Laplace Transforms

The Laplace transform turns a function of time into a function of a complex variable s:

F(s) = L{f(t)} = ∫₀^∞ e^−st f(t) dt

It turns differential equations into algebraic ones, which is why it is the main tool for circuits and control systems.

Table of transforms

Properties

• Linearity: L{af + bg} = aF + bG.
• First shifting (frequency shift): L{e^atf(t)} = F(s − a). So L{e^−2t sin 3t} = 3/((s + 2)² + 9).
• Second shifting (time shift): L{f(t − a)u(t − a)} = e^−asF(s).
• Derivatives: L{f′} = sF(s) − f(0); L{f″} = s²F(s) − sf(0) − f′(0).
• Integral: L{∫₀ᵗ f(τ) dτ} = F(s)/s.
• Multiplication by t: L{t f(t)} = −dF/ds.
• Convolution: L{f ∗ g} = F(s)G(s), with (f ∗ g)(t) = ∫₀ᵗ f(τ)g(t − τ) dτ.
• Initial value theorem: f(0⁺) = lim_s→∞ sF(s). Final value theorem: f(∞) = lim_s→0 sF(s), valid only if all poles of sF(s) are in the left half-plane.

Inverse transforms

Split F(s) into partial fractions and use the table. For example,

(s + 3)/((s + 1)(s + 2)) = 2/(s + 1) − 1/(s + 2), so f(t) = 2e^−t − e^−2t.

Solving differential equations

Transform both sides, put in the initial conditions, solve for Y(s), and invert.

Example. y′ + 2y = 4, y(0) = 1: sY − 1 + 2Y = 4/s gives Y = (s + 4)/(s(s + 2)) = 2/s − 1/(s + 2), so y = 2 − e^−2t.

Worked example

Check L{t e^−t} = 1/(s + 1)² numerically at s = 2, check the partial fraction expansion above, and the solution of y′ + 2y = 4.

import math

def laplace(f, s, T=60.0, n=600000):
    h = T / n
    return sum(math.exp(-s * (i + 0.5) * h) * f((i + 0.5) * h) for i in range(n)) * h

print(f"L{{t e^-t}} at s = 2: numerical {laplace(lambda t: t * math.exp(-t), 2):.6f}, formula {1 / 9:.6f}")
F = lambda s: (s + 3) / ((s + 1) * (s + 2))
P = lambda s: 2 / (s + 1) - 1 / (s + 2)
print("partial fractions agree:", all(abs(F(s) - P(s)) < 1e-12 for s in [0.5, 1, 3, 10]))
y = lambda t: 2 - math.exp(-2 * t)
dy = lambda t, h=1e-6: (y(t + h) - y(t - h)) / (2 * h)
print("y(0) =", y(0), " y' + 2y at t = 0.7:", round(dy(0.7) + 2 * y(0.7), 6))

L{t e^-t} at s = 2: numerical 0.111111, formula 0.111111
partial fractions agree: True
y(0) = 1.0  y' + 2y at t = 0.7: 4.0

Practice questions