Band-pass filter characteristic parameters and maximum gain frequency

Question

I think I have a big conceptual gap, I hope someone can clarify things to me.

The exercise asks the following:

1. The analytic expression of the gain \$A_v\; = \; \frac{V_o (j\omega)}{V_{in} (j\omega)}\$
2. The function of the circuit and its characteristic parameters
3. The analytic expression and value of the frequency of maximum gain \$ \omega_0\$
4. The behavior of the circuit relative to the two frequencies \$\omega = 0\$ and \$\omega \to \infty \$

I proceeded as follows:
I consider the circuit in the frequency domain first through the Laplace transform for easier calculations.
Because of negative feedback I have virtual ground at the op amp inputs and therefore I can write the following expressions:

\$KCL \; A) \; \frac{V_{in} (s) - V_A}{R_1} \; = \; \frac{V_A}{R_3} + sC_1 \left( V_A - V_o (s) \right) + sC_2 V_A \$

\$KCL \; B) \; sC_2 V_A \; = \; - \frac{V_o (s) }{R_2} \; therefore \; V_A \; = \; - \frac{1}{sC_2 R_2} V_o (s)\$

\$therefore \; V_o (s) \left( \frac{1}{sC_2 R_1 R_2} + \frac{1}{sC_2 R_2 R_3} + \frac{C_1}{C_2 R_2} + sC_1 + \frac{1}{R_2}\right) \; = \; - \frac{V_{in} (s)}{R_1}\$

\$therefore \; A_v (s) \; = \; \frac{V_o (s)}{V_{in} (s)} \; = \; - \frac{sC_2 R_2 R_3}{s^2 C_1 C_2 R_1 R_2 R_3 + sR_1 R_3 \left( C_1 + C_2 \right) + \left( R_1 + R_3 \right)}\$

Assuming \$ s = σ + j\omega \$ with \$ σ = 0 \$ , I can write:
\$A_v(j\omega) \; = \; - \frac{j\omega C_2 R_2 R_3}{\left( R_1 + R_3 \right) - \omega^2 C_1 C_2 R_1 R_2 R_3 + j\omega R_1 R_3 \left( C_1 + C_2 \right)}\$

Now, I sketched the Bode plots and since I have this transfer function with 1 zero and a pair of complex conjugate poles \$ A_v (s) \; = \; - \left( \frac{C_2 R_2 R_3}{R_1 + R_3} \right) \frac{s}{1 + s \frac{R_1 R_3 \left( C_1 + C_2 \right)}{R_1 + R_3} + s^2 \frac{C_1 C_2 R_1 R_2 R_3}{R_1 + R_3}} \$ , I think I can say it's a band-pass filter with natural frequency \$ \omega_n \; = \; \sqrt{\frac{R_1 + R_3}{C_1 C_2 R_1 R_2 R_3}} \$.

I stopped right here because I didn't understand how to derive the analytic expression of the maximum frequency gain so I asked my professor about it. He told me that the condition for maximum gain is that the transfer function has to be real.

Since the numerator is imaginary I have to make the denominator imaginary too so I can simplify the two \$ j \$ and that happens only when \$ \left( R_1 + R_3 \right) - \omega^2 C_1 C_2 R_1 R_2 R_3 \; = \; 0 \$ , therefore when \$ w \; = \; \sqrt{\frac{R_1 + R_3}{C_1 C_2 R_1 R_2 R_3}} \$ , but I don't understand the theory behind it..

(For the last point I drew the circuit respectively considering the capacitors as open and short circuit since:
\$with \; \omega = 0 \; then \; \frac{1}{j\omega C} \to \infty \$
\$with \; \omega \to \infty \; then \; \frac{1}{j\omega C} \to 0\$ )

Answer 1

Note that in the following, it's assumed that the coefficients are real-valued and not complex:

$$\begin{align*} G_s &= \frac{N_s}{D_s}=\frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0} \end{align*}$$

Now, for these purposes \$\sigma=0\$ (no spiraling effects) and the focus is on the rotation (\$j\omega\ge 0\$), so the substitution is made for \$s=j\omega\$.

$$\begin{align*} G_{j\,\omega} &=\underbrace{\overbrace{\left[\frac{a_2}{b_2}\vphantom{\frac{-\left(\frac{\omega}{\omega_{_0}}\right)^2}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}}\right]}^{\text{gain}}\cdot\left[\frac{-\left(\frac{\omega}{\omega_{_0}}\right)^2}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}\right]}_{\text{high-pass}}\\ &+\underbrace{\overbrace{\left[\frac{a_1}{b_1}\vphantom{\frac{-\left(\frac{\omega}{\omega_{_0}}\right)^2}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}}\right]}^{\text{gain}}\cdot\left[\frac{j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}\right]}_{\text{band-pass}}\\& + \underbrace{\overbrace{\left[\frac{a_0}{b_0}\vphantom{\frac{-\left(\frac{\omega}{\omega_{_0}}\right)^2}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}}\right]}^{\text{gain}}\cdot\left[\frac{1}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)}\right]}_{\text{low-pass}} \end{align*}$$

(To see the development of the above, read this.)

In the high-pass case, as \$\omega\to\infty\$, the denominator's magnitude approaches \$-\left(\frac{\omega}{\omega_{_0}}\right)^2\$ and so the right-side factor reduces to a maximum of 1. The gain is real, obviously.

In the low-pass case, as \$\omega\to 0\$, the denominator's magnitude approaches \$1\$ and so the right-side factor reduces to a maximum of 1. The gain again is real.

In the band-pass case, there are two components that reflect the magnitude -- the real and imaginary parts. But the denominator magnitude will always be larger than the numerator except in the case where \$\omega=\omega_{_0}\$. In that case, once again the right-side factor reduces to a maximum of 1. And once again, the gain is real.

Just look over the links I gave. Think over what I just wrote. You should be able to follow the professor's point.