Step 1 – Calculate $I_DQ$

$V_G = \frac{R_2}{( R_1+ R_2 )} x V_DD= \frac{110}{(910 + 110)} x 20$

∴ $V_G$ = 2.156 V

$R_S$ = 1.1k (given)

$V_S = I_DQx R_S = I_DQ $ x 1.1k

∴$V_GSQ = V_G - V_S = (2.156 - I_DQ1.1k) V$

Calculate $I_DQ$:

$I_DQ = I_DSS (1-\frac{V_GSQ}{V_p} )^2 = 10 mA (1+\frac{(2.156- I_DQ 1.1 k)}{3.5})^2$

$1.21x 10^6 x I_DQ^2 -13668.2 I_DQ-31.99 $= 0

$I_DQ$= 7.98 mA and $I_DQ$= 3.31 mA

To avoid ohmic region, Select $I_DQ$ = 3.31 mA

∴$V_GSQ$ = 2.156 - $I_DQ$1.1K = 2.156 – 3.31 mA x 1.1K = -1.485 V

Calculate value of $g_m$:

$g_m = g_mo [1- \frac{V_GS}{V_p} ]$

$g_mo = \frac{(2 I_DSS)}{V_p} = \frac{(2 x 10 mA)}{3.5}$ = 5.714 mS

$g_m = 5.714 mS [1+ \frac{(-1.485 )}{3.5}]$ = 3.2896

Obtain the value of $A_v$:

$A_v= -g_m R_D$ = -3.2896 x 2.2k = -7.237

Obtain $Z_i$ and $Z_o$:

$Z_i = R_1 || R_2$ = 98.137 kΩ

$Z_o = R_D$ = 2.2 kΩ