Question

An amplitude modulator has output (in Volts)
s(t) = A cos(400πt) + B cos(360πt) + B cos(440πt).
The carrier power normalized to 1 Ω resistance is 50 Watts. The ratio of the total sideband power to the total power is \( \frac{1}{9} \). The value of B (in Volts, rounded off to two decimal places) is ____.

Hint:

In amplitude modulation, calculate sideband power and total power using their respective relationships. The given power ratio helps in determining the unknown sideband amplitude.

Answer 1

Correct Answer: 1.49 - 2.5

Given the amplitude modulated signal:
s(t) = A cos(400πt) + B cos(360πt) + B cos(440πt).
The carrier power P_c equals 50 Watts. 
Since power is given by P = (A²/2R) and R = 1 Ω, P_c = A²/2.
Therefore, A²/2 = 50, so A = √(100) = 10 Volts.
The total sideband power to carrier power ratio is 1/9. Let P_sb be the total sideband power.
Thus: P_sb/P_total = 1/9.
We know P_total = P_c + P_sb, thus P_sb/ (50 + P_sb) = 1/9.
Let x = P_sb, solving gives: x/(50 + x) = 1/9. Multiplying both sides by 9(50+x) leads to:
9x = 50 + x => 8x = 50 => x = 6.25 Watts.
The power of each sideband component (360π, 440π) is B²/2, hence 2(B²/2) = 6.25.
So, B² = 6.25, B = √6.25 = 2.5.