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Question 3.SGPYQ.41: The Fourier transform of a signal h(t) is H(jω) = (2 cosω) (......

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

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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}

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