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please help me find the error in the following counter example.

Consider we take sinus with period of $2\pi$. We sample it many time, and more than 3. We make convolution with rectangle of height 1 and width $2\pi$.

We’ve got sine wave sampled many times without periodicity.

Now we make fourier transform of it. We shall get delta at omega 2pi and as we know, since we transformed a discrete signal, we shall get this delta every 2pi, alas at 4pi, 6pi, and so on. Meaning by that, that each multiplication by two of this signal in the time domain shall fit the discrete signal’s samples.

But it does not work for me.

Say we have 999 samples on the original sine on single period. Signal which is shortened by two will simply remain long enough not to fit between each of two samples in each of the halves of the signal because it works only if we have 3 samples (on the 0s). Only then the $4\pi$ sinus will lay down on on the samples, which are the zeroes of the sine wave.

What am I missing? Please help.

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  • $\begingroup$ This question is relatively hard to understand. To make it easier for people to help it would most probably be beneficial to improve on the description of your problem. That is, be more precise and specific: what exactly do you want to know, what have you tried? For example, why do you assume the sampled sine wave is not periodic? Doesn't periodicity depend on the ratio of the sine wave frequency and the sampling frequency? Also, in the second sentence, you are describing a period of $2\pi$ - further down the period is $4\pi$. $\endgroup$ – applesoup May 19 '20 at 19:15
  • $\begingroup$ Hi, @applesoup the ‘no periodicity’ is caused by low pass filter (convolution with rectangle) on the time domain. And low pass filter is the wrong name, but it just to clarify. Period of 2pi is the signal I sampled. But in frequency domain discrete signal of non periodic wave, after fourier transform turns to be periodic with period of 2pi. Which means that as I described, the 4pi sine shall match the samples of the first half of the 2pi sine signal. $\endgroup$ – Vitali Pom May 19 '20 at 19:51
  • $\begingroup$ Hi, Vitali (it’s me) - of course we shall not get any delta in the result. Low Pass Filter convolution won’t give any deltas, it will give a continuous signal, which in fact will be concatinated $sinc$ waves in a row. I will now punish myself and write an apologies letter to the Math World Wide Committete. Silly me. Thank you apple soup! $\endgroup$ – Vitali Pom May 20 '20 at 16:00

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