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I have the following signal for which amplitude and phase spectrums have to be computed:

enter image description here

This exercise also has a solution which begins by deriving the signal twice.

Next, they say the spectral density function for the given signal is:

$$X(j\omega)=\dfrac{X''(j\omega)}{(j\omega)^2} $$

In my attempt to figure out how did they get to that, I found the derivative property of the Fourier transform in the few pages of notes I have(there should be an F above the equivalency sign):

$$x''(t) \Leftrightarrow (j\omega)^2X(j\omega) $$

It looks quite a lot to what they have, but I do not know what do from here in order to reach their equation(formula). Also, can this property be applied to this signal or it only applies to non-periodic signals?

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  • $\begingroup$ It is not clear what you are asking, please try to be more concise and focus on a specific, answerable question. $\endgroup$ – PAK-9 Aug 28 '14 at 16:26
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I'll try to give you some clues so you'll be able to understand and solve this problem by yourself. First of all it's important to understand what they mean by $X''(j\omega)$. This looks like the second derivative of the function $X(j\omega)$, but it isn't! What they mean is the Fourier transform of the second derivative of $x(t)$, and this is exactly the correspondence you wrote down in your question:

$$X''(j\omega)=\mathcal{F}\{x''(t)\}=(j\omega)^2X(j\omega)\tag{1}$$

So if you want to compute $X(j\omega)$ you might as well compute $X''(j\omega)$ and divide by $(j\omega)^2$ (for $\omega\neq 0)$. You still have to figure out $X(0)$ because you can't compute it form Eq. (1). But since this is just the DC component of $x(t)$ it's usually quite easy to compute.

The reason why they chose to compute $X(j\omega)$ from $X''(j\omega)$ is because for the given function $x(t)$ it's much easier to compute the Fourier transform of the second derivative of $x(t)$. Since $x(t)$ is a piecewise linear function, the first derivative $x'(t)$ is a piecewise constant function, and the second derivative $x''(t)$ just has delta impulses at $t=0$ and $t=\tau$ (and at $t=nT$ and $t=nT+\tau$), i.e. where $x'(t)$ has jumps. Computing the Fourier coefficients or the Fourier transform is now easy because you just have to integrate delta impulses. After having obtained $X''(j\omega)$ in this way, you can compute $X(j\omega)$ (apart from $\omega=0$) by using (1).

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