# 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 $$(-1)^n?$$

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
$$H(f)=G(f)\left(1-e^{-j2\pi fT_0}\right)\tag{2}$$
• I think I have it: If $g(t)=\sum_{n=-\infty}^\infty\delta(t-2nT_0)$, then $g(t)-g(t-T_0)=\sum_{n=-\infty}^\infty\delta(t-2nT_0)-\sum_{n=-\infty}^\infty\delta(t-2(n+1)T_0)=h(t)$, and then we can calculate $H(f)$. So tricky! Thanks!! Jan 12, 2020 at 21:06
• @Gibbs: Almost there. There should be a minus sign on the RHS of your equation, and the argument of the delta impulse should be $(t-T_0-2nT_0)=(t-(2n+1)T_0)$. But you just need $g(t)$ and $G(f)$, and then you just use Eq. $(2)$. Jan 12, 2020 at 21:10