Partial Fractions
Factor the denominator and split the rational function into simpler fractions. Then use
\[\mathcal L^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}.\]
Inverse Laplace Transforms
Partial Fractions, Transform Differentiation and Shifting
Inverse Laplace transforms are developed through three IOE past-paper questions. Use partial fractions, differentiation of a transform and shifted standard forms.
UnitLaplace Transform
TopicInverse Laplace Transform
Question Type2 Mark Problems
Theory Required for These Problems
Differentiation of a Transform
If \(\mathcal L^{-1}\{F(s)\}=f(t)\), then
\[\mathcal L\{tf(t)\}=-F'(s).\]
This is useful when differentiating the given expression produces a standard transform.
First Shifting Theorem
If \(\mathcal L^{-1}\{F(s)\}=f(t)\), then
\[\mathcal L^{-1}\{F(s+a)\}=e^{-at}f(t).\]
Useful Standard Results
\[\mathcal L^{-1}\left\{\frac{1}{(s+a)^2}\right\}=te^{-at},\qquad\mathcal L^{-1}\left\{\frac{1}{(s+a)^3}\right\}=\frac{t^2}{2}e^{-at}.\]
\[\mathcal L^{-1}\left\{\frac{1}{(s+a)^2}\right\}=te^{-at}.\]\[\mathcal L^{-1}\left\{\frac{1}{(s+a)^3}\right\}=\frac{t^2}{2}e^{-at}.\]
Note: Rewrite the numerator in terms of the shifted expression \(s+a\) before taking the inverse transform.
Method used: Convert the given function into standard transform forms, apply the inverse transform to each term, and simplify the final expression.
Continue through the IOE Engineering Mathematics II Digital Workbook. This inverse laplace transforms page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
42
2 Marks
Inverse Transform by Partial Fractions
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 4(b) • 2 Marks
Find
\[\mathcal L^{-1}\left\{\frac{1}{s^2-3s+2}\right\}.\]
Solution
Partial fractionsFactor the denominator:
\[\frac{1}{s^2-3s+2}=\frac{1}{(s-1)(s-2)}.\]Let
\[\frac{1}{(s-1)(s-2)}=\frac{A}{s-1}+\frac{B}{s-2}.\]Therefore,
\[1=A(s-2)+B(s-1).\]Putting \(s=1\) gives \(A=-1\), and putting \(s=2\) gives \(B=1\).
Hence,
\[\frac{1}{s^2-3s+2}=-\frac{1}{s-1}+\frac{1}{s-2}.\]Taking inverse Laplace transforms,
\[\mathcal L^{-1}\left\{\frac{1}{s^2-3s+2}\right\}=-e^t+e^{2t}.\]\[e^{2t}-e^t\]
43
2 Marks
Inverse Transform of \(\cot^{-1}(s+1)\)
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 4(b) • 2 Marks
Find
\[\mathcal L^{-1}\{\cot^{-1}(s+1)\}.\]
Solution
Differentiation of the transformLet
\[F(s)=\cot^{-1}(s+1),\qquad\mathcal L^{-1}\{F(s)\}=f(t).\]
\[F(s)=\cot^{-1}(s+1).\]\[\mathcal L^{-1}\{F(s)\}=f(t).\]
Now,
\[F'(s)=-\frac{1}{1+(s+1)^2}.\]Therefore,
\[-F'(s)=\frac{1}{(s+1)^2+1}=\mathcal L\{e^{-t}\sin t\}.\]But
\[\mathcal L\{tf(t)\}=-F'(s).\]Hence,
\[tf(t)=e^{-t}\sin t.\]Therefore,
\[f(t)=\frac{e^{-t}\sin t}{t}.\]\[\mathcal L^{-1}\{\cot^{-1}(s+1)\}=\frac{e^{-t}\sin t}{t}\]
44
2 Marks
Inverse Transform Using a Shifted Numerator
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 4(b) • 2 Marks
Find
\[\mathcal L^{-1}\left\{\frac{s}{(s+2)^3}\right\}.\]
Solution
Rewrite the numerator and apply shiftingWrite
\[s=(s+2)-2.\]Therefore,
\[\frac{s}{(s+2)^3}=\frac{1}{(s+2)^2}-\frac{2}{(s+2)^3}.\]Using the standard results,
\[\mathcal L^{-1}\left\{\frac{1}{(s+a)^2}\right\}=te^{-at},\qquad\mathcal L^{-1}\left\{\frac{1}{(s+a)^3}\right\}=\frac{t^2}{2}e^{-at}.\]
\[\mathcal L^{-1}\left\{\frac{1}{(s+a)^2}\right\}=te^{-at}.\]\[\mathcal L^{-1}\left\{\frac{1}{(s+a)^3}\right\}=\frac{t^2}{2}e^{-at}.\]
Now,
\[\mathcal L^{-1}\left\{\frac{s}{(s+2)^3}\right\}=te^{-2t}-2\left(\frac{t^2}{2}e^{-2t}\right).\]
\[\mathcal L^{-1}\left\{\frac{s}{(s+2)^3}\right\}\]\[=te^{-2t}-2\left(\frac{t^2}{2}e^{-2t}\right).\]
Therefore,
\[\mathcal L^{-1}\left\{\frac{s}{(s+2)^3}\right\}=(t-t^2)e^{-2t}.\]\[(t-t^2)e^{-2t}\]