Electrical – Is this equation for transconductance correct

To find the input resistance, the procedure is to apply a test voltage \$v_t\$ to the input terminal (the gate). With the source node as our reference voltage \$v_t\$ is simply \$v_{gs}\$ (voltage from gate to source). The input resistance \$R_i\$ is then the ratio of the test voltage divided by the current \$i_t\$ supplied by the test voltage:

$$R_i = \frac{v_t}{i_t}$$

In this case \$i_t\$ is the current flowing from the gate node to the T junction (between the dependent current source and \$r_s\$).

The dependent current source \$i_s\$ in the MOSFET T-model is equal to \$g_m v_{gs}\$ (just like the hybrid-\$\pi\$ model). By Ohm's Law the current from gate to source (through \$r_s\$) is

$$i_{gs} = \frac{v_{gs}}{r_{s}} = \frac{v_{gs}}{1/g_m} = g_m v_{gs}$$

Since \$i_s = i_{gs}\$ it must be true by KCL that \$i_t = 0\$ no matter what the input voltage \$v_t\$ is. Therefore

$$R_i = \frac{v_t}{i_t} = \frac{v_t}{0} = \infty$$

The input resistance for the T-model is infinite, just as it is for the hybrid-\$\pi\$ model. It is not equal to \$1/g_m\$.

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.