CHAPTER 8 Discrete-Time Signals and Systems
8.2.1 Periodic and Aperiodic Signals
A discrete-time signalx[n]isperiodicif
n It is defined for all possible values ofn,−∞<n<∞.
n There is a positive integerN, the period ofx[n], such that
x[n+kN]=x[n] (8.2)
for any integerk.
Periodic discrete-time sinusoids, of periodN, are of the form x[n]=Acos
2πm N n+θ
−∞<n<∞ (8.3)
where the discrete frequency isω0=2πm/Nrad, for positive integersmandN, which are not divisible by each other, andθis the phase angle.
The definition of a discrete-time periodic signal is similar to that of continuous-time periodic signals, except for the period being an integer. That discrete-time sinusoids are of the given form can be easily shown: Shifting the sinusoid in Equation (8.3) by a multiplekof the periodN, we have
x[n+kN]=Acos 2πm
N (n+kN)+θ
=Acos 2πm
N n+2πmk+θ
=x[n]
since we add to the original angle a multiplemk(an integer) of 2π, which does not change the angle.
Remarks
n The units of the discrete frequency ω is radians. Moreover, discrete frequencies repeat every 2π (i.e., ω=ω+2πk for any integer k), and as such we only need to consider the range−π≤ω < π. This is in contrast with the analog frequency, which has rad/sec as units, and its range is from−∞to∞.
n If the frequency of a periodic sinusoid is
ω=2π Nm
for nondivisible integers m and N>0, the period is N. If the frequency of the sinusoid cannot be written like this, the discrete sinusoid is not periodic.
nExample 8.3
Consider the sinusoids
x1[n]=2 cos(πn−π/3)
x2[n]=3 sin(3πn+π/2) −∞<n<∞
8.2 Discrete-Time Signals 455
From their frequencies determine if these signals are periodic, and if so, determine their corresponding periods.
Solution
The frequency ofx1[n] can be written as
ω1=π= 2π 2
wherem=1 andN=2, so thatx1[n] is periodic of periodN1=2. Likewise, the frequency ofx2[n]
can be written as
ω2 =3π= 2π 2 3
wherem=3 andN=2, so thatx2[n] is also periodic of periodN2=2, which can be verified as follows:
x2[n+2]=3 sin(3π(n+2)+π/2)=3 sin(3πn+6π+π/2)=x[n]
n nExample 8.4
What is true for continuous-time sinusoids—that they are always periodic—is not true for discrete- time sinusoids. These sinusoids can be nonperiodic even if they result from uniformly sampling a continuous-time sinusoid. Consider the discrete signalx[n]=cos(n+π/4), which is obtained by sampling the analog sinusoidx(t)=cos(t+π/4)with a sampling periodTs=1 sec/sample. Isx[n]
periodic? If so, indicate its period. Otherwise, determine values of the sampling period, satisfying the Nyquist sampling rate condition, that when used in samplingx(t)result in periodic signals.
Solution
The sampled signalx[n]=x(t)|t=nTs=cos(n+π/4)has a discrete frequencyω=1 rad that cannot be expressed as 2πm/Nfor any integersmandNbecauseπis an irrational number. Sox[n] is not periodic.
Since the frequency of the continuous-time signalx(t)is=1 (rad/sec), then the sampling period, according to the Nyquist sampling rate condition, should be
Ts≤ π
=π
and for the sampled signalx(t)|t=nTs=cos(nTs+π/4)to be periodic of periodNor cos((n+N)Ts+π/4)=cos(nTs+π/4) is necessary that NTs=2kπ
for an integerk(i.e., a multiple of 2π). Thus,Ts=2kπ/N≤π satisfies the Nyquist sampling con- dition at the same time that it ensures the periodicity of the sampled signal. For instance, if we
wish to have a sinusoid with periodN=10, thenTs=0.2kπforkchosen so the Nyquist sampling rate condition is satisfied—that is,
0<Ts=kπ/5≤π so that 0<k≤5.
From these possible values forkwe choosek=1 and 3 so thatNandkare not divisible by each other and we get the desired periodN=10 (the values k=2 and 4 would give 5 as the period, andk=5 would give a period of 2 instead of 10). Indeed, if we letk=1, thenTs=0.2π satisfies the Nyquist sampling rate condition, and we obtain the sampled signal
x[n]=cos(0.2nπ+π/4)=cos 2π
10n+π 4
which according to its frequency is periodic of period 10. This is the same fork=3. n
When sampling an analog sinusoid
x(t)=Acos(0t+θ) −∞<t<∞ (8.4)
of periodT0=2π/0,0>0, we obtain aperiodic discrete sinusoid, x[n]=Acos(0Tsn+θ)=Acos
2πTs T0 n+θ
(8.5)
provided that
Ts T0 = m
N (8.6)
for positive integersNandm, which are not divisible by each other. To avoid frequency aliasing the sampling period should also satisfy
Ts≤ π
0 = T0
2 (8.7)
Indeed, sampling a continuous-time signalx(t)using as sampling periodTs, we obtain x[n]=Acos(0Tsn+θ)
=Acos 2πTs
T0
n+θ
where the discrete frequency isω0 =2πTs/T0. For this signal to be periodic we should be able to express this frequency as 2πm/Nfor nondivisible positive integersmandN. This requires that
T0
Ts = N m be a rational number, or that
mT0=NTs (8.8)
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 =T0 2
The sumz[n]=x[n]+y[n]of periodic signalsx[n]with periodN1, andy[n]with periodN2is periodic if the ratio of periods of the summands is rational—that is,
N2 N1 = p
q
wherepandqare integers not divisible by each other. If so, the period ofz[n]isqN2=pN1.
IfqN2=pN1, we then have that
z[n+pN1]=x[n+pN1]+y[n+pN1]
=x[n]+y[n+qN2]
=x[n]+y[n]=z[n]
sincepN1andqN2are 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 periodsN1=2,N2=3, andN3=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 sinceN2/N1 =3/2 is a rational number and 3 and 2 are non-divisible by each other, and its period isN4=3N1=2N2=6.
The signalz[n] is also periodic since
N4 N3 = 6
4 = 3 2
Its period isN=2N4=3N3=12. Thus,z[n] is periodic of period 12, indeed
z[n+12]=v[n+6N1]+w[n+4N2]+y[n+3N3]=v[n]+w[n]+y[n]=z[n]
n
nExample 8.6
Determine if the signal
x[n]= X∞ m=0
Xmcos(mω0n) ω0 =2π N0 is periodic, and if so, determine its period.
Solution
The signalx[n] consists of the sum of a constantX0and cosines of frequency mω0= 2πm
N0 m=1, 2,. . .
The periodicity ofx[n] depends on the periodicity of the cosines. According to the frequency of the cosines, they are periodic of periodN0. Thus,x[n] is periodic of periodN0. Indeed
x[n+N0]=
∞
X
m=0
Xmcos(mω0(n+N0))
=
∞
X
m=0
Xmcos(mω0n+2πm)=x[n]
n