ABSTRACT
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Engineers (or IEEE) Code of ethics is presented here to acquaint students with ethical behavior in engineering professions
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The z-transform is the discrete-time counterpart of the Laplace transform It puts at our disposal a compact tool for describing a broad variety of systems and their properties
Just as Laplace transform is useful in handling signals that do not have Fourier transform, the z-transform enables us to treat discrete-time signals that do not have discrete-time Fourier transforms (DTFT) (For example, the unit sequence u[n] does not have a DTFT) Also, just as the Laplace transform converts integrodifferential equations into algebraic equations, the z-transform converts difference equations into algebraic equations that are easier to manipulate and solve Therefore, techniques in this chapter parallel the techniques used in Chapter 3 on Laplace transform Although the properties of the z-transform are similar to those of the Laplace transform, there are some differences Like Laplace transform, the z-transform is applicable to systems with initial conditions
We begin by defining the z-properties and studying its important properties The z-transform of some common discrete-time signals is derived We present methods for finding the inverse z-transform Two important applications of the z-transform are discussed Finally, we consider using MATLAB® to find z-transform and its inverse
The z-transform is the generalization of the discrete-time Fourier transform (DTFT) We recall that from Chapter 6, the DTFT of a signal x[n] is given by
X x n e j n
( ) [ ]W = - W =-¥
Inserting a factor ρ−n in Equation 71 leads to
X x n e x n en j n
( ) [ ] [ ]( )W = =- - W =-¥
å år r (72)
If we let z = ρejΩ Then, Equation 72 becomes
X z x n z n
( ) [ ]= - =-¥
This is the two-sided (or bilateral) z-transform The one-sided (or unilateral) z-transform is
X z x n x n z n
( ) { [ ]} [ ]= = - =
(74)
Only the one-sided z-transform will be discussed in this chapter X(z) and x[n] constitute a z-transform pair
x n X z Z[ ] ( )¾ ®¾ (75)
Note the following: 1 From Equation 74, we notice that the z-transform of x[n] is a power series
of z−1 whose coefficients are x[n], that is,
X z x n z x x z x z x zn
( ) [ ] [ ] [ ] [ ] [ ]= = + + + +- =
1 2 30 1 2 3 (76)
In this power series, z −n can be interpreted as indicating the nth sampling instant
2 We also notice that the sum in Equation 74 may not converge for all values of z Each z-transform is associated with a region in the complex plane for which the transform exists We will discuss this further in the next section
3 Generally, z is a complex variable with z = ∑ + jΩ just as the Laplace transform variable s = σ + jω z as a complex number z = ρejΩ is illustrated in Figure 71
4 In our development of z-transforms, we will be using the formula for the geometric series
1 1 1
+ + + + = = -
åa a a a an n
(77)
which is convergent if and only if |a| < 1
Example 7.1
Example 7.2
Just like Laplace transform, the z-transform X(z) converges only for certain range of z known as the region of convergence (ROC) In order for the z-transform to be complete, its ROC must be specified The ROC specifies the region where the z-transform converges The ROC may be within a circle (such as |z| < |a|, as shown typically in Figure 72a), outside a circle (such as |z| > |a|, as typically shown in Figure 72b), or within an annulus (such as |a| < |z| < |b|, as shown typically in Figure 72c) An ROC does not include a pole because the z-transform does not converge at a pole
To illustrate how to find the ROC, let us consider the following examples
Example 7.3
X z a z a z
a
z
a
( ) = = + + + +- =
Example 7.4
We now consider some basic properties of the z-transform We use those properties to derive the z-transform of some signals and develop a z-transform table We first state each property, prove it, and then illustrate it with examples
Let x[n] and y[n] be two discrete-time signals with transforms X(z) and Y(z), respectively The z-transform of their linear combination is
Z ax n by n aX z bY z[ ] [ ] ( ) ( )+{ } = + (78)
where a and b are constants For example, from Example 73 and Practice Problem 73, we know that
a u n
z
z a z a u n
z
z zn [ ] , [ ] , | |«
- < «
- >and
1 1 (79)
By the linearity property,
2 3 2 3
1 1a u n u n z
z a
z
z z a zn [ ] [ ] , | | | | | |+ =
- +
- > >and (710)
This properties states that if X(z) is the z-transform of x[n], then
x n m z X zm[ ] ( )- « - (711)
assuming zero initial conditions To prove this, we use Equation 74
Z{ [ ]} [ ]x n m x n m z n n
- = - -
By changing variables k = n − m, we get
Z{ [ ]} [ ] [ ] ( )( )x n m x k z z x k z z X zm k m k
- = = =- + -
(712)
This assumes that the signal is causal (x[n] = 0 for n < 0) For example, when m = 1 or −1, we have
x n z X z
x n zX z
[ ] ( ) [ ] ( )
- «
+ «
-1
(713)
For nonzero initial conditions, the time delay property becomes
x n m z X z z x z x m x m m m[ ] ( ) [ ] [ ] [ ]- « + - + + - + + -- - + -1 11 1 (714)
If X(z) is the z-transform of x[n], then
a x n X z a
n [ ] « æ è ç
ö ø ÷ (715)
This implies that multiplication by an in the time domain corresponds to scaling in the frequency or z domain
By definition, the z-transform of the sequence anx[n] is
Z a x n a x n z x n z
a
X z a
[ ] [ ] [ ]{ } = = æ è ç
ö ø ÷
= æ è ç
ö ø ÷
(716)
This property can be used to find the z-transform of signals multiplied by an exponential signal For example,
e x n X e z j n jW - W«0 0[ ] ( ) (717)
We know that
cos( ) [ ] [ ]W = +( )W - Wn u n e e u nj n j n12 (718)
and that
u n
z
z [ ] «
-1 (719)
e u n
ze
ze
z
z e j n
W - W« - =
- [ ]
1 (720)
Similarly,
e u n
z
z e j n
W« - [ ] (721)
Combining Equations 720 and 721 gives
Z nu n z
z e
z
z e
z z
z zj j cos [ ] ( cos )
cos W{ } =
- +
- é ëê
ù ûú
= - W
- W +- W W 1 2 2 12
(722)
If x[n]↔X(z), then
x n X z
[ ]- « æ è ç
ö ø ÷
1 (723)
To prove this, we use Equation 74
Z{ [ ]} [ ]x n x n z n n
- = - -
Replacing −n with k,
Z{ [ ]} [ ] [ ]x n x k z x k
z X
z k
- = = æ è ç
ö ø ÷ =
æ è ç
ö ø ÷å å
-1 1 (724)
For example, from Equation 717,
u n
z
z [ ] «
-1
Applying Equation 724,
u n
z
z z [ ]- «
- =
- 1
1 1 1
1 /
/ (725)
The modulation property is very important in communication theory and digital signal processing If x[n]↔X(z), then
(cos ) [ ] ( ) ( )W « +éë ùû
W - Wn x n X e z X e zj j1 2
(726)
(sin ) [ ] ( ) ( )W « -éë ùû
W - Wn x n j X e z X e zj j 2
(727)
To prove Equations 726 and 727, we use Euler’s identity
cos [ ] [ ] [ ]W = +( )- W Wnx n e x n e x nj n j n12 (728)
sin [ ] [ ] [ ]W = -( )- W Wnx n j e x n e x nj n j n2 (729)
From Equation 717,
e x n X e z j n jW - W«0 0[ ] ( ) (730)
e x n X e z j n j-W W«0 0[ ] ( ) (731)
Applying Equation 730 and 731 on Equations 728 and 729 produces Equations 726 and 727 We can use this property to find the z-transforms of cosine and sine signals We already got the z-transform of cosine in Equation 722
Let y[n] be the summation or accumulation of the discrete-time signal x[n], with x[n] = 0 for n = −1, −2, −3,… Then
y n x k k
[ ] [ ]= =
(732)
The accumulation property states that
x k z z
X z k
[ ] ( ) =
å « - 0
1 (733)
It states that the z-transform of the sum of x[n] is equal to the z/(z − 1) times the z-transform of x[n]
To prove Equation 733, we write
y n x k x n k
[ ] [ ] [ ]= + =
(734)
This may be written as
y n y n x n[ ] [ ] [ ]= - +1 (735)
Taking the z-transform of both sides and applying the time-shifting property, we obtain
Y z
z Y z X z( ) ( ) ( )= +1 (736)
Solving for Y(z) yields
Y z
z X z z
z X z( ) ( ) ( )=
- =
-- 1
1 11 (737)
This property plays a very useful role in linear system theory It states that if
x n X z h n H z[ ] ( ) [ ] ( )« «and (738)
then their time convolution in the time domain is equivalent to their product in the frequency or z domain, that is,
y n x n h n Y z X z H z[ ] [ ]* [ ] ( ) ( ) ( )= « = (739)
Since the property applies to causal as well as noncausal signals, we prove it to the general case of noncausal signals, where the convolution sum ranges from −∞ to ∞ By definition,
Y z x n h n z x n h n k zn n k
( ) [ ]* [ ] [ ] [ ]= = - æ
è çç
ö
ø ÷÷
=
x k h n k z n nk
[ ] [ ]- æ
è çç
ö
ø ÷÷
(740)
We note that the term in brackets is the z-transform of the shifted signal h[n − k] By applying the time-shifting property, we get
Y z x k z H z x k z H z X z H zk k
( ) [ ] ( ) [ ] ( ) ( ) (= ( ) = æ
è çç
ö
ø ÷÷ =
The initial x[0] and the final value x[∞] are given by the initial and value theorems, which are stated as
x X z
z [ ] lim ( )0 =
®¥ (742)
x z X z
z [ ] lim( ) ( )¥ = -
11 (743)
assuming that x[∞] exists
To prove Equation 742, we write Equation 74 explicitly as
X z x x z x z x k z k( ) [ ] [ ] [ ] [ ]= + + + + +- - -0 1 21 2 (744)
As z→∞, the terms z−k →0 for all k > 0 Hence, Equation 744 becomes
lim ( ) [ ] z
X z x ®¥
= 0
To prove Equation 743, we apply the time-shifting property
Z{ [ ] [ ]} ( ) ( )x n x n z X z-- = - -1 1 1 (745)
The left-hand side of Equation 745 can be written as
{ [ ] [ ]} lim { [ ] [ ]}x n x n z x n x n z n
N n-- = - -
(746)
If we take the limit as z→1, combining Equations 745 and 746 produces
lim( ) ( ) lim { [ ] [ ]}
lim ( ) ( )
z X z x n x n
x N x
- = - -
= = ¥
1 1
(747)
The initial value theorem can be used to check the z-transform of a given signal The final value theorem gives the correct result only when x[∞] exists It can be shown that if there are poles on or outside the unit circle, except for a single pole at z = 1, the final value theorem does not apply
For example, let us find the initial and final values of a unit step function, u[n] We know that u[0] = 1 = u[∞] By the initial value theorem,
u U z z
zz z [ ] lim ( ) lim0
1 1= =
- æ è ç
ö ø ÷ =®¥ ®¥
(748)
By the final value theorem,
x z X z z
z
z
zz z [ ] lim( ) ( ) lim¥ = - = -
- æ è ç
ö ø ÷ =®
1 1 1
1 1 (749)
The theorems confirm what we know already
These properties of the z-transform are listed in Table 71, while Table 72 provides a collection of the z-transforms of common signals
Example 7.5
TABLE 7.1 Properties of the z-Transforms
TABLE 7.2 Some Common z-Transform Pairs
Example 7.6
Example 7.7
Example 7.8
Example 7.9
The problem of finding the inverse z-transform is the problem of finding x[n] for a given X(z) The direct way to do this is to evaluate the contour integral:
x n j X z z dz n[ ] ( )= -ò12 1p
G (750)
where Γ is a closed contour that encloses the origin and includes all poles of X(z) Γ is usually determined by the region of convergence of the summation in Equation 74 It encircles all singularities in the ROC This direct method of finding the inverse transformation is computationally involved and will not be pursued further We would rather use two other methods: long division expansion and partial fraction expansion
To find the inverse z-transform of X(z), we expand X(z) = N(z)/D(z) in a power series in z−1 by using long division The values of x[n] are taken as the coefficients of the series expansion The long division becomes tedious if more than few values are needed This major deficiency limits the usefulness of the long division approach
Example 7.10
Example 7.11
We use partial fraction for finding the inverse z-transform in the same way we used it to find the inverse Laplace transform in Section 34 In using partial fraction, a function X(z) that does not appear in the z-transform in Table 72 is expressed as a sum of functions that are listed in the table The inverse z-transform is then computed term by term using the appropriate entries in Table 72 We may also need to use the z-transform properties in Table 71 We will illustrate with examples for the cases of simple poles, repeated poles, and complex poles
Example 7.12
Example 7.13
Example 7.14
The z-transform is fundamentally important to digital signal processing, digital communications, and linear control systems In this section, we consider the use of the z-transform in the analysis of discrete-time linear systems We will apply the z-transform to two areas: linear difference equation and transfer function
Most discrete-time systems of practical interest can be described by linear difference equations We can use the z-transform to solve the difference equation for n ≥ 0 To do this we first take the z-transform of the two sides of the difference equation Next, we solve algebraically the transformed difference equation At this point, we can evaluate Y(z) and invert it using Table 72 to determine y[n]
To obtain the complete solution requires that we know the input signal x[n] For Nth order difference equation, we must also know N initial conditions Using the initial conditions, we obtain the z-transform of the delayed output signals as follows:
Z y n z Y z y
Z y n z Y z z y y
Z y
[ ] ( ) [ ]
[ ] ( ) [ ] [ ]
[
-{ } = + -
-{ } = + - + -
1 1
2 1 2
n z Y z z y z y y-{ } = + - + - + -- - -3 1 2 33 2 1] ( ) [ ] [ ] [ ]
(751)
In general,
Z y n m z Y z z y n zm m n
{ [ ] ( ) [ ]- = + -- - =
(752)
Similarly, the z-transforms of advanced output signals are
Z y n zY z zy
Z y n z Y z z y zy
Z y n
{ [ ]} ( ) [ ] { [ ]} ( ) [ ] [ ] { [ ]
+ = -
+ = - -
+
1 0
2 0 1
} ( ) [ ] [ ] [ ]= - - -z Y z z y z y zy3 3 20 1 2 (753)
In general,
Z y n m z Y z z y n zm m n
{ [ ] ( ) [ ]+ = - - =
(754)
We will illustrate this with an example
Example 7.15
The transfer function plays an important role in the analysis of discrete-time linear systems In this section, we will define the discrete-time transfer function and how to find the system impulse and step responses We will also learn how to use the transfer function to determine system performance characteristics such as stability and frequency response
The relationships between the system input x[n], output y[n], and impulse response and their z-transforms are given in Equation 739, namely,
y n x n h n Y z X z H z[ ] [ ]* [ ] ( ) ( ) ( )= « = (755)
From this, we obtain the transfer function H(z) as
H z Y z
X z ( ) ( )( )= (756)
This implies that the transfer function is the ratio of the z-transforms of the output and input We obtain the transfer function by either transforming the difference equation or transforming the system representation
When the system input is a unit impulse δ[n] (δ[n]↔1), the output of the system is equal to the transfer function H(z) Thus, the discrete-time system impulse response is given by
h n Z H z[ ] ( )= { }-1 (757)
When the input x[n] is a unit step u[n] so that X(z) = z/(z − 1), the corresponding output y[n] is called the step response.
