I solved for Vo in terms of Vi and I get the negative of the correct answer according to the book. Someone told me that the TA set the node to the left of Vo = to -Vo, and this explains why my answer is the negative, but I don't understand why the node adjacent to Vo is -Vo instead of just Vo. In this problem, gm = 2m / 1 ohm, and I get that Vo = -2 Vi, but the book gets that V0 = 2 Vi.
gm = 2m/Ω = 2mA/V Vo = -(gm*Vx)*2kΩ Vx = Vi/2 Vo = -((2m/Ω)*(Vi/2))*2kΩ Vo = -2Vi You're right, they're wrong. I have NO idea why they are putting Vo on the top-right node of the left hand circuit (that's what you are talking about, right?).
tAllan, the circuit shown equals the classical equivalent small-signal ac circuit diagram for a common emitter transistor amplifier. As you will know, this amplifier has inverting gain characteristics. Thus, for a positive input voltage, the output is negative, indeed.
Of course you can solve it without relying on this. You did. I did. He is just pointing out that this is a circuit that happens to match a common circuit that you will run into and, if you DID happen to know that, that it confirms that the output voltage is inverted relative to the input voltage.