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I need help with convolution sum in Labview

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I uploaded the wrong version, wait here it is corrected

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Message 11 of 27
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@altenbach can you see this final VI correctly?
It's in 2019 with all the datas

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Message 12 of 27
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Yes, this works and produces output. You say that the result is not as expected. What is the expected result?

 

(Sorry, I need to switch to my desktop later. my laptop screen is not big enough to efficiently look at this)

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Message 13 of 27
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I should be obtaining the same graph I obtain in MatLab (the output of waveform graph 3). The output of matlab is attached. 
You're really kind btw thank you so much for your help! 
I will probably look at your adjustments tomorrow morning (I'm in Italy), so feel free to do it when you have time 🙂

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Message 14 of 27
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I can attach the MatLab code that works and does the convolution correctly. 

 
This is the code that I wrote on Matlab (and gives the correct result) and I wanted to implement on LabView:
for i=1:1:200 (200 is Mtot*2-1, I wrote it as a constant on Matlab for simplicity)
l=200;
for k=1:1:200
if i-k>0
c(i)=c(i)+a(k)*b(i-k);
else
c(i)=c(i)+a(k)*b(l);
l=l-1;
end
end
end
This is my implementation of those lines on labview. The problem must be there, as every other part works perfectly (simulation, algorithm one and all the spectrum) or maybe where I mirrored and expanded the periodic spectrum of V and I.
gaiab03_0-1732904825083.png

 

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Message 15 of 27
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Sorry, I no longer do text based code*. Can you save your Matlab output for the current inputs to a simple text file?

 

*(I did text based code from 1973 to 1996, switched to LabVIEW 4.0 and never looed back since 😄 )

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Message 16 of 27
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The output is the graph attached in the previous message (It's in png format). 

Do you prefer another format for images? I can't turn an image into a txt. 

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Message 17 of 27
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Solution
Accepted by gaiab03

@gaiab03 wrote:

I should be obtaining the same graph I obtain in MatLab. The output of matlab is attached. 🙂


If you need to obtain same output as from given script,

 

 

Spoiler
clear all
a=zeros(200, 1);
a(2,1)=-150i;
a(4,1)=-100i;
a(6,1)=-50i;
a(102,1)=-150i;
a(104,1)=-100i;
a(106,1)=-50i;
a(100,1)=150i;
a(98,1)=100i;
a(96,1)=50i;
a(200,1)=150i;
a(198,1)=100i;
a(196,1)=50i;
b=zeros(200, 1);
b(2,1)=-250i;
b(4,1)=-100i;
b(102,1)=-250i;
b(104,1)=-100i;
b(100,1)=250i;
b(98,1)=100i;
b(200,1)=250i;
b(198,1)=100i;
c=zeros(200, 1);
for i=1:1:200
l=200;
for k=1:1:200
if i-k>0
c(i)=c(i)+a(k)*b(i-k);
else
c(i)=c(i)+a(k)*b(l);
l=l-1;
end
end
end

 
% Plot the histogram of abs(c)
figure(1);
bar(1:200, abs(c)); % Plot the absolute values of c
xlabel('Array Indexes');
ylabel('Amplitude');
title('Histogram of abs(c)');
grid on;

 

 

then it could be like this:

Matlab.png

 

The graphs are the same:

 

Screenshot 2024-11-30 00.25.26.png

 

And values are identical:

Screenshot 2024-11-30 00.32.28.png

Some issues may present, but hopefully the overall idea is clear more or less.

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Message 18 of 27
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Thank you so much!
You're really helped me a lot!
The solution works also on my project

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Message 19 of 27
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@Andrey_Dmitriev wrote:

@gaiab03 wrote:

I should be obtaining the same graph I obtain in MatLab. The output of matlab is attached. 🙂


If you need to obtain same output as from given script,

.


It is just a circular convolution and all you need is code the size of a postage stamp. The following gives the same result!

 

altenbach_0-1732982389424.png

 

I changed the cosmetics of the graph to show both as points. Modify as desired.

 

Of course the lingering question is if this is really what you need. 😄

 

Message 20 of 27
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