Relationship between fourier transform and fourier series

Question

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? infinite thanks!

Answer 1

Based on your expression, let's assume we're working with DFT.

Known variable:

$x(t) = A \cdot \sin(2 \pi f_0 t + \alpha)$

Discrete Fourier Transform (DTF)

$$X_k(\omega) = \sum^{N}_{n=0}{x(t) \cdot e^{\frac{-i2{\pi}kn} N}}$$

$$X_k(\omega) = \sum^{N}_{n=0}{A \cdot \sin(2 \pi f_0 t + \alpha) \cdot e^{\frac{-i2{\pi}kn} N}}$$

Because $\sin(x) = \frac{e^{ix} - e^{-ix}} {2i}$, we can substitute it to be:

$$X_k(\omega) = \sum^{N}_{n=0}{A \cdot \frac{e^{i(2 \pi f_0 t + \alpha)} - e^{-i(2 \pi f_0 t + \alpha)}} {2i} \cdot e^{\frac{-i2{\pi}kn} N}}$$

$$X_k(\omega) = \sum^{N}_{n=0}{A \cdot \frac{e^{i2 \pi f_0 t + i \alpha} - e^{-i2 \pi f_0 t - i \alpha}} {2i} \cdot e^{\frac{-i2{\pi}kn} N}}$$

$$X_k(\omega) = \sum^{N}_{n=0}{\frac{A}{2i} \cdot \left(e^{i2 \pi f_0 t + i \alpha} - e^{-(i2 \pi f_0 t + i \alpha)}\right) \cdot e^{\frac{-i2{\pi}kn} N}}$$

$$X_k(\omega) = \sum^{N}_{n=0}{\frac{A}{2i} \cdot \left(e^{i2 \pi f_0 t + i \alpha - \frac{i2{\pi}kn} N} - e^{-(i2 \pi f_0 t + i \alpha) - \frac{i2{\pi}kn} N}\right)}$$

$$X_k(\omega) = \sum^{N}_{n=0}{\frac{A}{2i} \cdot \left(e^{1 - \frac{i2{\pi}kn} N} - e^{-1 - \frac{i2{\pi}kn} N}\right)^{i2 \pi f_0 t + i \alpha}}$$

$$\therefore X_k(\omega) = \sum^{N}_{n=0}{\frac{A}{2i} \cdot \left(e^{1 - \frac{i2{\pi}kn} N} - e^{-(1 + \frac{i2{\pi}kn} N)}\right)^{i2 \pi f_0 t + i \alpha}}$$

From this point onward, I would let you to compare yourself with your formula because I don't have your full formula (and privilege to comment for clarification). For amplitude, it's quite easy if the magnitude in frequency domain ($X(\omega)$) was known:

$$X_k(\omega) = \sum^{N}_{n=0}{\frac{A}{2i} \cdot \left(e^{1 - \frac{i2{\pi}kn} N} - e^{-(1 + \frac{i2{\pi}kn} N)}\right)^{i2 \pi f_0 t + i \alpha}}$$

$$X_k(\omega) = A \cdot \sum^{N}_{n=0}{\frac{1}{2i} \cdot \left(e^{1 - \frac{i2{\pi}kn} N} - e^{-(1 + \frac{i2{\pi}kn} N)}\right)^{i2 \pi f_0 t + i \alpha}}$$

$$\therefore A = \frac{X_k(\omega)} {\sum^{N}_{n=0}{\frac{1}{2i} \cdot \left(e^{1 - \frac{i2{\pi}kn} N} - e^{-(1 + \frac{i2{\pi}kn} N)}\right)^{i2 \pi f_0 t + i \alpha}}}$$