Question

The signal x(t) is given as

$x\left(t\right)=\{\begin{array}{cc}\frac{3}{{ϵ}^3}(t-ϵ{)}^2& 0\le t\le ϵ\\ 0& otherwise\end{array}$

Select the correct statement about x(t).

• A. As $ϵ\to 0,x\left(t\right)$ becomes a unit step signal.
• B. None of these
• C. As $ϵ\to \infty ,x\left(t\right)$ becomes a unit impulse signal.
• D. As $ϵ\to 0,x\left(t\right)$ becomes a unit impulse signal.

Answer 1

The correct option is D As $ϵ\to 0,x\left(t\right)$ becomes a unit impulse signal.

We know that area under unit impulse signal is 1.

${\int }_{-\infty }^{\infty }x\left(t\right)dt={\int }_0^{ϵ}\frac{3}{{ϵ}^3}(t-ϵ{)}^2={\left[\frac{3}{{ϵ}^3}\frac{(t-ϵ{)}^3}{3}\right]}_0^{ϵ}={\left[\frac{(t-ϵ{)}^3}{{ϵ}^3}\right]}_0^{ϵ}$

$=\frac{(ϵ-ϵ{)}^3}{{ϵ}^3}-\frac{(-ϵ{)}^3}{{ϵ}^3}=1$

As $ϵ\to 0$ width of x(t) $\to$0 and height approached and its area remains at unity.
Therefore as $ϵ\to 0,x\left(t\right)$ becomes a unit impulse function.