Find fourier transform given the graph of a function

Question

Cheers, I am given the following graph for $x(t)$:

and I am asked to find the fourier tranform by using the integral property and knowing that $F(Π(t)) = sinc (\frac{\omega}{2\pi})$

The solution I saw stated that we can see that $x(t) = \int_{- \infty}^tΠ(ρ)dρ$, but I just can't see why that is, could that be explained?

Also I tried to go a different route, by finding it's equation. So I saw that : $x(t) = \begin{cases} 0, & \text{if $t < 0.5$ } \\ t + 0.5, & \text{if $t \in [-0.5,0.5]$} \\ 1, & \text{if $t>0.5$} \end{cases}$

Now I try using the differentiation property of the fourier transform. So I see that: $\frac{dx(t)}{dt} = \begin{cases} 0, & \text{if $t < 0.5$ } \\ 1, & \text{if $t \in [-0.5,0.5]$} \\ 0, & \text{if $t>0.5$} \end{cases}$ which I rewrite as: $u(t+0.5) - u(t-0.5)$, and then I differentiate again, so I get that $x''(t) = \delta(t + 0.5) - \delta(t-0.5)$ and now I get that:

$(j\omega)^2X(\omega) = e^{j\omega\frac{1}{2}} - e^{-j\omega\frac{1}{2}} \rightarrow X(\omega) = \frac{2j}{-\omega^2} \frac{e^{j\omega\frac{1}{2}} - e^{-j\omega\frac{1}{2}}}{2j} = \frac{2j}{-\omega^2} \sin(\frac{\omega}{2})$ which is not the answer that I should get which is:

Any help on If I am doing anything right? Thanks =)

Answer 1

Your method correctly produces the term $$\frac{1}{j\omega}\frac{\sin(\omega/2)}{\omega/2}$$ of the given solution, but you forget that by taking the derivative you lose information about any constant terms in $x(t)$. However, looking at the graph it's easy to figure out that the constant term in $x(t)$ is $\frac12$. The Fourier transform of that constant term is $\frac12\cdot 2\pi\delta(\omega)$, which gives you the missing term.

Concerning your first question, note that differentiating the given function resulted in a rectangle, so it shouldn't surprise you that $x(t)$ can be written as the integral of a rectangular function.