gm of this feedback common source transistor

How to calculate this Gm? How to remove the Vo item? Thanks.

So theoretically, if this stage were to be represented as an equivalent transconductor stage, then you need to short the output to gnd and see how much current flows through 'the short' , which in this case will be 1/Rf + gm.
But this will not be the complete picture. In your case, it will be an equivalent Gm (value given above) along with a parallel output impedance.

Thanks. It brings to Gm=(1/gm）-Rf.

In this application, it is in the Y parameter, Y21=I2/V1, with V2=0.

It is always hard to memorize to conditions of calculating certain parameters in the Z parameters, Y parameters, and Y parameters network. Like which parameter is open or short. Is there any intuitive way to memorize?

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Another question is if the output is short, then the drain and source of the transistor are short, leading to the parallel output impedance is also short?

To the question on different parameters, I don't know any intuitive method. I keep a printed copy of the table near me when I am trying to work on problems like this.
And to the next question, yes, if the output is shorted, it also shorts the output impedance of the device. And it also makes sense. 'Gm' is a measure of how much (change in) current you measure with respect to (change in) input voltage. If you did not short it and connected some other impedance, then you would see the impact of the variation in the output voltage on the current which is not what you are trying to measure. You are trying to replace the circuit by an equivalent Gm (and a parallel output impedance). The output impedance is what relates the variation in output voltage to the variation in the current (or in other words, Z22).

You don't really need to remember Y, Z, etc. parameters, especially for this simple circuit. Enough is to use superposition for finding the equivalent Gm and finding the output resistance to determine the gain using that equivalent Gm. Here is a quick calculation.




By the way, this shunt-shunt feedback is usually used with current input, not voltage. In other words - a TIA amplifier.

Thanks. the superposition helps