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Consider a Fourier series $x(t)=\sum_{n=-\infty}^{\infty} C_n e^{j n \omega_0 t}$. Let's assume $\mathrm{x}_1(\mathrm{t}) {\leftrightarrow} \mathrm{C}_{\mathrm{n}}, \mathrm{x}_2(\mathrm{t}){\leftrightarrow} \mathrm{D}_{\mathrm{n}}$. Then we know that $\mathrm{x}_1(\mathrm{t}) \cdot \mathrm{x}_2(\mathrm{t}) {\leftrightarrow} \sum_{\mathrm{k}=-\infty}^{\infty} \mathrm{C}_{\mathrm{k}} \mathrm{D}_{\mathrm{n}-\mathrm{k}}$. But how about $\mathrm{x}_1(\mathrm{t}) \cdot \frac{d}{dt}\mathrm{x}_2(\mathrm{t}) $ ? Is it $\mathrm{x}_1(\mathrm{t}) \cdot \frac{d}{dt}\mathrm{x}_2(\mathrm{t}) {\leftrightarrow} \sum_{\mathrm{k}=-\infty}^{\infty} \mathrm{C}_{\mathrm{k}}.(in\omega_0). \mathrm{D}_{\mathrm{n}-\mathrm{k}}$ ?

egc
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  • You seem to be using both $j$ and $i$ to mean the same thing. Anyway, you can use your formula but with $D_n$ replaced by $(in\omega_0)D_n$, which means that $D_{n-k}$ should be replaced by $(i(n-k)\omega_0)D_{n-k}$. So it would be $\sum_{\mathrm{k}=-\infty}^{\infty} \mathrm{C}{\mathrm{k}}.(i(n-k)\omega_0). \mathrm{D}{\mathrm{n}-\mathrm{k}}.$ – Jair Taylor Oct 16 '23 at 20:07
  • Thank you for your answer. – egc Oct 16 '23 at 22:19

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