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 $(-1)^n?$


1 Answer 1



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


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

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

  • $\begingroup$ 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!! $\endgroup$
    – Gibbs
    Jan 12, 2020 at 21:06
  • 1
    $\begingroup$ @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. $(2)$. $\endgroup$
    – Matt L.
    Jan 12, 2020 at 21:10

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