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Tally ERP user manual

The CDC’s FFT method is similar to that of the Parseval’s theorem, which only uses the square of the Fourier coefficients:
How do these two expressions differ in each case?

A:

The Fourier coefficients are the complex coefficients of the Fourier series decomposition for $f$ and its even and odd parts.

For the first formula, use the Parseval’s identity $\sum_{ -N}^N a_n \overline{a_n} = \sum_{ -\infty}^\infty |a_n|^2$. The Fourier series of $f$ can be written as $\sum_k a_k \mathrm{e}^{\mathrm{i} k \omega}$ where $a_k = \frac{2}{\pi} \int_{ -\pi}^{\pi} f \mathrm{e}^{ -\mathrm{i} k \omega} \mathrm{d} \omega$.
The last expression is the Fourier transform of the square of $f$. For this to make sense, we need some additional conditions on $f$. For instance, you have $f \in L^2$.

For the second formula, use the Parseval’s identity again $\sum_{ -N}^N a_n \overline{a_n} = \sum_{ -\infty}^\infty |a_n|^2$. The Fourier series of $f$ can be written as $\sum_k b_k \mathrm{e}^{\mathrm{i} k \omega}$ where $b_k = \frac{2}{\pi} \int_{ -\pi}^{\pi} f \mathrm{e}^{ -\mathrm{i} k \omega} \mathrm{d} \omega$.
Now, we square $\sum_k b_k \mathrm{e}^{\mathrm{i} k \omega}$. The Parseval’s identity tells us that this is equal to the square of the $L^2$ inner product of the functions $f$ and $1$: \$\sum_k b_k \overline{b_k} = \sum_{ -\infty}^\
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