Laplace transform

The Laplace transform of a function $f$ defined on $[0,\infty )$ is defined as an integral transform of a function $f(t)$ to a function of a complex variable $s=\sigma +j\omega$.

$$\mathcal{L}\{f(t)\}(s)=F(s)={\int }_{0^{-}}^{\infty }{e}^{-st}f(t)dt$$

The inverse Laplace transform reverses this process. The domain and range of $f(t)$ are in the set of real numbers, while the domain and range of $F(s)$ are in the complex plane ($s$-domain, $s$-plane, frequency domain), where $s$ has units of hertz.

The real part of $s$ is only defined for $\text{Re}(s)\ge {\sigma }^{\star }(x)$, where ${\sigma }^{\star }$ is the exponential class bounding growth rate. What this means is that any function $F(s)$ is only defined for values of $s$ that satisfy this property. Otherwise it’s undefined.

This means we can link $s=j\omega$ with the CTFT of a signal. However, if the region of definition of the Laplace transform is purely real (i.e., it doesn’t include the imaginary axis), we cannot obtain the CTFT from the Laplace transform (and the CTFT may not even exist).

Applications

We can use the Laplace transform to compute the output response of a causal LTI CT system with a right-sided input. This is used especially in circuit analysis.

Laplace transforms can be used to solve systems of ODEs, which show up often in linear circuits. In particular, phasor analysis is sometimes insufficient because it only deals with sinusoidal steady state signals. The Laplace transform is not restricted to signals of this type.

Many Laplace transforms in circuit analysis are rational functions. For linear circuits, they’re always rational, where the degree of the denominator is greater than that of the numerator. The poles and zeroes are the critical frequencies of $F(s)$. $F(s)$ always has an equal number of poles and zeroes (this may include $\pm \infty$). The Laplace domain in circuit analysis is also called the frequency domain.

In probability theory

We can use the Laplace transform if our random variable is continuous and non-negative (in the discrete non-negative case, we have the probability generating function).

$$X(s)=E\left(e^{-sX}\right)={\int }_0^{\infty }{f}_X(x)e^{-sx}dx$$

where the characteristic function is:

$${\Phi }_X(\omega )={\int }_{-\infty }^{\infty }{e}^{j\omega x}{f}_X(x)dx$$

and ${\Phi }_X(0)=1$. And where the moment theorem is:

$$E(X^n)=(-1{)}^n\frac{d^n}{d{s}^n}X(s){|}_{s=0}$$

Properties

The Laplace transform is a linear transform:

$$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f(t)\}+\beta \mathcal{L}\{g(t)\}$$

One of the most important properties is that differential equations (with constant coefficients) in terms of $t$ can become algebraic equations in terms of $s$ via the Laplace transform. For example:

$$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)$$ $$\mathcal{L}\{f^{\prime \prime }(t)\}=s^{2}F(s)-sf(0)-f^{\prime }(0)$$ $$\mathcal{L}\{f^n(t)\}=s^{n}F(s)-s^{n-1}f(0)-\cdots -s{f}^{n-2}(0)-f^{n-1}(0)$$

The change of scale theorem is:¹

$$\mathcal{L}\{f(at)\}=\frac{1}{a}F\left(\frac{s}{a}\right)$$

Table of transforms

The basic transforms are as followed. They can be derived with integration by parts.

$$\mathcal{L}\{1\}=\frac{1}{s}$$ $$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}};n\in \mathbb{N}$$

The below also works for $a\in \mathbb{C}$, provided $s>\text{Re}(a)$:

$$\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$$

The transforms of trigonometric functions can be derived with IBP or an equally valid approach using complex analysis. See this page.

$$\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2}$$ $$\mathcal{L}\{\cos kt\}=\frac{s}{s^2+k^2}$$ $$\mathcal{L}\{\cosh kt\}=\frac{s}{s^2-k^2}$$ $$\mathcal{L}\{\sinh kt\}=\frac{k}{s^2-k^2}$$

Special cases

The first and second translation theorems.

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

For functions multiplied by $t^n$. This is especially useful for ODEs.

$$\mathcal{L}\{t^{n}f(t)\}=(-1{)}^n\frac{d^n}{d{s}^n}F(s)$$

For the convolution of two functions:

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

For the integral of a function (i.e., we divide by $s$ in the $s$-domain):

$$\mathcal{L}\{{\int }_0^{t}f(\tau )d\tau \}=\frac{F(s)}{s}$$

For a periodic function where $f(t+T)=f(t)$, when we can integrate over one period:

$$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-sT}}{\int }_0^T{e}^{-st}f(t)dt$$

For the Dirac delta function:

$$\mathcal{L}\{\delta (t-t_0)\}=e^{-s{t}_0}$$

Definition

We must assume $f(t)=0\forall t<0$. The integral converges to a specific value in $\mathbb{C}$ is $f(t)$ is piecewise continuous (it has a finite number of discontinuities over any finite interval), and if $f(t)$ is of exponential order, where there exist two constants $K,a\in \mathbb{R}$ such that $|f(t)|<K{e}^{at}\forall t>0$

Note the above transforms for $e^{at}$ and $1$: this means the transforms aren’t defined for some values in $\mathbb{C}$. The Laplace transform has a uniqueness property (except for points where either function aren’t defined):

$$\mathcal{L}\{f(t)\}=F(s)⟷{\mathcal{L}}^{-1}\{F(s)\}=f(t)$$

In terms of units (especially when dealing with circuits), we have unit seconds ($V\cdot \text{sec}$). This is all fully consistent. As above, $s$ is in units of Hertz.

Limit properties

The initial value theorem states:

$$\underset{t\to 0^{+}}{\lim }f(t)=\underset{s\to \infty }{\lim }sF(s)$$

and is valid if the limits exist and $F(s)$ is a proper rational function (or equivalently continuous at 0).

The final value theorem states:

$$\underset{t\to \infty }{\lim }f(t)=\underset{s\to 0}{\lim }sF(s)$$

which is valid when the poles of $sF(s)$ are in the left half-plane.

Computer-aided

Through the Symbolic Math Toolbox, MATLAB can compute the Laplace transforms of symbolic functions. By default, the output is in terms of s and the independent variable is $x$ or $t$.

syms t, a, s;
f = exp(-a * t);
F = laplace(f, t, s) % function, independent variable, transform variable

For the Dirac delta function and the Heaviside function, we can:

syms t, s;
syms a positive; % only positive a
F_1 = laplace(dirac(t-a),t,s)
F_2 = laplace(heaviside(t-a),t,s)

Footnotes

1. Derivation here. We essentially assume $z=at$ and $dt=\frac{dz}{a}$. ↩