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I need to calculate the fourier transform of this $ t \cdot sin(t-3)+2 \cdot t \cdot cos(3t) \cdot rect(6t) $

is the following valid, based on the first fourier identity i've read in a book, and using linearity

$F\{ (-jt)x(t)\} \quad =\quad \frac { dX(ω) }{ dω } \quad \Longleftrightarrow \\ -j\quad F\{ tx(t)\} =\quad \quad \frac { dX(ω) }{ dω } \quad \Longleftrightarrow \\ F\{ tx(t)\} \quad =\quad j\frac { dX(ω) }{ dω } $

so i can make the problem easier? If not, what is mistaken, and how can i solve this problem? convolution?

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    $\begingroup$ It doesn't have a Fourier transform in the usual sense. Are you sure that was the question asked? Otherwise you need to deal with distributions. $\endgroup$
    – copper.hat
    Jun 10, 2014 at 16:26
  • $\begingroup$ to be exact the question was $ t \cdot sin(t-3)+2 \cdot t \cdot cos(3t) \cdot rect(6t)$, this update makes fourier possible? i thought it was a linear combination so it didn't matter $\endgroup$
    – Cris
    Jun 10, 2014 at 16:37
  • $\begingroup$ I guess you are using distributions (delta functions here). You have something like ${\cal F}(f \cdot g) = {\cal F} f * {\cal F} g$ where $*$ is convolution. You need to deal with the convolution of two delta functions. $\endgroup$
    – copper.hat
    Jun 10, 2014 at 16:54
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    $\begingroup$ yeah delta functions will be used for sure, but can i use the property above so i can make easier the calculation? (one convolution instead of 2) $\endgroup$
    – Cris
    Jun 10, 2014 at 17:10
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    $\begingroup$ You can, but you need to deal with derivatives of distributions... $\endgroup$
    – copper.hat
    Jun 10, 2014 at 17:18

1 Answer 1

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Yes, you can use these properties. For the first part $t\sin(t-3)$ just use

$$\sin(t-3)\Longleftrightarrow -je^{-3j\omega}\pi[\delta(\omega-1)-\delta(\omega+1)]=-j\pi[e^{-3j}\delta(\omega-1)-e^{3j}\delta(\omega+1)]$$

which after multiplication by $j$ and differentiation gives

$$t\sin(t-3)\Longleftrightarrow \pi[e^{-3j}\delta^{\prime}(\omega-1)-e^{3j}\delta^{\prime}(\omega+1)]$$

where $\delta^{\prime}(\omega)$ is the derivative of the distribution $\delta(\omega)$. In a similar way you can compute the Fourier transform of $2t\cos(3t)$, which you can then convolve with the Fourier transform of the rectangular function. Just note the following relation for convolution with the derivative of the Dirac impulse:

$$\delta^{\prime}(\omega-\omega_0)*X(\omega)=X^{\prime}(\omega-\omega_0)$$

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