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Find the fourier transform for signal in this picture (sorry for the bad quality) enter image description here

Could it be done like this? The signal is a sum of two triangular waves that are each delayed. $$x(t)=A\Lambda\left( \frac{t+T/2}{T}\right)-A\Lambda\left( \frac{t-T/2}{T}\right)$$ And the fourier transform for delayed signal $F\{x(t+t_d)\}=X(f)\cdot e^{-i2\pi ft_d}$. And the fourier transform for triangular pulse is defined $F\{ \Lambda(t/T)\}=T \operatorname{sinc}^2(\pi fT) $ \begin{align} F\{x(t)\}&=AT \operatorname{sinc}^2(\pi fT)\cdot e^{-i2\pi f (-\frac{T}{2})} -AT \operatorname{sinc}^2(\pi fT)\cdot e^{-i 2 \pi f \frac T2} \\ &=AT \operatorname{sinc}^2(\pi fT)\cdot (e^{i\pi f T} - e^{-i \pi f T}) \\ &=AT \operatorname{sinc}^2(\pi fT)\cdot2i\sin(\pi f T) \end{align}

Is this solution valid?

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    $\begingroup$ Absolutely. 15 characters are required in a comment. $\endgroup$ Commented Dec 8, 2014 at 21:30
  • $\begingroup$ Okey nice! I first tried to use time derivative properties of F-transform, but I figured it would be easier to do like this :) $\endgroup$
    – ELEC
    Commented Dec 8, 2014 at 21:58

1 Answer 1

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Use differentiation property twice

Differentiate $x(t)$ two times thus it will get converted into impulses

$$\frac{d^2x(t)}{dt^2} = \frac{2A}{T} \delta(t+T)- \frac{4A}{T} \delta(t+\frac{T}{2}) +\frac{4A}{T} \delta(t-\frac{T}{2}) -\frac{2A}{T} \delta(t-T)$$

Now take fourier transform on both sides

$$(j\omega)^2 X(\omega) = \frac{2A}{T} e^{-\omega T}- \frac{4A}{T} e^{-\omega\frac{T}{2}} +\frac{4A}{T} e^{\omega\frac{T}{2}} -\frac{2A}{T} e^{\omega T} $$

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