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I'm trying to find the Fourier Transform of: $$t^2e^{-2t}u(t)$$ I know that from the FT pairs table, $$ FT(te^{-2t})= \frac{1}{(2+jw)^2}$$ So I assume that we can just use the multiplication property and do $$\frac{1}{2\pi}\left(FT(t)*\frac{1}{(2+jw)^2}\right)$$ However when I try that, I do not get the correct answer, which should be $$\frac{2}{(2+jw)^3}$$ Am I missing some kind of trick I can apply here?

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Here we can use the property stating that $$\operatorname{FT}(t^nf(t))=j^n\frac{d^n}{dw^n}\operatorname{FT}(f(t))$$

In our case \$n=2, f(t)=e^{-2t}u(t)\$. So \$\operatorname{FT}(f(t))=\frac{1}{2+j\omega}\$. The second derivative of it would be \$\frac{2j^2}{(j\omega+2)^3}=\frac{-2}{(j\omega+2)^3}\$, and I will leave it to you to calculate. Multiplying it by \$j^n=-1\$, giving us \$\frac{2}{(j\omega+2)^3}\$ as expected.

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