## Dienstag, 22. November 2011

### EE4206/EE5806 : Digital Image Processing : 2-D DFT Symmetry Property 3

EE4206/EE5806 : Digital Image Processing : 2-D DFT Symmetry Property 3

Prove : If f(x,y) is a real function, the real part of its DFT is even and the imaginary part is odd;
similarly, if a DFT has real and imaginary parts that are even and odd respectively, then its IDFT is a real function.

ie., f(x,y) real <---> R(u,v) even; I(u,v) odd

steps :

1. F(u,v) is complex in general, so it can be expressed as the SUM of real and an imaginary part:

f(u,v) = R(u,v) + jI(u,v).

2. Then, F^*(u,v)= R(u,v) - jI(u,v).

3. Also, F(-u,-v) = R(-u,-v) + jI(-u,-v).

4. The Fourier transform of a real function f(x,y) is conjugate symmetric (even):

F^*(u,v) = F(-u,-v) (property 1)

5. Therefore, F^*(u,v) = R(-u,-v) + jI(-u,-v)

6. Then, F(u,v) = R(-u,-v) - jI(-u,-v)

7. from 1, F(u,v) = R (u,v) + jI(u,v)

8. from 6 and 7, therefore, R(u,v) = R(-u,-v) (even) and
I(u,v) = -I(-u,-v) (odd)

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Property 8

Prove : f(x,y) real and even <---> F(u,v) real and even

1. If f(x,y) is real we know that from property 3 that the real part of F(u,v) is even, so to prove property 8 all we need to do is show that if f(x,y) is real and even then the imaginary part of F(u,v) is 0, ie., I(u,v) = 0 ( ie., F is real)

2. use Euler's equation and the fact that cos(x) is an even function and sin(x) is an odd function.

3. even function X even function = even function

4. odd function X odd function = even function

5. even funcion X odd function = odd function

6. The product of two even or two odd functions is even, and that the product of an even and odd function is odd.

7. the only way that a discrete function can be odd is ifall its samples sum to zero.

excerpted from :
Digital Image Processing, 3rd., by Gonzalez and Woods, ch4, page 262-266.

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