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```# -*- coding: utf-8 -*-
# Talbot suggested that the Bromwich line be deformed into a contour that begins
# and ends in the left half plane, i.e., z → −∞ at both ends.
# Due to the exponential factor the integrand decays rapidly
# on such a contour. In such situations the trapezoidal rule converge
# extraordinarily rapidly.
# For example here we compute the inverse transform of F(s) = 1/(s+1) at t = 1
#
# >>> error = Talbot(1,24)-exp(-1)
# >>> error
#   (3.3306690738754696e-015+0j)
#
# Talbot method is very powerful here we see an error of 3.3e-015
# with only 24 function evaluations
#
# Created by Fernando Damian Nieuwveldt
# email:fdnieuwveldt@gmail.com
# Date : 25 October 2009
#
# Reference
# L.N.Trefethen, J.A.C.Weideman, and T.Schmelzer. Talbot quadratures
# and rational approximations. BIT. Numerical Mathematics,
# 46(3):653 670, 2006.

from cmath import *

def Talbot(t,N):

#   Initiate the stepsize
h = 2*pi/N;

#   Shift contour to the right in case there is a pole on the positive real axis : Note the contour will
#   not be optimal since it was originally devoloped for function with
#   singularities on the negative real axis
#   For example take F(s) = 1/(s-1), it has a pole at s = 1, the contour needs to be shifted with one
#   unit, i.e shift  = 1. But in the test example no shifting is necessary

shift = 0.0;
ans =   0.0;

if t == 0:
print "ERROR:   Inverse transform can not be calculated for t=0"
return ("Error");

#   The for loop is evaluating the Laplace inversion at each point theta which is based on the trapezoidal   rule
for k in range(0,N):
theta = -pi + (k+1./2)*h;
z = shift + N/t*(0.5017*theta*cot(0.6407*theta) - 0.6122 + 0.2645j*theta);
dz = N/t*(-0.5017*0.6407*theta*(csc(0.6407*theta)**2)+0.5017*cot(0.6407*theta)+0.2645j);
ans = ans + exp(z*t)*F(z)*dz;

return ((h/(2j*pi))*ans).real

#   Here is the Laplace function to be inverted, should be changed manually
def F(s):
return 1.0/(s+1.0)

#   Define the trig functions cot(phi) and csc(phi)
def cot(phi):
return 1.0/tan(phi)

def csc(phi):
return 1.0/sin(phi)

```