Please help me understand the solution to this graph

Question

I have a function

$$f(t) = \begin{cases} 0, & \text{0≤t<1} \\[2ex] 2, & \text{1≤t<2} \end{cases}$$

According to the solutions manual, the period is 2

this is an odd function so I add $$-f(-t)$$

And get

$$f_o(t) = \begin{cases} 0, & \text{0≤t<1} \\[2ex] 2, & \text{1≤t<2} \\[2ex] -2, & \text{-2≤t<-1} \\[2ex] 0, & \text{-1≤t<0} \end{cases}$$

I have trouble extending the graph. The period is 2, so initially I thought the graph of f(t) should repeat itself for every 2 units and the graph of -f(-t) should repeat itself for every 2pi, but thats not the case.

So if the function f(t) repeats itself every 2 units. The first period is from x=0 to x=2. And after x=2 the function should start again right? For instance y=0 is between x=0 and 1, y=2 is between x= 1 and 2. If the function repeats itself after every two units, that means the function should continue after x=2 and we should get the new lines, y=0 between x=2 and 3, and y= 2 between x= 3 and 4. But looking at the solution, that clearly isnt the case.

Can someone please explain whats going on?

Edit:

Problem:

Solution:

Answer 1

Thank you for the extra information.

a) You are asked to determine the period and the frequency. Based on the information you are given, the best choice is a period of 2.

The graph with period 2 should look like this:

b) You are asked to express the function f as a Fourier Sine Series. Because the sine function is an odd function, we pretend that your function is also odd!

Rotating your initial function about the origin gives us:

Taking this element and repeating it gives us the periodic function with period $T=4$ as plotted in the solutions manual.

Answer 2

The conditions you are given on the function are inconsistent. If you are told the function is odd and period $2$ you should only specify it over a range of $1$, such as $[0,1]$. You can use the fact that it is odd to extend that to $[-1,1]$, then use the fact that it is period $2$ to extend it over the whole real line. Alternately, you are given the values over $[0,2)$. You can extend it as period $2$, but it will not be odd. You can extend it to be odd, giving values over $(-2,2)$, and then extend it as a period $4$ function (though you will not have values for $2+4k$ if you do this.