differential amplifier operating point cmos
The two signals are generated with different common mode voltages by the circuitry that produced them. For now assume that the are produce by a signal generator.
When you say Vswing, I assume that you are referring to the output swing of the circuit. If we assume that the circuit is linear, We can approximate the output swing as follows. Vout = A*vin. where A = gm * (Rout || Rd) where gm is the transconductance of the transistor, and Rout is the output impedance. The output swing amplitude is the product of the input swing , the transconductance, and the effective output impedance of the circuit. You can change any of the variables listed to change vswing.
Clipping occurred on the 2nd input signal with lower common mode. This is because the common-source amplifier is an inverting structure. Lower input common mode voltage will produce a higher output common mode voltage as shown in the diagram.
you can check with KCL as well. as the input common mode voltage decreases. The current through the transistor, IDS, decreases since IDS is approx 1/2 W/L u COX (Vgs-Vt)^2. Now the output voltage is Vout = VDD-IDS*RD. As IDS decreases, Vout goes up.