Let ๐ฅ1(๐ก)=๐^(-t)๐ข(๐ก) and ๐ฅ2(๐ก)=๐ข(๐ก)−๐ข(๐ก−2), where ๐ข(⋅) denotes the unit step function...
Let ๐ฅ1(๐ก)=๐^(-t)๐ข(๐ก) and ๐ฅ2(๐ก)=๐ข(๐ก)−๐ข(๐ก−2), where ๐ข(⋅) denotes the unit step function. If ๐ฆ(๐ก) denotes the convolution of ๐ฅ1(๐ก) and ๐ฅ2(๐ก), then lim(x→0) ๐ฆ(๐ก) = _________ (rounded off to one decimal place).
Solution
Let
\[
x_1(t)=e^{-t}u(t)
\]
and
\[
x_2(t)=u(t)-u(t-2)
\]
where \(u(t)\) is the unit step function.
\[
y(t)=x_1(t)*x_2(t)
\]
Find the final value of \(y(t)\).
Laplace Transforms
\[
\mathcal{L}\{e^{-t}u(t)\}
=\frac{1}{s+1}
\]
\[
\mathcal{L}\{u(t)-u(t-2)\}
=\frac{1}{s}-\frac{e^{-2s}}{s}
=\frac{1-e^{-2s}}{s}
\]
Using Convolution Property
\[
Y(s)=X_1(s)X_2(s)
\]
\[
Y(s)
=
\frac{1}{s+1}
\cdot
\frac{1-e^{-2s}}{s}
\]
Applying Final Value Theorem
\[
\lim_{t\to\infty}y(t)
=
\lim_{s\to0}sY(s)
\]
\[
=
\lim_{s\to0}
s
\left(
\frac{1}{s+1}
\cdot
\frac{1-e^{-2s}}{s}
\right)
\]
\[
=
\lim_{s\to0}
\frac{1-e^{-2s}}{s+1}
\]
Using the limit
we obtain
\[
\boxed{\lim_{t\to\infty}y(t)=0}
\]