A sinusoidal wave $$y(t) = 40\sin(10 \times 10^6 \pi t)$$ is amplitude modulated by another sinusoidal wave $$x(t) = 20\sin(1000\pi t)$$. The amplitude of minimum frequency component of modulated signal is

We need to find the amplitude of the minimum frequency component of the amplitude modulated signal.

The carrier wave is $$y(t) = 40\sin(10 \times 10^6 \pi t)$$ and the modulating signal is $$x(t) = 20\sin(1000\pi t)$$.

Thus the carrier amplitude is $$A_c = 40$$ while the modulating amplitude is $$A_m = 20$$.

The modulation index is given by $$\mu = \frac{A_m}{A_c} = \frac{20}{40} = 0.5$$.

The carrier frequency can be calculated as $$f_c = \frac{10 \times 10^6 \pi}{2\pi} = 5 \times 10^6$$ Hz and the modulating frequency as $$f_m = \frac{1000\pi}{2\pi} = 500$$ Hz.

In an amplitude modulated signal there are three frequency components: the lower sideband at $$f_c - f_m = 5 \times 10^6 - 500$$ Hz with amplitude $$\frac{\mu A_c}{2}$$, the carrier at $$f_c = 5 \times 10^6$$ Hz with amplitude $$A_c$$, and the upper sideband at $$f_c + f_m = 5 \times 10^6 + 500$$ Hz with amplitude $$\frac{\mu A_c}{2}$$.

The minimum frequency among these is the lower sideband frequency $$f_c - f_m$$, and its amplitude is $$\frac{\mu A_c}{2} = \frac{0.5 \times 40}{2} = \frac{20}{2} = 10$$.

Hence, the correct answer is Option D.