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I encountered this in my notes:

$$\int^{\frac{1}{2}}_{-\frac{1}{2}}\mathcal{F}(f- f')df = 1\;\;\;\;\;(1)$$ where $$\mathcal{F}(f) = \frac{\sin^2(N\pi f)}{N \sin^2(\pi f)}$$

I know that $$\int^{\frac{1}{2}}_{-\frac{1}{2}}\mathcal{F}(f)df = 1$$

But I do not know how $(1)$ comes about. I mean how does it integrate to $1$? Please give me a hand here plz

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1 Answer 1

up vote 1 down vote accepted

$\mathcal{F}$ is a periodic function with periodicity of $1$. So it wouldn't matter which value you are using for $f'$.

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can frequency $f'$ be 0.5, 0.75 ...? OR does it have to take integer values ??? –  user1769197 Mar 31 '13 at 21:51
    
$f'$ can be any real values. –  Patrick Li Mar 31 '13 at 22:22
    
do u mind illustrating what you mean by "periodicity of 1" please ? –  user1769197 Mar 31 '13 at 22:29
1  
I suppose $N$ is an integer. Then $\frac{\sin^2(N\pi(x+1))}{\sin^2(\pi(x+1))} = \frac{\sin^2(N\pi x)}{\sin^2(\pi x)}$ –  Patrick Li Mar 31 '13 at 22:33

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