Signals and Systems - Electrical Engineering

524 C H A P T E R 9: The Z-Transform

For a real valueα=|α|ejω^0 forω 0 = 0 orπ,

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 , thenω 0 = 0 and the signal is less and less damped asα→∞.

n Whenα < 0 , thenω 0 =πand the signal is a modulated exponential that grows asα→−∞.

To compute the Z-transform ofx[n]=cos(ω 0 n+θ)u[n], we use Euler’s identity to writex[n] as

x[n]=

[

ej(ω^0 n+θ)

2

+

e−j(ω^0 n+θ)

2

]

u[n]

Applying the linearity property and using the above Z-transform whenα=ejω^0 and its conjugate

α∗=e−jω^0 , we get

X(z)=

1

2

[

ejθ

1 −ejω^0 z−^1

+

e−jθ

1 −e−jω^0 z−^1

]

=

1

2

[

2 cos(θ)−2 cos(ω 0 −θ)z−^1

1 −2 cos(ω 0 )z−^1 +z−^2

]

=

cos(θ)−cos(ω 0 −θ)z−^1

1 −2 cos(ω 0 )z−^1 +z−^2

(9.12)

ExpressingX(z)in terms of positive powers ofz, we get

X(z)=

z(zcos(θ)−cos(ω 0 −θ))

z^2 −2 cos(ω 0 )z+ 1

=

z(zcos(θ)−cos(ω 0 −θ))

(z−ejω^0 )(z−e−jω^0 )

(9.13)

which is valid for any value ofθ. Ifx[n]=cos(ω 0 n)u[n], thenθ=0 and the poles ofX(z)are a

complex conjugate pair on the unit circle at frequencyω 0 radians. The zeros are atz=0 andz=

cos(ω 0 ). Whenx[n]=sin(ω 0 n)u[n]=cos(ω 0 n−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(ω 0 +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(ω 0 −θ)/cos(θ).

For simplicity, we letθ=0. Ifω 0 =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 whenω 0 =0 andθ= 0

isx[n]=cos( 0 n)u[n]=u[n]. When the frequencyω 0 >0 the poles move along the unit circle from

the lowest(ω 0 =0 rad)to the highest(ω 0 =πrad)frequency.