Question 14.13: Design a bandpass filter in the form of Fig. 14.45 to pass f......

1. Define. The problem is clearly stated and the circuit to be used in the design is specified.
2. Present. We are asked to use the op amp circuit specified in Fig. 14.45 to design a bandpass filter. We are given the value of R to use (20 kΩ) . In addition, the frequency range of the signals to be passed is 250 Hz to 3 kHz.
3. Alternative. We will use the equations developed in Section 14.8.3 to obtain a solution. We will then use the resulting transfer function to validate the answer.
4. Attempt. Since ω_{1} = 1/RC_{2}   we obtain

C_{2} = \frac{1}{Rω_{1}} = \frac{1}{2πf_{1}R} = \frac{1}{2π  ×  250  ×  20  ×  10^{3}} = 31.83  nF

Similarly, since ω_{2} = 1/RC_{1}, 

C_{1} = \frac{1}{Rω_{2}} = \frac{1}{2πf_{1}R} = \frac{1}{2π  ×  3,000  ×  20  ×  10^{3}} = 2.65  nF

From Eq. (14.73),

K = \frac{R_{f}}{R_{i}} \frac{ω_{2}}{ω_{1}  +  ω_{2}}         (14.73)

\frac{R_{f}}{R_{i}} = K \frac{ω_{1}  +  ω_{2}}{ω_{2}} = K \frac{f_{1}  +  f_{2}}{f_{2}} = \frac{10(3,250)}{3,000} = 10.83

If we select R_{i} = 10  kΩ ,   then R_{f} = 10.83 R_{i} \simeq  108.3  k Ω .  

5. Evaluate. The output of the first op amp is given by

\frac{V_{i}  –  0}{20  k Ω }  + \frac{V_{1}  –  0}{20  k Ω } + \frac{s2.65  ×  10^{-9}  (V_{1}   –   0 )}{1}

= 0     →  V_{1} = – \frac{V_{i}}{1  +  5.3  ×  10^{-5}  s}

The output of the second op amp is given by

\frac{V_{1}   –  0}{20  kΩ  +  \frac{1}{s31.83  nF}} + \frac{V_{2}  –  0}{20  k Ω} = 0  →

V_{2} = – \frac{6.366  ×  10^{-4} s V_{1}}{1  +  6.366  ×  10^{-4}  s}

= \frac{6.366  ×  10^{-4}  s V_{i}}{(1  +  6.366  ×  10^{-4} s)(1  +  5.3  ×  10^{-5}  s)}

The output of the third op amp is given by

\frac{V_{2}  –  0}{10  k Ω} + \frac{V_{o}  –  0}{108.3  kΩ} = 0 → V_{o} = 10.83 V_{2} → j2π   ×  25°

V_{0} = – \frac{ 6.894  ×  10^{-3}  sV_{i}}{ (1  +  6.366  ×   10^{-4} s)(1   +   5.3  ×   10^{-5}  s)}

Let  j2π  ×  25°  and solve for the magnitude of V_{o}/V_{i}  

\frac{V_{o}}{V_{i}} = \frac{-j10.829}{(1  +  j1)(1)}

\left|V_{o}/V_{i}\right|= (0.7071)10.829, which is the lower corner frequency point . Let  s = j2π  ×  3000  = j18.849 k Ω We then get

\frac{V_{o}}{V_{i}} = \frac{-j129.94}{(1  +  j12)(1  +  j1)}

= \frac{129.94 \underline{/-90^\circ}}{(12.042\underline{/85.24^\circ})(1.4142 \underline{/45^\circ })} = (0.7071)10.791 \underline{/- 18.61^\circ}

Clearly this is the upper corner frequency and the answer checks.
6. Satisfactory? We have satisfactorily designed the circuit and can present the results as a solution to the problem.