CHAPTER 5 Signals, Systems, and Transforms
5.11 The Z Transform
The Z transform generalizes the DFT in roughly the same way as the Laplace transform generalizes the Fourier transform.
The integral in the Fourier transform can be viewed as the limiting sum of a series of unmodulated complex exponentials as their inter-frequency separation tends to zero, The Laplace transform generalizes this by taking the limiting sum over a series of modulated complex exponentials. In the same way, the DFT is a finite sum of unmodulated complex exponentials. The Z transform generalizes this to an infinite sum of modulated complex exponentials8. However, the notation of the Z transform is particularly simple because the complex transform variable, denoted z, is written in the a+jb form rather than the equiva- lent modulated complex exponential form.
We will consider only causal signals, that is, signals x[k] such that x[k] = 0 for k<0. We define the Z transform of such a sig- nal as the infinite sum . We denote this by
(EQ 69)
As with the Laplace transform, the infinite series defining the Z transform converges only for certain values of z, which define its region of convergence.
The inverse Z transform is given by:
(EQ 70)
8. A careful reader may note that we are simultaneously generalizing in two dimensions: from a finite sum to an infinite sum and from a sum of unmodulated complex exponentials to a sum of modulated complex exponentials. Indeed, there is an intermediate step, the discrete Laplace transform, which corresponds to a finite sum of modulated complex exponentials, that we have glossed over in this exposition.
ωF–k ωE–k ωN–k
ωN–k e–j
4π
---8 π 2---
⎝ ⎠–
cos⎛ ⎞ j π
---2
⎝ ⎠– sin⎛ ⎞
+ –j
= = =
ωE–k = ωN–2k = –1 ω–kF = ωE–2k = 1
1
8--- 4(( )–1 2( +6)) j 8---
⎝ ⎠–
⎛ ⎞((1+5)–1 3( +7))
+ 1
8---( )–4 j 8---
⎝ ⎠–
⎛ ⎞( )–4
+ –0.5+j0.5
= =
x k[ ]z–k
k=0
∞
∑
x k[ ]↔Z X z( ) x k[ ]z–k
k=0
∞
∑
=
x k[ ] 1 2πj
--- X z( )zk–1dz
C
∫°
=
DRAFT - Version 2 - The Z Transform
where C is a circle with its center at the origin of the z plane such that all values z such that (its poles) are inside this circle. As with the Laplace transform, the inverse is usually found from a table of standard transforms and the properties of the Z transform. Similarly, the concepts of poles and zeroes introduced for the Laplace transform continue to hold for the Z transform, as the next examples show.
EXAMPLE 27: Z TRANSFORMOFAUNITSTEP
Compute the Z transform of the discrete unit step signal defined by x[k] =1 for .
Solution: By definition, which converges when | | > 1, which is the region of conver- gence of this transform. Recall that z is complex, so this is the set of points outside a unit circle in the complex plane centered on the origin. Note that, unlike the Laplace transform, in the case of the Z transform, the region of convergence is expressed in terms of circular regions or their intersections, rather than half-planes.
EXAMPLE 28: Z TRANSFORMOFADISCRETEEXPONENTIAL
Compute the Z transform of the signal where a is a complex constant and .
Solution: By definition, . The series converges and the
transform is defined for the region where which is a circle of radius e-a centered at the origin. What if z = e-a? In this case, the denominator becomes zero and the transform’s value is infinite. This is called a pole of the system (the pole in the previous example was at 1). Intuitively, the series diverges when . We can ensure convergence by modulating this signal with the complex value z-1, but only when the absolute value of z is greater than e-a.
The values of z for which the transform vanishes are called the zeroes of the transform. This is illustrated by the next exam- ple.
EXAMPLE 29: Z TRANSFORMOFASINUSOID
Compute the Z transform of the discrete sinusoid .
Solution: We use Euler’s formula to rewrite the signal as . By definition,
, where, in the last step,
we used the result from the previous example, and we are assuming that . This reduces to , with the region of convergence (the entire z plane excluding a disk of radius 1). Note that the transform becomes infinite for (these are the poles of the transform) and is zero for z=0, which is the zero of the transform.
X z( )zk–1 = ∞
k≥0
X z( ) z–k
k=0
∞
∑
1---–1z–1 z z–1 ---= = = z
x k[ ] = e–ak k≥0
X z( ) e–akz–k
k=0
∞
∑
(e–az–1)kk=0
∞
∑
1–e1–az–1 --- zz–e–a ---
= = = =
e–az–1<1⇒ z >e–a
e–ak a<1
ωk [ ] cos
ejωk+e–jωk ---2
⎝ ⎠
⎛ ⎞
X z( ) ejωk+e–jωk ---2
⎝ ⎠
⎛ ⎞z–k
k=0
∞
∑
12--- ejωkz–kk=0
∞
∑
12--- e–jωkz–kk=0
∞
∑
+ 1
2--- 1 1–ejωz–1
--- 1 1–e–jωz–1 ---
⎝ + ⎠
⎛ ⎞
= = =
z >e–jω z2–zcosω z2–2zcosω+1 --- z > e–jω = 1
z = ±ejω
DRAFT - Version 2 -Relationship between Z and Laplace transform
163
It is important to keep track of the region of convergence of the transform. As with the Laplace transform, it can be shown that two time-domain functions that are completely different may have an identical transform, other than the region of con- vergence. The transform is unique only if the region is also specified. In general, the region of convergence of the Z trans- form is given by the annulus in the complex plane specified by .
5.11.1 Relationship between Z and Laplace transform
(This section can be skipped during a first reading) The Z transform is a compact representation of the Laplace transform of a discrete-time signal. We now discuss how the Z transform’s z auxiliary variable relates to the s auxiliary variable used in the Laplace transform.
Recall the following facts:
• The Laplace transform of a continuous signal is the Fourier transform of the signal after it has been modulated by the real exponential
• A discrete-time signal whose values are separated in time by T seconds can be viewed as the product of a corresponding continuous-time signal and an impulse train with impulses spaced T seconds apart
• The transform of a discrete-time impulse train with impulses spaced T seconds apart is an impulse train in the frequency domain with impulses spaced Hz apart
• The Fourier transform of the product of two signals is the convolution of their Fourier transforms
• The convolution of a signal with an impulse train replicates the signal
Given these facts, we see that the Laplace transform of a discrete-time signal with values spaced apart by T seconds results in infinitely replicating the transform of the signal modulated by the real exponential with a period of . Therefore, it is possible to fully describe the signal in the frequency domain considering only frequencies in the range . This corre- sponds to values of the Laplace variable that lie in band , as shown by the two dashed hori- zontal lines in Figure 14.
The Z transform denotes these values with the z variable, using the relationship . Representing z in the form , we see that
(EQ 71)
As a result of this mapping, lines parallel to the Y axis in the s plane correspond to circles in the z plane and lines parallel to the X axis in the s plane correspond to radial lines in the z plane. This is shown in Figure 14, where the two vertical lines marked by a diamond and a cross in the s plane are transformed to circles in the z plane and the two horizontal dashed lines marked A and B map to the same radial line in the z plane.
The Y axis in the s plane corresponds to the unit circle in the z plane. The left half-plane in the s domain corresponds to points within the unit circle in the z plane. Note that the vertical line marked with a diamond lies in the left half-plane of the s plane. Therefore, it lies within the unit circle in the z plane. Similarly, point X in the s plane, that lies in the left half-plane, is mapped to a point within the unit circle in the z plane, and point Y in the s plane, that lies in the right half-plane is mapped to a point outside the unit circle. Moreover, because Y has a larger value than X, it is rotated further in the anti-clockwise direction in the z plane than X.
α< z <β
e–σt
2π T---
e–σt 2π
T--- 0 π T---
±
s = σ+jω σ jπ
T---
– s σ jπ
T---
≤ ≤ +
z = esT = eσTejωT z = Aejθ
A = eσT θ = ωT
ω
DRAFT - Version 2 - The Z Transform
FIGURE 14. Mapping from the s plane to the z plane
EXAMPLE 30: MAPPINGFROMTHESPLANETOTHEZPLANE
What is the z value corresponding to the s value 3+j2 assuming that T = 1? Express this value in both polar and Cartesian coordinates.
Solution: The z value is given by . Using Euler’s formula, we can write this as .
5.11.2 Properties of the Z transform
The Z transform shares many of the properties of the Fourier and Laplace transforms, namely linearity, time-shifting, time- scaling, differentiation, and the convolution property, though in a slightly different form. We summarize some important properties in Table 5. For simplicity, we implicitly assume that signal values are spaced a unit time step apart, i.e., T =1.
Property Pre-condition Post condition Notes
Linearity a and b are arbitrary
constants and can be complex. This assumes that no pole-zero cancella- tions are involved.
Time shifting Shifting a signal a
steps forward multi- plies its transform by .
Scaling in the z domain
An expansion in the frequency scale compresses the time scale
σ jω
s plane π/T
-π/T σ2+jω2
●
z plane A
B
A
B
● σ1+jω1
● ●
θ=0 θ=π/2
θ=π
θ=3π/2
r=eσ2T,θ=ω2T
r=eσ1T,θ=ω1T X
Y
Y
X
♢
♢
✚
✚
♢
✚
Aejθ esT = e3+j2 = e3ej2 = 20.08ej2 20.08(cos( )2 +jsin( )2 ) = 20.08(–0.42+j(0.91)) = –8.36+j18.26
x1[ ]k ↔Z X1( )z α1< z <β1
x2[ ]k ↔Z X2( )z α2< z <β2
ax1[ ]k +bx2[ ]k ↔Z aX1( )z +bX2( )z
max(α1,α2)< z <min(β1,β2)
x k[ ]↔Z X z( ) α< z <β Z
x k[ –a]↔z–aX z( ) α< z <β
except z=0 if k>0 and z=∞ if k<0 z–a x k[ ]↔Z X z( ) α< z <β Z
akx k[ ] X z a---
⎝ ⎠⎛ ⎞
↔ (aα< z < aβ)
DRAFT - Version 2 -Properties of the Z transform
165
TABLE 5. Some properties of the Z transform
These properties, along with the a table of common transforms (Table 6) allow us to derive the transform of most common signals without having to derive them from first principles.
TABLE 6. Some standard Z transforms
EXAMPLE 31: SOLVINGASYSTEMUSINGTHE Z TRANSFORM
Time reversal Reversing the direc-
tion of a signal is equivalent to using
instead of z in the expression of the transform
Time-domain convolution
Convolution in the time domain corre- sponds to multipli- cation in the transform domain First differ-
ence
The first difference is equivalent to dif- ferentiation in the time domain
No. Signal x[k] X[z] Region of convergence
1 Delta or unit impulse 1 All z
2 Delayed delta
3 Unit step u[k]
4 Ramp ku[k]
5 Exponential
6 Sine
7 Cosine
8 Exponentially modu- lated sine
9 Exponentially modu- late cosine
Property Pre-condition Post condition Notes
x k[ ]↔Z X z( ) α< z <β Z
x[ ]–k ↔X z( –1) 1
β--- z 1 α---
< <
z–1
x1[ ]k ↔Z X1( )z α1< z <β1 x2[ ]k ↔Z X2( )z α2< z <β2
x1[ ]k ⊗x2[ ]k ↔Z X1( )Xz 2( )z
max(α1,α2)< z <min(β1,β2)
x k[ ]↔Z X z( ) α< z <β Z
x k[ ]–x k[ –1]↔(1–z–1)X z( ) max(α,0)< z <β
δ[ ]k δ[k–k0]
z–k0 z≠0
1 1–z–1
--- z >1 z–1
1–z–1
( )2
--- z >1
aku k[ ] 1
1–az–1
--- z >a ω0k
( )u k[ ]
sin z–1sin(ω0)
1–2z–1cos(ω0)+z–2 ---
z >1
ω0k
( )
cos u k[ ] 1–z–1cos(ω0) 1–2z–1cos(ω0)+z–2 ---
z >1
aksin(ω0k)u k[ ] az–1sin(ω0) 1–2az–1cos(ω0)+a2z–2 ---
z >a
akcos(ω0k)u k[ ] 1–az–1cos(ω0) 1–2z–1cos(ω0)+(a2z)–2 ---
z >a
DRAFT - Version 2 - Further Reading
Consider a discrete-time system such that the Z transform H(z) of its transfer function h[k] is . What is its response to
the input ?
Solution: Since , from Row 5 of Table 6, . Therefore, . Using
the method of partial fractions (i.e., writing the expression as and solving for a and b), we find that
. To find the inverse transform, recall that the Z transform is linear, so we only need to find the
inverse transform of each term in isolation. From Row 3 and Row 5 of Table 6, we get , which is a discrete unit step of height 2 to which is added a decaying discrete exponential.