Is fft2 in MATLAB unitary? Some differences happen I meet a problem when implementing fft2 in MATLAB. 

The question is I try to simulate the realistic measurements $Y = |FCXF^H|^2$ - the intensity of Fourier domain of object $X$, where $F$ denotes the Fourier transform matrix $C$ denotes $\{+1, -1\}$ matrix. 

But I want to replace the $C$ to realistic $D = \{0,1\}$ with $2D - 1 = C$. Therefore, after some simple calculations, the measuremnts $Y$ becomes $Y_{real} = 2(|FDXF^H|^2 + |F\bar{D}XF^H|^2) - |FXF^H|^2$, where $\bar{D}$ denotes the inversion of $D$. The calculations are from a paper. 

For original $Y$ and $C$ the algorithm could work. Therefore, I try to verify with $Y_{real}$. Strange things happen, although the intensity of Fourier domain of $Y_{real}$ looks similar with $Y$. They are totally different!!! 

I am very confused, the equations are right, but why they are different? I think the problem is from fft2 function. Could anyone tell me the principle of fft2? Thanks in advance!!!



Here is the code:

This implements the original algorithm with $C$:

Y = abs(fftshift(fft2((2*D-ones(n1,n2)).*x))).^2;


This implements the equation - realisitic measurement:

Y_real = 2*(  abs(fftshift(fft2(D.*x))).^2 + abs(fftshift(fft2( ( ones(n1,n2) - D ).*x ))).^2 ) - abs(fftshift(fft2(x))).^2;
The standard (conventional) definition of DFT (1D or 2D) is not unitary.

See for example the 1D standard (conventional) DFT pair as:

$$ X[k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} n k } $$

and 

$$ x[n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{j \frac{2\pi}{N} k n} $$

DFT is not unitary due to the fact that the forward and backward transforms are not symmetric (the scale $\frac{1}{N}$) 

Hence MATLAB's fft() and fft2() functions will not provide unitary transforms; i.e., they do not preserve energy.

However, you can define unitary version of DFT by distributing the scale as:

$$ X[k] = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} n k } $$

and 

$$ x[n] = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} X[k] e^{j \frac{2\pi}{N} k n} $$

Now this pair is unitary.