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i have a periodic signal $x[n] = \cos (\frac{2 \pi}{10}n)$.
I found this DTFT pair:

enter image description here

That's the only pair for a cosine function i found. But what is $\delta _{2\pi}$ ? Is it just a Dirac $\delta$ function for every period ?
And the $\pi$ before the brackets is just for a higher amplitude of the impulse ?

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madmax
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    Ultimately, the $\pi$ before the brackets is a convention. You can change how you define the Fourier transform, but then you'd also need to change the inverse transform.

    Normally, $\delta_{2\pi}$ denotes a periodic delta function, meaning that $\delta_{2\pi}(x) = \delta_{2\pi}(y) \Leftrightarrow x = y + 2 k \pi$ for some integer $k$. I can't really see if that should be the case here, but you should verify it.

     – Gerben May 25 '11 at 07:52
  • Well, it just says calculate the periodic signal $x[n]$. – madmax May 25 '11 at 08:35

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The DTFT is always $2\pi$-periodic, since changing the frequency by $2\pi$ doesn't change the phase at the integers. The cosine is a sum of two exponentials and has frequency components at $\theta_0$ and $-\theta_0$. The $\delta_{2\pi}$ function replicates these with period $2\pi$. As was already stated in the comments, the factor $\pi$ is a matter of convention; it gives the right normalization for the definition used e.g. in the Wikipedia article.

joriki
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