Signals and Systems - Electrical Engineering

76 C H A P T E R 1: Continuous-Time Signals

Solution

The reflection ofx(t)isx(−t)=cos(− 2 πt+θ). Then:

1. x(t)is even ifx(t)=x(−t)or

cos( 2 πt+θ)=cos(− 2 πt+θ)

=cos( 2 πt−θ)

orθ=−θorθ=0,π. Thus,x 1 (t)=cos( 2 πt)as well asx 2 (t)=cos( 2 πt+π)=−cos( 2 πt)are

even.

2. forx(t)to be odd, we need thatx(t)=−x(−t)or

cos( 2 πt+θ)=−cos(− 2 πt+θ)=cos(− 2 πt+θ±π)=cos( 2 πt−θ∓π)

which can be obtained withθ=−θ∓πorθ=∓π/2. Indeed, cos( 2 πt−π/ 2 )=sin( 2 πt)and

cos( 2 πt+π/ 2 )=−sin( 2 πt)are both odd. Thus,x 3 (t)=±sin( 2 πt)is odd.

Whenθ=π/4,x(t)=cos( 2 πt+π/ 4 )is neither even nor odd according to the above. n

nExample 1.7

Consider the signal

x(t)=

{

2 cos( 4 t) t> 0

0 otherwise

Find its even and odd decomposition. What would happen ifx( 0 )=2 instead of 0—that is, when

we define the sinusoid att=0? Explain.

Solution

The signalx(t)is neither even nor odd given that its values fort≤0 are zero. For its even–odd

decomposition, the even component is given by

xe(t)=0.5[x(t)+x(−t)]

=







cos( 4 t) t> 0

cos( 4 t) t< 0

0 t= 0

and the odd component is given by

xo(t)=0.5[x(t)−x(−t)]

=







cos( 4 t) t> 0

−cos( 4 t) t< 0

0 t= 0

which when added together become the given signal.