Question

Average power for the signal, $x\left(t\right)=\sum _{n=-\infty }^{\infty }{e}^{-(2t-n)}u(2t-n)$is __________(Answer upto 2 decimal places)

• A. 1.08

Answer 1

The correct option is A 1.08
For the given system to be periodic
$x(t+T)=x\left(t\right)$

$=\sum _{n=-\infty }^{\infty }{e}^{-(2t+2T-n)}u(2t+2T-n)$

2T - n = -m

2T =n -m

$T=\frac{n-m}{2}$

$MinimumvalueofT=\frac{1}{2}$

$x\left(t\right)=\sum _{n=-\infty }^0{e}^{-(2t-n)}0<t<\frac{1}{2}$

$=e^{-2t}\sum _{n=-\infty }^0{e}^n$

$x\left(t\right)=\frac{e^{-2t}}{1-e^{-1}}0<t<\frac{1}{2}$

$P_x=2\underset{0}{\overset{1/2}{\int }}\frac{e^{-4t}}{(1-e^{-1}{)}^2}dt$

$=\frac{2}{(1-e^{-1}{)}^2}\left(\frac{-1}{4}\right)e^{-4t}{|}_0^{1/2}$

$=\frac{-1}{2(1-e^{-1}{)}^2}(e^{-2}-1)$

$=\frac{(1+e^{-1})}{2(1-e^{-1})}=\frac{(e+1)}{2(e-1)}=1.08$