Question 24.5: Use the first shift theorem to find the function whose Fouri......

From the given result we have

${\mathcal{F}}\{u(t)\,\mathrm{e}^{-3t}\}={\frac{1}{3+\mathrm{j}\omega}}=F(\omega)$

Now

${\frac{1}{3+\mathrm{j}(\omega-2)}}=F(\omega-2)$

Therefore, from the first shift theorem with a = 2 we have

${\mathcal{F}}\{\mathrm{e}^{2j}u(t)\,\mathrm{e}^{-3t}\}={\frac{1}{3+\mathrm{j}(\omega-2)}}$

Consequently the function whose Fourier transform is $\frac{1}{3+\mathrm{j}(\omega-2)}\;\mathrm{is}\;u(t)\,\mathrm{e}^{-(3-2)j)t}.$