Fourier series of a square wave signal with a bias

Question

Given a $f(t)$ of the kind: $$f(t)=1, \{kt_0\le t\le kt_0+\tau\}$$ $$f(t)=a,\{kt_0+\tau\le t\le (k+1)t_0\}$$ with $a\lt 1$ what is the Fouries series development of f(t)? Thanks

Answer 1

$f$ is the sum of a constant and a multiple of a characteristic function of an interval,

$$f(t) = a + (1-a)\cdot \chi_{[0,\tau]}(t).$$

So the Fourier series of $f$ is the sum of the Fourier series of the two parts. The Fourier series of the constant is trivial if the basis contains a constant function, as is the case for the most common bases $\{e^{2\pi int/t_0}\}_{n\in\mathbb{Z}}$ and $\{\cos (2\pi nt/t_0)\}_{n\in\mathbb{N}} \cup \{\sin (2\pi nt/t_0)\}_{n\in\mathbb{Z}^+}$, both systems scaled to obtain an orthonormal basis.

So it remains to find the coefficients of the Fourier series of the characteristic function. For the exponential basis,

$$\begin{align} \int_0^{t_0} \chi_{[0,\tau]}(t) e^{-2\pi int/t_0}\,dt &= \int_0^\tau e^{-2\pi int/t_0}\,dt\\ &= \begin{cases} \qquad \tau &, n = 0\\ \frac{t_0}{2\pi in}\left(1-e^{-2\pi in\tau/t_0}\right) &, n \neq 0. \end{cases} \end{align}$$

For the trigonometric basis, we get

$$\begin{align} \int_0^{t_0} \chi_{[0,\tau]}(t)\sin (2\pi nt/t_0)\,dt &= \int_0^\tau \sin(2\pi nt/t_0)\,dt\\ &= \frac{t_0}{2\pi n}\left(1 - \cos (2\pi n\tau/t_0)\right), \end{align}$$

and

$$\begin{align} \int_0^{t_0} \chi_{[0,\tau]}(t)\cos (2\pi nt/t_0)\,dt &= \int_0^\tau \cos(2\pi nt/t_0)\,dt\\ &= \begin{cases} \qquad\tau &, n = 0\\ \frac{t_0}{2\pi n}\sin (2\pi n\tau/t_0) &, n \neq 0. \end{cases} \end{align}$$