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Explanation / Answer

as u see the base and emitter of Q3 are grounded so there is noVpi or Vpi = 0 or gm = 0.

so the net resistence at the collector of Q2 will ber02 = r03||r02 = r0/2.

for Q2 similarly the base will be grounded ac wise. if we assume v is the voltage at the collectror of Q1 in small signal model then we can have two equations from the Q1 and Q2 as

gm1*Vin + v/r01 + v/r_pi2 = 0--------------------------1

and

Vout = gm2*v*r02 ------------------2

getting the value of v from the first equation and putting it in the second equation we get

Av = Vout/Vin = - gm*gm*(r0*r_pi/(r0 + r_pi))*r0/2 assuming r0 and gm are same .

for calculating Rout we will sort the input voltage so then from the small signal model we can write,

Iout *(r0||r_pi) + (Iout + gm*Iout*(r0||r_pi))*r0/2 = Vout

so Rout = (r0||r_pi) + (1 + gm*(r0||r_pi))*r0/2

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