Question 24.12: Given that F{δ(t − a)} = e^−jωa find F{e^−jta}....

We have f(t) = δ(t − a),  F(\omega)=\mathrm{e}^{-\mathrm{j}\omega t}.  Applying the t–ω duality principle we find

f(-\omega)=\delta(-\omega-a)={\frac{1}{2\pi}}{\mathcal{F}}\{\mathrm{e}^{-\mathrm{j}ta}\}

Therefore

\begin{array}{c}{{{\mathcal{F}}\{\mathrm{e}^{-\mathrm{j}ta}\}=2\pi\delta(-\omega-a)}}\\ {{\phantom{\qquad}=2\pi\delta(-(\omega+a))}}\\ {{\quad}=2\pi\delta(\omega+a)}\end{array}

since δ(ω) is an even function.