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December 29th, 2016, 05:15 PM   #1
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e^-(j2PI(mx+ny)) = 1 ??? apparently I have forgotten some things

Hello all, I am not sure if this is the correct location for this thread but I am starting a co-op that deals with image processing in a book called "Feature Extraction & Image Processing" 2nd edition P 50. there is a part on the sixth from the last line that states that:
e^-(j2PI(mx+ny)) = 1 (since the term in brackets is always an integer and then the exponent is always an integer multiple of 2*PI)

where j is imaginary(the link below is the pdf for a context reference or you can simply google feature extractions & image processing pdf)

http://vlm1.uta.edu/~patjang/cvpr201...nd-edition.pdf

I do not understand how this equals 1.

Seems pretty elementary but I am not sure what I am overlooking.

Thanks
Chris
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December 29th, 2016, 05:16 PM   #2
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It would help if I included the URL:

http://vlm1.uta.edu/~patjang/cvpr201...nd-edition.pdf

Last edited by skipjack; December 30th, 2016 at 09:53 AM.
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December 29th, 2016, 05:24 PM   #3
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if $(m x + n y) \in \mathbb{Z}$ then yes

$e^{-j 2\pi(m x + n y)} = 1$
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December 29th, 2016, 05:39 PM   #4
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I am not doubting that it is the case, but I am unsure how the math works out. Could you explain the math a little bit or point me to a source that would? But yes, you are correct on the assumption that it is in summation notation.

In fact, it is a nested as follows:
FP_(u+mN,v+nN )=1/N ∑_(x=0)^(N-1) ∑_(y=0)^(N-1)〖P_(x,y) e^(-j(2π/N)(ux+vy) )*e^(-j2π(mx+ny)) 〗

I am not sure how to justify saying (maybe I don't know how to handle the imaginary portion):
e^(-j2π(mx+ny)) = 1

Sorry, I am not sure how to get an image in as I do not have a URL for my image.

Thanks for the help.

Last edited by skipjack; December 30th, 2016 at 09:55 AM.
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December 29th, 2016, 05:44 PM   #5
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Quote:
Originally Posted by ComputerEGR View Post
I am not doubting that it is the case, but I am unsure how the math works out. Could you explain the math a little bit or point me to a source that would? But yes, you are correct on the assumption that it is in summation notation.

In fact, it is a nested as follows:
FP_(u+mN,v+nN )=1/N ∑_(x=0)^(N-1) ∑_(y=0)^(N-1)〖P_(x,y) e^(-j(2π/N)(ux+vy) )*e^(-j2π(mx+ny)) 〗

I am not sure how to justify saying (maybe I don't know how to handle the imaginary portion):
e^(-j2π(mx+ny)) = 1

Sorry, I am not sure how to get an image in as I do not have a URL for my image.

Thanks for the help.
Euler's formula says

$e^{j x} = \cos(x) + j \sin(x)$

Applying that here, we get

$e^{-j 2 \pi(m x + n y)} = \cos(-2\pi(m x+ n y)) + j \sin(-2\pi(m x + n y)) = 1 + j0=1, \text{ when } (m x + n y) \in \mathbb{Z}$

Last edited by skipjack; December 30th, 2016 at 09:56 AM.
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December 29th, 2016, 05:46 PM   #6
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ohh crap, long day thank you very much.
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