Finding the Fourier Series of a step function

[Tony Hau]

Homework Statement

Given : $f(x) = \begin{cases} 0, & -\pi \lt x \lt 0 \\ 1, & 0 \lt x \lt \frac{\pi}{2} \\ 0, & \frac{\pi}{2} \lt x \lt \pi \end{cases}$,
Find the Fourier Series of $f(x)$.

Relevant Equations

$a_n =\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx)dx$
$b_n =\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)dx$

The answer in the textbook writes: $$ f(x) = \frac{1}{4} +\frac{1}{\pi}(\frac{\cos(x)}{1}-\frac{\cos(3x)}{3}+\frac{\cos(5x)}{5} \dots) + \frac{1}{\pi}(\frac{\sin(x)}{1}-\frac{2\sin(2x)}{2}+\frac{\sin(3x)}{3} + \frac{\sin(5x)}{5}\dots)$$

I am ok with the two trigonometric series in the answer. However, I don't understand where that $\frac{1}{4}$ comes from.
Since the formula for $a_0$ is $a_0 =\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} dx$, which gives $\frac{1}{2}$, isn't that constant equal to $\frac{1}{2}$ instead of $\frac{1}{4}$.

 

[etotheipi]

Since the formula for $a_0$ is $a_0 =\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} dx$, which gives $\frac{1}{2}$, isn't that constant equal to $\frac{1}{2}$ instead of $\frac{1}{4}$.

The first term is the series is $a_0/2$, not $a_0$.

 

[Tony Hau]

The first term is the series is $a_0/2$, not $a_0$.

Thanks. No joke but this thing has bothered me for the whole afternoon... :)

 

[etotheipi]

Thanks. No joke but this thing has bothered me for the whole afternoon... :)

That's annoying... but at least you will never forget it again!

 

[jbunniii]

Note that the constant term of the Fourier series is simply the average value of $f(x)$ over the interval of interest (which is $[-\pi,\pi]$ in this case). Since your $f(x)$ has value $1$ on one fourth of this interval (namely on $[0,\pi/2]$) and value 0 elsewhere, it's clear that the average value of $f(x)$ is $1/4$. This is a useful property to keep in mind when double-checking your calculation.

 

[Tony Hau]

Note that the constant term of the Fourier series is simply the average value of $f(x)$ over the interval of interest (which is $[-\pi,\pi]$ in this case). Since your $f(x)$ has value $1$ on one fourth of this interval (namely on $[0,\pi/2]$) and value 0 elsewhere, it's clear that the average value of $f(x)$ is $1/4$. This is a useful property to keep in mind when double-checking your calculation.

Thanks. This is a nice interpretation that is not mentioned in the book.