Examples of Laplace Transform
Sources:
- B. P. Lathi & Roger Green. (2018). Chapter 4: The Laplace Transform. Signal Processing and Linear Systems (3rd ed., pp. 338-345). Oxford University Press.
Heaviside’s Cover-up Method
My solutions involve Heaviside’s Cover-up Method, so it's heloful to understand it beforehead.
The cover-up method was introduced by Oliver Heaviside as a fast way to do a decomposition into partial fractions. This is an essential step in using the Laplace transform to solve differential equations, and this was more or less Heaviside's original motivation.
The cover-up method can be used to make a partial fractions decomposition of a rational function \(\frac{p(x)}{q(x)}\) whenever the denominator can be factored into distinct linear factors. We first show how the method works on a simple example, and then show why it works.
Example 1
Decompose \(\frac{x-7}{(x-1)(x+2)}\) into partial fractions.
Solution:
We know the answer will have the form \[ \frac{x-7}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} \text {. } \]
To determine \(A\) by the cover-up method, on the left-hand side we mentally remove (or cover up with a finger) the factor \(x-1\) associated with \(A\), and substitute \(x=1\) into what's left; this gives \(A\) : \[ \left.\frac{x-7}{(x+2)}\right|_{x=1}=\frac{1-7}{1+2}=-2=A . \]
Similarly, \(B\) is found by covering up the factor \(x+2\) on the left, and substituting \(x=-2\) into what's left. This gives \[ \left.\frac{x-7}{(x-1)}\right|_{x=-2}=\frac{-2-7}{-2-1}=3=B . \]
Thus, our answer is \[ \frac{x-7}{(x-1)(x+2)}=\frac{-2}{x-1}+\frac{3}{x+2} \text {. } \] Why does the method work? The reason is simple. The "right" way to determine \(A\) from equation (1) would be to multiply both sides by \((x-1)\); this would give \[ \frac{x-7}{(x+2)}=A+\frac{B}{x+2}(x-1) \text {. } \]
Now if we substitute \(x=1\), what we get is exactly equation (2), since the term on the right disappears. The cover-up method therefore is just any easy way of doing the calculation without going to the fuss of writing (4) — it's unnecessary to write the term containing \(B\) since it will become 0 .
In general, if the denominator of the rational function factors into the product of distinct linear factors: \[ \frac{p(x)}{\left(x-a_1\right)\left(x-a_2\right) \cdots\left(x-a_r\right)}=\frac{A_1}{x-a_1}+\ldots+\frac{A_r}{x-a_r}, \quad a_i \neq a_j, \] then \(A_i\) is found by covering up the factor \(x-a_i\) on the left, and setting \(x=a_i\) in the rest of the expression.
Examples for inverse Laplace Transform
Find the inverse unilateral Laplace transforms of (a) \(\frac{7 s-6}{s^2-s-6}\)
(a)
\[ X(s)=\frac{7 s-6}{(s+2)(s-3)}=\frac{k_1}{s+2}+\frac{k_2}{s-3} \]
To determine \(k_1\), corresponding to the term \((s+2)\), we cover up the term \((s+2)\) in \(X(s)\) and substitute \(s=-2\) (the value of \(s\) that makes \(s+2=0\) ) in the remaining expression: \[ k_1=\left.\frac{7 s-6}{(s+2)(s-3)}\right|_{s=-2}=\frac{-14-6}{-2-3}=4 \]
Similarly, to determine \(k_2\) corresponding to the term \((s-3)\), we cover up the term \((s-3)\) in \(X(s)\) and substitute \(s=3\) in the remaining expression \[ k_2=\left.\frac{7 s-6}{(s+2)(s-3)}\right|_{s=3}=\frac{21-6}{3+2}=3 \] Therefore, \[ X(s)=\frac{7 s-6}{(s+2)(s-3)}=\frac{4}{s+2}+\frac{3}{s-3} \]