Why is the continuous time Fourier series of DC signal an impulse?

In case of continuous time Fourier transform(CTFT), I can easily calculate the Fourier transform of DC signal by using Fourier duality or inverse CTFT. But I don't know how to calculate the continuous time Fourier series(CTFS) of DC signal. Some people say "One point in the frequency domain which includes the energy of all time should be an impulse." and many textbook says "the CTFS of DC signal is an impulse." But I think the inverse CTFS of an impulse is not DC signal by my calculation. I think I should consider a periodic function that converges to DC, calculate it's fourier series and then take the limit. But all my attempts failed. Who knows correct solution to this problem?

CTFS , $$a_k$$, of a periodic function $$x(t)$$ of period $$T$$ is :

$$a_k = \frac{1}{T} \int_{} x(t) e^{-j \frac{2\pi}{T} k t } dt \tag{1}$$

and the DC coefficient corresponds to setting $$k=0$$ above;

$$a_0 = \frac{1}{T} \int_{} x(t) dt \tag{2}$$

where the integration is along one period of $$x(t)$$. The DC coefficient $$a_0$$ cannot be an impulse; indeed the CTFS coefficients $$a_k$$ form a discrete sequence of finite amplitude by definition of their existance.

A signal must have finite energy per period in order to have a valid CTFS representation (Dirichlet conditions). Hence, the integral in Eq.2 should have a finite value; i.e., the integral converges, and thus cannot result in an impulse.

• Thank you for your answer. You are right. The CTFS of DC signal is not an impulse. Sep 25 '20 at 15:00
• Where is the "Accept an answer" icon? I can find only "This answer is useful" icon. I already click that icon, but the number 0 in the middle isn't increased. Sep 27 '20 at 17:21
• @kappysuper right now your user rep is not sufficient to upvote. Once you gain a few more points you can up vote. But even without a rep, you can still accept an answer to your question. Sep 27 '20 at 18:31
• @kappysuper the accpet tick is below the down vote button... Sep 27 '20 at 18:33