$\begin{aligned} &\ { Given, } X(t)=\sqrt{2} \sin (2 \pi t+\varphi) \\ &\ { Where, } \phi=[0,2 \pi] \end{aligned}$

So, the probability function is defined as

$\mathrm{f}(\phi)=\frac{1}{2 \pi}$

According to the question

The auto-correlation is

$\begin{aligned} &\mathrm{E} \ \left[\mathrm{x}\left(t_{1}\right) \mathrm{x}\left(\mathrm{t}_{2}\right)\right]=\int_{0}^{2 \pi} \mathrm{x}\left(t_{1}\right) \mathrm{x}\left(\mathrm{t}_{2}\right) \mathrm{f}_{\phi}(\phi) \mathrm{d} \phi \\ &\left.=\int_{0}^{2 \pi}\left[\sqrt{2} \sin 2 \pi \mathrm{t}_{1}+\phi\right)\right]\left[\sqrt{2} \sin \left(2 \pi t_{2}+\phi\right)\right] \frac{1}{2 \pi} \mathrm{d} \phi \\ &\left.\mathrm{E}\left[\mathrm{x}\left(t_{1}\right) \mathrm{x}\left(t_{2}\right)\right]=\frac{2}{2 \pi} \int_{0}^{2 \pi} \sin \left(2 \pi t_{1}+\phi\right) \sin 2 \pi \mathrm{t}_{2}+\phi\right) \mathrm{d} \phi \ldots \ldots \ldots \ldots . .(1) \end{aligned}$

From trigonometric relation

We know,

$\sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$

Thus using this property in equation (1) we get,

$\begin{aligned} &E\left[x\left(t_{1}\right) x\left(t_{2}\right)\right]=\frac{2}{2 \pi} \int_{0}^{2 \pi}\left[\cos \left(2 \pi t_{1}-2 \pi t_{2}\right)-\cos \left(2 \pi\left(t_{1}+t_{2}\right)+2 \phi\right] d \phi\right. \\ &=\frac{1}{2 \pi} \cos \left[2 \pi\left(t_{1}-t_{2}\right)\right] \times 2 \pi \\ &E\left[x\left(t_{1}\right) x\left(t_{2}\right)\right]=\cos \left[2 \pi\left(t_{1}-t_{2}\right)\right] \end{aligned}$

Hence, the correct option is (D).

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Amplitude Modulation

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