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

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!

• Typically, the coefficients in a Fourier series decay, so they couldn't be periodic and nonzero, see Magnitude of Fourier Series. Also, Parseval's identity wouldn't converge. – Michael Burr Jun 22 '15 at 15:53
• @MichaelBurr In classical analysis, it is true that the coefficients of a FS must decay in order to have convergence, in the theory of Generalized Functions, that is not required. Note the appearance of the Dirac Delta in the proposed answer. The Dirac Delta is not a classical function. – Mark Viola Jun 22 '15 at 15:58
• In your question, $b_k$ and $g_k$ are meant to be the same? – user225318 Jun 22 '15 at 20:58
• Fourier coefficients of functions decay, that's why cphiozn's guess has delta functions in there, so it is not a function. Which of course defeats the original assumption of a periodic continuous time signal... – GEdgar Jun 24 '15 at 13:45

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 $a_k$.