# 4.3 Finding the inverse laplace transform

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## Using transform tables

The inverse Laplace transform, given by

$x\left(t\right)=\frac{1}{2\pi j}{\int }_{\sigma -j\infty }^{\sigma +j\infty }X\left(s\right){e}^{st}ds$

can be found by directly evaluating the above integral. However since this requires a background in the theory of complex variables, which is beyond the scope of this book, we will not be directly evaluating the inverse Laplace transform. Instead, we will utilize the Laplace transform pairs and properties . Consider the following examples:

Example 3.1 Find the inverse Laplace transform of

$X\left(s\right)=\frac{{e}^{-10s}}{s+5}$

By looking at the table of Laplace transform properties we find that multiplication by ${e}^{-10s}$ corresponds to a time delay of 10 sec. Then from the table of Laplace transform pairs , we see that

$\frac{1}{s+5}$

corresponds to the Laplace transform of the exponential signal ${e}^{-5t}u\left(t\right)$ . Therefore we must have

$x\left(t\right)={e}^{-5\left(t-10\right)}u\left(t-10\right)$

Example 3.2 Find the inverse Laplace transform of

$X\left(s\right)=\frac{1}{{\left(s+2\right)}^{2}}$

First we note that from the table of Laplace transform pairs , the Laplace transform of $tu\left(t\right)$ is

$\frac{1}{{s}^{2}}$

Then using the $s$ -shift property in the table of Laplace transform properties gives

$x\left(t\right)=t{e}^{-2t}u\left(t\right)$

Also, the same answer may be arrived at by combining the Laplace transform of ${e}^{-2t}u\left(t\right)$ with the multiplication by $t$ property.

## Partial fraction expansions

Partial fraction expansions are useful when we can express the Laplace transform in the form of a rational function ,

$\begin{array}{cc}\hfill X\left(s\right)& =\frac{{b}_{q}{s}^{q}+{b}_{q-1}{s}^{q-1}+\cdots +{b}_{1}s+{b}_{0}}{{a}_{p}{s}^{p}+{a}_{p-1}{s}^{p-1}+\cdots +{a}_{1}s+{a}_{0}}\hfill \\ & =\frac{B\left(s\right)}{A\left(s\right)}\hfill \end{array}$

A rational function is a ratio of two polynomials. The numerator polynomial $B\left(s\right)$ has order $q$ , i.e., the largest power of $s$ in this polynomial is $q$ , while the denominator polynomial has order $p$ . The partial fraction expansion also requires that the Laplace transform be a proper rational function, which means that $q . Since $B\left(s\right)$ and $A\left(s\right)$ can be factored, we can write

$X\left(s\right)=\frac{\left(s-{\beta }_{1}\right)\left(s-{\beta }_{2}\right)\cdots \left(s-{\beta }_{q}\right)}{\left(s-{\alpha }_{1}\right)\left(s-{\alpha }_{2}\right)\cdots \left(s-{\alpha }_{p}\right)}$

The ${\beta }_{i},i=1,2,...,q$ are the roots of $B\left(s\right)$ , and are called the zeros of $X\left(s\right)$ . The roots of $A\left(s\right)$ , are ${\alpha }_{i},i=1,...,p$ and are called the poles of $X\left(s\right)$ . If we evaluate $X\left(s\right)$ at one of the zeros we get $X\left({\beta }_{i}\right)=0,i=1,...,q$ . Similarly, evaluating $X\left(s\right)$ at a pole gives The actual sign would need to be evaluated at some value of $s$ that is sufficiently close to the pole. $X\left({\alpha }_{i}\right)=±\infty ,i=1,...,p$ . The partial fraction expansion of a Laplace transform will usually involve relatively simple terms whose inverse Laplace transforms can be easily determined from a table of Laplace transforms. We must consider several different cases which depend on whether the poles are distinct.

## Distinct Poles:

When all of the poles are distinct (i.e. ${\alpha }_{i}\ne {\alpha }_{j},i\ne j$ ) then we can use the following partial fraction expansion:

$X\left(s\right)=\frac{{A}_{1}}{s-{\alpha }_{1}}+\frac{{A}_{2}}{s-{\alpha }_{2}}+\cdots +\frac{{A}_{p}}{s-{\alpha }_{p}}$

The coefficients, ${A}_{i},i=1,...,p$ can then be found using the following formula

${A}_{i}={\left(X,\left(s\right),\left(s-{\alpha }_{i}\right)|}_{s={\alpha }_{i}},i=1,...,p$

Equation [link] is easily derived by clearing fractions in [link] . The inverse Fourier transform of $X\left(s\right)$ can then be easily found since each of the terms in the right-hand side of [link] is the Laplace transform of an exponential signal. This method is called the cover up method .

Example 3.3 Find the inverse Laplace transform of

$\begin{array}{cc}\hfill X\left(s\right)& =\frac{2s-10}{{s}^{2}+3s+2}\hfill \\ & =\frac{2s-10}{\left(s+1\right)\left(s+2\right)}\hfill \end{array}$

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