Answer:

Step 1 - Calculate $I_{DQ}$

$V_G=\dfrac{R_2}{R_1+R_2} \times V_{DD}$

$V_G=\dfrac{110}{910+110} \times 20$

$\therefore V_G=2.156V$

Now , $V_S=I_{DQ}\times R_S=I_{DQ}\times 1.1$

$\therefore V_{GS}=(2.156-1.1I_{DQ})$

By schokley equation , 

$I_D=I_{DSS}[1-\dfrac{V_{GS}}{V_P}]^2$

$=10[1+\dfrac{2.156-1.1I_{D}}{3.5}]^2$

$1.21I_D^2-12.35I_D+25.60=0$

$\therefore I_D=7.31mA, \ I_D=2.89mA$

Now , $I_D\neq7.31mA$ because for this value of $I_D$ , $V_{DS}$comes out to be negative.

Hence , $ I_{DQ}=2.89mA$

Step 2 - Calculate $V_{GSQ}$

$\therefore V_{GSQ}=2.156-1.1I_{DQ}$

$\therefore V_{GSQ}=2.156-1.1 \times 2.89=-1.023V$

Step 3 - Obtain the value of $g_m$

$g_m=g_{mo}[1-\dfrac{V_{GS}}{V_P}]$

$g_{mo}=\dfrac{2 I_{DSS}}{V_p}=\dfrac{2 \times 10}{3.5}=5.71mS$

$g_m=5.71mS[1-\dfrac{-1.02}{-3.5}]=4.04mS$

Step 4 - Calculate $A_V, \ Z_i\ Z_0$

$\therefore A_V=-g_mR_D=-4.04 \times 10^{-3}\times 2.2 \times 10^3=-8.90$

$\therefore Z_i=R_1 \parallel R_2=910 \parallel110=98.13k \Omega$

$\therefore Z_0=R_D=2.2k\Omega$