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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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closed as unclear what you're asking by Dilip Sarwate, jojek, Deve, Paul R, PAK-9 Aug 28 '14 at 16:26

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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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