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Assuming the signal shown below : enter image description here

I have found an expression for fourier series coeffecients as the following: $$a_{k} = \frac{1}{5}+\frac{j}{5}\sin{\frac{2\pi}{5}k}$$

Which matches with what the books suggests as an answer. My confusion is this : why are they not purely imaginary ?

I tried to prove this property the following way: Signal is real and even: $$\implies x[n] = x^{*}[n] = -x[-n] $$
with * denoting the conjugate. And using properties of conjugation, time-reversal and linearity we get : $$a_{k} = a^{*}_{-k} = -a_{-k} $$

Then I concluded : $$a_{k} = -a_{-k} \implies a_{k}\ are\ odd\ in\ k$$ $$ a^{*}_{-k} = -a_{-k} \implies a_{k}\ are\ purely\ imaginary $$

What is wrong with all the above? Is any step in my reasoning wrong or is the solution to the problem incorrect ?

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  • $\begingroup$ What value should $x[0]$ have if the signal was odd? $\endgroup$
    – Matt L.
    Dec 28, 2022 at 10:53
  • $\begingroup$ @MattL. It should be zero, which is not the case here. $\endgroup$
    – Ait-Gacem Nabil
    Dec 28, 2022 at 11:51
  • $\begingroup$ On a second look now, I see that there was a mistake in the assumption of "the signal is odd". I don't know how that escaped me. But is the proof sound? $\endgroup$
    – Ait-Gacem Nabil
    Dec 28, 2022 at 11:53

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You are correct that a purely odd signal should have purely imaginary coefficients. Your signal, however, is not odd -- it is the sum of an odd part (most of the samples), and an even part (the nonzero sample at time = 0).

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