$x[t]=\cos^2[\dfrac {\pi t}2]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : DEC 2015

$x[t] = \cos^2 \dfrac {πt}2 $ $∴x[t] = \dfrac {1 + \cos 2 \dfrac π2 t}2 \hspace {1cm} \cos^2A =\dfrac { 1 + cos2A}2 \ ∴x[t] = \dfrac 12 + \dfrac12 \cos πt $ Compare the equation with $x[t] = A + A \cos (2πft) \ ∴2πft = πt \ ∴f = \dfrac 12 \ ∴T = 2$ Even part of the signal is given by $x_e(t) = \dfrac12 [x(t) + x(-t)] \ \text {Now,} x(-t) = \dfrac12 + \dfrac12 \cos (-πt) \ = \dfrac12 + \dfrac12 \cos πt \ ∴x_e(t) = \dfrac12 [\dfrac12 + \dfrac12 \cos πt + \dfrac12 + \dfrac12 \cos πt] \ ∴x_e(t) = \dfrac12 + \dfrac12 \cos πt $ Odd part of the signal is given by $x_o(t) = \dfrac12 [x(t) - x(-t)] \ = \dfrac12 [\dfrac12 + \dfrac12 \cos πt - \dfrac12 - \dfrac12 \cos πt] \ ∴x_o(t) = 0$

Even and odd components of the signal are shown in the figure below: