------,b c---------------------------
| | |
Rpi | |
| V gm Vbe Rc
|____e_____| |
| |
Re |
| |
------------G----------------------------------
I don't see Figure 8-1 but it's probably the same as the circuit I've tried to draw above above. The key to solving for the gain and input impedance is to write Kirchhoff's current law (KCL) at node e. This will give you...
Vbe/Rpi + gmVbe = Ve/Re
Kirchhoff's voltage law at the input side gives you
Vin = Vbe + Ve
Combining these two equations
Vin = Vbe (1 + Re/Rpi + gm Re)
At the output we have
Vout=Vc=gm Vbe Rc
From the last two equations, the small signal gain is
Vout/Vin = gmRc/ (1 + Re/Rpi + gm Re) = 400/(1+0.2+40)=9.53
To compute the input impedance, note that the input current is
Iin = Vbe/Rpi
so that
Iin = Vin/(Rpi(1 + Re/Rpi + gm Re))
and
Vin/Iin = Rpi + Re +gm Re Rpi =5K+1K+0.04*1K*5K=206K
If my circuit does not match Figure 8-1, you should be able to use the same method to derive the correct answer.