Suppose that the modulating signal is (t) = 2 cos(2πfm t) and the car

1.	Suppose that the modulating signal is (t) = 2 cos(2πfm t) and the carrier signal is xc (t) = AC cos(2πfc t).Which one of the following is a conventional AM signal without over-modulation? (a) (t) = Acm (t) cos(2πfc t) (b) (t) = Ac[1 + m(t)] cos(2πfc t) (c) (t) = Ac cos(2πfc t) +Ac /4m (t) cos(2πfc t) (d) (t) = Ac cos(2πfm t) cos(2πfc t) +Ac sin(2πfm t) sin(2πfc t)
Answer» (c) (t) = A_c cos(2πf_c t) +A_c /4m (t) cos(2πf_c t) Given Modulation signal (t) = 2 cos(2πf_m t), Carrier signal x_c(t) = AC cos(2πf_c t). Note that conventional AM is DSB – FC (DSB full carrier) . Standard Expression is given by e(t) = E_c(t) [1+m(t)] cos2_c t – I Option (b) is (t) = [1 + (t)] cos(2πf_c t) Comparing this expression with the standard one given equation (I) We get µ = 2 i.e. conventional AM with over modulation. Option (c) (t) = A_C cos(2πf_c t) +A_C /4 (t) cos(2πf_c t) (t) = cos(2πf_ct)[1+ 1/4 (2 cos(2πf_m t)) ] (t) =A_C cos(2πf_c t)[1+ 1 /2( cos(2πf_m t)) ] Here µ= 1/2 So, this represents conventional AM without over modulation.

Suppose that the modulating signal is (t) = 2 cos(2πfm t) and the carrier signal is xc (t) = AC cos(2πfc t).Which one of the following is a conventional AM signal without over-modulation? (a) (t) = Acm (t) cos(2πfc t) (b) (t) = Ac[1 + m(t)] cos(2πfc t) (c) (t) = Ac cos(2πfc t) +Ac /4m (t) cos(2πfc t) (d) (t) = Ac cos(2πfm t) cos(2πfc t) +Ac sin(2πfm t) sin(2πfc t)

Answer»

(c) (t) = A_c cos(2πf_c t) +A_c /4m (t) cos(2πf_c t)

Given Modulation signal (t) = 2 cos(2πf_m t), Carrier signal x_c(t) = AC cos(2πf_c t). Note that conventional AM is DSB – FC (DSB full carrier) .

Standard Expression is given by e(t) = E_c(t) [1+m(t)] cos2_c t – I

Option (b) is (t) = [1 + (t)] cos(2πf_c t)

Comparing this expression with the standard one given equation (I) We get µ = 2 i.e. conventional AM with over modulation.

Option (c) (t) = A_C cos(2πf_c t) +A_C /4 (t) cos(2πf_c t)

(t) = cos(2πf_ct)[1+ 1/4 (2 cos(2πf_m t)) ]

(t) =A_C cos(2πf_c t)[1+ 1 /2( cos(2πf_m t)) ]

Here µ= 1/2

So, this represents conventional AM without over modulation.