Series Formula for Inverse Laplace Transform related to The Telegrapher's (Heaviside's) Equations

Question

A problem related to The Telegrapher's (Heaviside's) Equations

$$V_x = -LI_t - RI$$ $$I_x = -CV_t - GV$$

gives rise to a Laplace Transform

$$Z(s)=\sqrt{\frac{R+s L}{G +s C}}$$

This problem has been discussed on Electrical Engineering Stack Exchange at Transmission Line Transient Response To Step Voltage Prior To Any Reflections.

My question is: what is the inverse Laplace Transform of the above transform? ($R$, $L$, $G$, and $C$ are constants). I suspect there may be an answer in terms of Bessel functions (and/or error functions?). However, I would be happy at this point to simply know a convergent series (or any convergent algorithm) which would allow numerical computation of the inverse Laplace Transform. Extra bonus if you can explain how the answer is found.

Answer 1

Original problem \begin{equation} f(t)=\mathcal{L}^{-1}\left\{\sqrt{\frac{R+s L}{G +s C}}\right\}. \end{equation} Cleaning up the notation by using Tyma's suggestion, \begin{align} f(t)&=\mathcal{L}^{-1}\left\{\sqrt{\frac{L}{C}}\frac{\sqrt{R/L+s}}{\sqrt{G/C+s}}\right\}\\ &=c\mathcal{L}^{-1}\left\{\frac{\sqrt{a+s}}{\sqrt{b+s}}\right\}. \end{align} with $c=\sqrt{L/C}$, $a=R/L$, and $b=G/C$.

The above inverse Laplace transform was discussed here. Reproducing the answer, \begin{align} \mathcal{L}^{-1}\left\{\sqrt{\dfrac{s+a}{s+b}}\right\}&=e^{-bt}\mathcal{L}^{-1}\sqrt{\dfrac{s+a-b}{s}}\\ &=e^{-bt}\left\{\dfrac{(a-b)e^\frac{(b-a)t}{2}}{2}\left[I_1\left(\dfrac{(a-b)t}{2}\right)+I_0\left(\dfrac{(a-b)t}{2}\right)\right]+\delta(t)\right\} \\ &=\frac{1}{2}(a-b)e^{-\frac{(a+b)t}{2}}\left[I_1\left(\dfrac{(a-b)t}{2}\right)+I_0\left(\dfrac{(a-b)t}{2}\right)\right]+\delta(t) \end{align} where we used the $t$-shift rule in the first line and http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv3.pdf in the second line.

Multiplying by $c$, we find the solution to the original problem, \begin{equation} f(t)=\frac{c}{2}(a-b)e^{-\frac{(a+b)t}{2}}\left[I_1\left(\dfrac{(a-b)t}{2}\right)+I_0\left(\dfrac{(a-b)t}{2}\right)\right]+c\delta(t) \end{equation}

For $a=1$, $b=3$, $c=1$, I plot the above function (green line), albeit without the delta function. Comparing to a numerical inversion of the original function (green dots), we see that we found the correct result. In red I do the same, changing $a$ to $a=3$.

Code:

import numpy as np
import matplotlib.pyplot as plt
import scipy.special as sp
from mpmath import *

def f(t,a,b,c):
    arg = (a-b)*t/2
    return c*(a-b)*np.exp(-(a+b)*t/2)*(sp.iv(1,arg)+sp.iv(0,arg))/2

def F(s,a,b,c):
    return c*mp.sqrt(a+s)/mp.sqrt(b+s)

mp.dps = 30
def fnum(t,a,b,c):
    ft = lambda s: F(s,a,b,c)
    return invertlaplace(ft,t,method='dehoog')

t  = np.linspace(0,3,100)
t2 = np.linspace(0.1,3,30)

a1=1
a2=3
b1=2
c1=1

fig, ax = plt.subplots()
ax.plot(t,[f(i,a1,b1,c1) for i in t],  'g',lw=2,label='analytical $a=1$')
ax.plot(t,[f(i,a2,b1,c1) for i in t],  'r',lw=2,label='analytical $a=3$')
ax.plot(t2,[fnum(i,a1,b1,c1) for i in t2],'g',lw=0,marker='o',label='numerical')
ax.plot(t2,[fnum(i,a2,b1,c1) for i in t2],'r',lw=0,marker='o',label='numerical')
ax.set(xlabel=r'$t$', ylabel=r'$f(t)$') 
ax.legend()