Question 20.3: Determine the output voltage waveform for the first two puls...

First, calculate the circuit time constant.

Obviously, the time constant is much longer than the input pulse width or the interval between pulses (notice that the input is not a square wave). In this case, the exponential formulas must be applied, and the analysis is relatively difficult. Follow the solution carefully.
1. Calculation for first pulse: Use the equation for an increasing exponential because C is charging. Note that V_{F} is 5 V, and t equals the pulse width of 10 μs. Therefore,

2. Calculation for interval between first and second pulse: Use the equation for a decreasing exponential because C is discharging. Note that V_{i} is 906 mV because C begins to discharge from this value at the end of the first pulse. The discharge time is 15 μs. Therefore,

This result is shown in Figure 20-20(b).
3. Calculation for second pulse: At the beginning of the second pulse, the output voltage is 671 mV. During the second pulse, the capacitor will again charge. In this case, it does not begin at zero volts. It already has 671 mV from the previous charge and discharge. To handle this situation, you must use the general exponential formula.

Using this equation, you can calculate the voltage across the capacitor at the end of the second pulse as follows:

This result is shown in Figure 20-20(c).
Notice that the output waveform builds up on successive input pulses. After approximately 5\tau, it will reach its steady state and will fluctuate between a constant maximum and a constant minimum, with an average equal to the average value of the input.
You can see this pattern by carrying the analysis in this example further.