Why does the solution for the voltage of this RLC circuit result in a maximum capacitor value? I have a circuit with an inductor (L), capacitor (C), and two resistors (R1=R2):



This represents an equivalent electrical circuit to a plectrum or hammer (cap=spring, inductor=mass) striking a string divided into two segments (R1 &amp; R2).

The system is energized by a current impulse at time 0. This DSP synthesis technique is explained more here: https://www.dsprelated.com/freebooks/pasp/String_Excitation.html

The problem I am having is that the solution for the voltage leads to a condition where there is a maximum capacitor value or it is unsolvable, and I am not sure why or what this represents.

The current across the parallel component (the "exciter") is described as follows:

$I(s) = I_C(s)+I_L(s)$

$I(s) = sCV_C(s) + \frac{1}{sL}V_L(s) + \frac{i_0}{s}$

$I(s) = \frac{CLs^2+1}{Ls} V_{exciter}(s) + \frac{i_0}{s}$

The current across each string is represented by:

$I(s) = V_R(s)/R$

Furthermore, if $R1=R2$, $V_R(s)$ can be approximated given that with the exception of the momentary impulse:

$V_{exciter}(s) + V_{R1}(s) + V_{R2}(s) = 0$

$V_{exciter}(s) + 2V_{R}(s) = 0$

$V_R(s) = \frac{-V_{exciter}(s)}{2}$

Therefore we can put these equations together as:

$\frac{-V_{exciter}(s)}{2R} = \frac{CLs^2+1}{Ls} V_{exciter}(s) + \frac{i_0}{s}$

$V_{exciter}(s) = \frac{-2i_0LR}{2CLRs^2+ Ls + 2R}$

To then convert this to the time domain, I need it in terms of: $\frac{1}{(s+a)(s+b)} = \frac{e^{-at}-e^{-bt}}{b-a}$

This is solved as:

$V_{exciter}(s) = \frac{-2i_0LR}{2CLR(s^2+s/(2CR)+1/(CL))}$

$V_{exciter}(s) = \frac{-i_0}{C}\frac{1}{(s+a)(s+b)}$

$a=\frac{-1}{4LR}\left(-\frac{L}{C}-\sqrt{\frac{L^{2}}{C^{2}}-\frac{16CLR^{2}}{C^2}}\right)$

$b=\frac{-1}{4LR}\left(-\frac{L}{C}+\sqrt{\frac{L^{2}}{C^{2}}-\frac{16LR^{2}}{C}}\right)$

This results in a final solution of:

$v_{exciter}(t) =\frac{-i_0}{C}\frac{e^{-at}-e^{-bt}}{b-a}$

The problem now is the square root term for $a$ and $b$. This creates a condition where in order to be solvable:

$L (L - 16 C R^2) &gt;= 0$

So at a given $L$ and $R$ value, $C$ must be $&lt; \frac{m}{16R^2}$ for this equation to be solvable or you otherwise end up with a square root of a negative which I am not sure how to deal with and get real results from.

Why does this happen and what does it mean? Is there some way to solve this equation differently so it will still work at any $C$ value?

It was suggested to me that above this capacitor threshold, the circuit may be going from exponential decay to sinusoidal.  But even still, should I not be able to work out an equation for the voltage that will still give me a value?
Why the fear of complex numbers? Just do a distinction of cases and generate solutions for the real and the complex case. $b-a$ in the denominator will always be real, as the imaginary parts will cancel out each other, so complex numbers will only be in the exponents, wich will lead to some nice sinusoidal functions, which is expected, as this is an oscillating system. The real case just represents the overdamped case with no oscillation. If you don't know what I'm talking about have a look at this wikipedia article:
https://en.wikipedia.org/wiki/Damping_ratio