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}.