Signals and Systems - Electrical Engineering

464 C H A P T E R 8: Discrete-Time Signals and Systems

Obtain an expression forx[n−k] for− 2 ≤n≤2 and determine in which direction it shifts asn

increases from−2 to 2.

Solution

Althoughx[n−k], as a function ofk, is reflected it is not clear if it is advanced or delayed asn

increases from−2 to 2. Ifn=0,

x[−k]=

{

−k − 3 ≤k≤ 0

0 otherwise

Forn6=0, we have that

x[n−k]=

{

n−k n− 3 ≤k≤n

0 otherwise

Asnincreases from−2 to 2 the supports ofx[n−k] move to the right. Forn=−2 the support of

x[− 2 −k] is− 5 ≤k≤−2, while forn=0 the support ofx[−k] is− 3 ≤k≤0, and forn=2 the

support ofx[2−k] is− 1 ≤k≤2, each shifted to the right. n

We can thus use the above to define even and odd signals and obtain a general decomposition of any

signal in terms of even and odd signals.

Even and odd discrete-time signals are defined as

x[n]iseven: ⇔ x[n]=x[−n] (8.12)

x[n]isodd: ⇔ x[n]=−x[−n] (8.13)

Any discrete-time signalx[n]can be represented as the sum of an even and an odd component,

x[n]=

1

2

(

x[n]+x[−n]

)

︸ ︷︷ ︸

xe[n]

+

1

2

(

x[n]−x[−n]

)

︸ ︷︷ ︸

xo[n]

=xe[n]+xo[n] (8.14)

The even and odd decomposition can be easily seen. The even componentxe[n]=0.5(x[n]+x[−n])

is even sincexe[−n]=0.5(x[−n]+x[n])equalsxe[n], and the odd componentxo[n]=0.5(x[n]−

x[−n])is odd sincexo[−n]=0.5(x[−n]−x[n])=−xo[n].

nExample 8.11

Find the even and the odd components of the discrete-time signal

x[n]=

{

4 −n 0 ≤n≤ 4

0 otherwise