When you halve the amplitude and double the frequency of the pulses, the average power will remain the same. So the correct answer is:
the average power will not change.
Here's why:
- Power: The average power of a periodic signal is defined as the average of the square of its instantaneous power over one period.
- Instantaneous power: The instantaneous power of a periodic signal is the product of its voltage and current at any given instant.
- Halving the amplitude: Reducing the amplitude by half will reduce the instantaneous power by a factor of four (since power is proportional to the square of voltage or current).
- Doubling the frequency: Doubling the frequency will double the number of periods per unit time.
These two effects work together to cancel each other out. Doubling the frequency doubles the rate at which we sum the instantaneous power, but each individual term in the sum is reduced by a factor of four. Therefore, the average power remains unchanged.
Here's an analogy: Imagine you're counting marbles falling into a bucket at a certain rate. Each marble represents the instantaneous power, and the bucket represents the average power. If you slow down the rate at which the marbles fall (halving the frequency), but also double the size of each marble (halving the amplitude), the number of marbles in the bucket at any given time will remain the same (the average power remains unchanged).