I came across this problem

Strange signal

I am supposed to find the Fourier transform of $g(t)$, but I am not able to find the analytical expression of such signal.

The teacher suggests that I should consider the inclined part in $\left[-T, 0\right]$ as the integral of a determined function.

It turns out it is the integral of some $\rm{rect}()$ function. However, the teacher noted that I should subtract from it the (???) step function.

Can you explain to me how to find the fourier transform of $g(t)$ more comprehensively?


I tried to solve it, and I came to this solution. Is it correct?

enter image description here


The simple answer is “Linearity”

The Fourier Transform is linear. $g(t)$ can be expressed as a sum of signals that have known, as in there are tables of transform pairs, in most books, probably also in your course notes, along with the shift and scale properties. Look at the tables you have and find entries that you can add to get $g(t))$. You already found rect(). Look for a triangle piece.

A full triangle is the convolution of 2 rects. A half triangle is a full triangle multiplied by a step.

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  • $\begingroup$ Care to expand? I know linearity and such, but I do not know how to express such function, and the logical process behind it. $\endgroup$ – Riccardo M. Pesce Dec 8 '17 at 17:03
  • $\begingroup$ Is it correct this way? $\endgroup$ – Riccardo M. Pesce Dec 9 '17 at 20:33

In addition to the linearity property in decompositing the $g(t)$ into simpler waveforms, you can also make use of the differentiation property which states that:

$$ x(t) \longleftrightarrow X(\omega) \implies x'(t) \longleftrightarrow j\omega X(\omega) $$

Taking the derivative of $g(t)$ is easy, resulting in two impulses and a shifted rectangular pulse. Their CTFTs are readily available; and hence keeping track of phase modifications and weights you can conclude $X(\omega)$ easily.

From a visual analysis of the given function $g(t)$, it can be deduced that the derivative function $$g'(t) = \delta(t+T) + \frac{1}{T} \text{p}_T(t+T/2) - 2 \delta(t-T)$$ where the pulse function $\text{p}_T(t)$ is defined as $$\text{p}(t) = \begin{cases} 1 &, \text{ for } -\frac{T}{2} < t < \frac{T}{2} \\ 0 &, \text{ otherwise } \\ \end{cases} $$

The CTFT of $g'(t)$ can be found from: $$ \mathcal{F} \{ g'(t) \} = \mathcal{F} \{ \delta(t+T) \} + \mathcal{F} \{ \frac{1}{T} \text{p}_T(t+T/2) \} + \mathcal{F} \{ - 2 \delta(t-T) \} $$

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  • $\begingroup$ Would you mind explaining more? I already now the formulae about the Fourier transform, including derivative and integral. What I do not understand is the excercise resolution, itself. $\endgroup$ – Riccardo M. Pesce Dec 8 '17 at 17:02
  • $\begingroup$ I solved it this way. $\endgroup$ – Riccardo M. Pesce Dec 9 '17 at 20:33

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