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.

 

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.

But can't you solve the circuit without knowing this with simple analysis? Is my analysis wrong?

 

But can't you solve the circuit without knowing this with simple analysis? Is my analysis wrong?

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.