A carrier wave $V_{C}(t)=160 \sin \left(2 \pi \times 10^{6} t\right)$ volts is made to vary between $V_{\max }=200 \mathrm{~V}$ and $V_{\text {min }}=120 \mathrm{~V}$ by a message signal $V_{m}(t)=A_{m} \sin \left(2 \pi \times 10^{3} t\right)$ volts. The peak voltage $A_{m}$ of the modulating signal is ______ V.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Semiconductors

A carrier wave

V_{C} (t) = 160 sin (2 π \times 1 0^{6} t)

volts is made to vary between

V_{m a x} = 200 V

and

V_{min} = 120 V

by a message signal

V_{m} (t) = A_{m} sin (2 π \times 1 0^{3} t)

volts. The peak voltage

A_{m}

of the modulating signal is ______ V.

Answer: 40

In amplitude modulation the envelope of the carrier swings between

V_{ma x} = V_{c} + A_{m}

and

V_{min} = V_{c} - A_{m}

. Subtracting these,

A_{m} = \frac{V _{ma x} - V _{min}}{2} = \frac{200 - 120}{2} = 40 V

Adding them instead returns

\frac{200 + 120}{2} = 160

V, which agrees with the stated carrier amplitude. The peak voltage of the modulating signal is

40

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer