Introduction to the z-transform

Overview  

    The z-transform is useful for the manipulation of discrete data sequences and has acquired a new significance in the formulation and analysis of discrete-time systems.  It is used extensively today in the areas of applied mathematics, digital signal processing, control theory, population science, economics.  These discrete models are solved with difference equations in a manner that is analogous to solving continuous models with differential equations.  The role played by the z-transform in the solution of difference equations corresponds to that played by the Laplace transforms in the solution of differential equations.

9.1  The  z-transform

    The function notation for sequences is used in the study and application of z-transforms.  Consider a function    defined for    that is sampled at times  ,  where    is the sampling period (or rate).   We can write the sample as a sequence using the notation  .  Without loss of generality we will set    and consider real sequences such as,  .  The definition of the z-transform involves an infinite series of the reciprocals  .  

Definition 9.1 (z-transform)  Given the sequence    the z-transform is defined as follows

(9-1)            ,

which is a series involving powers of  .  

Remark 9.1.  The z-transform is defined at points  where the Laurent series (9-1) converges.  The z-transform region of convergence (ROC) for the Laurent series is chosen to be

            ,    where    .

Remark 9.2.  The sequence notation    is used in mathematics to study difference equations and the function notation    is used by engineers for signal processing.  It's a good idea to know both notations.  

Remark 9.3.  In the applications, the sequence    will be used for inputs and the sequence    will be used for outputs.  We will also use the notations

            ,  and

            .  

Theorem 9.1 (Inverse z-transform)   Let    be the z-transform of the sequence    defined in the region  .  Then  is given by the formula

(9-2)            ,
      
where  is any positively oriented simple closed curve that lies in the region    and winds around the origin.  

Proof.

9.1.1  Admissible form of a z-transform

    Formulas for   do not arise in a vacuum.  In an introductory course they are expressed as linear combinations of z-transforms corresponding to elementary functions such as
  
                       . 

In Table 9.1, we will see that the z-transform of each function in    is a rational function of the complex variable  .  It can be shown that a linear combination of rational functions is a rational function.  Therefore, for the examples and applications considered in this book we can restrict the z-transforms to be rational functions.  This restriction is emphasized this in the following definition.

Definition 9.2 (Admissible z-transform)  Given the z-transform    we say that  is an admissible z-transform, provided that it is a rational function, that is

(9-3)            ,

where , are polynomials of degree , respectively.

    From our knowledge of rational functions, we see that an admissible z-transform is defined everywhere in the complex plane except at a finite number of isolated singularities that are poles and occur at the points where  .  The Laurent series expansion in  (9-1) can be obtained by a partial fraction manipulation and followed by geometric series expansions in powers of .  However, the signal feature of formula (9-3) is the calculation of the inverse z-transform via residues.  

Theorem 9.2 (Cauchy's Residue Theorem)  Let  D  be a simply connected domain, and let  C  be a simple closed positively oriented contour that lies in  D.  If  f(z)  is analytic inside  C  and on  C,  except at the points    that lie inside  C,  then

            .  

Proof.

Corollary 9.1 (Inverse z-transform)   Let    be the z-transform of the sequence .  Then  is given by the formula

            .

where    are the poles of  .  

Corollary 9.2 (Inverse z-transform)   Let    be the z-transform of the sequence.  If     has simple poles at the points    then   is given by the formula

            .  

Example 9.1.  Find the z-transform of the unit pulse or impulse sequence   .  

Solution 9.1.  This follows trivially from Equation (9-1)    

            .  

Explore Solution 9.1.

Example 9.2.  The z-transform of the unit-step sequence     is   .  

Solution 9.2.  From Equation (9-1)    

               

Explore Solution 9.2.

Example 9.3.  The z-transform of the sequence        is   .  

Solution 9.3.  From Definition 9.1

            .  

Explore Solution 9.3.

Example 9.4.  The z-transform of the exponential sequence      is   .  

Solution 9.4.  From Definition 9.1

            

Explore Solution 9.4.

9.1.2  Properties of the z-transform

    Given that      and   .  We have the following properties:
  
(i)    Linearity.            .

(ii)   Delay Shift.        .

(iii)  Advance Shift.        ,   or

                

(iv)   Multiplication by .    .  

Example 9.5 (a).   The z-transform of the sequence      is   .  

Example 9.5 (b).  The z-transform of the sequence      is   .  

Solution 9.5 (a).

                

Solution 9.5 (b).  This is left as an exercise for the reader.

Remark 9.4.  When using the residue theorem to compute inverse z-transforms, the complex form is preferred, i. e.

            

Explore Solution 9.5 (a).

9.1.3  Table of z-transforms

    We list the following table of z-transforms.  

It can also be used to find the inverse z-transform.       

Exploration

Theorem 9.3 (Residues at Poles)

(i)      If  has a simple pole at  ,  then the residue is 

            .

(ii)      If  has a pole of order  at  ,  then the residue is 

            .

(iii)      If  has a pole of order  at  ,  then the residue is 

            .  

Proof.

Subroutines for finding the inverse z-transform

Example 9.6.  Find the inverse z-transform   .   Use (a) series, (b) table of z-transforms, (c) residues.  

Solution 9.6.

Explore Solution 9.6.

    The following two theorems about z-transforms are useful in finding the solution to a difference equation.  

Theorem 9.4 (Shifted Sequences & Initial Conditions)  Define the sequence  and let    be its z-transform.  Then

    (i)          

    (ii)          

    (iii)           

Theorem 9.5 (Convolution)  Let    and    be sequences with z-transforms  , respectively.  Then

            
      
where the operation    is defined as the convolution sum  .  

Proof.

9.1.4  Properties of the z-transform

    The following properties of z-transforms listed in Table 9.2 are well known in the field of digital signal analysis.  

The reader will be asked to prove some of these properties in the exercises.      

Exploration

Example 9.7.  Given   .   Use convolution to show that the z-transform is  .  

Solution 9.7.

    Let both    be the unit step sequence, and both    and  .  Then
  
            ,

so that    is given by the convolution

            .

9.1.5  Application to signal processing

    Digital signal processing often involves the design of finite impulse response (FIR) filters.  A simple 3-point FIR filter can be described as
  
(9-4)            .

Here, we choose real coefficients  so that the homogeneous difference equation

(9-5)            

has solutions  .  That is, if the linear combination    is input on the right side of the FIR filter equation, the output  on the left side of the equation will be zero.  

    Applying the time delay property to the z-transforms of each term in (9-4), we obtain  .  Factoring, we get

(9-6)            ,  where

(9-7)            

represents the filter transfer function.  Now, in order for the filter to suppress the inputs , we must have

            

and an easy calculation reveals that

            ,   and

            .

A complete discussion of this process is given in Section 9.3 of this chapter.

Example 9.8.  (FIR filter design)  Use residues to find the inverse z-transform      of   .
Then, write down the FIR filter equation that suppresses .  

Solution 9.8.

Explore Solution 9.8.

9.1.6  First Order Difference Equations

    The solution of difference equations is analogous to the solution of differential equations.  Consider the first order homogeneous equation  

             

where  is a constant.  The following method is often used.Trial solution method.

    Use the trial solution   ,   and substitute it into the above equation and get   .  Then divide through by 

and simplify to obtain  .  The general solution to the difference equation is

            .

    Familiar models of difference equations are given in the table below.       

Exploration

9.1.7  Methods for Solving First Order Difference Equations

    Consider the first order linear constant coefficient difference equation (LCCDE)

              with the initial condition   .

Trial solution method.

    First, solve the homogeneous equation     and get   .   Then use a trial solution that is appropriate for the sequence  on the right side of the equation and solve to obtain a particular solution  .   Then the general solution is
      
            .

The shortcoming of this method is that an extensive list of appropriate trial solutions must be available.  Details can be found in difference equations textbooks.  We will emphasize techniques that use the z-transform.  

z-transform method.

(i)    Use the time forward property    and take the z-transform of each term and get

            
  
(ii)    Solve the equation in (i) for  .

(iii)    Use partial fractions to expand    in a sum of terms, and look up the inverse z-transform(s) using Table 1, to get

            

Residue method.

    Perform steps (i) and (ii) of the above z-transform method.  Then find the solution using the formula

(iii)            .

where    are the poles of   .  

Convolution method.

(i)    Solve the homogeneous equation      and get   .

(ii)    Use the transfer function   

    and construct the unit-sample response   .

(iii)    Construct the particular solution   ,  

    in convolution form   .

(iv)    The general solution to the nonhomogeneous difference equation is

            .

(v)     The constant    will produce the proper initial condition  .  Therefore,

             .

Remark 9.6.  The particular solution    obtained by using convolution has the initial condition

               

Example 9.9.  Solve the difference equation    with initial condition  .
9 (a).  Use the z-transform and Tables 9.1 - 9.2 to find the solution.
9 (b).  Use residues to find the solution.

Solution 9.9.

Explore Solution 9.9.

Example 9.10.  Solve the difference equation    with initial condition  .
9.10 (a).  Use the z-transform and Tables 9.1 - 9.2 to find the solution.
9.10 (b).  Use residues to find the solution.

Solution 9.10.

Explore Solution 9.10.

Example 9.11.  Given the repeated dosage drug level model   with the initial condition  . 
9.11 (a).  Use the trial solution method.
9.11 (b).  Use z-transforms to find the solution.
9.11 (c).  Use residues to find the solution.
9.11 (d).  Use convolution to find the solution.

Solution 9.11.

    An illustration of the dosage model using the parameters  and initial condition   is shown in Figure 1 below.

            Figure 9.1.  The solution to  with .