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[solved] Show that i(t)=(n+1)sin 10t


Active member
Jan 26, 2012
$\displaystyle i(t)=\sum_{n=0}^\infty(-1)^nu\left(t-\frac{n\pi}{10}\right)\sin\left[10\left(t-\frac{n\pi}{10}\right)\right]$, where $\displaystyle u(t)$ is the unit step function.

Show that $\displaystyle i(t)=(n+1)\sin 10t$ if $\displaystyle\frac{n\pi}{10}<t<\frac{(n+1)\pi}{10}$.

---------- Post added 03-05-2012 at 12:21 AM ---------- Previous post was 03-04-2012 at 11:47 PM ----------

Solved it, don't bother.
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