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 &quot;One point in the frequency domain which includes the energy of all time should be an impulse.&quot; and many textbook says &quot;the CTFS of DC signal is an impulse.&quot; 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 :
$$ a_k = \frac{1}{T} \int_{&lt;T&gt;} x(t) e^{-j \frac{2\pi}{T} k t } dt \tag{1} $$
and the DC coefficient corresponds to setting $k=0$ above;
$$ a_0 = \frac{1}{T} \int_{&lt;T&gt;} x(t) dt \tag{2} $$
where the integration is along one period of $x(t)$. The DC coefficient cannot be an impulse, (indeed the CTFS coefficients $a_k$ form a discrete sequence, in $k$, of finite amplitude by definition of its existance.)
A signal must have finite energy per period (aka finite avg power) in order to have a valid CTFS (due to Dirichlet conditions) representation, hence the above integral in Eq.2 should be finite; i.e., the integral converges, and cannot result in an impulse.