Laplace Transform Calculator

The Laplace transform of f(t) = e^(at) is F(s) = 1/(s - a), with region of convergence Re(s) > a. For example, L{e^(3t)} = 1/(s - 3) for s > 3. This is one of the most commonly used transform pairs in circuit analysis and control theory, where exponential decay or growth models natural system responses.

L{sin(bt)} = b/(s^2 + b^2) with ROC Re(s) > 0. For sin(5t), substitute b = 5 to get F(s) = 5/(s^2 + 25). This pair appears frequently when analyzing oscillatory systems like RLC circuits or mechanical spring-mass systems where the natural response involves sinusoidal functions.

The Laplace transform uses a complex variable s = sigma + j*omega and integrates from 0 to infinity, making it ideal for causal systems and transient analysis. The Fourier transform uses purely imaginary j*omega and integrates from negative infinity to infinity, suited for steady-state frequency analysis. Setting sigma = 0 in the Laplace transform gives the Fourier transform for signals that satisfy convergence conditions.

The ROC defines the values of Re(s) for which the Laplace integral converges to a finite value. For L{e^(at)} = 1/(s - a), the ROC is Re(s) > a. If a = 3, the transform only exists for s > 3. The ROC is critical for determining system stability: a causal and stable system must have its ROC include the j*omega axis (Re(s) = 0).

The Laplace transform converts derivatives into algebraic terms: L{f'(t)} = sF(s) - f(0) and L{f''(t)} = s^2F(s) - sf(0) - f'(0). This turns a differential equation into an algebraic equation in s, which you solve for F(s), then apply the inverse Laplace transform to get f(t). For example, solving y'' + 4y = 0 with y(0) = 1 gives Y(s) = s/(s^2 + 4), so y(t) = cos(2t).

L{t^n} = n!/(s^(n+1)) with ROC Re(s) > 0, where n is a non-negative integer. For t^3: F(s) = 3!/(s^4) = 6/s^4. For t^1 (simply t): F(s) = 1/s^2. This generalizes to the gamma function for non-integer n: L{t^a} = Gamma(a+1)/s^(a+1).

The essential pairs to memorize:

L{1} = 1/s (s > 0)

L{t^n} = n!/s^(n+1) (s > 0)

L{e^(at)} = 1/(s-a) (s > a)

L{sin(bt)} = b/(s^2+b^2) (s > 0)

L{cos(bt)} = s/(s^2+b^2) (s > 0)

L{e^(at)sin(bt)} = b/((s-a)^2+b^2) (s > a)

L{t*e^(at)} = 1/(s-a)^2 (s > a)

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