What if the Fourier series of a periodic function also has periodic coefficients $a_k$

Question

If given that $x(t)$ is a periodic continuous time signal, with periodic $T$. It can be expressed by the Fourier series, i.e. $x(t)=\sum\limits_{k=-\infty}^{+\infty}\,a_k\cdot e^{j k \frac{2 \pi}{T}t}$. What I would like to prove is that if the coefficients $a_k$ also have period $N$, i.e. $a_{k+N}=a_k$, then there exists some periodic sequence $b_k$ so that $x(t)=\sum\limits_{k=-\infty}^{+\infty}\,g_k\cdot \delta(t-kT/N)$. I tried to write $x(t)$ as the concatenated sums like $x(t) = \sum\limits_{k=1}^N\left[\sum\limits_{n=-\infty}^{+\infty}\,e^{j(k+nN)\frac{2\pi}{T}t}\right]$, but not able to go further any more. I can't thank more if some could give some hints!

Answer 1

If the Fourier coefficients $a_k$ are periodic with period $N$, the $T$-periodic signal $x(t)$ can be written as

$$\begin{align}x(t)&=\sum_{k=-\infty}^{\infty}a_ke^{j2\pi k t/T}\\&=\sum_{l=-\infty}^{\infty}\sum_{k=0}^{N-1}a_ke^{j(k+lN)2\pi t/T}\\&= \sum_{k=0}^{N-1}a_ke^{j2\pi kt/T}\sum_{l=-\infty}^{\infty}e^{j2\pi lNt/T}\tag{1}\end{align}$$

Note that the last sum in (1) is the Fourier series of a Dirac comb:

$$\sum_{l=-\infty}^{\infty}e^{j2\pi lNt/T}=\frac{T}{N}\sum_{m=-\infty}^{\infty}\delta\left(t-m\frac{T}{N}\right)\tag{2}$$

Combining (1) and (2) gives

$$\begin{align}x(t)&=\frac{T}{N}\sum_{k=0}^{N-1}a_ke^{j2\pi kt/T}\sum_{m=-\infty}^{\infty}\delta\left(t-m\frac{T}{N}\right)\\ &=\sum_{m=-\infty}^{\infty}\underbrace{\left(\frac{T}{N}\sum_{k=0}^{N-1}a_ke^{j2\pi km/N}\right)}_{b_m}\delta\left(t-m\frac{T}{N}\right) \tag{3}\end{align}$$

which has the desired form. Note that the coefficients $b_m$ are basically the discrete Fourier series coefficients of $a_k$.