Use the complex inversion formula to calculate the inverse Laplace transform $f(t)$ of the following Laplace transform: $$F_L (s) = \frac{1}{(s+2)(s^2 +4)}.$$ When the region of convergence is: \begin{align}(1)& \quad Re(s)<-2;\\(2)&\quad -2<Re(s)<0;\\(3)&\quad Re(s)>0.\end{align}


Here is an explanation of the complex inversion formula. Plugging the function into the formula:

$$f(t) = \frac{1}{j2\pi} \int^{\sigma + j \infty}_{\sigma - j \infty} \frac{e^{st}}{(s+2)(s^2+4)}\ ds \tag{1}$$

  • So, how do I need to choose $\sigma$?
  • And how do I evaluate this for each of the three regions?

P. S. I tried to solve this without the complex inversion formula, just to see what the answer should look like. I started out by expanding using partial fractions as: \begin{align} \frac{1}{(s+2)(s^2 +4)}&= \frac{1}{8} \left( \frac{1}{s+2} + \frac{1}{s^2+4} \right)\\ &=\frac{1}{8} \left( \frac{1}{s+2} + \frac{1}{(s+2j)(s-2j)} \right)\\ &=\frac{1}{8} \left( \frac{1}{s+2} + \frac{j}{4(s+2j)} - \frac{j}{4(s-2j)} \right). \end{align}

Looking at a Laplace transform table, $\frac{1}{s-a} \leftrightarrow e^{at},$ so

$$f(t) = \frac{1}{8} \left( e^{-2t} + \frac{j}{4} \left(e^{-2jt} + e^{2jt}\right) \right).$$

  • Is this correct?
  • If so, how can I get to this using the complex inversion formula?

1 Answer 1


In engineering practice, the complex inversion integral is hardly ever used. As an engineer, you will almost exclusively need to invert rational functions, and this can be done by partial fraction expansion and elementary inversions. So first I'll show you how to obtain the inverse Laplace transform by partial fraction expansion, then I'll explain the evaluation of the inversion integral using Cauchy's residue theorem.

You have an error in the partial fraction expansion. Furthermore, you don't need to split up the complex pole pair. I would rewrite the Laplace transform like this:


with $A=\frac18$, $B=-\frac18$, and $C=\frac14$. The terms on the right-hand side of $(1)$ are elementary Laplace transforms. Now you just have to consider the different regions of convergence (ROC):

$$\begin{align}\frac{1}{s+2}&\Longleftrightarrow e^{-2t}u(t),&\quad \text{Re}\{s\}>-2\\\frac{1}{s+2}&\Longleftrightarrow -e^{-2t}u(-t),&\quad \text{Re}\{s\}<-2\\ \frac{s}{s^2+4}&\Longleftrightarrow \cos(2t)u(t),&\quad\text{Re}\{s\}>0\\ \frac{s}{s^2+4}&\Longleftrightarrow -\cos(2t)u(-t),&\quad\text{Re}\{s\}<0\\ \frac{1}{s^2+4}&\Longleftrightarrow \frac12\sin(2t)u(t),&\quad\text{Re}\{s\}>0\\ \frac{1}{s^2+4}&\Longleftrightarrow -\frac12\sin(2t)u(-t),&\quad\text{Re}\{s\}<0 \end{align}$$

So for the ROC $\text{Re}\{s\}<-2$ you get the anti-causal signal $$f(t)=\frac18\left[-e^{-2t}+\cos(2t)-\sin(2t)\right]u(-t)\tag{2}$$

For the ROC $-2<\text{Re}\{s\}<0$ you get the two-sided signal


And, finally, for the ROC $\text{Re}\{s\}>0$ you get the causal signal $$f(t)=\frac18\left[e^{-2t}-\cos(2t)+\sin(2t)\right]u(t)\tag{4}$$

If you need to use the inversion formula then it is very helpful to know Cauchy's residue theorem, which says that

$$\frac{1}{2\pi j}\oint_Cf(s)ds=\sum_kR_k\tag{5}$$

where $f(s)$ is analytic with finitely many poles, $C$ is a positively oriented closed curve, and $R_k$ are the residues of the poles inside $C$. It can be shown that the inversion integral equals a contour integral if the curve $C$ is chosen appropriately:

$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds= \frac{1}{2\pi j}\oint_CF(s)e^{st}ds\tag{6}$$

In the case of a rational function $F(s)$ the curve $C$ is chosen as a Bromwich contour, as shown here in Fig.2. The straight line part of the curve is the actual integration path we're interested in. The contribution from the circular part of $C$ approaches zero. Depending on the chosen ROC, we have to choose the position of the straight line (i.e., the value of $\sigma)$ differently. For ROC $\text{Re}\{s\}<-2$ (i.e., the anti-causal solution), the straight line is anywhere to the left of the left-most pole, and so the Bromwich contour enclosing all poles is negatively oriented, which results in a sign change:

$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds=-\sum_kR_k,\quad\sigma<-2,\quad t<0\tag{7}$$

where $R_k$ are the residues corresponding to the poles of the function $f(s)=F(s)e^{st}$. For ROC $\text{Re}\{s\}>0$ (i.e., the causal solution), the straight line is anywhere to the right of the right-most pole, and the Bromwich contour enclosing all poles is positively oriented. Consequently, we have

$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds=\sum_kR_k,\quad\sigma>0,\quad t>0\tag{8}$$

For the two-side solution with ROC $-2<\text{Re}\{s\}<0$ we need to choose two curves with the straight line inside the ROC. One encloses the pole to its left at $s=-2$, so the curve is positively oriented, and the other one encloses the two poles at $s=2j$ and $s=-2j$ to the right of the straight line, so it is negatively oriented (which adds a negative sign to the corresponding residues). Let $R_1$ be the residue corresponding to the pole at $s=-2$, and let $R_2$ and $R_3$ be the residues of the two poles at $\pm 2j$, respectively. The inversion integral is then given by:

$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds=\begin{cases}R_1,&t>0\\-R_2-R_3,&t<0\end{cases},\quad-2<\sigma<0\tag{9}$$

What remains is the computation of the residues. The residue at pole $p_k$ is given by $$R_k=\lim_{s\rightarrow p_k}(s-p_k)f(s)\tag{10}$$

With $f(s)=F(s)e^{st}$ we get for $p_1=-2$

$$R_1=\lim_{s\rightarrow -2}(s+2)F(s)e^{st}=\lim_{s\rightarrow -2}\frac{e^{st}}{s^2+4}=\frac{e^{-2t}}{8}\tag{11}$$

In a similar manner the other residues are obtained:

$$\begin{align}R_2&=-\frac{e^{2jt}}{8}\frac{1+j}{2}\\ R_3&=-\frac{e^{-2jt}}{8}\frac{1-j}{2}\end{align}\tag{12}$$

Now $(7)-(9)$ can be evaluated, and the results are of course the same as $(2)-(4)$.

  • $\begingroup$ Thank you very much. But the question is specifically asking to use the complex inversion formula. So if I have $$f(t) = \frac{1}{j2 \pi} \int^{\sigma + j \infty}_{\sigma - j \infty} \frac{e^{st}}{(s+2)(s^2+4)} \ ds,$$ for instance for the region of absolute convergence $Re\{ s \} < -2,$ how exactly do I choose $\sigma$? The thing we start with is real, so the answer we end up with should be real as well. $\endgroup$
    – Merin
    Commented May 10, 2016 at 21:37
  • 1
    $\begingroup$ @Merin: I've added the evaluation of the complex integral. $\endgroup$
    – Matt L.
    Commented May 11, 2016 at 8:22

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