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I need to prove the following DFT for a periodic 𝑠𝑖𝑛𝑐 sequence or ideal lowpass filter:

$\sqrt{\frac{p}{m}}\frac{sinc(\frac{\pi np}{m})}{sinc(\frac{\pi p}{m})} \stackrel{DFT}{\longleftrightarrow} \begin{cases} \sqrt{\frac{m}{p}},\: if \: |k-\frac{m}{2}|\ge \frac{p-1}{2}\\ 0, \: otherwise \end{cases}$

But actually, I do not know how to do it. Maybe someone gives me at least a hint?

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Hints:

1) Definition of the unnormalized DFT:

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

2) Geometric Summation Formula:

$$ \sum_{n=0}^{N-1} r^n = \frac{1-r^N}{1-r} $$

3) Exponential form of Sine:

$$ \sin( \theta ) = \frac{ e^{i\theta} - e^{-i\theta} }{ 2i } $$

4) Definition of unnormalized sinc:

$$ \operatorname{sinc}(x) = \frac{\sin(x)}{x} $$

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  • $\begingroup$ How can geometric summation formula help? I rewrite all sinc functions using formula 3 and 4 and added it to the formula 1, but I got stuck at this point, I cannot figure out how to rewrite this bug formula with exponents. $\endgroup$ – darktealeaf May 31 at 11:24
  • $\begingroup$ @darktealeaf Do the reverse first, which is taking the DFT (or iDFT) of a rectangle function. You will find very similar math here: dsprelated.com/showarticle/1038.php $\endgroup$ – Cedron Dawg May 31 at 12:00
  • $\begingroup$ So, got this: $x(n) = \frac{1}{N} \sum\limits_{k=0}^{N-1} \sqrt{\frac{m}{p}} e^{\frac{i2\pi kn}{N}} = \frac{1}{N} \sqrt{\frac{m}{p}} \frac{1 - e^{i2\pi n}}{1 - e^{\frac{i2\pi n}{N}}} = \frac{1}{N} \sqrt{\frac{m}{p}} \frac{e^{i\pi n}}{e^{\frac{i\pi n}{N}}} \frac{e^{-i\pi n} - e^{i\pi n}}{e^{\frac{-i\pi n}{N}} - e^{\frac{i\pi n}{N}}} = -\frac{1}{N} \sqrt{\frac{m}{p}} e^{i\pi n - \frac{i\pi n}{N}} \frac{sin(\pi n)}{sin(\frac{\pi n}{N})} = -\sqrt{\frac{m}{p}} e^{i\pi n - \frac{i\pi n}{N}} \frac{sinc(\pi n)}{sinc(\frac{\pi n}{N})}$ But still did not reach the correct formula. What do I do wrong? $\endgroup$ – darktealeaf May 31 at 14:51
  • $\begingroup$ @darktealeaf Your case statement needs to be translated into summation intervals. I just noticed the inequality is the opposite of a rectangle function, so that means two intervals or you can work with the complement. $\endgroup$ – Cedron Dawg May 31 at 15:11

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