9.4 One-Sided Z-Transform 525The Z-transform pairs of a cosine and a sine are, respectively,cos(ω 0 n)u[n] ⇔z(z−cos(ω 0 ))
(z−ejω^0 )(z−e−jω^0 )ROC :|z|> 1 (9.14)sin(ω 0 n)u[n] ⇔
zsin(ω 0 )
(z−ejω^0 )(z−e−jω^0 )ROC :|z|> 1 (9.15)The Z-transforms for these sinusoids have identical poles 1 e±jω^0 , but different zeros. The frequency of the
sinusoid increases from the lowest(ω 0 =0 rad)to the highest(ω 0 =πrad)) as the poles move along the unit
circle from 1 to− 1 in its lower and upper parts.Consider then the signalx[n]=rncos(ω 0 n+θ)u[n], which is a combination of the above cases. As
before, the signal is expressed as a linear combination
x[n]=[
ejθ(rejω^0 )n
2+
e−jθ(re−jω^0 )n
2]
u[n]and it can be shown that its Z-transform is
X(z)=z(zcos(θ)−rcos(ω 0 −θ))
(z−rejω^0 )(z−re−jω^0 )(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 ofrdecreases 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 pairrncos(ω 0 n+θ)u[n] ⇔
z(zcos(θ)−rcos(ω 0 −θ))
(z−rejω^0 )(z−re−jω^0 )(9.17)shows how complex conjugate pairs of poles inside the unit circle represent the damping indicated by the
radiusrand the frequency given byω 0 in radians.Double poles are related to the derivative ofX(z)or to the multiplication of the signal byn. If
X(z)=∑∞
n= 0x[n]z−nits derivative with respect tozis
dX(z)
dz=
∑∞
n= 0x[n]dz−n
dz=−z−^1∑∞
n= 0nx[n]z−n