My question is similar to this one and this one, but neither answers address my concern.
Suppose you have a signal x(n) = {-1,2,-3,2,-1} where we can assume x(0) = -3. So this is an even signal.
How do you find the phase of the DFT of x(n)?
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$\begingroup$ What you show is not an even signal, did you mean x(n) = {1,2,-3,2,1}? $\endgroup$ – Dan Boschen Mar 21 '20 at 3:28
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$\begingroup$ This post should help you: dsp.stackexchange.com/questions/38544/… $\endgroup$ – Dan Boschen Mar 21 '20 at 3:29
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$\begingroup$ Sorry my mistake. Now it's even $\endgroup$ – user1068636 Mar 21 '20 at 3:30
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$\begingroup$ @DanBoschen - I am not sure I see which part of the post is relevant to this question. So the answer to this question is $\pi$, but I'm not sure I see why. $\endgroup$ – user1068636 Mar 21 '20 at 3:33
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1$\begingroup$ No problem, this was very helpful $\endgroup$ – user1068636 Mar 21 '20 at 3:51
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Here is a hint that will help you:
The DFT is cyclical in time and in frequency. For the sequence given by
$$x(n) = [-1,2,-3,2,-1]$$
With x(0) = -3 would be solved using the standard DFT equation that starts at n=0 using
$$x(n) = [-3, 2, -1,-1, 2]$$
From that you can solve for the DFT and then determine easily for each result what it's phase is.
answered Mar 21 '20 at 3:44
