Assuming $x(t)$ is a periodic function of period $T$ and having the Fourier transform $X(\omega)$, it is required to calculate the Fourier transform of the signal $x(t)+x(t-T)$. Since x(t-T) is equal to x(t) the Fourier transform should simply be $2X(\omega)$ but if we use the time-shifting property of the Fourier transform the answer should also be $X(\omega)+e^{-j\omega T} X(\omega)$. But how come I am getting two different answers.

Actually, I am confused about the concepts involving the Fourier transform of periodic signals and while practicing I came up with this weird thing and it's getting more confusing, and chances are it may not make sense to some people but I don't know why two fully applicable things giving different conclusions.


1 Answer 1


It's important to realize that a $T$-periodic function has a discrete frequency spectrum with contributions at integer multiples of $\omega_0=2\pi/T$. Consequently, the spectrum $X(\omega)$ has the form

$$X(\omega)=\sum_kc_k\delta\left(\omega-\frac{2\pi k}{T}\right)\tag{1}$$

with constants $c_k$, which are just scaled versions of the Fourier coefficients of $x(t)$.

When multiplied by $e^{-j\omega T}$ one obtains contributions

$$e^{-j\omega T}\delta\left(\omega-\frac{2\pi k}{T}\right)= e^{-j2\pi k}\delta\left(\omega-\frac{2\pi k}{T}\right)\tag{2}$$

because for any $f(\omega)$ that is continuous at $\omega_k$ we have $f(\omega)\delta(\omega-\omega_k)=f(\omega_k)\delta(\omega-\omega_k)$. Furthermore, since $e^{-j2\pi k}=1$ the result follows, i.e.,

$$e^{-j\omega T}X(\omega)=X(\omega)\tag{3}$$

  • $\begingroup$ Thanks, I now realize that the Fourier transform of periodic sequences has much more profound implications than any general Fourier transform because of its discrete nature in terms of the impulse/delta train. $\endgroup$ Commented Sep 30, 2021 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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