input-impedance
Please see images of question and solution. Please see my working out using the small-signal model
I'm really confused about how they went about solving the input impedance of this circuit? Did they use DC biasing or the small-signal model to work out the input impedance?
In general case the situations look like this
Now let as try to find a input resistance.
Rin = Vin/Iin
In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R
Rin = Vin/Iin = R/(1 - A)
As you can see we have a "minus" sign. We get the "plus" sign only when our amplifier gain is negative (inverting amplify)
Rin = R/(1 - (-A)) = R/(1+|A|)
The problem in your reasoning is that you do not show the complete path of the signal. More specific the impedance level of the signal.
You are right in that you cannot have both a high impedance and a low noise. If you want low noise you must keep the impedance low. Simple as that.
In the two circuits you have drawn it is unclear what the impedance is of the source which you use to feed a signal to your amplifier. Assuming that the AC coupling capacitors are large and that this source impedance is low (for example: 50 ohms) then the noise will be low !
Why ? Because the noise generated by the 100 Mohm DC bias resistors will be shorted by the AC coupling capacitors and that low source impedance. So in this situation the effective signal impedance (at a certain frequency) is much lower than 100 Mohm. Resulting in a low noise.
If the 50 ohms source impedance was not there that noise current would multiply by the 100 Mohm of the resistor itself resulting in a high noise level.
You can do calculations on this more easily by considering the noise current generated by the 100 Mohm resistors. That current will be multiplied by the signal source impedance (for example, 50 ohms) resulting in a small noise voltage !
So the circuit on the right is no better than your circuit on the left. Carefully read how they measured that low noise and try to figure out what the impedance level of the input signal was. I guarantee you that they will have used a source impedance such that the noise of the 100 Mohm DC biasing resistors can be neglected (a very low source impedance, they might even have shorted/grounded the inputs !). In that circuit the noise of the FETs should be dominant as these should determine the lowest possible noise level (at least in a properly designed amplifier).
Best Answer
Small signal is all about change. Input impedance merely means the ratio of the change in input voltage at the point by some source and the change in current obtained by it. You could wiggle a current through the node too and measure the change in voltage at the node.
If you pass a small current di into the node (causing the total current to become I+di), since di = gm.dvgs (gm is transconductance), the change in voltage developed is 1/gm times di. Thus, the answer would be dv/di = 1/gm. The resistor appears as a resistor to ground (a small current change would occur through the resistor since other voltages are fixed giving the impression of resistance to the current source that caused the di) and thus zin = 1/gm || r.