A carrier wave of 10 MHz and peak voltage of 14 V is amplitude modulated by a sinusoidal wave of 5 kHz and amp

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Q. A carrier wave of 10 MHz and peak voltage of 14 V is amplitude modulated by a sinusoidal wave of 5 kHz and amplitude 6 V. Write the equation of the AM wave. What is the bandwidth of the modulated signal?

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Concept understanding — Amplitude Modulation and the AM Wave Equation

Amplitude modulation (AM) is the process in which the amplitude of a high-frequency carrier is varied in accordance with the instantaneous value of the modulating signal, while the carrier's frequency and phase are held constant. The rapid carrier oscillation then sits inside a slowly varying outline called the envelope, whose shape is identical to the original message — which is precisely why the message can later be recovered from the envelope.

To describe it mathematically, take the modulating signal $v_m = V_m \sin(\omega_m t)$ and the carrier $v_C = V_C \sin(\omega_c t)$ , with $f_c \gg f_m$ . During modulation the carrier amplitude is no longer constant but becomes

A = V_C + v_m = V_C\left(1 + \tfrac{V_m}{V_C}\sin\omega_m t\right) = V_C(1 + m_a \sin\omega_m t)

so the instantaneous AM voltage is

v_{AM} = V_C(1 + m_a \sin\omega_m t)\sin(\omega_c t)

where $m_a = V_m/V_C$ is the modulation index. Expanding with a product-to-sum identity shows the AM wave is the sum of three components — the carrier at $f_c$ , a lower sideband at $f_c-f_m$ and an upper sideband at $f_c+f_m$ :

v_{AM} = V_C \sin(\omega_c t) + \tfrac{m_a V_C}{2}\cos(\omega_c-\omega_m)t - \tfrac{m_a V_C}{2}\cos(\omega_c+\omega_m)t

Notably, the modulated wave contains no term at the original modulating frequency $f_m$ . Being able to write and interpret this equation is a core skill for the II PUC Electronics board exam.

[!TLDR]

With $m_a = 6/14 = 0.43$ , the AM equation is $v_{AM} = 14(1 + 0.43\sin\omega_m t)\sin\omega_c t$ and the bandwidth is $2f_m = 10$ kHz.

A single-tone AM wave is written in the standard form

v_{AM} = V_C\left(1 + m_a\sin\omega_m t\right)\sin\omega_c t

where $m_a = V_m/V_C$ is the modulation index, $\omega_m = 2\pi f_m$ is the modulating angular frequency and $\omega_c = 2\pi f_c$ is the carrier angular frequency. The AM wave contains three frequencies — the carrier and the two side frequencies at $f_c \pm f_m$ — so the band of frequencies it occupies (its bandwidth) stretches from the lower to the upper side frequency, giving $BW = 2f_m$ .

Given $V_C = 14$ V, $V_m = 6$ V, $f_c = 10$ MHz and $f_m = 5$ kHz:

m_a = \frac{V_m}{V_C} = \frac{6}{14} = 0.43

\omega_m = 2\pi f_m = 2\pi\times5\times10^3 = 10\pi\times10^3\ \text{rad/s}

\omega_c = 2\pi f_c = 2\pi\times10\times10^6 = 20\pi\times10^6\ \text{rad/s}

v_{AM} = 14\big(1 + 0.43\sin(10\pi\times10^3\,t)\big)\sin(20\pi\times10^6\,t)

BW = 2f_m = 2\times5\ \text{kHz} = 10\ \text{kHz}

[!ANSWER]

$v_{AM} = 14\big(1 + 0.43\sin(10\pi\times10^3\,t)\big)\sin(20\pi\times10^6\,t)$ ; bandwidth $= 10$ kHz

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Q.A carrier wave of 10 MHz and peak voltage of 14 V is amplitude modulated by a sinusoidal wave of 5 kHz and amplitude 6 V. Write the equation of the AM wave. What is the bandwidth of the modulated signal?

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