Consider the following circuit. The switch closes at \$t=0\$. The voltage source is \$V_0e^{-t/t_0}\$. We assume to op is ideal, so there's no voltage difference between the input terminals and no current through the input.
Approach 1
For \$t > 0\$ we have immediately that \$V=\frac{R}{R+R}V_0e^{-t/t_0}=\frac{1}{2}V_0e^{-t/t_0}\$.
Let \$i_L\$ be the current through the inductor from \$V_{out}\$ to \$V\$. We see that \$i_L = V/R = \frac{V_0}{2R}e^{-t/t_0}\$. We have \$V_{out}-V=L\frac{di_L}{dt}=-\frac{V_0L}{2Rt_0}e^{-t/t_0}\$, and we find \$V_{out}=V-\frac{V_0L}{2Rt_0}e^{-t/t_0}=\frac{V_0}{2}\left(1-\frac{L}{t_0R}\right)e^{-t/t_0}.\$
Approach 2
Let's analyse the circuit in the Laplace domain. We thus let the voltage source be \$V_0e^{-t/t_0}H(t)\$ in the time domain, where \$H(t)\$ is the step function. The Laplace transform of this is \$V_0\frac{1}{s+1/t_0}\$.
In the same way as before, we find \$V(s)=\frac{V_0}{2}\frac{1}{s+1/t_0}\$. We also have \$V_{out}-V=\frac{sL}{sL+R}V_{out}\$, whence \$V_{out}=\frac{sL+R}{R}V=\frac{V_0L}{2R}\frac{s+R/L}{s+1/t_0}=\frac{V_0L}{2R}\left(1+\frac{R/L-1/t_0}{s+1/t_0} \right).\$
Taking the inverse gives \$V_{out} = \frac{V_0L}{2R}\delta(t)+\frac{V_0}{2}\left(1-\frac{L}{t_0R}\right)e^{-t/t_0}.\$
Questions
Apparently, we pick up a delta-distribution with the second approach. Obviously, this cannot occur in a real circuit, so I assume the problem somehow lies in our assumptions about ideality of the OP?
Is it then in general true that we cannot analyse transient behaviour of ideal OP circuits in the Laplace domain? (Or rather, if we do so, we have to neglect all non-physical terms?).
simulate this circuit – Schematic created using CircuitLab
Best Answer
You have several problems.
You are using derivatives on something that is discontinuous. That usually tends not to work. Also, the opamp's output voltage will go to infinity at t=0. I will now explain...
So, after the two rather useless resistors which divide this by 2, when \$ t=0^- \$ which is a tiny instant before t=0, we have \$ V= 0 \$. Then at \$ t=0 \$ the switch closes and we have \$V=V_0/2\$, thus V is discontinuous.
Our nice perfect opamp must adjust its output voltage to keep its two inputs at the same potential. So, at t=0 the output of the opamp will go up. However, there is a perfect inductor in the feedback loop. And the current in an inductor cannot change instantly. And the opamp's negative input voltage V is \$ R i_L \$. And at t=0, \$ i_L =0 \$.
Thus, the voltage at the two inputs of the opamp is not equal. IN+ is at Vo/2 and IN- is at 0V. So, the perfect opamp model stops working, but you got other problems. For the ideal opamp, there is only one choice... its output voltage instantly jumps to +infinity.
Now, at t=0 voltage across the inductor which is \$ V_{inductor} = L\frac{di_L}{dt}\$ is thus +infinity.
If we break enough math, spill the pieces over the floor, then roll over it forward and reverse with a tank, which you did, we can thus integrate that and conclude that \$ \int V_{inductor} = L i_L\$ and therefore right after t=0, give or take, more or less, \$ i_L = 1/L \int +\infty = V_o/2R\$ ...
Tada! Done.
Now, your problem is you didn't notice this, so your you got the wrong result in "Approach 1". No blame, we all make mistakes, and the ideal opamp model really invites mistakes too, since any deviation from ideal operating point creates impossible conditions.
I didn't check the Laplace Transform calculations, but let's examine the circuit again. Let's throw away the switch and both resistors, and use the "V" node as input. We now have a bog-standard non-inverting amplifier. Its gain is:
\$ G = 1+ Ls/R \$
Now, as is obvious from this equation, G goes to infinity as frequency goes up. This kinda works like a differentiator. But it is given a discontinuous signal as an input. Thus it derivates a non-derivable input. Hence your Laplace results have a delta.
At infinitely high frequency, the inductor's impedance is infinite, thus it is an open circuit, and we can remove it from the schematic: on discontinuous inputs, the opamp no longer has feedback, so the ideal opamp model cannot apply.
In real life, opamps are not infinitely fast, therefore the phase lag caused by the inductor in the feedback loop will turn the opamp into an oscillator.
So this circuit is a trap ;)