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Find out even and odd component of the following signal

x[t]=cos2[πt2]

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : DEC 2015

1 Answer
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x[t]=cos2πt2 ∴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:

enter image description here

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