Inverse Laplace Transform Calculator
Enter a Laplace-domain function F(s) as a sum of simple rational terms.
Supported forms (combined by +/-):
A/(s-a)^n, A/s^n, (k1*(s-a)+k0)/((s-a)^2+w^2), (k1*s+k0)/(s^2+w^2), and general quadratics (k1*s+k0)/(a2*s^2+a1*s+a0).
s for Laplace variable (or change below), ^ for powers, and parentheses.
Examples: 1/(s-2), s/(s^2+9), (s+2)/(s^2+4s+5).
s).
t).
f(t) = 𝓛⁻¹{F(s)} (formatted by math.js, not LaTeX).
Results
Time-domain Function f(t)
What is an Inverse Laplace Transform Calculator?
An Inverse Laplace Transform Calculator converts a function of the complex variable \(s\), typically denoted \(F(s)\), back into a causal time-domain function \(f(t)\). If \(\mathcal{L}\{f(t)\}=F(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t)\,dt\), then the inverse operation returns \(f(t)=\mathcal{L}^{-1}\{F(s)\}\). This is essential for solving linear differential equations, analyzing control systems, circuits, and signals. In practice, we use linearity, partial-fraction decomposition, shift theorems, and the convolution theorem to reconstruct \(f(t)\) efficiently.
About the Inverse Laplace Transform Calculator
This calculator accepts symbolic \(F(s)\) (and parameters) and returns \(f(t)\) with step-by-step reasoning. It leverages table lookups for common pairs, applies partial fractions when rational functions appear, and uses theorems for shifts and products. Repeated poles are handled via known patterns:
For responsiveness and crisp math on all screens, render formulas with MathJax; for numeric evaluations (parameters, plotting samples) use math.js. Configure MathJax with container-width line breaks so equations wrap neatly on mobile while preserving the same formulas shown below.
How to Use this Inverse Laplace Transform Calculator
- Enter \(F(s)\) and specify constants (e.g., poles, delays, shifts).
- Choose method: table/partial fractions, convolution, or shift rules; the tool combines methods automatically when needed.
- Review intermediate steps and the final \(f(t)=\mathcal{L}^{-1}\{F(s)\}\), including any unit-step \(u(t)\) factors for delays.
- Optionally verify by forward transform or by substituting into your differential equation.
Examples (using the same formulas)
Example 1 — Partial fractions:
\(F(s)=\dfrac{s+3}{(s+1)(s+2)}=\dfrac{1}{s+1}-\dfrac{1}{s+2}\ \Rightarrow\
f(t)=e^{-t}-e^{-2t}.\)
Example 2 — Repeated pole:
\(F(s)=\dfrac{2}{(s-3)^2}\ \Rightarrow\ f(t)=2\,t\,e^{3t}u(t)\) by
\(\mathcal{L}^{-1}\{1/(s-a)^2\}=t\,e^{at}u(t)\).
Example 3 — Time delay:
\(F(s)=\dfrac{e^{-4s}}{s+2}\ \Rightarrow\ f(t)=u(t-4)\,e^{-2(t-4)}\) using
\(\mathcal{L}^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)\) and
\(\mathcal{L}^{-1}\{1/(s+2)\}=e^{-2t}.\)
Example 4 — Product / convolution:
\(F(s)=\dfrac{1}{s}\cdot\dfrac{1}{s+2}\ \Rightarrow\
f(t)=(1 * e^{-2t})(t)=\displaystyle\int_0^t e^{-2(t-\tau)}\,d\tau=\frac{1-e^{-2t}}{2}.\)
FAQs
Q1: What’s the difference between Laplace and inverse Laplace?
Laplace maps \(f(t)\) to \(F(s)\); inverse Laplace reconstructs \(f(t)\) from \(F(s)\).
Q2: How do I handle non-rational \(F(s)\)?
Use tables, series expansions, or residues; many common forms (e.g., exponentials, s-shifts) have closed-form inverses.
Q3: What if there’s a time delay?
A factor \(e^{-as}\) implies \(u(t-a)f(t-a)\) in time; include the step function explicitly.
Q4: Can this solve ODEs?
Yes. Transform the ODE, solve algebraically for \(Y(s)\), then apply the inverse to get \(y(t)\).
Q5: Are repeated poles difficult?
No. Use \(\mathcal{L}^{-1}\{1/(s-a)^n\}=t^{n-1}e^{at}/(n-1)!\) and combine with linearity and shifts.