ECE 580 Mock Final Solutions
1.


2. KCL at C1: -gmi.Vi -gm2.V1+gm4.Vo = sc1.V1
KCL at C2: -gm3.V1 = sC2.Vo
Ho(s) = Vo/Vi = gm1.gm3 / (s2.C1.C2 + s.gm2.C2 + gm3.gm4)
H1(s) = V1/Vi = (-sC2 / gm3) Ho(s) = - sC2.gm1 / (s2.C1.C2 + s.gm2.C2 + gm3.gm4)


ECE 580 Mock Final Solutions
3. By interreciprocity, Vo = - ∑ I’k Vk . Using Bashkow’s method, we assume I’1 = 1 A,
and also to simplify the calculations, R = 1 (this doesn’t change the voltage gain).
Then Va’ = 2, I2’ = 2, Ib’ = 3, Vb’ = 5 I3’ = 5, Ic’ = 8, Vc’ = 13, I4’ = 13, Id’ = 21.
Hence, the scale factor is -1/21, and Vo = (1/21) [ 1x1 + 2x1 +5x2 + 13x2] = 39/21 ~
1.857 V.




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