Hi everyone!
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 \( A_v\; = \; \frac{V_o (jw)}{V_{in} (jw)} \)
2. The function of the circuit and its characteristic parameters
3. The analytic expression and value of the frequency of maximum gain \( w_0\)
4. The behavior of the circuit relative to the two frequencies \(w = 0\) and \(w \to \infty \)

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 \( s = σ + jw \) with \( σ = 0 \) , I can write:

\(
A_v(jw) \; = \; - \frac{jwC_2 R_2 R_3}{\left( R_1 + R_3 \right) - w^2 C_1 C_2 R_1 R_2 R_3 + jw R_1 R_3 \left( C_1 + C_2 \right)}
\)

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 \( w_n \; = \; \sqrt{\frac{R_1 + R_3}{C_1 C_2 R_1 R_2 R_3}} \).

I stopped right here because I didn't understand how to derive the analytic expression of the maximum frequency gain so I went to my professor to ask him 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 \( \left( R_1 + R_3 \right) - w^2 C_1 C_2 R_1 R_2 R_3 \; = \; 0 \) , 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
\(
with \; w = 0 \; then \; \frac{1}{jwC} \to \infty \\
\\
with \; w \to \infty \; then \; \frac{1}{jwC} \to 0
\))

 

Hi everyone!
I think I have a big conceptual gap, I hope someone can clarify things to me.
View attachment 295694

The exercise asks the following:

1. The analytic expression of the gain \( A_v\; = \; \frac{V_o (jw)}{V_{in} (jw)} \)
2. The function of the circuit and its characteristic parameters
3. The analytic expression and value of the frequency of maximum gain \( w_0\)
4. The behavior of the circuit relative to the two frequencies \(w = 0\) and \(w \to \infty \)

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 \( s = σ + jw \) with \( σ = 0 \) , I can write:

\(
A_v(jw) \; = \; - \frac{jwC_2 R_2 R_3}{\left( R_1 + R_3 \right) - w^2 C_1 C_2 R_1 R_2 R_3 + jw R_1 R_3 \left( C_1 + C_2 \right)}
\)

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 \( w_n \; = \; \sqrt{\frac{R_1 + R_3}{C_1 C_2 R_1 R_2 R_3}} \).

I stopped right here because I didn't understand how to derive the analytic expression of the maximum frequency gain so I went to my professor to ask him 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 \( \left( R_1 + R_3 \right) - w^2 C_1 C_2 R_1 R_2 R_3 \; = \; 0 \) , 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
\(
with \; w = 0 \; then \; \frac{1}{jwC} \to \infty \\
\\
with \; w \to \infty \; then \; \frac{1}{jwC} \to 0
\))

I didn't check all your calculations.

To find the frequency at which the gain is a maximum you have to do the following:
1. Derive an expression for the absolute value of the gain.
2. Take the derivative of this expression and set it equal to zero to find where its maximum is.

 

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

The answer is quite simple (no need to differentiate the magnitude function):

* Question: Under which condition has the transfer function a maximum in magnitude?
* Answer: When the denominator assumes its minimum. This is the case when both real parts cancel each other (at the pole frequency). Now the denominator is pure imaginary - and the numerator as well.
* Consequence: The transfer function is real at the pole frequency (the phase crosses zero) and it assumes its maximum at this frequency. Therefore, the mid-frequency (center frequency) of the bandpass is identical to the pole frequency.

 

When the denominator assumes its minimum. This is the case when both real parts cancel each other (at the pole frequency). Now the denominator is pure imaginary - and the numerator as well.

I have problems grasping the math behind this reasoning. How can you say that the denominator is at its minimum in that condition? Can't it also tend to \( 0 \)?

 

I have problems grasping the math behind this reasoning. How can you say that the denominator is at its minimum in that condition? Can't it also tend to \( 0 \)?

Try it. And you will see that there is no frequency which makes the real as well as imginary part of the denominator equal to zero.
As I have shown - there is one single frequency (pole frequency) which makes the real part equal to zero. And as everybody can see - the imaginary part is not zero. So, do you see any chance for another frequency which makes both parts equal to zero?

 

I have problems grasping the math behind this reasoning. How can you say that the denominator is at its minimum in that condition? Can't it also tend to \( 0 \)?

I also do not understand the reasoning:
1. The denominator is a complex number. How do you decide if one complex number is smaller than another?
2. In the case of real-valued functions the statement would be true if the numerator is constant. In this case it is not and also depends on w.

 

How do you decide if one complex number is smaller than another?

I think that the reasoning is:

1. A complex number \( x+jy = 0\) only when both \( x \) and \( y \) are \( 0 \).
2. Since there is no value of \( \omega \) for which \( x \) and \( y \) equal \( 0 \) simultaneously, your best bet is trying to cancel only one of them off.
3. \( y = 0 \) only for \( \omega = 0 \) but that also ends up making the gain null.
4. \( x = 0 \) only when \( \omega = \omega_n \).

Therefore the gain is maximum when the denominator is minimum and that happens only when \( \omega = \omega_n \), which also happens to be when the gain is purely real.