discrete fourier transform in 2D

Besy should be the formulas for convertion but I haven´t found them.

Hello

Is that a more reliable program than matlab?

S D

I was out walking to buy some milk for tomorrow. Found two ideas. One was to change the m-1 or the n-1 to m respectively n. Beacase the answer was wery close but had the following look.

Firs column right. column two=equal to the last and the rest was in reversed order.

SO

the picture was

1 5 4 3 2 if 5×5 matrix

1 4 3 2 if 4×4

So i realised that I was close to the solution.

I got the idea to change the rows with the columns in the fft, had tried it before but now I did want to do it on two places at the same time. this wery intuitive idea turned out the be rigth. Have not realised why jet. That is the work for tommorrow.

Today I travelled in to the university and looked for books on this at several places, but couldn´t find any. So during the luch brake I got the basic idea in my head. sketched it on the bus and programmed it this evening. Found the error made beacase I didn´t have the full theory just new that 2d discrtet could be done in two steps first as a a 1d at each row and then a new to the result.

I´m happy to have solved it, but without any rigourous theory, bad.

S D

The changes are marked with many !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

clear
format short
%This is the matrix to do fft on
tal=[13 3 55 42
3 5 7 9
4 2 6 88
2 4 6 7]

%This is the make the complex coefficients
koff=fft(tal’);

%I want later to do the fourier transform on a and b real cofficients, they will be %aranged so that I do the FFT on the coefficients with the same frequency
koff=koff; TAKEEN AWAY THE TRANSPOSE !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ACTION ONE OF TWO

%formulas for calculating the real from the complex
konjkoff=conj(koff);
ak=koff+konjkoff;
bk=i*(koff-konjkoff);

%found out that the discrete real coefficients should be half of what the
%continuous was, when calculating them from the complex
ak=ak/2;
bk=bk/2;

%1D fourier transform over each column with real coefficients of the same frequenscy
koffa=fft(ak’);!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!I put in the transpose here. The difference is that now the fft is ower different frequencies. I think. havent analysed it jet.

koffb=fft(bk’);
konjkoffa=conj(koffa);
konjkoffb=conj(koffb);

%for the one start with an a
aa=(koffa+konjkoffa)/2;
ab=(i*(koffa-konjkoffa))/2;

%for the one start with an b
ba=(koffb+konjkoffb)/2;
bb=(i*(koffb-konjkoffb))/2;

%summa=zeros(4,4);
%n and m are the place in the matrix for reconstructing tal from the real
%coefficinets
for n=1:4
for m=1:4

%m=1
%n=3
summa=0;
%k and p are the different frequencies
for k=1:4
for p=1:4
summa=summa+aa(p,k)*cos((2*pi*(k-1)*(n-1))/4)*cos((2*pi*(p-1)*(m-1))/4)…
+ab(p,k)*cos((2*pi*(k-1)*(n-1))/4)*sin((2*pi*(p-1)*(m-1))/4)…
+ba(p,k)*sin((2*pi*(k-1)*(n-1))/4)*cos((2*pi*(p-1)*(m-1))/4)…
+bb(p,k)*sin((2*pi*(k-1)*(n-1))/4)*sin((2*pi*(p-1)*(m-1))/4);

end
end

svar(m,n)=summa/16;

end
end
svar

Hello

The error was actually that the transpose with a complex matrix is another thing than the transpose of a real one.

And the second fourier transform work over equal frequencies.

S D