From Eq. (8.31), one can deduce that the transfer function of the matched filter is given by \(H(f)=\sin \left(\pi \tau_{0} fight) /\left(\pi \tau_{0} fight)\). Show that

Equation (8.31)
\[
\int_{-\frac{1}{2 \tau_{0}}}^{\frac{1}{2 \tau_{0}}} H(f) \quad d f=\frac{1}{2 \tau_{0}}
\]

x(0;fa)= sin tofa