The easiest way is to search for language translators that translates from language to other.

Prof. S. EL-Rabaie

thanks but I can not find any software that converts V.B to Matlab.

If you are comfortable with matlab then VB is the easiest language for programming (vice versa).  I am not a fan of translators (never tried ).

You have the code in VB(non- GUI part) . It will/might be tedious but you can do it if the logic  is clear.

First of all look @ the data/data structures

https://msdn.microsoft.com/en-CA/library/wak0wfyt.aspx

Eg: Dim numbers(4)  As Integer => Declaration of an array to store Integer data and size of the array is 4.

Next look for program constructs

Then functions and regroup them based on your convenience (say if you do not want you can insert the matlab constructs)

Not very fun but you have to be diligent.

thanks dear Aparna. hit-or-miss the logic is about FFt (fast fourier transform) which is used for a structural dynamics problem and also you may dealing it for signal processing in Electrical Engineering.

Yep ... Only signal and image processing Saeed , let me know if you need any help

I prefer to use 'fft' build-in function in Matlab for solving the following problem (see Fig.1). the formulation is shown in Fig. 2. In my matlab code Force Fourier Coeff. and half of Response Fourier Coeff. are calculated  successfully (Fig. 3) but Disp.Real and Disp.Imag are not matched with results of the example (Fig. 4). Where do I mistake??

My Matlab code:

clc

clear all

a=[0 -6.25 -9.5 -8.25 -6 -3.5 -1 4.5 4 1.5 0 0 0 0 0 0];

FFC=fft(a/16)

T=1;

k=200;

mass=0.259;

w=(k/mass)^0.5;

wbar=2*pi/T;

kessi=0.15;

r=wbar/w;

rn=[0 r 2*r 3*r 4*r 5*r 6*r 7*r 8*r 9*r 10*r 11*r 12*r 13*r 14*r 15*r];

RFC=FFC./(k*(1-rn.^2+2*kessi*i.*rn))

Disp=fft(RFC)

Will go thru it , give me some time Saeed...

thanks a lot.

Saeed,

I have some weird #s , is the FFT in VB giving you different values.

If yes then fft of matlab should be checked , I changed individual computing the values change.

sqr_rn=rn.^2;

for k=1:16

tmp_dt= 2*kessi*i*rn;

tmp_denom= k*(1-sqr_rn+tmp_dt) ;

end

RFC=FFC./tmp_denom

RFC and hence the fft() computation change (numerical values !!) 

My suggestion would be : in VB get the fft() values and then we have to look @ numerically and validate before giving it to built-in fft of matlab !!

dear Aparna

I really appreciate your concern.

In fact, the results of Fig. 4 (achieved from a reliable book) are obtained by VB code and we can use them to validate our results.

Ok ... I was wondering , FFC coefficients are from the fft so your RFC is the fft of the fft ? This seems to be more of computational glitch .

Yes. Really, it is complex. but the formulation enforce us to use FFC in calculation of RFC. Of course the FFC can be corrected manually (because of symmetry around  N/2+1=9). but the main problem are Disp.Real and Disp.Imag.

Saeed , are still on the issue or were you able to fix it !!

From the Displacement and FFC equation , they are conjugates aren't they ??

so I was thinking , let me add to the confusion (just kidding) , just check if these are #s you want :)

Disp=ifft(FFC./(k*(1-(rn.^2)+(2*kessi*i.*rn))))

Disp =

Columns 1 through 6

-0.0001 + 0.0007i -0.0005 + 0.0015i -0.0019 + 0.0011i -0.0021 - 0.0006i -0.0010 - 0.0013i -0.0005 - 0.0008i

Columns 7 through 12

-0.0007 - 0.0009i -0.0002 - 0.0015i 0.0010 - 0.0005i 0.0004 + 0.0009i -0.0008 + 0.0005i -0.0005 - 0.0003i

Columns 13 through 16

0.0001 + 0.0000i -0.0002 + 0.0005i -0.0005 + 0.0003i -0.0003 + 0.0002i

I tried splitting magnitude and phase ...

Hi. Thanks for your companionship.

Recently I fix it by adding a loop that create symmetric values about point N/2+1=9 for calculating RFC. Then displacements calculated successfully by using "Disp=ifft(RFC)*16".

but I can not realize why RFC which is obtained from Matlab function (fft) is not symmetric?

Thank you, the fft algorithm is a bit complicated in terms of implementation.

If you have observed lot of approximation will change the values and fft bit reversal may result in #%$^