$(i) x (t) = Ae^{-αt}u(t), α > 0\ (ii) x[n] = u[n]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 04

Year : MAY 2015

$$(i) x(t) = Ae^{-αt} u(t) α > 0$$

This signal is non-periodic and of infinite duration. It can be energy signal.

Therefore calculating its energy

Energy is given by the equation,

$E = ∫\limits_{-∞}^∞|x(t)|2 dt \\ = ∫\limits_0^∞ [A e^{-∝t}]^2 dt\hspace {1cm} \text {Since} \space u(t) = 1 \text {for} 0 ≤ t ≤ ∞ \\ = A^2 ∫\limits_0^∞e^{-2∝t} dt \\ = A^2 ( [\dfrac {e^{-2∝t}}{-2∝}]^∞- [\dfrac {e^{-2∝t}}{-2∝}]^0) \\ ∴E =\dfrac { A^2}{2α}$

The energy is finite and non-zero, hence the given signal is energy signal with $E =\dfrac {A^2}{2α}.$

$$(ii) x(n) = u(n)$$

This signal is periodic since u(n) repeats after every sample and of infinite duration . Hence it may be power signal.

Therefore calculating power

Power is given by the equation,

$P = \lim\limits_{N→ ∞}\dfrac 1{2N+1} ∑\limits_{n= -N}^N|x(n)|^2 \\ = \lim\limits_{N→ ∞}\dfrac 1{2N+1} ∑\limits_{n= 0}^N|(1)|^2\hspace {1cm} \text {Since} u(n) = 1 \text {for} 0 ≤ n ≤ ∞ \\ \text {Here} ∑\limits_{n= 0}^N |(1)|^2 \text { means} 1 + 1 + 1 + 1………. \text {For} n = 0\space \space to\space \space N. \\ \text {In other words,} 1 + 1 + 1 + 1………. (N + 1)\text { times} = (N + 1) $

Therefore above equation will be,

$P =\lim\limits_{N→ ∞}\dfrac 1{2N+1} (N + 1) \\ = \lim\limits_{N→ ∞} \dfrac {N+1}{2N+1}\\ = \lim\limits_{N→ ∞}\dfrac {1+ \dfrac 1N}{2+\dfrac 1N} \\ ∴P =\dfrac 12 $

The power is finite and non-zero, hence unit step function is power signal with $P = \dfrac 12 .$