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December 29th, 2016, 05:15 PM  #1 
Newbie Joined: Dec 2016 From: Michigan Posts: 5 Thanks: 0  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 coop 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...ndedition.pdf I do not understand how this equals 1. Seems pretty elementary but I am not sure what I am overlooking. Thanks Chris 
December 29th, 2016, 05:16 PM  #2 
Newbie Joined: Dec 2016 From: Michigan Posts: 5 Thanks: 0  Last edited by skipjack; December 30th, 2016 at 09:53 AM. 
December 29th, 2016, 05:24 PM  #3 
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637 
if $(m x + n y) \in \mathbb{Z}$ then yes $e^{j 2\pi(m x + n y)} = 1$ 
December 29th, 2016, 05:39 PM  #4 
Newbie Joined: Dec 2016 From: Michigan Posts: 5 Thanks: 0 
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)^(N1) ∑_(y=0)^(N1)〖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. 
December 29th, 2016, 05:44 PM  #5  
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637  Quote:
$e^{j x} = \cos(x) + j \sin(x)$ 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.  
December 29th, 2016, 05:46 PM  #6 
Newbie Joined: Dec 2016 From: Michigan Posts: 5 Thanks: 0 
ohh crap, long day thank you very much.


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apparently, ej2pimx, forgotten, things 
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