["*"] represents convolution operator]
Laplace Transform of Convolution Signal
We need to find the Laplace Transform of the convolution $x(t) = u(t - 2) * (t u(t))$
Using the Laplace Transform property for convolution, $\mathcal{L}\{f(t) * g(t)\} = F(s) G(s)$.
Let $f(t) = u(t - 2)$ and $g(t) = t u(t)$.
Transform of $u(t-2)$ Signal
The Laplace Transform of a time-delayed function $f(t-a)u(t-a)$ is $e^{-as} F(s)$, where $F(s) = \mathcal{L}\{f(t)\}$.
For $u(t-2)$, we have $a = 2$ and $f(t) = u(t)$. Since $\mathcal{L}\{u(t)\} = \frac{1}{s}$,
$\mathcal{L}\{u(t - 2)\} = e^{-2s} \left(\frac{1}{s}\right) = \frac{e^{-2s}}{s}$. This is $F(s)$.
Transform of $t u(t)$ Signal
The Laplace Transform of $t^n u(t)$ is $\frac{n!}{s^{n+1}}$.
For $t u(t)$, $n=1$, so $\mathcal{L}\{t u(t)\} = \frac{1!}{s^{1+1}} = \frac{1}{s^2}$. This is $G(s)$.
Convolution Transform Result
Applying the convolution property:
$X(s) = F(s) G(s) = \left(\frac{e^{-2s}}{s}\right) \left(\frac{1}{s^2}\right)$
$X(s) = \frac{e^{-2s}}{s^3}$
Final Result Match
The calculated Laplace Transform is $\frac{e^{-2s}}{s^3}$, which corresponds to Option 4.
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
