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Find the Energy of the signal x(n) = 0.5*u(n) + 8nu(-n-1)

Mumbai University > Computer Engineering > Sem 7 > Digital Signal Processing

Marks: 5 Marks

Year: Dec 2015

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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 $n = -m \therefore$,

$$E_x=\dfrac14+0+\sum_1^∞|64^{-m} |$$

$$\therefore,E_x=\therefore14+0+\sum_1^∞|\dfrac{1}{64}^m |$$

$$\sum_{m=1}^{m=∞}a^m=\dfrac{a}{1-a}$$

$$\therefore,E_x=\dfrac{1}{4}+0+ \dfrac{(1⁄64)}{(1-1⁄64)}$$

$$\therefore,E_x=\dfrac{1}{4}+0+ \dfrac{1}{63} = \dfrac{67}{252}$$

Result: Energy of the given signal x(n) is, $E_x= \dfrac{67}{252}$

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