The problem asks for the amplitude $L$ of the modulating components in an amplitude modulated (AM) signal, given the carrier power, load resistance, and transmission efficiency.

Signal Analysis and Power Calculation

The AM signal is given by:

The load resistance is $R = 1\ \Omega$. The carrier frequency is $\omega_c = 300\pi$, corresponding to a carrier amplitude $A_c = K$. The sideband frequencies are $\omega_1 = 240\pi$ and $\omega_2 = 360\pi$. The amplitudes of these sideband components are both $L$. The average power of a sinusoidal signal $A \cos(\omega t)$ across a resistance $R$ is $P = \frac{A^2}{2R}$.

• Carrier Power ($P_c$): The power associated with the carrier term $K \cos(300\pi t)$ is $P_c = \frac{K^2}{2R}$. Given $P_c = 50\text{ W}$ and $R = 1\ \Omega$: $50 = \frac{K^2}{2 \times 1} \implies K^2 = 100 \implies K = 10\text{ V}$
• Sideband Power ($P_{SB}$): The power associated with the two sideband terms are $P_{s1} = \frac{L^2}{2R}$ and $P_{s2} = \frac{L^2}{2R}$. The total sideband power is $P_{SB} = P_{s1} + P_{s2} = \frac{L^2}{2R} + \frac{L^2}{2R} = \frac{L^2}{R}$. With $R=1\ \Omega$, $P_{SB} = L^2$.
• Total Power ($P_T$): Since the frequency components (carrier and sidebands) are distinct sinusoids, the total power is the sum of the individual powers: $P_T = P_c + P_{SB} = 50 + L^2$

Efficiency Calculation

The efficiency ($\eta$) of the transmitter is defined as the ratio of the sideband power to the total power:

Given $\eta = 60\% = 0.6$:

Solving for Amplitude L

Rearrange the efficiency equation to solve for $L$:

1. Multiply both sides by $(50 + L^2)$: $0.6 (50 + L^2) = L^2$
2. Distribute $0.6$: $30 + 0.6 L^2 = L^2$
3. Subtract $0.6 L^2$ from both sides: $30 = L^2 - 0.6 L^2$ $30 = 0.4 L^2$
4. Solve for $L^2$: $L^2 = \frac{30}{0.4} = \frac{300}{4} = 75$
5. Take the square root to find $L$: $L = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
6. Calculate the numerical value: $L \approx 5 \times 1.732 = 8.66\text{ V}$

The value of $L$ is approximately $8.66$ V.