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Let $$x(t) = A\sin(2 \pi f_0 t + \alpha)$$ its Fourier transform is given by $$ X(\omega) = \frac{A \pi}{i}(e^{ia}\delta(\omega-2\pi f_0) - e^{-ia}\delta(w+2\pi f_0)). $$ the Fourier series complex representation of a $T$-periodic is : $$x(t) = \sum_{n=-\infty}^\infty c_n e^{(2 i \pi n)/T \cdot t}$$ thus its Fourier transform is $X(\omega) =2 \pi \sum_{n=-\infty}^\infty c_n \delta(\omega - 2 \pi n/T)$ now here is my question, whats the expression of $c_n$ by identification with the first Fourier transform of the first signal previously, here is what i did : since $\exists k = \omega \cdot T/(2 \pi)$ then the Fourier transform becomes $$X(\omega) = 2 \pi c_k = \frac{A \pi}{i}(e^{ia}\delta(\omega-2\pi f_0) - e^{-ia}\delta(w+2\pi f_0)) \iff \begin{align} c_k = \frac{A}{2i}(e^{ia}\delta(\omega-2\pi f_0) - e^{-ia}\delta(w+2\pi f_0)) \end{align}$$ is this valid? if not whats the problem? if it is then how can i determine the amplitude and frequency spectrum of this signal?

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    $\begingroup$ Look at all them $2\pi$ factors floating around in there. No better example for why we should be using ordinary frequency, $f$, rather than angular frequency, $\omega$ in the Fourier Transform. It put's the $2\pi$ where it belongs. In the exponent. $\endgroup$ Nov 17, 2023 at 3:46
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    $\begingroup$ i really dont understand the hint or the point you trying to transmit , can you elaborate more ? @robertbristow-johnson $\endgroup$ Nov 17, 2023 at 7:22
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    $\begingroup$ As you've shown yourself, the coefficients $c_k$ are the weights of the Dirac impulses, so how can they themselves be Dirac impulses? And where is the index $k$ in your expression for $c_k$? You really just need to equate the Fourier transform of the sinusoid with the Fourier transform of a periodic signal in terms of $c_k$. $\endgroup$
    – Matt L.
    Nov 17, 2023 at 9:23

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Transform looks right, but the logic afterwards needs some correction.

$$ X(\omega) = -i \pi A \left(e^{i \alpha } \delta \left(\omega -2 \pi f_0\right)-e^{-i \alpha } \delta \left(2 \pi f_0+\omega \right)\right) $$ you made a slight mistake with this equality: $$ X(\omega) = 2 \pi \sum^{\infty}_{n=-\infty} c_n \delta ( \omega - 2 \pi n f_0 ) = 2 \pi \sum^{1}_{n=-1} c_n \delta ( \omega - 2 \pi n f_0 ) $$ which simplifies to: $$ X(\omega) = 2 \pi c_{-1} \delta(\omega - 2 \pi (-1) f_0) + 0 + 2 \pi c_{1} \delta(\omega - 2 \pi (1) f_0) = \pi \left(2 c_{-1} \delta(\omega + 2 \pi f_0) + 2c_{1} \delta(\omega - 2 \pi f_0) \right) $$ so we can equate to our original Fourier Transform $$ \pi \left( 2 c_{-1} \delta(\omega + 2 \pi f_0) + 2c_{1} \delta(\omega - 2 \pi f_0) \right) = -i \pi A \left( e^{i \alpha } \delta \left(\omega -2 \pi f_0\right)-e^{-i \alpha } \delta \left(2 \pi f_0+\omega \right)\right) $$ $$ \pi \left(2 c_{-1} \delta(\omega + 2 \pi f_0) + 2c_{1} \delta(\omega - 2 \pi f_0) \right) = \pi \left( -i A e^{i \alpha } \delta \left(\omega -2 \pi f_0\right) + i A e^{-i \alpha } \delta \left(2 \pi f_0+\omega \right) \right) $$ so by inspection then $$ c_{-1} = \frac{i A e^{-i \alpha}}{2} \qquad c_1= \frac{-iAe^{-i \alpha}}{2}. $$ Just note that depending how you define your Fourier Transform (there are a few conventions) your coefficients can change.

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