I am unable to confirm the following. If a signal is odd, then does the following hold? $$a_k = -a_{-k}$$ If a signal is even, then: $$a_k = a_{-k}$$ If the aforementioned qualities hold, then is the following true? If a signal is odd and the period is $2k$, then $a_k = -a_{-k} = 0$.


Your first two affirmations hold.

Let $f$ be $p$-periodic and $f \in L^1[-p/2,p/2]$. Then the Fourier coefficients of $f$ are $$ \hat{f}(k) = \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi k t/p} \,dt $$ If $f$ is odd then the change of variable $u=-t$ gives (we can keep the interval of integration $[-p/2,p/2]$ by $p$-periodicity) \begin{align} \hat{f}(k) &= -\frac{1}{p} \int_{-p/2}^{p/2} (-f(-t)) e^{i2\pi k t/p} \,dt \\ &= -\frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi (-k) t/p} \,dt \\ &= -\hat{f}(-k) \end{align}

If $f$ is even then the same change of variable gives \begin{align} \hat{f}(k) &= \frac{1}{p} \int_{-p/2}^{p/2} f(-t) e^{i2\pi k t/p} \,dt \\ &= \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi (-k) t/p} \,dt \\ &= \hat{f}(-k) \end{align}

I don't think your third affirmation holds.

Take $f(t):=\sin(\pi t)$. Then $f$ is odd and of period $2\cdot1$. Yet, we can calculate $\hat{f}(1)=-i/2 \neq 0$.

  • $\begingroup$ But logically speaking, if the period is 2k, then, the value at -1 and 1 should be the same. But them being odd requires that the signs of the values are opposite. Hence, the only value that satisfies both is 0, right? Or am I erring in my logic? $\endgroup$ – Jonathan Feb 16 '17 at 8:40
  • $\begingroup$ @Christian Hmmm yes I think you are correct, my example is not an odd signal. Let me rethink this. $\endgroup$ – NeedForHelp Feb 16 '17 at 8:43
  • $\begingroup$ @Christian Do you agree with my new counter-example? $\endgroup$ – NeedForHelp Feb 16 '17 at 8:49
  • $\begingroup$ Oh, I see. Just to confirm, the notation with caret sign over the function signifies the Fourier coefficient at k=1 right? Not the actual function itself? $\endgroup$ – Jonathan Feb 16 '17 at 9:04
  • 1
    $\begingroup$ @Christian The Fourier coefficients will satisfy the odd property $\hat{f}(k)=-\hat{f}(-k)$, as I showed. However they won't be periodic in general. That's why it doesn't work. $\endgroup$ – NeedForHelp Feb 16 '17 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.