Derive an expression for amplitude modulated wave.

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Q. Derive an expression for amplitude modulated wave.

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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]

The AM wave is $e_{AM}=E_c(1+m_a\cos\omega_m t)\cos\omega_c t$ , which expands into a carrier term plus an upper and a lower side band.

Let the carrier be $e_c = E_c\cos\omega_c t$ and the modulating signal be $e_m = E_m\cos\omega_m t$ , where $E_c$ and $E_m$ are the peak amplitudes and $\omega_c \gg \omega_m$ .

In amplitude modulation the instantaneous amplitude of the carrier is varied about $E_c$ in proportion to $e_m$ . Hence the amplitude of the modulated wave is

A = E_c + e_m = E_c + E_m\cos\omega_m t = E_c\left(1 + \frac{E_m}{E_c}\cos\omega_m t\right).

Defining the modulation index $m_a = E_m/E_c$ ,

A = E_c(1 + m_a\cos\omega_m t).

The AM wave is this varying amplitude multiplying the carrier oscillation:

e_{AM} = A\cos\omega_c t = E_c(1 + m_a\cos\omega_m t)\cos\omega_c t.

Expanding,

e_{AM} = E_c\cos\omega_c t + m_aE_c\cos\omega_m t\,\cos\omega_c t.

Using $\cos A\cos B = \tfrac{1}{2}[\cos(A+B)+\cos(A-B)]$ ,

e_{AM} = E_c\cos\omega_c t + \frac{m_aE_c}{2}\cos(\omega_c+\omega_m)t + \frac{m_aE_c}{2}\cos(\omega_c-\omega_m)t.

Thus the AM wave contains three frequency components: the carrier at $f_c$ with amplitude $E_c$ , the upper side band at $f_c+f_m$ and the lower side band at $f_c-f_m$ , each of amplitude $m_aE_c/2$ . The bandwidth is $2f_m$ .

[!ANSWER]

$e_{AM} = E_c(1+m_a\cos\omega_m t)\cos\omega_c t = E_c\cos\omega_c t + \dfrac{m_aE_c}{2}\cos(\omega_c+\omega_m)t + \dfrac{m_aE_c}{2}\cos(\omega_c-\omega_m)t$ ; the wave consists of the carrier and the upper and lower side bands.

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Q.Derive an expression for amplitude modulated wave.

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