Laplace Transform

Definition

The Laplace transform of a signal $x(t)$ is defined as follows:

$$X(s)\triangleq {\int }_{-\infty }^{+\infty }x(t)e^{-st}\mathrm{d}t$$

It should be particularly noted that this is a function with the independent variable $s$, where $s$ is the complex variable in the exponent of $e$. The complex variable $s$ can generally be written as $s=\sigma +j\omega$, where $\sigma$ and $\omega$ are its real and imaginary parts, respectively. For convenience, the Laplace transform is often expressed in the form of an operator $\mathcal{L}\{x(t)\}$, and the transformation relationship between $x(t)$ and $X(s)$ is denoted as:

$$x(t)\stackrel{\mathcal{L}}{\leftrightarrow }X(s)$$

When $s=j\omega$, this equation becomes:

$$X(j\omega )={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt$$

This is the Fourier transform of $x(t)$, i.e.,

$$X(s){|}_{s=j\omega }=\mathcal{F}\{x(t)\}$$

When the complex variable $s$ is not purely imaginary, the Laplace transform and the Fourier transform also have a direct relationship. To see this, express $s$ in as $s=\sigma +j\omega$, then we have:

$$X(\sigma +j\omega )={\int }_{-\infty }^{+\infty }x(t)e^{-(\sigma +j\omega )t}dt$$

or

$$X(\sigma +j\omega )={\int }_{-\infty }^{+\infty }[x(t)e^{-\sigma t}]e^{-j\omega t}dt$$

We can view the right-hand side as the Fourier transform of $x(t)e^{-\sigma t}$. This means that the Laplace transform of $x(t)$ can be seen as the Fourier transform of $x(t)$ multiplied by a real exponential signal. This real exponential $e^{-\sigma t}$ can be decaying or growing over time, depending on whether $\sigma$ is positive or negative.

Region of Convergence

The Region of Convergence (ROC) defines the range of values for the complex variable $s$ for which the Laplace transform integral converges. Without specifying the ROC, the Laplace transform is incomplete and often non-unique.

The ROC is a region in the complex $s$-plane, and is typically a vertical strip in the $s$-plane, extending from ${\sigma }_1$ to ${\sigma }_2$ (i.e., ${\sigma }_1<\mathcal{R}e(s)<{\sigma }_2$). This can sometimes be a left half-plane ($\mathcal{R}e(s)<{\sigma }_2$), a right half-plane ($\mathcal{R}e(s)>{\sigma }_1$), or the entire $s$-plane.

Pole-Zero Plot

The Laplace transform of a LTI system's impulse response, $h(t)$, results in the transfer function, $H(s)$, expressed as a ratio of two polynomials in the complex variable $s$

$$H(s)=\frac{N(s)}{D(s)}$$

Where:

• $s$ is the complex frequency, $s=\sigma +j\omega$

• $N(s)$ is the numerator polynomial.
• $D(s)$ is the denominator polynomial. Poles and Zeros:
• Zeros: The roots of the numerator polynomial, $N(s)$, are called zeros of the transfer function. When $s$ equals a zero, the transfer function $H(s)$ evaluates to zero. Mathematically, zeros are the values of $s$ where $N(s)=0$.
• Poles: The roots of the denominator polynomial, $D(s)$, are called poles of the transfer function. When $s$ equals a pole, the transfer function $H(s)$ becomes infinite (or undefined). Mathematically, poles are the values of $s$ where $D(s)=0$.

Note

There can not be poles in the region of convergence

Pole-zero Plot: A pole-zero plot is a two-dimensional graph on the complex s-plane where:

• The horizontal axis represents the real part of $s(\sigma )$.
• The vertical axis represents the imaginary part of $s(j\omega )$.
• Poles are marked with an "x".
• Zeros are marked with an "o".

Note

In general, for an LTI system to be stable (BIBO stable), the ROC of its Laplace transform must necessarily include the entire imaginary axis ($s=j\omega$).

And if the system is causal, then the system is stable if and only if all the poles are at the left side of $j\omega$, i.e., all the poles have a negative real part

Inverse Transform

For $H(s)$ with the form $N(s)/D(s)$, we can decompose the complex rational function $H(s)$ into a sum of simpler fractions, each of which has a known inverse Laplace transform.

The inverse transform is of the form $\exp (at)$ if the ROC is to the right of the pole $s=a$, and $-\exp (at)$ if the ROC is to the left of the pole $s=a$.

• Causal Systems: If the system is known to be causal, the ROC is always to the right of the rightmost pole. This dictates that we choose the inverse transform pair $\exp (at)u(t)$ for each pole at $s=a$.
• Anti-Causal Systems: If the system is anti-causal, ROC will be a left half plane and the inverse transform is $-\exp (at)u(-t)$ for a pole at $s=a$.

Example Let's say we have $H(s)=1/((s+1)(s-2))$

1. Partial Fraction Expansion: $H(s)=A/(s+1)+B/(s-2)$ Solving for A and B yields $A=-1/3$ and $B=1/3$, so that $H(s)=(-1/3)/(s+1)+(1/3)/(s-2)$
2. Poles and Possible ROCs:
   • Poles at $s=-1$ and $s=2$. These poles divide the s-plane in to three regions
     • Region 1 : $\mathcal{R}e(s)<-1$.
     • Region 2 : $-1<\mathcal{R}e(s)<2$.
     • Region 3 : $\mathcal{R}e(s)>2$.
3. Inverse Transforms based on ROC:
   • If ROC is Re{s} > 2 (Causal System):
     • $h(t)=(-1/3)e^{(-t)}u(t)+(1/3)e^{(2t)}u(t)$
   • If ROC is -1 < Re{s} < 2 (Non-causal System):
     • $h(t)=(1/3)e^{(-t)}u(t)-(1/3)e^{(2t)}u(-t)$
   • If ROC is Re{s} < -1 (Anti-causal System)
     • $h(t)=(1/3)e^{(-t)}u(-t)-(1/3)e^{(2t)}u(-t)$

Notice how the time-domain signal $h(t)$ changes drastically depending on the chosen ROC.

Properties

Property	Signal	Transform	ROC
Linear	$a{x}_1(t)+b{x}_2(t)$	$a{X}_1(s)+b{X}_2(s)$	At least $R_1\cap R_2$
Time shift	$x(t-t_0)$	$\exp (-s{t}_0)X(s)$	$R$
$s$-plane shift	$\exp (s_{0}t)x(t)$	$X(s-s_0)$	Shift of $R$. i.e. $s\in R^{\prime }⟺s-s_0\in R$
Time scale	$x(at)$	$\frac{1}{\mathrm{abs}(a)}X\left(\frac{s}{a}\right)$	$\frac{s}{a}\in R⟺s\in R^{\prime }$
Conjunction	$x^{\ast }(t)$	$X^{\ast }(s^{\ast })$	$R$
Convolution	$x_1(t)\ast x_2(t)$	$X_1(s)X_2(s)$	At least $R_1\cap R_2$
Time differentiate	$\frac{\mathrm{d}}{\mathrm{d}t}x(t)$	$sX(s)$	At least $R$
$s$-plane differentiate	$-tx(t)$	$\frac{\mathrm{d}}{\mathrm{d}s}X(s)$	$R$
Time integral	${\int }_{-\infty }^{t}x(\tau )\mathrm{d}\tau$	$\frac{1}{s}X(s)$	At least $R\cap \left\{\mathcal{R}e(s)>0\right\}$

Examples

Note

The $F(s)$ in the following table is unilateral Laplace transform. For bilateral case, you should add $u(t)$ terms.

Time-Domain Function $f(t)$	Laplace Transform $F(s)$	ROC
$1$	$\frac{1}{s}$	$\mathcal{R}e(s)>0$
$t$	$\frac{1}{s^2}$	$\mathcal{R}e(s)>0$
$t^n$	$\frac{n!}{s^{n+1}}$	$\mathcal{R}e(s)>0$
$e^{-at}$	$\frac{1}{s+a}$	$\mathcal{R}e(s)>-a$
$t{e}^{-at}$	$\frac{1}{(s+a{)}^2}$	$\mathcal{R}e(s)>-a$
$t^n{e}^{-at}$	$\frac{n!}{(s+a{)}^{n+1}}$	$\mathcal{R}e(s)>0$
$\sin (\omega t)$	$\frac{\omega }{s^2+{\omega }^2}$	$\mathcal{R}e(s)>0$
$\cos (\omega t)$	$\frac{s}{s^2+{\omega }^2}$	$\mathcal{R}e(s)>0$
$e^{-at}\sin (\omega t)$	$\frac{\omega }{(s+a{)}^2+{\omega }^2}$	$\mathcal{R}e(s)>-a$
$e^{-at}\cos (\omega t)$	$\frac{s+a}{(s+a{)}^2+{\omega }^2}$	$\mathcal{R}e(s)>-a$
$u(t-a)$	$\frac{e^{-as}}{s}$	$\mathcal{R}e(s)>0$
$\delta (t-a)$	$e^{-as}$	All $s$-plane