A carrier wave VC (t) = 160 sin(2p × 106t) volts is made to very between Vmax = 200V and Vmin = 120V by a message signal Vm(t) = Am sin(2p ×103t) volts. The peak voltage Am of the modulating single is_________V.

μ = A m A c = V 1 − V 2 V 1 + V 2 = 2 0 0 − 1 2 0 2 0 0 + 1 2 0 = 1 4 ⇒ Modulation indexHere VC (t) = 160 sin ( 2 π × 1 0 6 t ) V a n d V m ( t ) = A m s i n ( 2 π × 1 0 3 t ) V , s o <!- [if gte mso 9]><xml> </xml><! [endif]-> A m = A c 4 = 1 6 0 4 = 4 0 V <!- [if gte mso 9]><xml> </xml><! [endif]->

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A carrier wave V_C (t) = 160 sin(2p × 10⁶t) volts is made to very between V_max = 200V and V_min = 120V by a message signal V_m(t) = A_m sin(2p ×10³t) volts. The peak voltage A_m of the modulating single is_________V.

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alok kumar singh | Contributor-Level 10

2 months ago

$μ = \frac{A_{m}}{A_{c}} = \frac{V_{1} - V_{2}}{V_{1} + V_{2}} = \frac{200 - 120}{200 + 120} = \frac{1}{4} \Rightarrow$ Modulation index

Here V_C (t) = 160 sin $(2 π \times 1 0^{6} t) V a n d V_{m} (t) = A_{m} s i n (2 π \times 1 0^{3} t) V, s o$

$A_{m} = \frac{A_{c}}{4} = \frac{160}{4} = 40 V$

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A carrier wave V_C (t) = 160 sin(2p × 10⁶t) volts is made to very between V_max = 200V and V_min = 120V by a message signal V_m(t) = A_m sin(2p ×10³t) volts. The peak voltage A_m of the modulating single is_________V.

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