iLaplace 1.1

A minimal Python interface for computing inverse Laplace transforms using Talbot’s method — designed as a clean and practical wrapper around sympy & mpmath

iLaplace

A Simple Interface for Powerful Math

iLaplace is a lightweight Python wrapper around advanced symbolic and numerical math tools (sympy, numpy & mpmath) that provides a clean and intuitive interface for computing inverse Laplace transforms using Talbot's method.

I created this library to simplify the process of using the inverse Laplace transform, aiming to provide a MATLAB-like experience within Python. This means the output of my code in Python will be compared against a sample calculation in MathWorks MATLAB.

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Why This Library?

If you've ever tried to compute inverse Laplace transforms numerically in Python, you've probably noticed:

• mpmath.invertlaplace() is powerful but raw and low-level.
• sympy gives symbolic transforms, but not numerical answers.
• Combining them can be verbose and repetitive.

In short: You focus on the math, we handle the machinery.

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MATLAB vs. Python Comparison

Here is a sample calculation in MATLAB and its equivalent in Python using this library.

MATLAB Example

syms s t
X = (s+3)/((s+1)^2 + (s^2+9));
V = vpa(ilaplace(X, s, t));
t0   = 5.2643;
Volt = double(subs(V, t, t0));
disp(Volt)

The output in MATLAB is -0.020100091042301.

Python Equivalent with iLaplace

import iLaplace as il
import sympy as sp
t, s = sp.symbols('t s')
X = (s+3)/((s+1)**2 + (s**2+9))
t0 = 5.2643
Answer = il.inverse_laplace(X, t0)
print(Answer)

And the output is -0.020100091042300372.

##See more examples:

2nd Example:

import iLaplace as il
import sympy as sp
import numpy as np
s, t = sp.symbols('s t')
t0 = 5.2643
X = (3) / ((s**2) +  (3**2))
Answer = il.inverse_laplace(X, t0)
print(Answer)

the output is -0.08483425637752962.

MATLAB Example

syms s t
format long
X= 3 / (s^2 + 3^2);
V = vpa(ilaplace(X, s, t));
t0   = 5.2643;
Volt = double(subs(V, t, t0));
disp(Volt)

and The output in MATLAB is -0.084834643101466.

3rd Example:

import iLaplace as il
import sympy as sp
t, s = sp.symbols('t s')
t0 = 5.2643
X = ((s + 3) / ((s + 3)**2 + 25)
              + 1 / (s + 2)**3
              + sp.log((s + 10) / s))
Answer = il.inverse_laplace(X, t0)
print(Answer)

the output is 0.19032962256026703.

MATLAB Example

syms s t
X = (s + 3) / ((s + 3)^2 + 25) ...
    + 1 / (s + 2)^3 ...
    + log((s + 10) / s);  
V = vpa(ilaplace(X, s, t));
t0   = 5.2643;
Volt = double(subs(V, t, t0));
disp(Volt)

and The output in MATLAB is 0.190329629407706.

4th Example:

import iLaplace as il
import sympy as sp
t, s = sp.symbols('t s')
t0 = 5.2643
#######################################
X1 = ((s + 3) / ((s + 3)**2 + 25)
              + 1 / (s + 2)**3
              + sp.log((s + 10) / s))
X2 = (s+3)/((s+1)**2 + (s**2+9))
X3 = (3) / ((s**2) +  (3**2))
#######################################
X= X1 + X2 + X3
Answer = il.inverse_laplace(X, t0)
print(Answer)

the output is 0.08539527514043704.

MATLAB Example

syms s t
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
X1 = (s + 3) / ((s + 3)^2 + 25) ...
    + 1 / (s + 2)^3 ...
    + log((s + 10) / s);
X2 = (s+3)/((s+1)^2 + (s^2+9));
X3 = 3 / (s^2 + 3^2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
X = X1 + X2 + X3;
V = vpa(ilaplace(X, s, t));
t0   = 5.2643;
Volt = double(subs(V, t, t0));
disp(Volt)

and The output in MATLAB is 0.085394895263940.

License and Attributions

This project is licensed under the MIT License.

It internally utilizes components from the following libraries: mpmath & sympy.

All third-party libraries retain their original licenses and attributions.