Signals and Systems - Electrical Engineering

8.2 Discrete-Time Signals 457

which says that a period(m= 1 )or several periods(m> 1 )should be divided intoN>0 segments of
durationTsseconds. If the condition in Equation (8.6) is not satisfied, then the discretized sinusoid
is not periodic. To avoid frequency aliasing the sampling period should be chosen so that

Ts≤

π

 0

=

T 0

2

The sumz[n]=x[n]+y[n]of periodic signalsx[n]with periodN 1 , andy[n]with periodN 2 is periodic if the

ratio of periods of the summands is rational—that is,

N 2

N 1

=

p

q

wherepandqare integers not divisible by each other. If so, the period ofz[n]isqN 2 =pN 1.

IfqN 2 =pN 1 , we then have that

z[n+pN 1 ]=x[n+pN 1 ]+y[n+pN 1 ]

=x[n]+y[n+qN 2 ]

=x[n]+y[n]=z[n]

sincepN 1 andqN 2 are multiples of the periods ofx[n] andy[n].

nExample 8.5

The signal

z[n]=v[n]+w[n]+y[n]

is the sum of three periodic signalsv[n],w[n], andy[n] of periodsN 1 =2,N 2 =3, andN 3 =4,

respectively. Determine ifz[n] is periodic, and if so, determine its period.

Solution

Letx[n]=v[n]+w[n], so thatz[n]=x[n]+y[n]. The signalx[n] is periodic sinceN 2 /N 1 = 3 /2 is a

rational number and 3 and 2 are non-divisible by each other, and its period isN 4 = 3 N 1 = 2 N 2 =6.

The signalz[n] is also periodic since

N 4

N 3

=

6

4

=

3

2

Its period isN= 2 N 4 = 3 N 3 =12. Thus,z[n] is periodic of period 12, indeed

z[n+12]=v[n+ 6 N 1 ]+w[n+ 4 N 2 ]+y[n+ 3 N 3 ]=v[n]+w[n]+y[n]=z[n]

n