CTFS analysis and synthesis equations

The Fourier series representation of a periodic signal expresses it as a sum (linear combination) of harmonically related sinusoids or complex exponentials, whose frequencies are integer multiples of the fundamental frequency $\omega_0 = 2\pi/T$. It shows which frequencies the signal contains and how strongly.

Synthesis equation

Each $e^{jk\omega_0 t}$, $k = 0, \pm1, \pm2, \dots$, is periodic with period $T$, so their combination is also periodic with $T$:

$$x(t) = \sum_{k=-\infty}^{\infty}a_k\,e^{jk\omega_0 t}$$

This builds (synthesises) $x(t)$ from its harmonics; $a_k$ are the Fourier series coefficients.

Analysis equation

Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period:

$$\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt$$

For $k \ne n$, $e^{j(k-n)\omega_0 t}$ completes $(k - n)$ whole cycles in $T$, so

$$\int_0^T e^{j(k-n)\omega_0 t}dt = \int_0^T\cos((k-n)\omega_0 t)dt + j\int_0^T\sin((k-n)\omega_0 t)dt = 0$$

For $k = n$ the integrand is 1 and the integral is $T$. So only $k = n$ survives:

$$\int_0^T x(t)e^{-jn\omega_0 t}dt = T a_n$$ $$a_k = \frac{1}{T}\int_T x(t)\,e^{-jk\omega_0 t}\,dt$$

This extracts (analyses) the amount of the $k$th harmonic in $x(t)$; the integral may be taken over any interval of length $T$.

Summary

Example: $x(t) = \sin\omega_0 t = \frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omega_0 t}$, so $a_1 = \frac{1}{2j}$, $a_{-1} = -\frac{1}{2j}$, all others 0.