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I got this on my physics class but I post it here since it relate to math more enter image description here

Here, the explanation is $$s(t)e^{-j\frac{2\pi lt}{T}}=\sum c_ke^{j\frac{2\pi kt}{T}}e^{-j\frac{2\pi lt}{T}} \\ \implies\int s(t)e^{-j\frac{2\pi lt}{T}}=\int\sum c_ke^{j\frac{2\pi kt}{T}}e^{-j\frac{2\pi lt}{T}} $$ At this point, my lecturer said "over $e^{j\frac{2\pi kt}{T}}e^{-j\frac{2\pi lt}{T}}$ the only time you get anything is when $k=l$ so what happens, all the term in this sum disappear except for the $l$ for whatever $l$ is

So the following result is $$c_l=\frac1T \int_0^T s(t)e^{-j\frac{2\pi lt}{T}}dt$$

I don't understand why is this. Can someone explain it for me.

Gerry Myerson
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aukxn
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  • Integrating a complex exponential over a full period gives zero, unless the complex exponential is constant --- and that only happens when $k=l$ and the two exponentials multiply to give 1. – Gerry Myerson Mar 29 '14 at 05:47
  • but in $\int s(t)e^{-j\frac{2\pi lt}{T}}=\int\sum c_ke^{j\frac{2\pi kt}{T}}e^{-j\frac{2\pi lt}{T}} $, there is no $dt$ or $dl$ so how do you now what is integrated – aukxn Mar 29 '14 at 05:52
  • In your picture, there's a $dt$, so I assume that's what's meant. – Gerry Myerson Mar 29 '14 at 05:53

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