Timeline for Finding Discrete Fourier Transform (DFT) for different DFT size
Current License: CC BY-SA 4.0
16 events
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| Nov 18, 2023 at 12:28 | comment | added | nexxterp | Thanks for your answer, but it was suggested us to find it as a piecewise function. This is why at the beginning I write Y[2m] and Y[2m+1]. I find the even part, ie., if k is even, then Y[k]=X[k/2], but if k is odd then Y[k] is what in terms of X[k], I'm trying to find this. | |
| Nov 17, 2023 at 21:04 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 20:52 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 20:30 | comment | added | Jdip | That's it! See my edited answer to tie it all up together. Feel free to compute the convolution, but I don't think it's necessary, Also take a look at this similar question. If you're satisfied, please don't forget to accept/upvote this answer | |
| Nov 17, 2023 at 20:27 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 18:47 | comment | added | nexxterp | I did it in my edited version of question, but still I couldn't see how we can move from here to find DFT. | |
| Nov 17, 2023 at 18:00 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 17:54 | comment | added | Jdip | That's right, but you went a step too far. As you work your way through these, please edit your question with what you're doing so I can help correct you. I'll give you a couple more hints, and as you work your way through the problem I'll help. | |
| Nov 17, 2023 at 17:54 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 11:40 | comment | added | nexxterp | By using the DFT formula I found it as (1-(-1)^k)/(1-exp(-jπk/N)), but I could not visualize how I can use this result. | |
| Nov 17, 2023 at 9:54 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 9:49 | comment | added | Jdip | 1. Forget about Y[2m] and Y[2m+1]. 2. You do not have to compute the convolution of $X[k]$ and $\mathcal{F}\left\{e^{j\pi k n/N}\right\}$. Just try to compute $\mathcal{F}\left\{e^{j\pi k n/N}\right\}$ and we’ll go from there. | |
| Nov 17, 2023 at 9:47 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 8:14 | comment | added | nexxterp | I unfortunately could not do it. For Y[2m], I got the desired result but I cannot get it for Y[2m+1]=X[in terms of m]. Even though I take the fourier transform of exponential you write, I do not think its convolution with X[k] will be easy. | |
| Nov 17, 2023 at 7:58 | history | edited | Jdip | CC BY-SA 4.0 |
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| Nov 17, 2023 at 7:06 | history | answered | Jdip | CC BY-SA 4.0 |