Homogeneous Difference Equations


9.2  Second Order Homogeneous Difference Equations

    Before proceeding with the
z-transform method, we mention a heuristic method based on substitution of a trial solution.  Consider the second order homogeneous linear constant-coefficient difference equation (HLCCDE)

(9-8)            [Graphics:Images/ZTransformDEMod_gr_1.gif],  

[Graphics:Images/ZTransformDEMod_gr_2.gif] are constants.  Using the trial solution  [Graphics:Images/ZTransformDEMod_gr_3.gif],  and direct substitution into (9-8) produces the equation  [Graphics:Images/ZTransformDEMod_gr_4.gif],  and dividing through by [Graphics:Images/ZTransformDEMod_gr_5.gif] produces the characteristic polynomial  [Graphics:Images/ZTransformDEMod_gr_6.gif]  and characteristic equation

(9-9)            [Graphics:Images/ZTransformDEMod_gr_7.gif].

There are three types of solutions to (9-8) which are determined by the nature of the roots in (9-9).  

    Case (i)  If   [Graphics:Images/ZTransformDEMod_gr_8.gif]  then we have real distinct roots   [Graphics:Images/ZTransformDEMod_gr_9.gif]  and  [Graphics:Images/ZTransformDEMod_gr_10.gif],  and  

(9-10)            [Graphics:Images/ZTransformDEMod_gr_11.gif].  

    Case (ii)  If   [Graphics:Images/ZTransformDEMod_gr_12.gif]   then we have a double real root   [Graphics:Images/ZTransformDEMod_gr_13.gif],  and  

(9-11)            [Graphics:Images/ZTransformDEMod_gr_14.gif].  

    Case (iii)  If   [Graphics:Images/ZTransformDEMod_gr_15.gif]   then we have complex roots   [Graphics:Images/ZTransformDEMod_gr_16.gif]  and  [Graphics:Images/ZTransformDEMod_gr_17.gif],  and    

(9-12)            [Graphics:Images/ZTransformDEMod_gr_18.gif].    
    The solution for case (iii) can also be written as the following linear combination
(9-13)            [Graphics:Images/ZTransformDEMod_gr_19.gif],  
            where  [Graphics:Images/ZTransformDEMod_gr_20.gif]  and  [Graphics:Images/ZTransformDEMod_gr_21.gif].  
Caution.  Be sure to use the value of arctan that lies in the range  [Graphics:Images/ZTransformDEMod_gr_22.gif].


Remark About Stability

    Stability depends on the location of the roots of the characteristic polynomial.  Without loss if generality, if  [Graphics:Images/ZTransformDEMod_gr_23.gif]  then both roots lie inside the unit circle and the solutions are asymptotically stable and tends to zero as  [Graphics:Images/ZTransformDEMod_gr_24.gif].   If  [Graphics:Images/ZTransformDEMod_gr_25.gif] and [Graphics:Images/ZTransformDEMod_gr_26.gif]  then a root lies on the unit circle and the solutions are stable.  If  [Graphics:Images/ZTransformDEMod_gr_27.gif]  then there is an unstable solution.  If  [Graphics:Images/ZTransformDEMod_gr_28.gif]  then at least one root lies outside the unit circle and there is an unstable solution.  


    The following subroutine uses the characteristic equation to construct solutions to a second order homogeneous difference equation.

Mathematica Subroutine (Solution of a Difference Equation).


Example 9.12.  Solve  [Graphics:Images/ZTransformDEMod_gr_30.gif]   with  [Graphics:Images/ZTransformDEMod_gr_31.gif].  


Solution 9.12.

Explore Solution 9.12.


Example 9.13.  Solve  [Graphics:Images/ZTransformDEMod_gr_51.gif]   with  [Graphics:Images/ZTransformDEMod_gr_52.gif].  


Solution 9.13.

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Example 9.14.  Solve  [Graphics:Images/ZTransformDEMod_gr_74.gif]   with  [Graphics:Images/ZTransformDEMod_gr_75.gif].  



Solution 9.14.

Explore Solution 9.14.



Higher Order Difference Equations

he general form of a [Graphics:Images/ZTransformDEMod_gr_104.gif] order linear constant coefficient difference equation (LCCDE), is  
(9-14)            [Graphics:Images/ZTransformDEMod_gr_105.gif]

[Graphics:Images/ZTransformDEMod_gr_106.gif] and [Graphics:Images/ZTransformDEMod_gr_107.gif]. The sequence [Graphics:Images/ZTransformDEMod_gr_108.gif] is given and the sequence [Graphics:Images/ZTransformDEMod_gr_109.gif] is output.  The integer  [Graphics:Images/ZTransformDEMod_gr_110.gif]  is the order of the difference equation.  The compact form of writing this difference equation is

(9-15)            [Graphics:Images/ZTransformDEMod_gr_111.gif].  

This formula can be expressed in a recursive form:

(9-16)            [Graphics:Images/ZTransformDEMod_gr_112.gif].  

    This form of the LCCDE explicitly shows that the present output  [Graphics:Images/ZTransformDEMod_gr_113.gif]  is a function of the past output values  [Graphics:Images/ZTransformDEMod_gr_114.gif],  for  [Graphics:Images/ZTransformDEMod_gr_115.gif]; and the present input  [Graphics:Images/ZTransformDEMod_gr_116.gif],  and the previous inputs  [Graphics:Images/ZTransformDEMod_gr_117.gif]  for  [Graphics:Images/ZTransformDEMod_gr_118.gif].  

    Now we would like to emphasize the method of z-transforms for solving difference equations.  Applying the linearity and time delay shift property of the z-transform to equation (9-15), we obtain

(9-17)            [Graphics:Images/ZTransformDEMod_gr_119.gif].

This can be rearranged as  [Graphics:Images/ZTransformDEMod_gr_120.gif]  and then solved for the quotient [Graphics:Images/ZTransformDEMod_gr_121.gif].  The sequence  [Graphics:Images/ZTransformDEMod_gr_122.gif] can be used to construct a particular solution to (9-14), i.e.  [Graphics:Images/ZTransformDEMod_gr_123.gif].  This solution can be expressed using the convolution sum, as follows:

(9-18)            [Graphics:Images/ZTransformDEMod_gr_124.gif].  

Remark.  This particular solution   [Graphics:Images/ZTransformDEMod_gr_125.gif] does not involve initial conditions for (9-14).  We will illustrate how to use convolution at the end of this section.  


Difference Equations with Initial Conditions

    Often times a difference equation involves only one input on the right hand side of (9-14) and we write

then we could shift the index and use the form


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

(9-19)            [Graphics:Images/ZTransformDEMod_gr_128.gif]

with the initial conditions  [Graphics:Images/ZTransformDEMod_gr_129.gif]  (and implicitly we have  [Graphics:Images/ZTransformDEMod_gr_130.gif]).

    Step (i)  Using the time forward properties  
        and take the z-transform of each term and get the equation  

(9-20)            [Graphics:Images/ZTransformDEMod_gr_134.gif].
    Step (ii)  Solve equation (9-20) for  [Graphics:Images/ZTransformDEMod_gr_135.gif].

    Step (iii)  Use partial fractions to expand  [Graphics:Images/ZTransformDEMod_gr_136.gif]  in a sum of terms, and look up the inverse z-transform(s) using Table 1, to get the solution  


    Step (iv)  Alternate calculation using residues.  Perform steps (i) and (ii) listed above.  Then find  [Graphics:Images/ZTransformDEMod_gr_138.gif]  using residues  


        where  [Graphics:Images/ZTransformDEMod_gr_140.gif]  are the poles of  [Graphics:Images/ZTransformDEMod_gr_141.gif].  

    Remark  The function  [Graphics:Images/ZTransformDEMod_gr_142.gif]  has real coefficients.  Hence, if , and if [Graphics:Images/ZTransformDEMod_gr_143.gif] and [Graphics:Images/ZTransformDEMod_gr_144.gif] are poles, then we can use the computational fact:

(9-21)             [Graphics:Images/ZTransformDEMod_gr_145.gif]

    We now show how to obtain answers to Examples 9.12-9.14 using z-transform methods.


    The following subroutine uses z-transforms to construct solutions to a second order homogeneous difference equation.

Mathematica Subroutines (Solution of a Difference Equation).


Example 9.15 (a).  Use z-transform methods to solve  [Graphics:Images/ZTransformDEMod_gr_148.gif]   with  [Graphics:Images/ZTransformDEMod_gr_149.gif].  

9.15 (b).  Use z-transform methods to solve  [Graphics:Images/ZTransformDEMod_gr_150.gif]   with  [Graphics:Images/ZTransformDEMod_gr_151.gif].  


Solution 9.15 (a).

Solution 9.15 (b).

Explore Solution 9.15 (a).

Explore Solution 9.15 (b).


Example 9.16 (a).  Use z-transform methods to solve  [Graphics:Images/ZTransformDEMod_gr_214.gif]   with  [Graphics:Images/ZTransformDEMod_gr_215.gif].  

9.16 (b).  Use z-transform methods to solve  [Graphics:Images/ZTransformDEMod_gr_216.gif]   with  [Graphics:Images/ZTransformDEMod_gr_217.gif].  


Solution 9.16 (a).

Solution 9.16 (b).

Explore Solution 9.16(a).

Explore Solution 9.16 (b).


Example 9.17 (a).  Use z-transform methods to solve  [Graphics:Images/ZTransformDEMod_gr_280.gif]   with  [Graphics:Images/ZTransformDEMod_gr_281.gif].  

9.17 (b).  Use z-transform methods to solve  [Graphics:Images/ZTransformDEMod_gr_282.gif]   with  [Graphics:Images/ZTransformDEMod_gr_283.gif].  


Solution 9.17 (a).

Solution 9.17 (b).

Explore Solution 9.17 (a).

Explore Solution  9.17 (b).


Example 9.18.  Solve  [Graphics:Images/ZTransformDEMod_gr_354.gif]   with  [Graphics:Images/ZTransformDEMod_gr_355.gif].  


Solution 9.18.

Explore Solution 9.18.


Example 9.19.  Solve  [Graphics:Images/ZTransformDEMod_gr_400.gif]   with  [Graphics:Images/ZTransformDEMod_gr_401.gif].  


Solution 9.19.

Explore Solution 9.19.



Convolution for Solving a Non-homogeneous Equation

(i)    Solve the homogeneous equation   [Graphics:Images/ZTransformDEMod_gr_446.gif]   and get  [Graphics:Images/ZTransformDEMod_gr_447.gif].  

(ii)    Use the transfer function   [Graphics:Images/ZTransformDEMod_gr_448.gif].    

(iii)    Construct the particular solution using convolution  

    [Graphics:Images/ZTransformDEMod_gr_449.gif],   or


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



Example 9.20 (a).  Find the general solution to  [Graphics:Images/ZTransformDEMod_gr_452.gif].

9.20 (b).  Find the general solution to  [Graphics:Images/ZTransformDEMod_gr_453.gif].   


            Figure 9.2.  A typical solution to  [Graphics:Images/ZTransformDEMod_gr_455.gif].  


            Figure 9.3.  A typical solution to  [Graphics:Images/ZTransformDEMod_gr_457.gif].  

Solution 9.20 (a).

Solution 9.20 (b).

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Explore Solution 9.20 (b).



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(c) 2006 John H. Mathews, Russell W. Howell