For AM modulated signal given by i = 10 + K1 Vi + K2 Vi2

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For AM modulated signal given by i = 10 + K₁ V_i + K₂ V_i²
where V_i = A_c cos ω _ct + m (t),
uses square law demodulator, find μ given that
A_c = 1V, A_m = 0·5 V, K₁ = 2 and K2 = 0·2

1. 0.5
2. 0.2
3. 1
4. 0.4

Correct Option: B

We know that AM modulated wave represented by equation
x(t)=A_c (1 + μ cos ω_mt) cos ω_ct
s(t)=A_c cos ω_ct + μ·A_c cos ω_mt cos ω_ct …(A)
given that,
i _AM(t) = 10 + K₁V₁ + K₂V² _i
i_AM(t) = 10 + K₁A_c cos ω_ct + K₁A_m cos ω_mt + K₂A²_c cos²ω_ct + K²₂A²_m cos² ω_mt 2K₂A_cA_m cos ω_ct cos ω_mt …(B)
In order to calculate, modulation index μ.
Compare the coefficient of cos ωmt cos ωct in equation (A) and (B) we get
μA_c = 2K₂A_cA_m
or μ = 2K₂A_m = 2 × 0·2 × 0·5= 0·20 (˙.˙ K₂ = 0·2)
Hence alternative (B) is the correct choice.