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I think I must be missing something obvious, but I can't for the life of me see what it is. The discrete version of Parseval's theorem can be written like this:

$$\sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{n=0}^{N-1} |X[k]|^2 $$

Now, say you've got some function in time, like $x = \sin(\omega t)$. Depending on how large your value of $N$, the LHS could be arbitrarily large.

The FT of this function is a delta function at $\omega$. Everywhere else is zero, so your RHS is given by $1/N$. So how on earth does one side equal the other?! It seems to me like you've got a summation one one side and an average value on the other...

I must be missing something obvious!

Ned Yoxall
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  • Good point. I'll put it up there. – Ned Yoxall May 17 '13 at 08:35
  • You can get displayed equations by using double dollar signs instead of single dollar signs. That makes fractions and sums with limits come out less squashed. (It also centres the equation.) There's an edit link underneath the question. – joriki May 17 '13 at 10:43

1 Answers1

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It seems that you're used to a normalization convention that's different from the one used by the Wikipedia article from which you quoted the theorem. With the convention used in the article,

$$ X_k = \sum_{n=0}^{N-1} x_n \cdot\mathrm e^{-\mathrm i 2 \pi k n / N}\;, $$

the problem you describe doesn't occur. In your example, exactly one of the $X_k$ is non-zero, and it's proportional to $N$, so the left-hand side and the right-hand side are both proportional to $N$.

joriki
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