Inverse Laplace Transform Calculation
The problem requires finding the inverse Laplace transform of the function $F(s)=\frac{1}{s(s+1)}$.
Function Analysis
The given function is $F(s)=\frac{1}{s(s+1)}$. To find the inverse Laplace transform, we can use the method of Partial Fraction Decomposition.
Partial Fraction Decomposition
We express $F(s)$ as a sum of simpler fractions:
$ \frac{1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} $
To find the constants A and B, we clear the denominators:
$ 1 = A(s+1) + Bs $
Let $s=0$: $ 1 = A(0+1) + B(0) \implies 1 = A $
Let $s=-1$: $ 1 = A(-1+1) + B(-1) \implies 1 = -B \implies B = -1 $
Substituting the values of A and B back, we get:
$ F(s) = \frac{1}{s} - \frac{1}{s+1} $
Applying Inverse Laplace Transform
Now, we find the inverse Laplace transform of each term using standard pairs:
- $ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 $
- $ \mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\} = e^{-at} $. For $a=1$, $ \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} = e^{-t} $.
Therefore, the inverse Laplace transform $f(t)$ is:
$ f(t) = \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} $
$ f(t) = 1 - e^{-t} $
Conclusion
The inverse Laplace transform of $F(s)=\frac{1}{s(s+1)}$ is $f(t)=1-e^{-t}$.
Important Questions from Laplace Transform
Which of the following is the final value of the impulse response of the system whose transfer function is
(2s + 1)/(s 4 + 8s 3 + 16s 2 + s)
Find the Laplace transform for the following time domain.
y(t) = -2te -t + 4e -t - 4e -2t
Match List I with List II
List – I
List – II
f(t)
F(S)
A.
e -at
I.
\(\rm \frac{s}{s^2+ \omega^2}\)
B.
te at
II.
\(\rm \frac{\omega}{s^2+ \omega^2}\)
C.
sinωt
III.
\(\rm \frac{1}{(s- a)^2}\)
D.
cosωt
IV.
\(\rm \frac{1}{(s+ a)}\)
Choose the correct answer from the options given below:
The Laplace transform of sin h (at) is
The unilateral Laplace transform of f(t) is \(\frac{1}{s^2+s+1}\). The unilateral Laplace transform of t f(t) is
