Using the superposition theorem to solve for the output voltage

Question

I'm working on diodes again and thought I'd challenge myself to hopefully learn a bit more. I've tried solving this problem before, but the superposition theorem didn't come to mind then, but I think it'll be a good approach in this case. Either way, here's the problem:

We are given three input voltages \$u_1,u_2,u_3\$ whose voltages over time are described on the right. We've also assumed that \$R_1\$ is much smaller than \$R_2\$. Also, the diodes are ideal. We basically want to find \$U_{out}\$. So using the superposition principle, we can redraw the circuit for three different scenarios:

So, we zero all the other input voltages expect the first one. So diode \$S_1\$ is conducting current if and only if \$V_{R_1} > V_1\$. Also, we have to have that \$ E > V_{R_1}\$ otherwise we wont get current from anywhere flowing through the diode. But that just means that \$ V_1 < E \$. This only happens when \$V_1 = 0\$ from the diagram, but since the diode is ideal, no voltage drop over the diode exists, so \$V_{R_1} = 0 V\$ and therefore the output voltage \$U_o = U_{out} = 0 V\$ for \$ V_1 < E \$.

If the diode is not conducting any current, then \$V_1 > V_{R_1} = U_{o}\$ and \$E =V_{R_1}\$, so together, \$V_1 > E\$ which never happens according to the diagram. Therefore, the diode \$S_1\$ is always conducting, and \$V_1 = U_{out}\$ in our first case.

I'm unsure of my second argument. I also tried to use the same argument for the case when \$u_2\$ and \$u_3\$ but I couldn't really get anywhere. The final idea is to add these voltages contributed to our output voltage in a diagram. Any tips would be greatly appreciated.

Thank you once again.

Answer 1

It seems that you should simplify the problem by combining U2 and U3 into what I've called U4: -

Then, because of these words: -

We've also assumed that R1 is much smaller than R2

We can make another simplification: -

Can you run with it from here?

Using the superposition theorem to solve for the output voltage

To be a good EE means that one of your goals is to read circuit diagrams proficiently and, recognize the important things. This is what my answer is about; seeing simplifications and capitalizing on those simplifications. No need to use math to bludgeon your way to a solution until you have made the obvious simplifications. And, in this example, no need to use math at all; just your eyes and your brain. Final picture: -