Fourier Series Coefficient of a given signal

Question

$$ {\rm x}\left(t\right) = \sum_{k = -\infty}^{\infty}\left[\delta\left(t-\dfrac{k}{3}\right) + \delta\left(t-\dfrac{2k}{3}\right)\right] $$

I need to find the Fourier series coefficient of x(t). I know that

$$ a_{k} = \frac{1}{T}\int_{0}^{T}{\rm x}\left(t\right)\,{\rm e}^{-{\rm i}kw_{0}t}\,{\rm d}t $$

but I couldn't substitute my signal into this formula. Need some help, thanks.

Answer 1

Thanks to "impulse train", I guess I find the answer.

If $x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)$

$a_k = \frac{1}{T} \int_{-T/2}^{T/2} \delta(t) e^{-jkw_0t} dt = \frac{1}{T} $

Therefore in my question, $$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-\dfrac{k}{3}) + \delta(t-\dfrac{2k}{3})$$

$$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-\dfrac{k}{3})+\sum_{k=-\infty}^{\infty}\delta(t-\dfrac{2k}{3})$$

T = 1/3 for the first part and T = 2/3 for the second part. It follows that: $$a_k = 3 \int_{-1/6}^{1/6} \delta(t) e^{-jkw_0t} dt + \frac{3}{2} \int_{-2/6}^{2/6} \delta(t) e^{-jkw_0t} dt $$ $$a_k =3 + \frac{3}{2} = \frac{9}{2}$$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{% {\rm x}\left(t\right) = \sum_{k = -\infty}^{\infty}\left[\delta\left(t-\dfrac{k}{3}\right) + \delta\left(t-\dfrac{2k}{3}\right)\right]}$

$$ {\rm x}\left(t + T\right) = \sum_{k = -\infty}^{\infty}\left\{% \delta\left(t - {k - 3T \over 3}\right) + \delta\left(t - {2\bracks{k - 3T/2} \over 3}\right) \right\} $$ which clearly shows that $T = 2$: ${\rm x}\pars{t}$ is a $\it\mbox{periodic function}$ of $\it\mbox{period}\ T = 2$. In addition, $\left.\mbox{a}\right)~{\rm x}\pars{t}$ is an $\it\mbox{even}$ function of $t$ and $\left.\mbox{b}\right)~{\rm x}\pars{t} \in {\mathbb R}\,,\ \forall t$. It means that $$ {\rm x}\pars{t} = \sum_{n = 0}^{\infty}a_{n}\cos\pars{\omega_{n} t} \,,\quad\omega_{n} = n\,{2\pi \over T} = n\pi\quad\mbox{and}\quad a_{n} \in {\mathbb R}\,,\ \forall\ n = 0, 1, 2, \ldots $$ $$ \int_{-1^{-}}^{1^{+}}{\rm x}\pars{t}\cos\pars{\omega_{n}t}\,\dd t = \sum_{m = 0}^{\infty}a_{m}\ \overbrace{\quad% 2\int_{0}^{1^{+}}\cos\pars{\omega_{n}t}\cos\pars{\omega_{m}t}\,\dd t\quad} ^{\ds{\delta_{nm}}} = a_{n} $$

\begin{align} a_{n} &= \sum_{k = -\infty}^{\infty}\int_{-1^{-}}^{1^{+}}\delta\pars{t - {k \over 3}} \cos\pars{n\pi t}\,\dd t + \sum_{k = -\infty}^{\infty}\int_{-1^{-}}^{1^{+}}\delta\pars{t - {2k \over 3}} \cos\pars{n\pi t}\,\dd t \\[3mm]&= \sum_{k = -\infty}^{\infty}\Theta\pars{3 - \verts{k}} \cos\pars{n\pi\,{k \over 3}} + \sum_{k = -\infty}^{\infty} \Theta\pars{{3 \over 2} - \verts{k}}\cos\pars{n\pi\,{2k \over 3}} \\[3mm]&=\braces{% 1 + 2\bracks{\cos\pars{n\,{\pi \over 3}} + \cos\pars{n\,{2\pi \over 3}} + \cos\pars{\pi n}}} + \bracks{1 + 2\cos\pars{n\,{2\pi \over 3}}} \end{align}

$$\color{#0000ff}{\large% a_{n}= 2\braces{% \bracks{1 + \cos\pars{n\pi}} + \cos\pars{n\pi \over 3} + 2\cos\pars{2n\pi \over 3}}} $$