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Find the Energy of the signal x(n) = 0.5*u(n) + 8nu(-n-1)
1 Answer
| written 9.0 years ago by |
The total energy of a sequence x[n] is defined by,
$$E_x=\sum_{n=-∞}^∞|x[n]|^2$$
$$\therefore E_x=\sum_{-∞}^∞\bigg|\dfrac{u(n)}{2}+8^n u(-n-1)\bigg|^2$$
$$\therefore E_x=\sum_{-∞}^∞\bigg|\dfrac{u^2 (n)}{4}\bigg|+\sum_{-∞}^∞|2*u(n)*8^n u(-n-1)| +\sum_{-∞}^∞|8^n u(-n-1)|^2$$
We know, $u(n) = 1 ; n≥0 \\ = 0 ; n\lt0$
$\therefore,$ u(-n-1) $= 1 ; n\lt1 \\ = 0 ; n≥0$
$$\therefore E_x=\dfrac14 \sum_0^∞|u^2 |+0+\sum_{-∞}^{-1}|8^2n |$$
Substituting $n …