Question
Download Solution PDFThe inverse Laplace transform of \(\frac{1}{{\left( {s + 1} \right)\left( {s - 2} \right)}}\) is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The Laplace transform of a general exponential signal is given by:
\(L[e^{-at}]\longleftrightarrow \frac{1}{s+a}\)
where 'a' is any positive integer.
Calculation:
Given:
\(\frac{1}{{\left( {s + 1} \right)\left( {s - 2} \right)}}\)
\(\frac{1}{{\left( {s + 1} \right)\left( {s - 2} \right)}} = \frac{A}{{\left( {s - 2} \right)}} + \frac{B}{{\left( {s + 1} \right)}} = \frac{1}{3}\left\{ {\frac{1}{{s - 2}} + \frac{{ - 1}}{{s + 1}}} \right\}\)
\({L^{ - 1}}\left( {\frac{1}{{\left( {s + 1} \right)\left( {s - 2} \right)}}} \right) = {L^{ - 1}}\left\{ {\frac{1}{3}\left( {\frac{1}{{s - 2}} - \frac{1}{{s + 1}}} \right)} \right\} = \frac{{{e^{2t}} - {e^{ - t}}}}{3}\)
Additional Information
Some common inverse Laplace transforms are:
|
F(s) |
ROC |
f(t) |
|
1 |
All s |
δ (t) |
|
\(\frac{1}{s}\) |
Re (s) > 0 |
u(t) |
|
\(\frac{1}{{{s^2}}}\) |
Re (s) > 0 |
t |
|
\(\frac{{n!}}{{{s^{n + 1}}}}\) |
Re (s) > 0 |
tn |
|
\(\frac{1}{{s + a}}\) |
Re (s) > -a |
e-at |
|
\(\frac{1}{{{{\left( {s + a} \right)}^2}}}\) |
Re (s) > -a |
t e-at |
|
\(\frac{{n!}}{{{{\left( {s + a} \right)}^n}}}\) |
Re (s) > -a |
tn e-at |
|
\(\frac{a}{{{s^2} + {a^2}}}\) |
Re (s) > 0 |
sin at |
|
\(\frac{s}{{{s^2} + {a^2}}}\) |
Re (s) > 0 |
cos at |
Last updated on May 9, 2025
-> The BPSC Lecturere DV will be held on 28th May 2025.
-> The written exam took place on 20th March 2025.
-> BPSC Lecturer notification was released for 6 vacancies.
-> The BPSC recruits eligible candidates for the post of Lecturer for various disciplines in Government Colleges across the state.
-> Candidates must attempt the BPSC Lecturer EC Mock Tests. Candidates can also download and practice from the BPSC Lecturer previous year papers.