Question

Problem #1 (100pts) Consider the circuit shown in the Figure of three simpler circuits (stages). 1) Without solving for the transfer function find how each stage behaves versus frequency o 2) Using this information find how the total circuit behaves versus frequency 2) Find the transfer function for each stage 3) Find the transfer function of the total circuit 4) Plot the |H(jo) 2 as a cascade Av R4 R1 ww Vin Vout R3 Ov Stage II Stage II Stage I Assume: R,-R3 10K R2=R 40K C1-1pF C2=1nF

Answer 1

Solution:

1) The first stage is a high pass filter circuit because at low frequency capacitor is open and does not allow the input and at high frequency capacitor is shorted and output is equal to input. It passes high frequency signals and attenuates low frequency signals.

Second stage is a DC amplifier with gain is equal to (1+R4/R3)

The third stage is a low pass filter circuit because at high frequency capacitor is open and allow the input and at low frequency capacitor is shorted and output is equal to zero. It passes low frequency signals and attenuates high frequency signals.

2) The total response is band pass response that will allow only signal frequencies between higher cutoff frequency fH and lower cutoff frequency fL.

3) Transfer function of each stage is found as,

Let output of first stage is taken as V1.

By voltage divider rule,

Capacitive reactance = Xc = 1/jwC1

R1 1in R(1/jwC1)

jwCIRI (1jwC1R) Vin

Second stage, the transfer function is given by,

Let V2 be output voltage of second stage, and assume that op-amp is assumed to be ideal,

R4 1+ 22 R3

Transfer function of third stage is found as,

By voltage divider rule,

Capacitive reactance = Xc = 1/jwC2

1/jwC2 R2(1/jwC2) Vout 2

Vout (1jwC2R2)

4) The magnitude of the transfer function is given by,

Vaut Veut Vin I+wCR H(tw) +jw R2 + HG VIA (GR IGR) R4 (wc, R,) |+ The esponse i plod as. iCeueny) 2 R, C