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Can anyone tell me why the Fourier series coefficient of $$p(t) = \sum_{n=-\inf}^{\inf}\delta(t-nT)$$ is 1/T? $$\\$$ My derivation shows that it should be$$a_k=1/T \int_{0}^{T}p(t)e^{-j2\pi kt/T}dt\\ = 1/T \int_{0}^{T}{\sum_{n=-\inf}^{\inf}\delta(t-nT)}e^{-j2\pi kt/T}dt\\=1/T\sum_{n=-\inf}^{\inf}\int_{0}^{T}\delta(t-nT)e^{-j2\pi kt/T}dt\\=1/T\sum_{n=-\inf}^{\inf}e^{-j2\pi knT/T}\\=1/T\sum_{n=-\inf}^{\inf}1$$.

Therefore, how come $a_K$ is 1/T?

Bernard
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ytj_banana
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    The integral of $\delta(x-x_0)$ is $1$ if the bounds of integration include $x_0$. But if not, then the integral vanishes. As such, you need to confirm whether $t=nT$ for $0<t< T$. – Semiclassical Jan 21 '21 at 23:46
  • will this mean the summation will be gone and only n=0 makes the integral equals to 1. – ytj_banana Jan 21 '21 at 23:51
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    Pretty much, yes. There is a small awkwardness: the bounds of integration coincide with $n=0,1$ cases. Perhaps someone else can clarify this point. – Semiclassical Jan 22 '21 at 00:16
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    Technically speaking, integrating a the Dirac delta centered on one of the interval bounds is not well defined. To resolve this, just integrate between $-\frac T 2$ and $\frac T 2$. – Stefan Lafon Jan 22 '21 at 01:26

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