Laplace Transform Calculator

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Laplace transform of e^(2t)*sin(3t)
Laplace transform of t^2
inverse Laplace of 1/(s^2 + 4)
inverse Laplace of s/((s-1)(s+2))

What is the Laplace Transform?

The Laplace transform converts a function of time f(t)f(t) into a function of complex frequency F(s)F(s):

F(s)=L{f(t)}=∫0∞e−stf(t) dtF(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\,dt

The transform is defined for ss in some right half-plane Re⁡(s)>σ\operatorname{Re}(s) > \sigma where the integral converges.

Why this is useful: Laplace converts differentiation into multiplication by ss, turning linear ODEs with constant coefficients into algebraic equations in ss. You solve the algebra, then take the inverse Laplace transform to get the answer in the time domain.

Laplace transforms also handle discontinuous and impulsive inputs (step functions, Dirac deltas) elegantly, which makes them indispensable in control theory, signal processing, and electrical engineering.

How to Compute Laplace Transforms

Basic Transform Pairs

Memorize the core table:

f(t)f(t)F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}
111s\dfrac{1}{s}
tnt^nn!sn+1\dfrac{n!}{s^{n+1}}
eate^{at}1s−a\dfrac{1}{s - a}
sin⁡(ωt)\sin(\omega t)ωs2+ω2\dfrac{\omega}{s^2 + \omega^2}
cos⁡(ωt)\cos(\omega t)ss2+ω2\dfrac{s}{s^2 + \omega^2}
u(t−a)u(t - a) (step)e−ass\dfrac{e^{-as}}{s}
δ(t−a)\delta(t - a)e−ase^{-as}

Key Properties

Linearity:

L{af(t)+bg(t)}=aF(s)+bG(s)\mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)

First Shifting (s-shift):

L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at} f(t)\} = F(s - a)

This is how eatsin⁡(ωt)→ω(s−a)2+ω2e^{at}\sin(\omega t) \to \frac{\omega}{(s-a)^2 + \omega^2}.

Differentiation in the tt-domain:

L{f′(t)}=sF(s)−f(0)\mathcal{L}\{f'(t)\} = sF(s) - f(0)

L{f′′(t)}=s2F(s)−sf(0)−f′(0)\mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0) - f'(0)

This is what converts ODEs into algebra: derivatives become polynomials in ss multiplied by F(s)F(s), with initial conditions baked in.

Multiplication by tt:

L{tf(t)}=−F′(s)\mathcal{L}\{t f(t)\} = -F'(s)

Inverse Laplace Transform

Given F(s)F(s), find f(t)f(t) such that L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s). Standard techniques:

  1. Partial fractions: decompose F(s)F(s) into simple rational pieces matching the table.
  2. Completing the square: for 1s2+bs+c\frac{1}{s^2 + bs + c} shapes, rewrite as 1(s−a)2+ω2\frac{1}{(s - a)^2 + \omega^2} to match the shifted sine table entry.
  3. Look up and combine using linearity.

Solving ODEs with Laplace

For y′′+3y′+2y=e−ty'' + 3y' + 2y = e^{-t}, y(0)=0,y′(0)=1y(0) = 0, y'(0) = 1:

  1. Apply Laplace: s2Y−s⋅0−1+3(sY−0)+2Y=1s+1s^2 Y - s \cdot 0 - 1 + 3(sY - 0) + 2Y = \frac{1}{s+1}
  2. Solve for YY: Y(s2+3s+2)=1+1s+1Y(s^2 + 3s + 2) = 1 + \frac{1}{s+1}, so Y=s+2(s+1)(s2+3s+2)=1(s+1)2Y = \frac{s + 2}{(s+1)(s^2+3s+2)} = \frac{1}{(s+1)^2} (after simplification).
  3. Invert: y(t)=te−ty(t) = t e^{-t}.

Clean and mechanical — the same problem with variation of parameters takes twice the work.

Common Mistakes to Avoid

  • Forgetting initial conditions: L{f′}=sF(s)−f(0)\mathcal{L}\{f'\} = sF(s) - f(0). Skipping f(0)f(0) is the single most common error.
  • Wrong sign in s-shift: L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s - a), not F(s+a)F(s + a). Sign matters.
  • Mishandling discontinuities: For step inputs, use the unit-step function u(t−a)u(t-a) and the time-shift theorem L{u(t−a)f(t−a)}=e−asF(s)\mathcal{L}\{u(t-a)f(t-a)\} = e^{-as}F(s).
  • Inverse transform without partial fractions: 1(s−1)(s+2)\frac{1}{(s-1)(s+2)} does not invert directly — decompose first.
  • Confusing F(s)F(s) with L−1{F}\mathcal{L}^{-1}\{F\}: F(s)F(s) is the transform, f(t)f(t) is the original. Always end ODE problems back in the time domain.

Examples

Step 1: Use the rule L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at} f(t)\} = F(s - a) with f(t)=tf(t) = t, a=2a = 2
Step 2: L{t}=1/s2\mathcal{L}\{t\} = 1/s^2, so F(s)=1/s2F(s) = 1/s^2
Step 3: Apply s-shift: L{te2t}=1/(s−2)2\mathcal{L}\{t e^{2t}\} = 1/(s-2)^2
Answer: 1(s−2)2\dfrac{1}{(s - 2)^2}

Step 1: Compare to table: L{sin⁡(ωt)}=ω/(s2+ω2)\mathcal{L}\{\sin(\omega t)\} = \omega / (s^2 + \omega^2)
Step 2: Here ω2=4\omega^2 = 4 so ω=2\omega = 2
Step 3: Adjust constants: 1s2+4=12⋅2s2+4\frac{1}{s^2+4} = \frac{1}{2} \cdot \frac{2}{s^2+4}
Step 4: Therefore L−1{1/(s2+4)}=12sin⁡(2t)\mathcal{L}^{-1}\{1/(s^2+4)\} = \frac{1}{2}\sin(2t)
Answer: 12sin⁡(2t)\dfrac{1}{2}\sin(2t)

Step 1: Partial fractions: s(s−1)(s+2)=As−1+Bs+2\frac{s}{(s-1)(s+2)} = \frac{A}{s-1} + \frac{B}{s+2}
Step 2: Multiply out: s=A(s+2)+B(s−1)s = A(s+2) + B(s-1)
Step 3: Set s=1s = 1: 1=3A1 = 3A, so A=1/3A = 1/3
Step 4: Set s=−2s = -2: −2=−3B-2 = -3B, so B=2/3B = 2/3
Step 5: Invert each piece: 13et+23e−2t\frac{1}{3}e^t + \frac{2}{3}e^{-2t}
Answer: 13et+23e−2t\dfrac{1}{3}e^t + \dfrac{2}{3}e^{-2t}

Frequently Asked Questions

The Laplace transform exists when the integral ∫₀^∞ e^(-st)f(t) dt converges. This typically requires f to grow no faster than exponentially as t → ∞, and Re(s) to exceed the function's exponential order.

The Laplace transform integrates over [0, ∞) with kernel e^(-st) where s is complex; it handles initial-value problems and exponentially growing inputs. The Fourier transform integrates over (-∞, ∞) with kernel e^(-iωt); it handles steady-state frequency content of functions that decay at infinity.

Because ℒ{f'} = sF(s) - f(0), differentiation in t becomes multiplication by s in the s-domain. A linear ODE with constant coefficients becomes a polynomial equation in s, which you solve algebraically.

For rational F(s) with degree of numerator less than degree of denominator, yes — using partial fractions and the standard table. For non-rational F(s), the inverse may require contour integration (Bromwich integral) or have no closed form.

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