In this section we will use the linearity property of the Z-transform to connect the behavior of the signal with the poles of its Z-transform.
9.4 One-Sided Z-Transform 523
The Z-transform is a linear transformation, meaning that
Z(ax[n]+by[n])=aZ(x[n])+bZ(y[n]) (9.10) for signalsx[n]andy[n]and constantsaandb.
To illustrate the linearity property as well as the connection between the signal and the poles of its Z-transform, consider the signalx[n]=αnu[n] for real or complex values α. Its Z-transform will be used to compute the Z-transform of the following signals:
n x[n]=cos(ω0n+θ)u[n] for frequency 0≤ω0≤πand phaseθ.
n x[n]=αncos(ω0n+θ)u[n] for frequency 0≤ω0≤πand phaseθ.
Show how the poles of the corresponding Z-transform connect with the signals.
The Z-transform of the causal signalx[n]=αnu[n] is
X(z)=
∞
X
n=0
αnz−n=
∞
X
n=0
(αz−1)n= 1
1−αz−1 = z
z−α ROC: |z|>|α| (9.11) Using the last expression in Equation (9.11) the zero ofX(z)isz=0 and its pole isz=α, since the first value makesX(0)=0 and the second makesX(α)→ ∞. Forαreal, be it positive or negative, the region of convergence is the same, but the poles are located in different places. See Figure 9.2 for α <0.
Ifα=1 the signalx[n]=u[n] is constant forn≥0 and the pole ofX(z)is at z=1ej0 (the radius isr=1 and the lowest discrete frequencyω=0 rad). On the other hand, whenα= −1 the signal is x[n]=(−1)nu[n], which varies from sample to sample for n≥0; its Z-transform has a pole at z= −1=1ejπ (a radiusr=1 and the highest discrete frequencyω=π rad). As we move the pole toward the center of thez-plane (i.e.,|α| →0), the corresponding signal decays exponentially for 0<
α <1, and is a modulated exponential of|α|n(−1)nu[n]= |α|ncos(πn)u[n] for −1< α <0. When
|α|>1 the signal becomes either a growing exponential(α >1)or a growing modulated exponential (α <−1).
FIGURE 9.2
Region of convergence (shaded area) ofX(z)with a pole at z=α,α <0(same ROC if pole is atz= −α).
α×
−1 1
z-plane
For a real valueα= |α|ejω0forω0=0orπ,
x[n]=αnu[n] ⇔ X(z)= 1
1−αz−1 = z
z−α ROC:|z|>|α|
and the location of the pole ofX(z)determines the behavior of the signal:
n Whenα >0, thenω0=0and the signal is less and less damped asα→ ∞.
n Whenα <0, thenω0=πand the signal is a modulated exponential that grows asα→ −∞. To compute the Z-transform ofx[n]=cos(ω0n+θ)u[n], we use Euler’s identity to writex[n] as
x[n]=
"
ej(ω0n+θ)
2 +e−j(ω0n+θ) 2
# u[n]
Applying the linearity property and using the above Z-transform whenα=ejω0 and its conjugate α∗=e−jω0, we get
X(z)= 1 2
"
ejθ
1−ejω0z−1 + e−jθ 1−e−jω0z−1
#
= 1 2
2 cos(θ)−2 cos(ω0−θ)z−1 1−2 cos(ω0)z−1+z−2
= cos(θ)−cos(ω0−θ)z−1
1−2 cos(ω0)z−1+z−2 (9.12)
ExpressingX(z)in terms of positive powers ofz, we get
X(z)= z(zcos(θ)−cos(ω0−θ))
z2−2 cos(ω0)z+1 = z(zcos(θ)−cos(ω0−θ))
(z−ejω0)(z−e−jω0) (9.13) which is valid for any value of θ. Ifx[n]=cos(ω0n)u[n], then θ=0 and the poles of X(z) are a complex conjugate pair on the unit circle at frequencyω0 radians. The zeros are at z=0 andz= cos(ω0). Whenx[n]=sin(ω0n)u[n]=cos(ω0n−π/2)u[n], thenθ= −π/2 and the poles are at the same location as those for the cosine, but the zeros are atz=0 andz=cos(ω0+π/2)/cos(π/2)→
∞, so there is only one finite zero at zero. For any other value ofθ, the poles are located in the same place but there is a zero atz=0 and another atz=cos(ω0−θ)/cos(θ).
For simplicity, we letθ=0. Ifω0=0, one of the double poles atz=1 is canceled by one of the zeros atz=1, resulting in the poles and the zeros ofZ([u[n]). Indeed, the signal whenω0=0 andθ=0 isx[n]=cos(0n)u[n]=u[n]. When the frequencyω0>0 the poles move along the unit circle from the lowest(ω0=0 rad)to the highest(ω0 =π rad)frequency.
9.4 One-Sided Z-Transform 525
The Z-transform pairs of a cosine and a sine are, respectively, cos(ω0n)u[n] ⇔ z(z−cos(ω0))
(z−ejω0)(z−e−jω0) ROC :|z|>1 (9.14) sin(ω0n)u[n] ⇔ zsin(ω0)
(z−ejω0)(z−e−jω0) ROC :|z|>1 (9.15) The Z-transforms for these sinusoids have identical poles1e±jω0, but different zeros. The frequency of the sinusoid increases from the lowest(ω0=0 rad)to the highest(ω0=πrad)) as the poles move along the unit circle from1to−1in its lower and upper parts.
Consider then the signalx[n]=rncos(ω0n+θ)u[n], which is a combination of the above cases. As before, the signal is expressed as a linear combination
x[n]=
"
ejθ(rejω0)n
2 +e−jθ(re−jω0)n 2
# u[n]
and it can be shown that its Z-transform is
X(z)=z(zcos(θ)−rcos(ω0−θ))
(z−rejω0)(z−re−jω0) (9.16) The Z-transform of a sinusoid is a special case of the above (i.e., whenr=1). It also becomes clear that as the value ofr decreases toward zero, the exponential in the signal decays faster, and that wheneverr>1, the exponential in the signal grows making the signal unbound.
The Z-transform pair
rncos(ω0n+θ)u[n] ⇔ z(zcos(θ)−rcos(ω0−θ))
(z−rejω0)(z−re−jω0) (9.17) shows how complex conjugate pairs of poles inside the unit circle represent the damping indicated by the radiusrand the frequency given byω0in radians.
Double poles are related to the derivative ofX(z)or to the multiplication of the signal byn. If X(z)=
X∞ n=0
x[n]z−n its derivative with respect tozis
dX(z) dz =
∞
X
n=0
x[n]dz−n dz
= −z−1
∞
X
n=0
nx[n]z−n
Or the pair
nx[n]u[n] ⇔ −zdX(z)
dz (9.18)
For instance, ifX(z)=1/(1−αz−1)=z/(z−α), we find that dX(z)
dz = − α
(z−α)2 That is, the pair
nαnu[n] ⇔ αz (z−α)2 indicates that double poles correspond to multiplication ofx[n]byn.
The above shows that the location of the poles ofX(z)provides basic information about the signal x[n]. This is illustrated in Figure 9.3, where we display the signal and its corresponding poles.