Question

An amplitude modulated voltage signal \(x(t)\) drives a load of \(1\,\Omega\).
\[ x(t) = K\cos(300\pi t) + L\cos(240\pi t) + L\cos(360\pi t) \]
If the efficiency is \(60\%\) and the carrier power is \(50\) W, the value of \(L\) is ______ V.

• A. \(8.66\)
• B. \(1.22\)
• C. \(0.81\)
• D. \(17.32\)

Hint:

Split the terms into one carrier and two equal sidebands, use \(P_c=K^2/2R\) and \(\eta=P_{sb}/(P_c+P_{sb})\) to get \(P_{sb}\), then \(P_{sb}=L^2\) since \(R=1\,\Omega\).

Answer 1

The Correct Option is A

Step 1: Identify the carrier and the sideband terms.
Write the three cosine terms in terms of their angular frequencies. The carrier angular frequency is \(300\pi\), and the other two, \(240\pi\) and \(360\pi\), sit symmetrically \(60\pi\) below and above it. So
\[ K\cos(300\pi t)\ \text{is the carrier},\qquad L\cos(240\pi t),\ L\cos(360\pi t)\ \text{are the lower and upper sidebands}. \]
This is the standard structure of an amplitude modulated signal: one carrier term plus two sideband terms of equal amplitude \(L\).
Step 2: Find the carrier amplitude K from the given carrier power.
For a sinusoid of amplitude \(K\) across a resistance \(R\), the average power is
\[ P_c = \frac{K^2}{2R} \]
Here \(R=1\,\Omega\) and \(P_c=50\) W, so
\[ 50 = \frac{K^2}{2\times1}\ \Rightarrow\ K^2=100\ \Rightarrow\ K=10\ \text{V} \]
(We do not actually need the numeric value of \(K\) to get \(L\), but it confirms the setup is consistent.)
Step 3: Use the efficiency definition to find the total sideband power.
The efficiency of an AM signal is the fraction of the total transmitted power that is carried in the sidebands, since only the sidebands carry the information:
\[ \eta = \frac{P_{sb}}{P_c+P_{sb}} \]
Substituting \(\eta=0.6\) and \(P_c=50\),
\[ 0.6 = \frac{P_{sb}}{50+P_{sb}} \]
Step 4: Solve for the sideband power.
\[ 0.6(50+P_{sb}) = P_{sb} \]
\[ 30 + 0.6P_{sb} = P_{sb} \]
\[ 30 = 0.4P_{sb} \]
\[ P_{sb} = 75\ \text{W} \]
Step 5: Relate the sideband power to L.
Each sideband is a sinusoid of amplitude \(L\) across \(R=1\,\Omega\), so each carries power \(L^2/(2R)=L^2/2\). There are two sidebands (lower and upper), so
\[ P_{sb} = 2\times\frac{L^2}{2} = L^2 \]
Step 6: Solve for L.
\[ L^2 = 75\ \Rightarrow\ L = \sqrt{75} = 5\sqrt{3} \approx 8.66\ \text{V} \]
Step 7: Check why the other options fail.
A quick way to test each option is to plug it back into \(P_{sb}=L^2\) and see what efficiency it gives. For (B) \(1.22\): \(P_{sb}\approx1.49\) W, giving \(\eta\approx1.49/51.49\approx2.9\%\), far below \(60\%\). For (C) \(0.81\): \(P_{sb}\approx0.66\) W, giving \(\eta\approx1.3\%\), also far too low. For (D) \(17.32\): \(P_{sb}\approx300\) W, giving \(\eta\approx300/350\approx85.7\%\), too high. Only \(L=8.66\) V reproduces the stated \(60\%\) efficiency exactly.
Final Answer:
\[ \boxed{L \approx 8.66\ \text{V}} \]