Current will be defined by high impedance and voltage by low impedance. So M2 decides the current. Current in the branch will be gm2*vin. Voltage will be defined by low impedance, so M1 defines the voltage at Vo. Current through M1 is gm2*vin, so VGS of M2 should be gm2*vin/gm1. vo comes out to be vin*(1-(gm2/gm1)). Gain will be 1-(gn2/gm1).
My approach: Considering only M1, it is a source follower. So vo1=vin. Considering only M2, it is a common source. So vo2=-gm2*(1/gm1)vin. By superposition, vo=vo1+vo2=(1-gm2/gm1) or, vo/vin=1-gm2/gm1=(gm1-gm2)/gm1
for the 3rd circuit can we think it like this_ at small signal the current source will offer infinite resistance and will be open circuited .so Gm=gm/1+gmRs where Rs in infinite so Gm=0. And gain Av=-GmRd=0 ??
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So my approach was that, the drain current is same for both the MOSFETs. Assuming gm is same for both, and as Vg of both is same, then for current to be same, Vs should be equal for both the MOSFETs. Thus Vo becomes 0. Yeah if we assume gm1 & gm2 are different, then superposition approach yields better result.
