Exercise 5.4 A common gate amplifier

a)

An answer:
The gate current (for this model) is zero which does not require any (equivalent) small signal component. The drain current depends only on the gate-source voltage which can be modelled with a voltage controlled current source delivering $i_d=g_m\cdot v_{\mathit{gs}}$.

b)

An answer:

Which can be simplified into:

This small signal equivalent shows that the gate is connected to ground AND that both the input signal port AND the output port have this ground in common.

c)

An answer:

$$\begin{array}{rcll}& r_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & i_{\mathit{in}}=\frac{v_{\mathit{in}}}{R_S}-g_m\cdot v_{\mathit{gs}}& \\ & & v_{\mathit{gs}}=v_{\mathit{in}}& \\ & & i_{\mathit{in}}=\left(\frac{1}{R_S}+g_m\right)\cdot v_{\mathit{in}}& \\ & r_{\mathit{in}}& =R_S∕∕\frac{1}{g_m}& \end{array}$$

d)

An answer:
The derivation is pretty much the same as for question (c), $r_{\mathit{out}}=R_D$.