# Fourier transform of shifted periodic function

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.

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.,

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

• 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. Sep 30 '21 at 1:55