3
$\begingroup$

Find Laplace Transform of:$$F(s)=\frac{1}{s^4(s^2+1)}$$

It was the Bonus point question in my exam. I solved it with this Lemma :

Let $F(s)=\mathcal{L}\{f(t)\}$, we have $\frac{F(s)}{s}=\mathcal{L}\{\int_o^tf(x)dx\}$ and I evaluate 4 integrals and I get $\frac{t^3}{6}+\sin(t)-t$,correctly.

Here is my Question:

1.Is There any other method to evaluate it?(By the way I could solve it by using partial fractions. and convolution.)

2.In such questions that We can find Inverse Laplace Transform directly wtih respect to $t$ parameter or Solve it with Convolution. which method is better and worth more mathematically ? I mean when we use Convolution We have integral in our final answer.(I think writing the answer directly It has only variable $t$ is more propriety and better in math.)

Thanks in advance!

$\endgroup$
8
  • $\begingroup$ Covnolution theorem gives the same answer of course $\endgroup$ Commented Apr 27, 2020 at 0:22
  • $\begingroup$ @Aryadeva I dont Understand the answer contains integral $\endgroup$
    – Aligator
    Commented Apr 27, 2020 at 0:25
  • $\begingroup$ You can evaluate the integral of course $\endgroup$ Commented Apr 27, 2020 at 0:26
  • $\begingroup$ Can you please show it to me? then why text books doesnt evaluate integral at this point? Maybe it is not elementary? $\endgroup$
    – Aligator
    Commented Apr 27, 2020 at 0:29
  • $\begingroup$ It's of course elementary and gives the same answer as yours $\endgroup$ Commented Apr 27, 2020 at 0:40

4 Answers 4

2
$\begingroup$

If you want to find the inverse Laplace transform of $\frac{1}{s^4(s^2+1)}$ you just have to notice that there is a pole of order $4$ at the origin and simple poles at $\pm i$, so for some constants $$ \frac{1}{s^4(s^2+1)} = \frac{A}{s^4}+\frac{B}{s^3}+\frac{C}{s^2}+\frac{D}{s}+\frac{E}{s-i}+\frac{F}{s+i} $$ holds and it is very simple to compute $\mathcal{L}^{-1}$ of any term in the RHS. In order to find such constants you may notice that $$ \frac{1}{s^4(s^2+1)}=\frac{1}{s^2}\cdot\left(\frac{1}{s^2}-\frac{1}{s^2+1}\right)=\frac{1}{s^4}-\frac{1}{s^2}+\frac{1}{s^2+1} $$ so $(\mathcal{L}^{-1} F)(t)$ is given by $\frac{t^3}{6}-t+\sin(t)$.

$\endgroup$
3
  • $\begingroup$ Did you realized that $$\frac{1}{s^2(s^2+1)}=\frac{1}{s^2}-\frac{1}{s^2+1}$$observational? $\endgroup$
    – Aligator
    Commented Apr 27, 2020 at 6:33
  • 2
    $\begingroup$ @Soheil: just an instance of $\frac{1}{m(m+1)}=\frac{1}{m}-\frac{1}{m+1}$, reminiscent of Mengoli's series. $\endgroup$ Commented Apr 27, 2020 at 6:37
  • 1
    $\begingroup$ It is amazing you used this trick twice! $\endgroup$
    – Aligator
    Commented Apr 27, 2020 at 6:48
1
$\begingroup$

By the theorem of convolution you have : $$f(t)=I=\dfrac {1}{3!}\int_0^t\sin(t-\tau) \tau ^3 d\tau$$ You can evaluate this integral of course. Integrate by part.You will get the same answer as yours.

A first integration $$-6I=\int_0^t\sin(\tau-t) \tau ^3 d\tau$$ $$-6I=-\cos(\tau-t)\tau^3\bigg |_0^t+ \int_0^t-\cos(\tau-t)3\tau ^2 d\tau$$ $$-6I=-t^3+ 3\int_0^t\cos(\tau-t)\tau ^2 d\tau$$ $$-6I=-t^3- 6\int_0^t\sin(\tau-t)\tau d\tau$$ $$-6I=-t^3+ 6t+6\int_0^t\cos(\tau-t) d\tau$$ $$-6I=-t^3+ 6t-6\sin(t)$$ Finally : $$ \boxed {f(t)=\dfrac {t^3}6-t+\sin(t)}$$

$\endgroup$
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\mrm{f}\pars{t}\right\vert_{\, t\ >\ 0} & = \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {1 \over s^{4}\pars{s^{2} +1}}\,\expo{ts}\,{\dd s \over 2\pi\ic} \\[3mm]\ & = 2\pi\ic\bracks{{\expo{t\ic} \over 2\ic} + {\expo{t\pars{-\ic}} \over -2\ic}}\,{1 \over 2\pi\ic} + 2\pi\ic\,{1 \over 3!}\lim_{s \to 0} \partiald[3]{}{s}\pars{{\expo{ts} \over s^{2} +1}\,{1 \over 2\pi\ic}} \\[3mm]\ & = \sin\pars{t} + {1 \over 6}\lim_{s \to 0} \partiald[3]{}{s}\braces{\bracks{1 + ts + {1 \over 2}\pars{ts}^{2} + {1 \over 6}\pars{ts}^{3}}\pars{1 - s^{2}}} \\[3mm]\ & = \sin\pars{t} + {1 \over 6}\lim_{s \to 0} \partiald[3]{}{s}\pars{-ts^{3} + {1 \over 6}\,t^{3}s^{3}} \\[3mm] & = \bbx{\sin\pars{t} + {1 \over 6}\,t^{3} - t} \end{align}

$\endgroup$
1
$\begingroup$

An easy approach to split the fraction $F(s)=\frac{1}{s^4(s^2+1)}$ by partial fraction: $$\frac{1}{s^4(s^2+1)} = \frac{A}{s^4}+\frac{B}{s^3}+\frac{C}{s^2}+\frac{D}{s}+\frac{Es+F}{s^2+1}$$ Because $F(s)$ is even function and $F(s)=F(-s)$ so $B=D=E=0$: $$\frac{1}{s^4(s^2+1)} = \frac{A}{s^4}+\frac{C}{s^2}+\frac{F}{s^2+1}$$ $A=1,C=-1,F=1$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .