The even and odd components of the signal x(t) = e^-2t cos t are respectively.

Concept:

if x(t) is given signal then,

Even signal of x(t) = \(\frac{x(t)+x(-t)}{2}\)

&

odd signal of x(t) =\(\frac{x(t)-x(-t)}{2}\)

Solution:

Here,

x(t) = e-2t cos t 

Even signal of x(t) = \(\frac{x(t)+x(-t)}{2}\)

\(=\frac{e^{-2t} \space cos(t)+e^{-2(-t)} \space cos(-t)}{2}\)

\(=\frac{e^{-2t} \space cos(t)+e^{2t} \space cos(t)}{2}\)

\(= cos(t)\frac{(e^{-2t} \space+e^{2t}) }{2}\)

\(= cos(t)cosh(2t)\)

&

odd signal of x(t) =\(\frac{x(t)-x(-t)}{2}\)

\(=\frac{e^{-2t} \space cos(t)-e^{-2(-t)} \space cos(-t)}{2}\)

\(=\frac{e^{-2t} \space cos(t)-e^{2t} \space cos(t)}{2}\)

\(= cos(t)\frac{(e^{-2t} -e^{2t}) }{2}\)

\(=- cos(t)\frac{(e^{2t} -e^{-2t}) }{2}\)

\(= -cos(t)sinh(2t)\)