Odd components in these signals

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Find the even and odd components for each of the following signals:

i) x(t) = exp^(-2t) sin (2t)

ii) x(t) = 1 + t + t³ + t^5

iii) x(t) = 1 + sin(t) + t cos(t) + t² sin(t) cost(t)
 

Re: Signals

Even part of x(t) = 0.5 [ x(t) + x (-t) ]

and

Odd part of x(t) = 0.5 [ x(t) - x (-t) ]

Use those formulas, to find even and odd component of your signals.


For example

(ii) For x (t) = 1 + t + t^3 + t^5,

Even part of x(t) will be 0.5 {[ 1 + t + t^3 + t^5] + [1 + (-t) + (-t)^3 + (-t)^5]} = 1
Odd part of x(t) will be 0.5{[ 1 + t + t^3 + t^5] - [1 + (-t) + (-t)^3 + (-t)^5]} = t + t^3 + t^5
 

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    sky_tm

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Signals

Thanks for the part (ii) answer. Can you show part (i) and (iii) as well?
 

Re: Signals

Anwers to (i) & (iii)

(i) Even part = 0.5[exp^(-2t) sin (2t) + exp^(2t) sin (-2t)]
Odd part = 0.5[exp^(-2t) sin (2t) - exp^(2t) sin (-2t)]

(iii) Even part = 0.5[{1 + sin(t) + t cos(t) + t² sin(t) cos(t)} + {1 + sin(-t) - t cos(-t) + t² sin(-t) cos(-t)}]

Odd part = 0.5[{1 + sin(t) + t cos(t) + t² sin(t) cos(t)} - {1 + sin(-t) - t cos(-t) + t² sin(-t) cos(-t)}]

where sin(jw) = [exp^(jw) + exp^(-jw)]/2
and cos(jw) = [exp^(jw) - exp^(-jw)]/2

Further Eulers theorem can be used
exp^(jt)= cos(t) + jsin(t)

then they can easily be simplified !!!
 

Re: Signals

Note that by looking at the signal itself we can find the even and odd parts of it.

For example
(ii)

\[x(t)=1+t+t^3+t^5\]. In this except \[1\], which is even function, the remaining part \[t+t^3+t^5\] is odd function.

(iii) Same is the case for this too. Note that except
\[1\], which is even function, the remaining part \[\sin(t)+t \cos(t)+t^2 \sin(t)\cos(t)\] is odd function.

(i) For this type of functions we have to evaluate the formulas:
\[x_e(t)=\frac{x(t)+x(-t)}{2}\] and
\[x_o(t)=\frac{x(t)-x(-t)}{2}\].

For problems(ii) and (iii) even after applying these results, we get the same answers. The answers we get above is by inspection.

thnx

purna!
 
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Re: Signals

A further simplification of problem ii would lead to

even part= sin2t * sinh2t
odd part= sin2t * cosh2t

Added after 1 minutes:

sorry.. the above solution i've mentioned is for problem (i)
 

Signals

you can read the book, Signals and Systems by Oppenhiem
 

Re: Signals

Conclusion:

i) \[x_e (t) = - \sin (2t)*\sinh (2t)\]
\[x_o (t) = \sin (2t)*\cosh (2t)\]

ii) \[x_e (t) = 1\]
\[x_o (t) = t + t^3 + t^5\]

iii) \[x_e (t) = 1 + t^2 \sin t\cos t\]
\[x_o (t) = \sin t + t\cos t\]
 

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