The Fourier transform of a signal h(t) is H(jω) = (2 cosω) (sin 2ω)/w. The value of h(0) is
(a) \frac{1}{4} (b) \frac{1}{2} (c) 1 (d) 2
Given that
F[h(t)]=H(j \omega)=\frac{(2 \cos \omega)(\sin 2 \omega)}{\omega}
Inverse Fourier transform of sinc function is a rectangular function
\begin{aligned} & F^{-1}[H(j \omega)]=h(t)=h_1(t)+h_2(t) \\ & h(0)=h_1(0)+h_2(0)=\frac{1}{2}+\frac{1}{2}=1 \end{aligned}