Inverse Laplace Transform problem

In summary, the mistake in the conversation was that the inverse Laplace transform of the second fraction was not calculated correctly, resulting in an incorrect final answer. By correctly applying the inverse Laplace transform and combining it with the inverse Laplace transform of the first fraction, the correct answer was obtained, matching the result given by Mathematica.
  • #1
Dustinsfl
2,281
5
I can't seem to part of an inverse Laplace transform correct.

\begin{align*}
f(t) &= \frac{6}{5}\mathcal{L}^{-1}\bigg\{\frac{1}{s + 2}\bigg\} +
\frac{3}{5}\mathcal{L}^{-1}\bigg\{\frac{3s - 1}
{s^2 + 5s + 11}\bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}
- \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} - \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{s^2\big|_{s\to s + \frac{5}{2}} + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}e^{-5/2t}
\cos\bigg(\frac{\sqrt{19}}{2}t\bigg) - \frac{6}{5\sqrt{19}}
e^{-5/2t}\sin\bigg(\frac{\sqrt{19}}{2}t\bigg)
\end{align*}

When I check this with Mathematica, I should have \(\frac{17}{5\sqrt{19}}e^{-5/2t}\sin\left(\frac{\sqrt{19}}{2}t\right)\).

What am I doing wrong?
 
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  • #2
dwsmith said:
I can't seem to part of an inverse Laplace transform correct.

\begin{align*}
f(t) &= \frac{6}{5}\mathcal{L}^{-1}\bigg\{\frac{1}{s + 2}\bigg\} +
\frac{3}{5}\mathcal{L}^{-1}\bigg\{\frac{3s - 1}
{s^2 + 5s + 11}\bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}
- \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} - \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{s^2\big|_{s\to s + \frac{5}{2}} + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}e^{-5/2t}
\cos\bigg(\frac{\sqrt{19}}{2}t\bigg) - \frac{6}{5\sqrt{19}}
e^{-5/2t}\sin\bigg(\frac{\sqrt{19}}{2}t\bigg)
\end{align*}

When I check this with Mathematica, I should have \(\frac{17}{5\sqrt{19}}e^{-5/2t}\sin\left(\frac{\sqrt{19}}{2}t\right)\).

What am I doing wrong?

Careful!

\[\frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} \neq \frac{9}{5} e^{-5/2 t}\cos\Bigg(\frac{\sqrt{19}}{2}t\Bigg)\]

The result that you want is actually

\[\frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s\color{red}{\big|_{s\to s+\frac{5}{2}}}}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} = \frac{9}{5} e^{-5/2 t}\cos\Bigg(\frac{\sqrt{19}}{2}t\Bigg)\]

Let's start over with that second fraction; I would do things this way:

\[\begin{aligned}\frac{3}{5}\mathcal{L}^{-1}\Bigg\{\frac{3s-1}{s^2+5s+11}\Bigg\} &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{\frac{s-\frac{1}{3}}{\left(s+\frac{5}{2}\right)^2+\frac{19}{4}}\Bigg\}\\ &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{\frac{s+\frac{5}{2} - \frac{17}{6}}{\left(s+\frac{5}{2} \right)^2 + \frac{19}{4} } \Bigg\}\\ &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{ \frac{s+\frac{5}{2}}{\left(s+\frac{5}{2}\right)^2 + \frac{19}{4}}\Bigg\} - \frac{51}{10}\mathcal{L}^{-1}\Bigg\{ \frac{1}{\left(s+\frac{5}{2}\right)^2 + \frac{19}{4}} \Bigg\}\\ &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{ \frac{s\big|_{s\to s+\frac{5}{2}}}{ s^2\big|_{s\to s+\frac{5}{2}} + \frac{19}{4}}\Bigg\} - \frac{51}{5\sqrt{19}}\mathcal{L}^{-1}\Bigg\{ \frac{\frac{\sqrt{19}}{2} }{s^2\big|_{s\to s+ \frac{5}{2}} + \frac{19}{4}} \Bigg\} \\ &= \frac{9}{5} e^{-5/2 t} \cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - \frac{51}{5\sqrt{19}}e^{-5/2 t} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \end{aligned}\]

Putting this together with the inverse Laplace Transform of the first fraction gives us

\[\begin{aligned} f(t) &= \frac{6}{5}e^{-2t} + \frac{9}{5} e^{-5/2 t} \cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - \frac{51}{5\sqrt{19}}e^{-5/2 t} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \\ &= \frac{3}{5}e^{-5/2 t}\Bigg( 2e^{t/2} + 3\cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - \frac{17}{\sqrt{19}} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \Bigg)\\ &= \frac{3}{95}e^{-5/2 t}\Bigg( 38e^{t/2} + 57\cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - 17\sqrt{19} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \Bigg)\end{aligned}\]

Which matches the answer given to me by Mathematica:

MHB_InverseLaplace.png

I hope this clarifies where you made your mistake! (Smile)
 

Related to Inverse Laplace Transform problem

1. What is an inverse Laplace Transform?

The inverse Laplace Transform is a mathematical operation that takes a function in the complex frequency domain and transforms it into a function in the time domain. It is the inverse of the Laplace Transform, which transforms a function in the time domain into the complex frequency domain.

2. Why is the inverse Laplace Transform important?

The inverse Laplace Transform is important in many areas of science and engineering, particularly in the fields of control systems, signal processing, and circuit analysis. It allows us to analyze systems in the time domain, which is often more intuitive and practical than working in the complex frequency domain.

3. What is the process for solving an inverse Laplace Transform problem?

The process for solving an inverse Laplace Transform problem involves first identifying the Laplace Transform of the function in question, then using tables or other methods to find the inverse Laplace Transform. This may involve partial fraction decomposition, completing the square, or other techniques.

4. What are some common applications of the inverse Laplace Transform?

The inverse Laplace Transform has many applications in science and engineering, including analyzing and designing control systems, solving differential equations, and analyzing the behavior of electrical circuits. It is also used in signal processing to recover signals from their frequency domain representation.

5. Are there any limitations or challenges to using the inverse Laplace Transform?

Yes, there are limitations and challenges to using the inverse Laplace Transform. One challenge is that not all functions have a Laplace Transform, so the inverse Laplace Transform cannot be applied in these cases. Additionally, the inverse Laplace Transform can be difficult to compute analytically, so numerical methods may be necessary in some cases. Finally, the inverse Laplace Transform may not always give a unique solution, as some functions may have multiple Laplace Transforms.

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