Topic: Communication Systems
What is the peak to average power ratio for the signal $x(t)=A \sin(\omega t)$ with 50% duty cycle?
- A. 0 dB
- B. 1 dB
- C. 3 dB
- D. 6 dB
Answer: D. 6 dB
Explanation: For a continuous sine wave, the peak power is $A^2$ and the average power is $A^2/2$, yielding a Peak to Average Power Ratio (PAPR) of 2 (or 3 dB). A 50% duty cycle means the signal is only "on" half the time. This halves the average power to $A^2/4$ while the peak power remains $A^2$. The new PAPR is $\frac{A^2}{A^2/4} = 4$. Converting to decibels, $10 \log_{10}(4) \approx 6$ dB.