I could finally talk to someone who's responsible for the exercise. It turned out that A1's input terminals had been erroneously interchanged.

The conclusions about the above incorrect circuit are as follows:

• One should immediately recognize the positive feedback loop (and thus the faulty design).
• As there is positive feedback, A1 won't be able to establish virtual ground at its inverting input (Node A).
• One consequence is that it's futile to try to establish a transfer function.
• Furthermore one can't create a state space model of the circuit (which I tried) because some voltages/variable have a value of +/- infinity which leads to absurd calculations.

The correct circuit would have to look like this:

^{simulate this circuit – Schematic created using CircuitLab}

Here the results of a simple DC analysis and the transfer function as well as the state space model should be consistent (couldn't check it yet, though).

It's worth to note that for determining the type of feedback one can count the number of inversions in the path of the feedback path, starting from the input terminal up to the input terminal again.

In this case there is (no inversion from Ve1+ and Vout) and (one inversion from Vout to V2). This makes the total count of inversions odd, which is negative feedback - as opposed to the faulty circuit above.

As for the original questions (this is my interpretation again, so no guarantee about correctness):

i) Strictly speaking the official answer is correct (\$A_v=-\infty\$), because it's futile to calculate a transfer function based on the wrong assumption of virtual GND at Ve1-e, but the circuit doesn't make sense anyway.

ii) As @Chu has pointed out, a magnitude can't be negative. The absolute value of my expression would hence be \$\infty\$ again.

iii) The DC analysis should indeed be consistent with a (correctly derived) transfer function. One should then use a step function \$\mathscr{L}\{\sigma(t)\} = \frac{1}{s}\$ as input and then calculate the inverse Laplace transform of the resulting expression.

iv) If the transfer function were correct, its phase would be constant -90°, because if we denote \$H(s)=A_v(s)\$ then \$H(i \omega) = \frac{1}{i \omega R_1C_1} = 0 - i \frac{1}{\omega R_1C_1}\$

$$\varphi(H(i \omega))=\arctan \left( \frac{\Im\{H(i \omega)\}}{\Re\{H(i \omega)\}} \right) = \arctan \left( \frac{\frac{-1}{\omega R_1C_1}}{0} \right) = \arctan (-\infty) = -90°$$

Wolfram Alpha does nice Bode plots for arbitrary transfer functions, of course also for this one.

I could finally talk to someone who's responsible for the exercise. It turned out that A1's input terminals had been erroneously interchanged.

The conclusions about the above incorrect circuit are as follows:

• One should immediately recognize the positive feedback loop (and thus the faulty design).
• As there is positive feedback, A1 won't be able to establish virtual ground at its inverting input (Node A).
• One consequence is that it's futile to try to establish a transfer function.
• Furthermore one can't create a state space model of the circuit (which I tried) because some voltages/variable have a value of +/- infinity which leads to absurd calculations.

The correct circuit would have to look like this:

^{simulate this circuit – Schematic created using CircuitLab}

Here the results of a simple DC analysis and the transfer function as well as the state space model should be consistent (couldn't check it yet, though).

It's worth to note that for determining the type of feedback one can count the number of inversions in the path of the feedback path, starting from the input terminal up to the input terminal again.

In this case there is (no inversion from Ve1+ and Vout) and (one inversion from Vout to V2). This makes the total count of inversions odd, which is negative feedback - as opposed to the faulty circuit above.

As for the original questions (this is my interpretation again, so no guarantee about correctness):

i) Strictly speaking the official answer is correct (\$A_v=-\infty\$), because it's futile to calculate a transfer function based on the wrong assumption of virtual GND at Ve1-e, but the circuit doesn't make sense anyway.

ii) As @Chu has pointed out, a magnitude can't be negative. The absolute value of my expression would hence be \$\infty\$ again.

iii) The DC analysis should indeed be consistent with a (correctly derived) transfer function. One should then use a step function \$\mathscr{L}\{\sigma(t)\} = \frac{1}{s}\$ as input and then calculate the inverse Laplace transform of the resulting expression.

iv) If the transfer function were correct, its phase would be constant -90°, because if we denote \$H(s)=A_v(s)\$ then \$H(i \omega) = \frac{1}{i \omega R_1C_1} = 0 - i \frac{1}{\omega R_1C_1}\$

$$\varphi(H(i \omega))=\arctan \left( \frac{\Im\{H(i \omega)\}}{\Re\{H(i \omega)\}} \right) = \arctan \left( \frac{\frac{-1}{\omega R_1C_1}}{0} \right) = \arctan (-\infty) = -90°$$

Wolfram Alpha creates nice Bode plots for arbitrary transfer functions, of course also for this one.