We can also analyze the circuit by using the superposition theorem since all the elements here are linear and time-invariant. With that in mind. The effect of 1V at Vx (with Vin=0) is 0.5V. Then the effect of Vin at Vx (with 1V source shorted out) is a simple HPF response for a square wave input, which is well known and can be analyzed for all combinations of RC w.r.t T. Finally the response will be the HPF (C & R/2) shifted up by 0.5V. This is just another way to look at the circuit and quickly figure out what happens, but before doing that it is important to know what happens in each cycle as shown in the video. Both approaches are important!
We can use superposition right? So first we can ignore the dc voltage and get the charging-discharging curve and then shift that graph upward by 0.5V because of the DC Source.
The discharging waveform shape for even cycles need to change. |dv/dt| is going to be higher at the beginning - just like you have shown for the odd cycles.
Your curve from 1.5v to 0.5v is wrong. The slew rate starts high, then slows to 0 as it fully charges. Your curve shape from -0.5v to +0.5v is correct.
The square wave consists of a DC + Odd Sine Harmonics . So we can solve this by first considering the DC component along with 1V D.C. Source . Then we can consider Odd Sine Harmonics and use Sinusoidal Steady Analysis for the same .
check out any circuit analysis textbook in the chapter where it deals with natural and step response. I found Electric circuits by Nilsson and Riedel very fascinating. You will encounter this kind of problems if you solve end of chapter problems. I strongly recommend you to solve problems from chapters 8, 12, and 13.
