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

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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\$ )

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  • \$\begingroup\$ To find the gain at \$\omega\$ you need to divide the numerator s-term and the denominator s- term as well. \$A_V = \frac{C_2 R_2 R_3}{R_1 R_3( C_1 + C_2)} = \frac{R_2}{R_1} \frac{C_2}{C_1 + C_2} \$ \$\endgroup\$
    – G36
    Commented Jun 5, 2023 at 18:51
  • \$\begingroup\$ Ritter, is "but I don't understand the theory behind it.." (where it is from "He told me that the condition for maximum gain is that the transfer function has to be real") your question? I did find one discussion about gain here that is backed up with more development by the same author here. But whether those help you or don't, I'm not sure. You'll need to look and decide. If not, then I think you may at least be in a better position to sharpen your question here. \$\endgroup\$ Commented Jun 5, 2023 at 22:16
  • \$\begingroup\$ Ritter, also I wrote something on the multifeedback bandpass here, recently. \$\endgroup\$ Commented Jun 5, 2023 at 22:21
  • \$\begingroup\$ @G36 I need to find the maximum gain frequency, not the maximum gain. \$\endgroup\$
    – RitterTree
    Commented Jun 6, 2023 at 8:11
  • \$\begingroup\$ @periblepsis I'm sorry for not being clear, yes, that's my question. I don't understand why the transfer function has to be real in order for its absolute value to represent the maximum gain. \$\endgroup\$
    – RitterTree
    Commented Jun 6, 2023 at 8:16

1 Answer 1

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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.

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  • \$\begingroup\$ I read your post, it gave me beautiful insight, thanks! Writing the numerator as an expression dependent on the denominator to isolate the compound behaviors of the filter transfer function was pretty sweet. Therefore, in my case, can I say that, given \$A_v(j\omega) \; = \; - \frac{C_2 R_2}{R_1 \left( C_1 + C_2 \right)} \frac{j\omega R_1 R_3 \left( C_1 + C_2 \right)}{\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)}\$, in order for the right hand side contribution of the gain to be 1 (its maximum) \$\omega\$ has to be equal to \$\omega_n\$ ? \$\endgroup\$
    – RitterTree
    Commented Jun 10, 2023 at 9:06

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