# Laplace Transform and Fourier Transform of a function

I have this transfer function: $$h(t)= -\frac{1}{16}te^{-2t}$$ and the Laplace Transform is: $$H(s) = \frac{-\frac{1}{16}}{(s+2)^{2}}$$ I know that to find the Fourier Transform, I would just replace s with jw $$H(j\omega) = \frac{-\frac{1}{16}}{(j\omega+2)^{2}}$$ But, the Fourier Transform table says: $$tf(t) \Leftrightarrow j\frac{d}{d\omega}\hat{f}(\omega)$$ And the Fourier transform of $$e^{-2t} \Leftrightarrow \frac{1}{j\omega + 2}$$

$$-\frac{1}{16}te^{-2t} \Leftrightarrow \frac{j\frac{1}{16}}{(j\omega + 2)^{2}}$$

So, the final answer is not the same as $H(j\omega)$!

Do you see any thing wrong here? Please guide me!

When you differentiate, there is another $j$ in the numerator giving $j^2$.
Also, your $h(t)$ should be $$h(t)= -\frac{1}{16}te^{-2t} ~{\mathbf u(t)}$$