Prerequisite:

Definition and Region of Convergence

Definition. The Laplace transform of a function $f: [0, \infty) \to \mathbb{R}$ is

$$\mathcal{L}{f(t)}(s) = F(s) = \int_0^\infty f(t),e^{-st},dt$$

provided the integral converges. Here $s \in \mathbb{C}$ is a complex frequency variable.

Region of Convergence (ROC). The integral converges absolutely for all $s$ with $\operatorname{Re}(s) > \sigma_c$, where $\sigma_c$ is the abscissa of convergence. For example, $f(t) = e^{at}$ gives $F(s) = 1/(s-a)$ with ROC $\operatorname{Re}(s) > a$.

Geometrically, the ROC is a right half-plane in the complex $s$-plane:

        Im(s)
          |
          |   ROC: Re(s) > sigma_c
          |   /
          |  /
----------+--/----------------------> Re(s)
          | sigma_c
          |
          |
The Laplace transform is analytic in its ROC.

Transforms of Standard Functions

$f(t)$ $F(s) = \mathcal{L}{f}$ ROC
$1$ $\dfrac{1}{s}$ $\operatorname{Re}(s) > 0$
$t^n$ $\dfrac{n!}{s^{n+1}}$ $\operatorname{Re}(s) > 0$
$e^{at}$ $\dfrac{1}{s - a}$ $\operatorname{Re}(s) > a$
$\sin(\omega t)$ $\dfrac{\omega}{s^2 + \omega^2}$ $\operatorname{Re}(s) > 0$
$\cos(\omega t)$ $\dfrac{s}{s^2 + \omega^2}$ $\operatorname{Re}(s) > 0$
$u(t)$ (unit step) $\dfrac{1}{s}$ $\operatorname{Re}(s) > 0$
$\delta(t)$ (Dirac delta) $1$ all $s$

The $\sin$ and $\cos$ entries follow from computing $\mathcal{L}{e^{i\omega t}} = 1/(s - i\omega)$ and taking real and imaginary parts.

Properties

Linearity. $\mathcal{L}{af + bg} = aF(s) + bG(s)$.

Time Shift. If $f$ is shifted by $a \geq 0$:

$$\mathcal{L}{f(t - a),u(t - a)} = e^{-as},F(s).$$

Frequency Shift. Multiplying by an exponential shifts $s$:

$$\mathcal{L}{e^{at},f(t)} = F(s - a).$$

Convolution Theorem. The Laplace transform converts convolution to multiplication:

$$\mathcal{L}{(f * g)(t)} = F(s),G(s)$$

where $(f * g)(t) = \int_0^t f(\tau),g(t - \tau),d\tau$.

Proof sketch. Substitute the definition of convolution into the Laplace integral, interchange the order of integration (justified by absolute convergence), and collect the exponential factors to get $F(s)G(s)$.

Differentiation Rules

Theorem. If $f$ is differentiable and $\mathcal{L}{f} = F(s)$, then

$$\mathcal{L}{f'(t)} = s,F(s) - f(0).$$

Proof. Integrate by parts: $\int_0^\infty f'(t)e^{-st},dt = [f(t)e^{-st}]_0^\infty + s\int_0^\infty f(t)e^{-st},dt = -f(0) + sF(s)$.

Applying the rule repeatedly:

$$\mathcal{L}{f''(t)} = s^2 F(s) - s,f(0) - f'(0).$$

This is the key property that makes Laplace transforms useful for ODEs: differentiation becomes multiplication by $s$, reducing an ODE to an algebraic equation.

Solving ODEs with the Laplace Transform

Strategy. Given $ay'' + by' + cy = g(t)$ with initial conditions $y(0) = y_0$, $y'(0) = v_0$:

  1. Apply $\mathcal{L}$ to both sides.
  2. Use the differentiation rules to express $\mathcal{L}{y''}$ and $\mathcal{L}{y'}$ in terms of $Y(s) = \mathcal{L}{y}$.
  3. Solve algebraically for $Y(s)$.
  4. Invert to find $y(t) = \mathcal{L}^{-1}{Y(s)}$.

Example. Solve $y'' + y = 0$, $y(0) = 1$, $y'(0) = 0$.

Transform: $s^2 Y - s\cdot 1 - 0 + Y = 0 \implies (s^2 + 1)Y = s \implies Y(s) = \frac{s}{s^2 + 1}$.

Inverting: $y(t) = \cos(t)$.

Partial Fractions for Inverse Laplace

When $Y(s)$ is a ratio of polynomials, decompose into partial fractions before inverting. For example:

$$Y(s) = \frac{2s + 3}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}.$$

Solving: $A = 1$, $B = 1$, so $y(t) = e^{-t} + e^{-2t}$.

Complex poles contribute damped sinusoids: $\frac{1}{(s+a)^2 + \omega^2}$ inverts to $\frac{1}{\omega}e^{-at}\sin(\omega t)$.

Transfer Functions and Control Theory

In control theory, a linear time-invariant (LTI) system relates input $x(t)$ to output $y(t)$ via a differential equation. Taking the Laplace transform (assuming zero initial conditions) gives

$$Y(s) = H(s),X(s)$$

where the transfer function is $H(s) = Y(s)/X(s)$.

The transfer function is a rational function $H(s) = N(s)/D(s)$. Its poles are the roots of $D(s)$ and its zeros are the roots of $N(s)$.

Stability Condition. An LTI system is bounded-input bounded-output (BIBO) stable if and only if all poles of $H(s)$ lie in the open left half-plane $\operatorname{Re}(s) < 0$.

        Im(s)
          |
    X  X  |  stable
    X  X  |  region
----------+-----------> Re(s)
    X  X  |
    X  X  |
          |
Poles (X) in left half-plane => stable system.
Poles on imaginary axis => marginally stable.
Poles in right half-plane => unstable.

The step response $y(t) = \mathcal{L}^{-1}{H(s)/s}$ reveals how the system responds to a sudden input - fundamental in control design (PID controllers, Bode plots, Nyquist criterion).

Initial and Final Value Theorems

Initial Value Theorem. If $f$ and $f'$ are Laplace-transformable:

$$\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} s,F(s).$$

Final Value Theorem. If $f$ has a finite limit as $t \to \infty$, and all poles of $sF(s)$ are in the left half-plane:

$$\lim_{t \to \infty} f(t) = \lim_{s \to 0} s,F(s).$$

These let you extract limiting behavior directly from $F(s)$ without inverting the transform - useful for checking steady-state values in control systems.

Connection to the Fourier Transform

The Fourier transform is

$$\mathcal{F}{f}(\omega) = \int_{-\infty}^\infty f(t),e^{-i\omega t},dt.$$

For causal functions ($f(t) = 0$ for $t < 0$), evaluating the Laplace transform on the imaginary axis $s = i\omega$ gives the Fourier transform:

$$\mathcal{L}{f}(i\omega) = \int_0^\infty f(t),e^{-i\omega t},dt = \mathcal{F}{f}(\omega).$$

The Laplace transform is thus a generalization: it adds a real damping factor $e^{-\sigma t}$ that ensures convergence even when $f$ grows. The Fourier transform lives on the boundary of the ROC.

This explains why poles in the left half-plane correspond to stable behavior: the Fourier transform converges on the imaginary axis, and a left-half-plane pole at $s = -a + i\omega_0$ contributes a damped oscillation $e^{-at}\cos(\omega_0 t)$ that decays.

Examples

The Z-transform is the discrete analogue of the Laplace transform. For a sequence ${x[n]}$:

$$X(z) = \sum_{n=0}^\infty x[n],z^{-n}.$$

The substitution $z = e^{sT}$ (with $T$ the sampling period) links the two: poles in the left half-plane of the $s$-domain map to poles inside the unit circle $|z| < 1$ in the $z$-domain - the stability condition for discrete-time systems.

In digital signal processing, the transfer function $H(z)$ characterizes filters. The poles and zeros of $H(z)$ on the unit circle determine the frequency response $H(e^{i\omega})$ - the discrete-time analogue of evaluating $H(s)$ on the imaginary axis.

In machine learning, convolution layers implement the convolution theorem: a 1D convolution in time is pointwise multiplication in frequency. The Laplace/Fourier framework justifies why frequency-domain analysis of neural networks (e.g., NTK spectral analysis) works.

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