Find the inverse Laplace Transform from the function below. F(s)=frac(7)((s+11)(s+12))

texelaare 2021-09-28 Answered
Find the inverse Laplace Transform from the function below
\(\displaystyle{F}{\left({s}\right)}={\frac{{{7}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Fatema Sutton
Answered 2021-09-29 Author has 6053 answers

Step 1
We have to find inverse laplace transformation of given function.
Step 2
Now,
\(\displaystyle{F}{\left({s}\right)}={\frac{{{7}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}\)
Now,
\(\displaystyle{F}{\left({s}\right)}={7}{\left({\frac{{{1}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}\right)}\)
\(\displaystyle{F}{\left({s}\right)}={7}{\left({\frac{{{1}}}{{{s}+{11}}}}-{\frac{{{1}}}{{{s}+{12}}}}\right)}\)
Therefore,
\(\displaystyle{L}^{{-{1}}}{F}={L}^{{-{1}}}{\left[{\frac{{{7}}}{{{s}+{11}}}}-{\frac{{{7}}}{{{s}+{12}}}}\right]}\)
\(\displaystyle={L}^{{-{1}}}{\left[{\frac{{{7}}}{{{s}+{11}}}}\right]}-{L}^{{-{1}}}{\left[{\frac{{{7}}}{{{s}+{12}}}}\right]}\)
\(\displaystyle={7}{e}^{{-{11}{t}}}-{7}{e}^{{-{12}{t}}}\)
Hence,
\(\displaystyle{L}^{{-{1}}}{\left({\frac{{{7}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}\right)}={7}{e}^{{-{11}{t}}}-{7}{e}^{{-{12}{t}}}\)

Have a similar question?
Ask An Expert
9
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-09-08
Use the transforms in the table below to find the inverse Laplace transform of the following function.
\(\displaystyle{F}{\left({s}\right)}={\frac{{{5}}}{{{s}^{{4}}}}}\)
asked 2021-09-19
Find the inverse Laplace transform of the following transfer function:
\(\displaystyle{\frac{{{Y}{\left({s}\right)}}}{{{U}{\left({s}\right)}}}}={\frac{{{50}}}{{{\left({s}+{7}\right)}^{{2}}+{25}}}}\)
Select one:
a)\(\displaystyle{f{{\left({t}\right)}}}={10}{e}^{{-{7}{t}}}{\sin{{\left({5}{t}\right)}}}\)
b)\(\displaystyle{f{{\left({t}\right)}}}={10}{e}^{{{7}{t}}}{\sin{{\left({5}{t}\right)}}}\)
c)\(\displaystyle{f{{\left({t}\right)}}}={50}{e}^{{-{7}{t}}}{\sin{{\left({5}{t}\right)}}}\)
d)\(\displaystyle{f{{\left({t}\right)}}}={2}{e}^{{-{7}{t}}}{\sin{{\left({5}{t}\right)}}}\)
asked 2021-09-18

determine the inverse Laplace transform of the function.
\(L^{-1}\left\{R(s)\right\}=L^{-1}\left\{\frac{7}{(s+3)(s-3)}\right\}\)

asked 2021-09-25

Find the inverse Laplace transform \(f(t)=L^{-1}\left\{F(s)\right\}\) of the function \(\displaystyle{F}{\left({s}\right)}={\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\)
Use h(t-c) for the Heaviside function h_c(t) if necessary.
\(f(t)=L^{-1}\left\{\frac{8e^{-4s}}{s^2+64}\right\}=\)

asked 2021-09-27
Find the inverse Laplace transform of the function:
\(\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}\)
asked 2021-09-19
find the inverse Laplace transform of the given function.
\(\displaystyle{F}{\left({s}\right)}={\frac{{{3}!}}{{{\left({s}-{2}\right)}^{{4}}}}}\)
asked 2021-09-18
to find the inverse Laplace transform of the given function.
\(\displaystyle{F}{\left({s}\right)}={\frac{{{2}^{{{n}+{1}}}{n}!}}{{{s}^{{{n}+{1}}}}}}\)

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...