scale is also very strange & disturbing.

  Would appreciate any corrections & explanations so that I trust the convolution of the other wave forms of entry than just the step function.

  OK, I think I have it working now.  Your premise on the effect of sampling on the derivative is not the issue.  Does it affect what the FREQ of levy is the basis of time of convolution.  As the convolution product is not continuous but discrete the length of the array should be taken into account & the sampling frequency must be consistent with this length of array as well as 1 second corresponds to 1 second.  If sampling freq is 2 kHz & the length of the array is 1000 then to get the correct time base by a factor of 2 must be taken into account.  In addition, to take account of the DC, shift of the ZERO gain factor must be added to the convolved signal to get the correct size.

  Thanks for making me think more deeply.

• Synthetesis of frequencies with filtered signals (Convolution)

  Hello

  I am a new user of DIAdem and I analysis of the signal.

  To deal with my purchases, I use several tools alreaydy implement in DIAdem. But I want to do a convolution of the signal number to make a spectrum of synthesis.

  In fact, I want to filter several frequencies on a signal and put them in a same spectrum, think a convolution of each of the signals.

  How can I do that with DIAdem?

  Best regards

  Pascal GRAIGNIC

  Hi Pascal,.

  I'm no expert in acoustic analysis, but it seems to me that you could simply add together the resulting chains of each single frequency filter digital function of the results of the ANALYSES.  The result of the calculation of digital filter in tiara is a curve of time series that contains only the frequency components with a size proportional to the contribution of the relative amplitude of frequency of such component (s) to the curve of time series (complete) original.  There is a function of multichannel totaling in the palette of the ANALYSIS of "basic mathematics".

  Brad Turpin
  Tiara Product Support Engineer
  National Instruments

• convolution

  HII

  I am facing problem with 1 d convolution. I'm convolving two table 1 d of size 1024 but the result I get is of size 2047.How to get rid of this or how to remove extra specific samples which is the addition of convolution.

  I have attached the filter that I use with my VI.

  Thank you

  Parag

  It is an oversight the convolution 1 d without a 'format' entry, the 2D versions do. (please vote for my idea to have this corrected. (Thank you!)

  All you have to do is take the subset of the appropriate chart to cut output down to the same size.

  While you're there, you should also learn some very basic coding skills. All you have to do is autoindex on the array of names. This eliminates the indexing operation and you either it wire to N. N will be determined automatically based on the size of the array. What happens if the first iteration of the loop FOR occurs before the 'Names' terminal is even written? Other terminals such as 'File Options' and 'Number' was before the loop, because they will be hopefully no change during the execution of the loop.  Playback of a local variable is ineffective and can lead to race conditions. ('Names', 'Image' and 'Path'!) Delete local variables and thread just beyond the boundary of the loop.

• Need help with the Convolution

  I don't understand exactly what is happening with my convolution. I am trying to convolution my data with one {+ 1, -1} by wavelets, downsample by 2 and convolution again.

  The problem I have is that if the values of N data go in, he goes out with N + 1 values. When I created the VI to treat a large data file by using a loop, the question becomes much larger.

  How to solve the problem with my convolution? What should I do after the convolution to remove excess?

  As I said, a convolution with 1-1 is just a simple derived. I don't know why you would jump through all these hoops inflamed of advanced signal processing, if what you need is a difference in pairs. Someone test you?

  The attached amendment shows what I mean. As you can see, the results are the same!

  

• How ro provide Convolution in the frequency domain

  Hello world

  I have a problem with application a convolution of the image and kernel smoothing in LabView. The idea behind this is that convolution is a multiplication by the area of the Fourier transform. So firstly I want to apply convolution by standart IMAQ convoluted VI, then, to do essentially the same thing, but do it manually FFT of the image and the kernel and then their multiplication by then apply inverse fourier transform and observe the resulting image. Finally, I get two different results, but they should be the same, I can't get the idea why.

  two copies

• Leak memory in the Convolution

  The VI attached shows a memory leak in the Convolution routine in LabVIEW 8.6.1.  There is no leakage of memory in 8.2.1 - I have no other versions to test it on.

  Cheers ~ Greg

  Hi Tim,.

  Given that we develop are more for LabVIEW 8.6, this will be something that you need a solution to workaround or update for 2009.  The leak occurs only when you set the size of "Dimension X".  And he was set for 2009.

  See you soon,.

• convolution property of the Fourier transform

  
  

  The cited web page does not mention the word "linear convolution", but if you just want to do without any fourier transforms, you could multiply and sum up in a loop as in the attached demo. It gives the same result as the convolution stock VI (in ~ 10-15). Well, there are other ways to do it, of course.

  (There are a number of things wrong with your VI, so I'm not sure what you want at the end. For example, you are graphics of complex tables in a simple graph, then you lose the imaginary part. You also have a small wait in the loop, otherwise the VI consumes all CPU, the same recalculating over and over again. Ideally the loop should turn when an entry changes).

• Ideal no RC Impulse Response vague question the quadrature

  Salvation OR team,

  I have a little trouble understand why the amplitude of the response of my simulated square wave signal, when passed through a non-ideal answer RC filter, is incorrect. Qualitatively, the form of the answer seems correct, but the range is wide.

  I'm generating a square wave (where you specify the width of the frequency and pulse), the amplitude is set to 1 and the offset is also 1 for the square wave goes from 0 to 1...

  Then, I build the core of impulse response indeed RC ((1/RC) * exp(-t/RC)) using the formula of wave Vi, and finally, I use the continuous VI of Convolution to convolution of the signal with the impulse response of RC square wave to generate the output signal...

  For example, this is explained in the following link:

  http://Web.CECS.PDX.edu/~ece2xx/ECE222/slides/ConvolutionIntegralx4.PDF

  Although, I think the author may be typo in the amplitude of kernel (I think it should be 1/RC and not RC) but the rest seems fine.

  In the attached VI, use a frequency of 300 Hz, pulse width of 0.01 ms and RC 380 Hz bandwidth Signal.

  RC 380 Hz bandwidth would be a time constant RC of ~ 2.6 MS, when a square wave with a pulse width unit< than="" 2.6ms,="" the="" resultant="" response="" signal should="" have="" an="" amplitude="" of="" 1="" (matching="" the="" unit="" square="" wave="" signal="" here)="" and="" fall="" off="" appropriately="" as="" e^(-t/rc)="">

  This seems OK when running the simulation. However, when I increase the width of the impulse to say 2.6 ms, (which should show a refill of the RC circuit to an amplitude of ~0.5 and then discharged as e ^(-t/RC), the amplitude is too high...

  Have I a gross error in my understanding here, or maybe I use one of these VI incorrectly?

  Thank you for your help and time in advance!

  

  

  In fact, I made a mistake in my interpretation of the correct amplitude in both cases, I presented. I was still unable to resolve the discrepancy here, but thanks to this post, I managed to solve the problem. In this case, we need to 'normalise' the core of convolution to the 'sum' of the elements in the data of y of the signal of the nucleus resulting. I will post the solution here in case someone else runs into this problem.

  

  

• Imaging of double precision of uint8

  Hello

  I am writing a program that manipulates image processing image gradients. Since negative gradients come into picture, I need to convert the image in double precision type uint8. How can I do this? I tried to use mathscript and although the double precision is obtained using the double function I need to leave mathscript to perform other operations such as filtering. How can I transfer this image of double precision outdoors and filter with it?

  Hello

  If you are satisfied with the float data type, why not just use the convolution operator?

  For example (using a kernel of Sobel, you can choose your own of course):

  

  Best regards

  K

• time scale is not shown on my chart

  Hello

  This may seem a very stupid question, but I was stuck on this for the last 2 hours and I could not find an answer anywhere else on the forum, so here goes. The timestamp (dt) on my graph will not appear. It shows the whole numbers of samples and does not care what is my sampling rate (dt). I read somewhere that it could be because "the timestamp is ignored", this option is grayed out, so is not applicable here.

  I added a picture of the code where the waveform is created (if it is trivial). Any thoughts?

  Hello

  I just noticed that the point of constraint comes from my stream. Attaching a normal array it makes the point of disappearing, and then the chart ignores my dt. So the problem must be with my bitstream.

  EDIT: For some reason, wiring of the table that is normal to the curve, remove and then rewiring my flow for the chart has solved my problem... I have no idea what happened, but now the sampling frequency is taken into account. (the point of constraint moved to my block of convolution, now, which indicates a problem with my the filter coefficients, but it doesn't seem to work now).

• table tricks

  Hello

  I am building a derivation from an array, I came with the attached VI example, I don't know if it's intensive for my computer, I'm going to get a picture of 107 x 51 every 50 ms and want to derive from channels of them when you give values between-1 a 1, it is possible that I have 4 times a x 0.25 and channel - 1...

  Someone knows a better solution?

  Kind regards

  Thijs

  Ignore the comment of Convolution, you don't need to be so complicated. You need just A x B.vi as shown below, in the range of linear algebra.

  

• filter of signal distortion

  Hello everyone. I'm trying to filter a signal using filter Butterworth that VI included with LabView. The problem I have is that it is a signal in time real get sampled at a certain speed with a certain block of samples. The filter seems to do its job properly for all samples in the middle of the block, but it doesn't for the first part.

  I have attached two picture to show the difference with and without the filter. It is a sine wave.

  I think that the problem comes from the Convolution of the input signal filter and distort the signal in this way. I was hoping people can say this is the problem and help me to find a way to a solution.

  I can show you my code as well, but to me, that this is a general problem with filtering, rather than my code. If you think I'm wrong please tell me.

  David.

  
  

• parallel configuration for-lopp is not possible

  Hello

  I have problems running a parallel loop. The Task Manager indicates that 8% workload, showing that only one of the 12 runs. I need to process the image of a repertoire of about 20,000 (U16, grayscale) tiff images

  First of all, I thought that it is a problem of Vision-screws, because they use pointers as a reference image. Pointers would cause problems in a parallel loop. That's why I rewrote briefly the morphological filter screws without using the screw of the vision-Toolbox. but it does not help. The program still works with the workload of 8%, indicating only 1 core is active.

  A test showed that labview is able to run a task with all 12 hearts. It is not a problem in windows.

  Maybe someone has an idea why this is not working so far. The main program is called "StreifenWegmachenAlleBilderv11Multicore". All other screws are sub-vis. At present, it would take about 1 week to process all the images. I want to accelerate to a night operation.

  Greetings

  b

  Yamaeda wrote:

  GaussFilterOhneVision is not environment, which will effectively stop the parallelization.

  Yes, the missing reentrancy is the main problem. You could even inline it.

  I also get a faster result of more than 2 x in parallelizing the outer circle, instead of the inside OF the loop. (~ 1.2seconds on my 16 core Xeon)

  Have you tried the 2D convolution directly. Probably would save you some effort of programming. Even the padding integrated

  In addition, your generation of the Gaussian kernel is a bit complicated. You can use the outer product of a 1-d Gaussian. The results agree breast of ~ 1e-18 (see extract, probably not relevant after conversion back U16)