Hi guys i have to find the fourier transform of the convolution: $$ sinc(t/2T)*\sum\limits_{n-\infty}^{+\infty} (-1)^{n}\delta(t - nT) $$
i was thinking of express the summatory as : $$\sum\limits_{n-\infty}^{+\infty} (-1)^{n}\delta(t - nT) = \mathrm{III}_{2T}(t) -\mathrm{III}_{T}(t-1)$$ so we have:
$$ sinc(t/2T)*[\mathrm{III}_{2T}(t) -\mathrm{III}_{T}(t-1)]$$
So the convolution in the fourier domain became a product: $$ rect(f2T)\mathrm{III}_{1/2T}(t)-2rect(f2T)\mathrm{III}_{1/T}(t)e^{ipif} $$
Well is this a good result and how i can find the energy from this formula ?
$rect(f2T)\mathrm{III}_{1/2T}(t)$ is a single Dirac delta so you are almost done. I think you meant $2\mathrm{III}_{2T}(t) - \mathrm{III}_{T}(t)$ where $$\mathrm{III}_T(t)=\sum_n \delta(t-nT)=|T|^{-1} \mathrm{III}_1( t/|T|)$$ whose $e^{-2i\pi f t}$ Fourier transform is $$\mathrm{III}_1( |T| f)=|T|^{-1}\mathrm{III}_{1/|T|}( f)$$
Alternatively you could replace $\sum\limits_{n-\infty}^{+\infty} (-1)^{n}\delta(t - nT)$ by its Fourier series (which converges only in the sense of distributions) and change the order of summation obtaining a series of $sinc \ast e^{iat} = \widehat{sinc}(a) e^{iat}$.
Your function is $2T$ periodic, so with the energy you meant $\int_0^{2T} |g(t)|^2dt$ ?