Laplace Transform and Fourier Transform of a function

Question

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!

Answer 1

When you differentiate, there is another $j$ in the numerator giving $j^2$.

Also, your $h(t)$ should be $$ h(t)= -\frac{1}{16}te^{-2t} ~{\mathbf u(t)} $$