A carrier wave $$V_C(t) = 160 \sin(2\pi \times 10^6 t)$$ volts is made to vary between $$V_{max} = 200$$ V and $$V_{min} = 120$$ V by a message signal $$V_m(t) = A_m \sin(2\pi \times 10^3 t)$$ volts. The peak voltage $$A_m$$ of the modulating signal is ___.

In amplitude modulation, the carrier wave $$V_C(t) = 160\sin(2\pi \times 10^6 t)$$ volts has a carrier amplitude $$A_c = 160$$ V. The modulating signal $$V_m(t) = A_m\sin(2\pi \times 10^3 t)$$ varies the amplitude between $$V_{max} = 200$$ V and $$V_{min} = 120$$ V.

In AM modulation, the instantaneous amplitude oscillates between $$A_c + A_m$$ and $$A_c - A_m$$. Therefore:

$$V_{max} = A_c + A_m = 160 + A_m = 200 \implies A_m = 40 \text{ V}$$

$$V_{min} = A_c - A_m = 160 - A_m = 120 \implies A_m = 40 \text{ V}$$

Both conditions consistently give $$A_m = 40$$ V.

The peak voltage of the modulating signal is $$\boxed{40}$$ V.