Consider the two continuous-time signals defined below:

$x_1\left(t\right)=\{\begin{array}{c}\left|t\right|,-1\le t\le 1\\ 0,otherwise\end{array},x_2\left(t\right)=\{\begin{array}{c}1-\left|t\right|,-1\le t\le 1\\ 0,otherwise\end{array}$

These signals are sampled with a sampling period of T = 0.25 seconds to obtain discrete time signals x₁[n] and x₂[n], respectively. Which one of the following statements is true?

Concept:

Energy of the discrete time signal x(n) is,

                         $E=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x\left(n\right)\right|}^2$

Energy of the continuous time signal x(t) is,

                         $E=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left(x\left(t\right)\right)}^{2}dt$

Explanation:

$x_1\left(t\right)=\{\begin{array}{cc}\left|t\right|& -1\le t\le 1\\ 0& otherwise\end{array}$

After sampling the above signal for T = 0.25 sec, the discrete signal will be

Energy of $x_1\left(n\right)=E_1=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x_1\left(n\right)\right|}^2$

= 0² + 2 [0.25² + 0.5² + 0.75² + 1²]

= 3.75

$x_2\left(t\right)=\{\begin{array}{cc}1-\left|t\right|& -1\le t\le 1\\ 0& otherwise\end{array}$

After sampling the above signal for T = 0.25 sec, the discrete signal will be

Energy of $x_2\left(n\right)=E_2=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x_2\left(n\right)\right|}^2$

= 1² + 2 [0.75² + 0.5² + 0.25²]

= 2.75

⇒ E₁ > E₂