**1 Answer**

written 7.9 years ago by | • modified 5.7 years ago |

(i) $x(t) = 0.9e^{-3t} u(t)$

$E = ∫\limits_{-∞}^∞|x(t) |^2 dt $

$ =∫\limits_{-∞}^∞|0.9e^{-3t} u(t) |^2 dt \\ = 0.81∫\limits_0^∞e^{-6t} dt \\ = 0.81[\dfrac {e^{-6t}}{-6}]^{\infty}_0 \\ = 0.81/6 \\ E= 0.135J$

Since the signal is not periodic hence it’s not a power signal but it is a energy signal.

**(ii)** $ x[n] = u[n]$

The signal is periodic because u(n) repeats after every sample and of infinite duration.

Hence it is a Power Signal

$P = \lim\limits_{N→∞} \dfrac 1{2N+1} ∑\limits_{-N}^N|x[n]|^2 \\ = \lim\limits_{N→∞} \dfrac 1{2N+1} ∑\limits_{-N}^N|u[n]|^2 \\ = \lim\limits_{N→∞} \dfrac 1{2N+1} \\ ∑\limits_0^N(1)^2$

Since $∑\limits_0^N(1)^2 =1+1+1+⋯∞ = (N+1) \\ P = \lim\limits_{N→∞} \dfrac 1{2N+1}(N+1) \\ =\lim\limits_{N→∞} \dfrac { N(1+\dfrac 1N)}{N(2+\dfrac 1N)} = 0.5 W. \\ P=0.5W$