You haven't done anything wrong. The first stage of your circuit is an integrator, and when you integrate a step function (your input) you get a ramp. Since your input never ends, the ramp never ends and the final value is infinity.

Of course, this only holds for an ideal op-amp. In a real op-amp, output voltages would be limited to the rail voltage.

In case you're wondering, I've never seen the output voltage limitations accounted for in equations. You just have to introduce a nonlinear element into the model (a voltage limiter) and then all of your linear equations are essentially out the window. In other words, the linear equations you're used to are all well and good as long as the output of any of your stages doesn't exceed the rail voltage - as it will in this case.

Strictly speaking, the circuit has 5 nodes, at points labelled a, b, c, d and f. If you use the modified nodal analysis to solve the circuit, you'll apply KCL at all nodes but one (usualy the reference one) to end up with a system of lineal independent equations. So, you'll apply KCL at nodes a, b, c and d to determine the (initially) unknown voltages at these nodes.

However, this is modified by the presence of voltage sources connected to ground at nodes a and c. Due to the presence of these voltage sources, the voltages at nodes a and c are not unknowns, so you have only 2 "real" unknowns (voltages at b and d), and you have to write only 2 KCL. If you have to choose, you would like to write KCL at b and d, because writing the KCL at a and c involves the currents flowing through the voltage sources, and you don't know how to express these currents in terms of the node voltages, so you avoid writing KCL equations at nodes having voltage sources connected to ground.

So finally you have to write the KCL at b and d to solve the circuit, and you can also get rid of the KCL at d if you add the impedances of the resistor and the capacitor together to have 10-10j Ohm.

So your first equation (KCL) is ok, and all you have to do from that is to express the currents in terms of node voltages and impedances:

$$I_1+I_2−I_3−I_4=0$$

$$I_1 = \frac{E_1-V_b}{-10j}$$

$$I_2 = \frac{E_2-V_b}{+20j}$$

$$I_3 = \frac{V_b}{10-10j}$$

$$I_4 = \frac{V_b}{10}$$

And if you substitute these currents in the first KCL and solve for Vb you obtain:

$$V_b = \frac{1-j}{1-3j} * (2E_1-E_2) = \frac{1}{2-j} * (2E_1-E_2)$$

Knowing Vb (and E1 and E2) allows you to easily determine all other circuit variable.

## Best Answer

Jan, I won't bother with much (because you can pound out equations as well as the next one) except to say that I wrote the following:

So, for example, I find the impedance seen by \$V_n\$ as:

$$R=\frac{V_n}{I_4}+\frac{R_4+R_5}{\left[\frac{R_4}{R_1\mid\mid R_2}+\frac{R_4+R_5}{R_3}\right]}$$

KCL just works so far as I can tell. I guess I'm not sure what the question might be?