Basic 2d DFT - interpreting the coefficients

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elegysix
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Thanks for any help! I'm trying to understand the coefficients of a 2d DFT.

say we've got this matrix, f(a,b)

[itex] \left( \begin{array}{ccc}<br /> 9 & 1 & 9 \\<br /> 9 & 1 & 9 \\<br /> 9 & 1 & 9 \end{array} \right)[/itex]

I used wolfram alpha's function, Fourier{f(a,b)}
and the transform comes back as

[itex] \left( \begin{array}{ccc}<br /> 19 & 4-6.93i & 4+6.93i \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 0 \end{array} \right)[/itex]the characteristic polynomial is [itex]19x^{2} - x^{3}[/itex]I know the coeffecient at (0,0) is an average of something, but what?
what are the other two coefficients in the top row?

If I do a matrix with a frequency in both directions, I get coefficients in the first column as well. What do they represent?

Is there enough information here to determine a function z(x,y) that approximates f(a,b)? (like a sum of sines and cosines)

three eigenvectors are given as well, if needed

thanks for your help!
 
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Would a moderator move this to the calculus section please? 126 views and no responses yet :(

thanks
 
I checked it out, but I don't think it answers my questions. At least not in a way that I comprehend.

How are the coefficients related to frequencies?
 
The 1-D DFT is the dot product of the signal with a vector containing a complex sinusoid which oscillates over the indices at a variable frequency, according to the given formula.

http://reference.wolfram.com/mathematica/ref/Files/Fourier.en/3.gif

If s was 1, we'd get the zero frequency because s-1 = 0 and e^(2∏i0) = 1 (a constant).

We haven't defined a sampling rate. So, I suppose I could call (s-1) the frequency. Since (r-1)/n would be from 0 to (n-1)/n (evenly spaced sampling in the interval 0 to 1), the number of oscillations of the complex sinusoid from r = 1 to n is the oscillations of a sinusoid with frequency (s-1) over a domain of length one.
 
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