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Q.A 10 MHz sinusoidal carrier wave of amplitude 10 mV is modulated by a 5 kHz sinusoidal audio signal wave of amplitude 6 mV. Find the frequency components of the resultant modulated wave and their amplitudes.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 5mImportance★★★★★
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Amplitude modulation of a 10 MHz, 10 mV carrier by a 5 kHz, 6 mV audio tone produces a carrier component and two side-band components, whose frequencies and amplitudes follow from the standard AM expression.

Given data

Carrier frequency fc=10 MHzf_c = 10\text{ MHz}, carrier amplitude Ac=10 mVA_c = 10\text{ mV}

Modulating (audio) frequency fm=5 kHz=0.005 MHzf_m = 5\text{ kHz} = 0.005\text{ MHz}, modulating amplitude Am=6 mVA_m = 6\text{ mV}

Expression for the AM wave

An amplitude modulated wave is ec(t)=Ac(1+μsin⁡ωmt)sin⁡ωcte_c(t) = A_c(1+\mu\sin\omega_m t)\sin\omega_c t, where μ=Am/Ac\mu = A_m/A_c is the modulation index. Expanding using the product-to-sum identity sin⁡ωct⋅sin⁡ωmt=12[cos⁡(ωc−ωm)t−cos⁡(ωc+ωm)t]\sin\omega_c t \cdot \sin\omega_m t = \tfrac{1}{2}[\cos(\omega_c-\omega_m)t - \cos(\omega_c+\omega_m)t]:

ec(t)=Acsin⁡ωct+μAc2cos⁡(ωc−ωm)t−μAc2cos⁡(ωc+ωm)te_c(t) = A_c\sin\omega_c t + \frac{\mu A_c}{2}\cos(\omega_c-\omega_m)t - \frac{\mu A_c}{2}\cos(\omega_c+\omega_m)t

Since μAc=Am\mu A_c = A_m, this shows the modulated wave is the sum of three sinusoids: the unchanged carrier of amplitude AcA_c, a lower-side-band term at frequency (fc−fm)(f_c-f_m) of amplitude Am/2A_m/2, and an upper-side-band term at frequency (fc+fm)(f_c+f_m) of amplitude Am/2A_m/2.

Frequency components

Carrier: fc=10 MHzf_c = 10\text{ MHz}, amplitude =Ac=10 mV= A_c = 10\text{ mV}

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