Unilateral Laplace Transform

1Definition and Region of Convergence

This lecture explains how the unilateral Laplace transform converts differentiation into algebraic operations in $s$s while retaining the initial values of a linear differential equation.

Let $s=\sigma +i\omega \in \mathbb{C}$s=\sigma+i\omega\in\mathbb C. The unilateral Laplace transform of $f:[0,\mathrm{\infty })\to \mathbb{C}$f:[0,\infty)\to\mathbb C is

for those $s$s at which the integral converges. Suppose that $f$f is piecewise continuous on every finite interval and that, for some $M>0$M>0, $a\in \mathbb{R}$a\in\mathbb R, and $T\text{[PARSE ERROR: Undefined("Command("ge")")]}0$T\ge0,

Then $f$f is said to be of exponential order $a$a, and the transform converges absolutely for $\mathrm{Re}s>a$\operatorname{Re}s>a. A transform must be specified together with its region of convergence (ROC), the set of $s$s at which it converges. For a unilateral transform the ROC is normally a right half-plane. The inequality $\mathrm{Re}s>a$\operatorname{Re}s>a is the range guaranteed by the stated hypothesis; the actual ROC may be larger.

2Differentiation Rule and Initial Values

Assume that $f$f is absolutely continuous on every finite interval and that $f$f and $f^{\prime }$f' have Laplace transforms on a common right half-plane. Integration by parts, together with the vanishing boundary term at infinity, gives

where $f(0+)$f(0+) is the right-hand limit. If the required higher derivatives satisfy the same conditions, then

Thus the differentiation rule does not merely replace differentiation by multiplication by $s$s; it retains initial values as boundary terms.

Transforming the initial-value problem

gives $(s+1)F\left(s\right)=1$(s+1)F(s)=1 and hence $F\left(s\right)=1/(s+1)$F(s)=1/(s+1). Direct integration also gives

The uniqueness theorem therefore yields $f\left(t\right)=e^{-t}$f(t)=e^{-t}.

3Uniqueness and Inversion

If two piecewise-continuous functions of exponential order have the same Laplace transform on a common right half-plane, then they agree at every point where both are continuous. In this sense the inverse Laplace transform $\text{[PARSE ERROR: Undefined("Command("mathcal")")]}{L}^{-1}$\mathcal L^{-1} is unique. Computations normally use transform pairs, partial fractions, shift rules, and convolution.

Uniqueness can be reduced to uniqueness of the Fourier transform. Fix $\sigma$\sigma in the common half-plane and extend $e^{-\sigma t}\left(f\right(t)-g(t\left)\right)$e^{-\sigma t}(f(t)-g(t)) by zero for $t<0$t<0. The Fourier transform of this $L^1$L^1 function vanishes for every $\omega$\omega. Fourier uniqueness gives $f=g$f=g almost everywhere, and piecewise continuity then gives equality at continuity points.

If $F$F satisfies suitable analyticity and growth conditions and the vertical line $\mathrm{Re}s=\gamma$\operatorname{Re}s=\gamma lies inside the ROC and to the right of every singularity, the Bromwich integral

represents the inverse transform at continuity points. Its use requires convergence hypotheses; it cannot be applied formally to an arbitrary complex function.

4Relation to the Fourier Transform

With $s=\sigma +i\omega$s=\sigma+i\omega, the value $\text{[PARSE ERROR: Undefined("Command("mathcal")")]}L\left\{f\right\}(\sigma +i\omega )$\mathcal L\{f\}(\sigma+i\omega) is the Fourier transform of the damped signal $e^{-\sigma t}f\left(t\right)$e^{-\sigma t}f(t) extended by zero for $t<0$t<0. Choosing sufficiently large $\sigma$\sigma suppresses exponential growth. Identification with the Fourier transform on $\sigma =0$\sigma=0 is valid only when the imaginary axis belongs to the ROC and the Fourier integral converges.

5Poles, the ROC, and Control

The function $1/(s+1)$1/(s+1) has a pole at $s=-1$s=-1, and the ROC of $e^{-t}$e^{-t} defined for $t\text{[PARSE ERROR: Undefined("Command("ge")")]}0$t\ge0 is $\mathrm{Re}s>-1$\operatorname{Re}s>-1. Because the unilateral transform fixes the time domain to $t\text{[PARSE ERROR: Undefined("Command("ge")")]}0$t\ge0, the transform uniquely determines a function within the class stated above. The ROC accompanies the transform expression as its convergence and analyticity domain, but it is not independent information that selects a time direction. Using different ROCs for the same rational expression to distinguish causal and anticausal signals belongs to the bilateral Laplace transform, which also treats $t<0$t<0.

For a zero-initial-state linear time-invariant system, the ratio $G\left(s\right)=Y\left(s\right)/U\left(s\right)$G(s)=Y(s)/U(s) of output to input is the transfer function. Control engineering uses the Laplace transform to convert differential equations into algebraic equations and to analyze response and stability through poles. Nonzero initial values appear separately as boundary terms in the differentiation rule, so a transfer function alone does not describe the initial-state response.

Transform contract

• Treat a transform expression together with its ROC.
• Apply the differentiation rule on a common half-plane where the transforms of the function and its derivatives exist.
• Invert by using known transform pairs and uniqueness, or an inversion formula whose hypotheses have been verified.