Derive trigonometric Fourier series (CT)

A periodic signal $x(t)$ with period $T$ ($\omega_0 = 2\pi/T$) that meets the Dirichlet conditions can be written as a dc term plus sums of cosines and sines at multiples of $\omega_0$. This is the trigonometric Fourier series:

$$x(t) = a_0 + \sum_{n=1}^{\infty}\left[a_n\cos n\omega_0 t + b_n \sin n\omega_0 t\right]$$

Orthogonality relations (over one period $T$)

For integers $m, n \ge 1$:

$$\begin{aligned} &\int_T \cos m\omega_0 t\,dt = \int_T \sin m\omega_0 t\,dt = 0 \\ &\int_T \cos m\omega_0 t\,\cos n\omega_0 t\,dt = \begin{cases} T/2, & m = n \\ 0, & m \ne n \end{cases} \\ &\int_T \sin m\omega_0 t\,\sin n\omega_0 t\,dt = \begin{cases} T/2, & m = n \\ 0, & m \ne n \end{cases} \\ &\int_T \sin m\omega_0 t\,\cos n\omega_0 t\,dt = 0 \quad \text{for all } m, n \end{aligned}$$

Finding $a_0$

Integrate both sides over one period. Every sine and cosine term integrates to zero:

$$\int_T x(t)\,dt = a_0 T \quad\Rightarrow\quad a_0 = \frac{1}{T}\int_T x(t)\,dt$$

Finding $a_n$

Multiply both sides by $\cos m\omega_0 t$ and integrate over $T$. Only the term with $n = m$ in the cosine sum survives:

$$\int_T x(t)\cos m\omega_0 t\,dt = a_m\frac{T}{2} \quad\Rightarrow\quad a_n = \frac{2}{T}\int_T x(t)\cos n\omega_0 t\,dt$$

Finding $b_n$

Multiply by $\sin m\omega_0 t$ and integrate; only the $n = m$ sine term survives:

$$\int_T x(t)\sin m\omega_0 t\,dt = b_m\frac{T}{2} \quad\Rightarrow\quad b_n = \frac{2}{T}\int_T x(t)\sin n\omega_0 t\,dt$$

Summary and notes

• Even $x(t)$: $b_n = 0$ (cosine terms only). Odd $x(t)$: $a_0 = a_n = 0$ (sine terms only).
• Compact (polar) form: $x(t) = C_0 + \sum_{n=1}^{\infty}C_n\cos(n\omega_0 t + \theta_n)$ with $C_0 = a_0$, $C_n = \sqrt{a_n^2 + b_n^2}$, $\theta_n = -\tan^{-1}(b_n/a_n)$.
• Link to exponential form: $c_0 = a_0$, $c_n = \frac{a_n - jb_n}{2}$, $c_{-n} = \frac{a_n + jb_n}{2}$.