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(Q.1Q.3) An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V.
Q. The modulation index is
20 + 4sin(500πt) = 20(1 + 0.2sin(500πt)), hence the modulation index is 0.2.
Pc = 20×20/2 or 200 W. Pt = Pc(1 + 0.2×0.2/4) or 204 W.
Pt – Pc = 204 – 200 = 4W.
An AM broadcast station operates at its maximum allowed total output of 50 kW with 80% modulation. The power in the intelligence part is
Pi = Pt – Pc = 50 – 37.88 kW or 12.12 kW.
The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.The modulation index is
400/326 = 1 + (α^{2})/2, therefore α is 0.68.
A modulating signal is amplified by a 80% efficiency amplifier before being combined with a 20 kW carrier to generate an AM signal. The required DC input power to the amplifier, for the system to operate at 100% modulation, would be
Pi = Pt Pc = 30 – 20 = 10 kW. DC input = 10/0.8 or 12.5 kW.
A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is
Pt = Pc (1 + 0.49/2).
A carrier is simultaneously modulated by two sine waves with modulation indices of 0.4 and 0.3. The resultant modulation index will be
α^{2} = 0.3^{2} + 0.4^{2} = 0.5^{2} or α = 0.5.
In a DSBSC system with 100% modulation, the power saving is
This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%.
A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively. The total radiated power is
The required answer is 1 (1 + 0.4^{2} + 0.3^{2}) or 11.25 kW.
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11 videos57 docs108 tests
