I have to find the Fourier transform of this function:
I wrote that :
$x(t) = \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)$ where $k \in Z$
It looks like a convolution but the $t$ should be $\in Z$
Can I do this (?)
$X(\omega) = $
$\int_{-\infty}^{\infty}\sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)e^{j\omega t}dt$ $=\sum_{k= -\infty}^{\infty}\int_{-\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)e^{j\omega t}dt = \sum_{k= -\infty}^{\infty}sin(\frac{2\pi k}{3})\int_{-\infty}^{\infty}e^{j\omega t}\delta(t-k)dt$
And the Fourier tranform of : $\delta(t-k) \rightarrow e^{j\omega k}$
\
So : $X(\omega) = \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})e^{j\omega k}$
And : $\sin(\frac{2\pi k}{3}) =\frac{1}{2j}(e^{j\frac{2\pi k}{3}} - e^{-j\frac{2\pi k}{3}})$
$X(\omega) = \sum_{k= -\infty}^{\infty}\frac{1}{2}( e^{k(j\frac{2\pi}{3} + j\omega)} - e^{k(-j\frac{2\pi k}{3}+ j\omega)})$ but how do I proceeed?
update**
I thought another way :
this signal is periodic so I may can express it with fourier - analysis :
$ c_n = \begin{cases}
\delta(t-n) & n = 3k + 1 \\
-\delta(t-n) & n = 3k +2 \\
0 & n = 3k + 3
\end{cases}$ where $k \in Z$
Then we can express $x(t)$ as :
$x(t) = \begin{cases}
\sum_{n=-\infty}^{\infty}\delta(t-n)& n =3k+1 \\
\sum_{n=-\infty}^{\infty}-\delta(t-n)& n=3k+2 \\
0 & n = 3k + 3
\end{cases} \rightarrow x(t) = \sum_{n=-\infty}^{\infty}\delta(t-nT -1) -\delta(t-nT -2)$
In my table with fourier transforms (provided by my uni) there this property:
$\sum_{n=-\infty}^{\infty}\delta(t-nT) \leftrightarrow \frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta(\omega - n\omega_0)$
$\\ \\$
So :
$X(\omega) =\sum_{n=-\infty}^{\infty}e^{-j\omega}\delta(\omega -\omega_0)-e^{-j\omega 2}\delta(\omega -\omega_0) = \sum_{n=-\infty}^{\infty}e^{-j\omega}\delta(\omega -\omega_0)e^{-j\omega}(1 - e^{-j\omega}) $
And $e^{-j\omega}(1 - e^{-j\omega}) = e^{-j\omega}e^{\frac{-j\omega}{2}}(e^{\frac{j\omega}{2}} - e^{\frac{-j\omega}{2}} ) = e^{\frac{-3j\omega}{2}}(2jsin(\frac{\omega}{2}))$
And
$X(\omega) =
\frac{2\pi}{T} e^{\frac{-3j\omega}{2}}(2jsin(\frac{\omega}{2}))\sum_{n=-\infty}^{\infty}\delta(\omega -\omega_0)
$
First of all, due to the rule $f(x)\delta(x-a)=f(a)\delta(x-a)$:
$$x(t) = \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)= \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi t}{3})\delta(t-k)$$
Hereafter, I advise you to use the formalism of Dirac comb(s):
$$x(t)=\sin(\frac{2\pi t}{3})\underbrace{\sum_{k= -\infty}^{\infty}\delta(t-k)}_{\text{unit Dirac comb }\text{Ш}}\tag{1}$$
where, more generally:
$$\displaystyle \operatorname {\text{Ш}} _{T}(t)\ \triangleq \ \sum _{k=-\infty }^{\infty }\delta (t-kT)={\frac {1}{T}}\operatorname {\text{Ш}} \left({\frac {t}{T}}\right)$$
with the property that:
$$\displaystyle \operatorname {\text{Ш}} _{T}(t){ \ \ \stackrel {\mathcal {F}} {\longleftrightarrow }} \ \ {\frac {\sqrt {2\pi }}{T}}\operatorname {\text{Ш}} _{\frac {2\pi }{T}}(\omega ) \ \ \text{or } \ {\frac {1}{\sqrt {2\pi }}}\sum _{n=-\infty }^{\infty }e^{-i\omega nT}\tag{2}$$
(the last expression making the link with Fourier series, but we haven't to use it here).
Therefore, using (2), the Fourier Transform of product (1) is convolution product
$$(\mathcal{F}x)(\omega)=-i \pi (\delta(\omega-\frac{2\pi}{3})-\delta(\omega+\frac{2\pi}{3})) \color{red}{\star} \sqrt {2\pi } \sum _{k=-\infty }^{\infty }\delta (\omega-2 \pi k)$$
It remains to take all the convolution products, an easy task using the shifting property of delta distribution:
$$\delta(\omega-a) \star \delta(\omega-b)=\delta(\omega-(a+b)).$$
Remark 1: Formula (2) is rather natural: a Dirac comb with quite spaced "teeth" will be transformed into a Dirac comb with quite clenched "teeth".
Remark 2: I see in your computation that you come back to the definition of Fourier Transform (as an integral). Why not, but in virtually all practical cases, we only need tables and formulas.