Fourier transform of $\sum_{n=-\infty}^\infty(-1)^n\delta(t-nT_0)$ Given $x(t)$ and $h(t)=\sum_{n=-\infty}^\infty(-1)^n\delta(t-nT_0)$, I have to compute $Y(f)$, where $y(t)=x(t)h(t)$. I have thought about using that, in this case, $Y(f)=X(f)*H(f)$. I know that $\mathscr{F}(\sum_{n=-\infty}^\infty\delta(t-nT_0))=T_0^{-1}\sum_{n=-\infty}^\infty\delta(t-nf_0)$, but how can I deal with that $(-1)^n?$
HINT:

Note that the given $h(t)$ can be written as

$$h(t)=g(t)-g(t-T_0)\tag{1}$$

with some $g(t)$ the Fourier transform $G(f)$ of which you know. So from $(1)$ you then get

$$H(f)=G(f)\left(1-e^{-j2\pi fT_0}\right)\tag{2}$$