Q.A carrier wave with an RMS voltage of 3 V and a frequency of 1.5 MHz is modulated by sinusoidal wave with a frequency of 500 Hz and amplitude of 2 V RMS. Write the equation for the resulting signal.

Because the AM wave equation $v_{AM} = (V_C + V_m\sin\omega_m t)\sin\omega_c t$ is an instantaneous-voltage expression, the rms values must first be converted to peak amplitudes, $V = \sqrt{2}\,V_{rms}$.

[!ANSWER]

$v_{AM} = (4.24 + 2.82\sin(2\pi\times500\,t))\sin(3\pi\times10^6\,t)$ V

[!TLDR]

Converting the rms values to peaks gives $V_C = 4.24$ V and $V_m = 2.82$ V, so $v_{AM} = (4.24 + 2.82\sin(2\pi\times500\,t))\sin(3\pi\times10^6\,t)$ V.

The AM wave is written in the form $v_{AM} = (V_C + V_m\sin\omega_m t)\sin\omega_c t$, where $V_C$ and $V_m$ are the peak (not rms) amplitudes of the carrier and modulating signals. Since the data are given as rms values, each is first converted to its peak using $V_{peak} = \sqrt{2}\,V_{rms}$.

Given $V_{C(rms)} = 3$ V, $V_{m(rms)} = 2$ V, $f_c = 1.5$ MHz and $f_m = 500$ Hz:

[!NOTE] The textbook's own working correctly converts the rms values to the peak amplitudes $V_C = 4.24$ V and $V_m = 2.82$ V, but its printed final line then writes the equation using the rms figures 3 and 2. The instantaneous AM equation must use the peak amplitudes, so $(4.24 + 2.82\sin\ldots)$ is the correct form.

[!ANSWER]

$v_{AM} = (4.24 + 2.82\sin(2\pi\times500\,t))\sin(3\pi\times10^6\,t)$ V