100 Vcc R1 Rstg 47k R2 C2 C1 60p Cui = 20pF Vcc = 15V Rsig C1 Q1 RL vsig The circuit shows an NPN common emitter amplifier with a diode-connected biasing scheme. The component values are as shown on...

100 Vcc R1 Rstg 47k R2 C2 C1 60p Cui = 20pF Vcc = 15V Rsig C1 Q1 RL vsig The circuit shows an NPN common emitter amplifier with a diode-connected biasing scheme. The component values are as shown on the right. From the component values we can calculate that 9m 38.1mA/V. You don't need to calculate the bias point a. [2 pts] Draw a small signal model applicable at the low-frequency end of the passband near the low-frequency cutoff fi. Neglect any capacitances that will only matter at the high-frequency end of the passband. b. [2 pts] Apply the Miller approximation to the model to obtain a new model that is unilateral (i.e c. 12 pts] Use your answer from (b) to estimate fi, stating and justifying any approximations being d. [1 pt] where the input and output sides are not connected). Indicate the value of any new components introduced to your model by applying Miller's theorem. made ] Draw a small signal model applicable at the high-freqeuncy end of the passband near the high-frequency cutoff f. Neglect any capacitances that will only matter at the low-frequency end of the passband. You can assume r Rsig. e. [3 pts] Use the method of open-circuit time constants to calculate f