2303 Scaling and Shifting

There is a very useful class of shortcuts which allows us to use the known Fourier series of a function $f(t)$ to get the series for a function related to $f(t)$ by shifts and scale changes. We illustrate this technique with a collection of examples of related functions.

We let $sq(t)$ be the standard odd, period $2\pi$ square wave. $$sq(t)=\begin{cases} -1 &\text{for }-\pi\leq t<0\\ 1 &\text{for }0\leq t <\pi \end{cases}\tag{1}$$

We already know the Fourier series for $sq(t)$. It is $$sq(t)=\frac{4}{\pi}(\sin t+\frac{1}{3}\sin 3t+\frac{1}{5}\sin 5t+\cdots)=\frac{4}{\pi}\sum_{n \text{ odd}}\frac{\sin nt}{t}\tag{2}$$

Shifting and Scaling in the Vertical Direction

Example 1. (Shifting) Find the Fourier series of the function $f_1(t)$ whose graph is shown.

Solution. The graph in Figure 1 is simply the graph in Figure 0 shifted upwards one unit. That is, $f_1(t) = 1 + sq(t).$ Therefore $$f_1(t)=1+\frac{4}{\pi}\sum_{n \text{ odd}}\frac{\sin nt}{t}$$

Example 2. (Scaling) Let $f_2(t) = 2 sq(t).$ Sketch its graph and find its Fourier series.
Solution.

The Fourier series of $f_2(t)$ comes from that of $sq(t)$ by multiplying by 2. $$f_2(t)=\frac{8}{\pi}\sum_{n \text{ odd}}\frac{\sin nt}{t}$$

Example 3. We can combine shifting and scaling along the vertical axis. Let $f_3(t)$ be the function shown in Figure 3. Write it in terms of $sq(t)$ and find its Fourier series.

Solution. $$f_3(t)=\frac{1}{2}(1+sq(t))=\frac{1}{2}+\frac{2}{\pi}\sum_{n \text{ odd}}\frac{\sin nt}{t}$$

Scaling and Shifting in $t$

Example 4. (Scaling in time) Find the Fourier series of the function $f_4(t)$ whose graph is shown.

InFigure 4 the point marked 1 on the $t$-axis corresponds with the point marked $\pi$ in Figure 0. This shows that $f_4(t) = sq(\pi t)$ and therefore we replace $t$ by $\pi t$ in the Fourier series of $sq(t)$. $$f_4(t)=\frac{4}{\pi}\sum_{n \text{ odd}}\frac{\sin n\pi t}{t}$$

Example 5. (Shifting in time) Let $f_5(t) = sq(t + \pi /2)$. Graph this function and find its Fourier series.
Solution. We have $f_5(t)$ is $sq(t)$ shifted to the left by $\pi/2$. Therefore $$f_5(t)=\frac{4}{\pi}(\sin (t+\pi/2)+\frac{1}{3}\sin (3t+3\pi/2)+\cdots)=\frac{4}{\pi}(\cos t-\frac{\cos 3t}{3}+\cdots$$
Notice that $f_5(t)$ is even, and so must have only cosine terms in its series, which is in fact confirmed by the simplified form above.