Topic: Communication Systems
A message $m(t)$ bandlimited to the frequency $f_{m}$ has a power of $P_{m}$. The power of the output signal in the figure is
(The figure shows $m(t)\cos\omega_0t$ multiplied by $\cos(\omega_0t + \theta)$, followed by an ideal low pass filter with a cutoff of $f_m$ and passband gain of 1).
- A. $\frac{(P_m \cos \theta)}{2}$
- B. $\frac{(P_m \sin² \theta)}{2}$
- C. $\frac{P_m}{4}$
- D. $\frac{P_m \cos² \theta}{4}$
Answer: D. $\frac{P_m \cos² \theta}{4}$
Explanation: The output of the multiplier is $A(t) = m(t)\cos(\omega_0 t) \cdot \cos(\omega_0 t + \theta) = \frac{m(t)}{2}[\cos(\theta) + \cos(2\omega_0 t + \theta)]$. After passing through the ideal low pass filter with a cut-off frequency $f_m$, the high-frequency term is removed. The output signal is $y(t) = \frac{m(t)}{2}\cos(\theta)$. The power of $y(t)$ is $\frac{\text{Power of } m(t) \times \cos²(\theta)}{4} = \frac{P_m \cos²(\theta)}{4}$.