Exercise 4.  Solve the homogeneous difference equations.  

Exercise 4 (b).  [Graphics:Images/ZTransformDEModHome_gr_773.gif]   with   [Graphics:Images/ZTransformDEModHome_gr_774.gif].   
                          Hint.  Get  [Graphics:Images/ZTransformDEModHome_gr_775.gif].  

Solution 4 (b).

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/ZTransformDEModHome_gr_884.gif].  

Remark.   It is our goal to learn Z-transforms and use residues.  

Solution.   

Method 1.
  The characteristic equation  

                    [Graphics:../Images/ZTransformDEModHome_gr_885.gif]  

has roots [Graphics:../Images/ZTransformDEModHome_gr_886.gif].  

The general solution is  

                    [Graphics:../Images/ZTransformDEModHome_gr_887.gif].  

Solve the linear system  

                    [Graphics:../Images/ZTransformDEModHome_gr_888.gif]  

and get  [Graphics:../Images/ZTransformDEModHome_gr_889.gif]  and  [Graphics:../Images/ZTransformDEModHome_gr_890.gif].

Therefore,  

                    [Graphics:../Images/ZTransformDEModHome_gr_891.gif].

Method 2.  Take the z-transform of both sides and use the initial conditions   [Graphics:../Images/ZTransformDEModHome_gr_892.gif]:  

                    [Graphics:../Images/ZTransformDEModHome_gr_893.gif],  

and then get  

                    [Graphics:../Images/ZTransformDEModHome_gr_894.gif].  

New solve for  [Graphics:../Images/ZTransformDEModHome_gr_895.gif]  and obtain:  

                    [Graphics:../Images/ZTransformDEModHome_gr_896.gif].  

Using Tables.   Using Table 9.1 and the formulas   [Graphics:../Images/ZTransformDEModHome_gr_897.gif]   we get:  

                    [Graphics:../Images/ZTransformDEModHome_gr_898.gif]  

Remark.  The details for the partial fraction expansion are at the bottom of the page.

Therefore,  

                    [Graphics:../Images/ZTransformDEModHome_gr_899.gif].

We are done.   

Alternative Solution Using Tables.   Using ordinary partial fractions and Table 9.1  

and   
[Graphics:../Images/ZTransformDEModHome_gr_900.gif]   and   [Graphics:../Images/ZTransformDEModHome_gr_901.gif]   with   [Graphics:../Images/ZTransformDEModHome_gr_902.gif]   we get:  

                    [Graphics:../Images/ZTransformDEModHome_gr_903.gif]  

        Since   [Graphics:../Images/ZTransformDEModHome_gr_904.gif] and   [Graphics:../Images/ZTransformDEModHome_gr_905.gif]we can write  

                    [Graphics:../Images/ZTransformDEModHome_gr_906.gif]   and   

                    [Graphics:../Images/ZTransformDEModHome_gr_907.gif]   for   [Graphics:../Images/ZTransformDEModHome_gr_908.gif].  

Therefore, the solution has the following form:

                    [Graphics:../Images/ZTransformDEModHome_gr_909.gif]   

We are done.   

Aside.  The commands for the ordinary partial fraction expansion are:  

[Graphics:../Images/ZTransformDEModHome_gr_910.gif]

[Graphics:../Images/ZTransformDEModHome_gr_911.gif]


[Graphics:../Images/ZTransformDEModHome_gr_912.gif]

[Graphics:../Images/ZTransformDEModHome_gr_913.gif]

        Here   [Graphics:../Images/ZTransformDEModHome_gr_914.gif]   and   [Graphics:../Images/ZTransformDEModHome_gr_915.gif],   and we can corroborate this solution.

 

[Graphics:../Images/ZTransformDEModHome_gr_916.gif]

[Graphics:../Images/ZTransformDEModHome_gr_917.gif]
[Graphics:../Images/ZTransformDEModHome_gr_918.gif]

We are really done.   

Using Residues.   Calculate the residue of  [Graphics:../Images/ZTransformDEModHome_gr_919.gif]  at the pole   [Graphics:../Images/ZTransformDEModHome_gr_920.gif].  

                    [Graphics:../Images/ZTransformDEModHome_gr_921.gif]  

Therefore,  

                    [Graphics:../Images/ZTransformDEModHome_gr_922.gif]  

We are really really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ZTransformDEModHome_gr_923.gif]

[Graphics:../Images/ZTransformDEModHome_gr_924.gif]


[Graphics:../Images/ZTransformDEModHome_gr_925.gif]

[Graphics:../Images/ZTransformDEModHome_gr_926.gif]


[Graphics:../Images/ZTransformDEModHome_gr_927.gif]

[Graphics:../Images/ZTransformDEModHome_gr_928.gif]


[Graphics:../Images/ZTransformDEModHome_gr_929.gif]

[Graphics:../Images/ZTransformDEModHome_gr_930.gif]


[Graphics:../Images/ZTransformDEModHome_gr_931.gif]

[Graphics:../Images/ZTransformDEModHome_gr_932.gif]


[Graphics:../Images/ZTransformDEModHome_gr_933.gif]

[Graphics:../Images/ZTransformDEModHome_gr_934.gif]


[Graphics:../Images/ZTransformDEModHome_gr_935.gif]

[Graphics:../Images/ZTransformDEModHome_gr_936.gif]


[Graphics:../Images/ZTransformDEModHome_gr_937.gif]

[Graphics:../Images/ZTransformDEModHome_gr_938.gif]

Aside.  The Maple commands are similar  

[Graphics:../Images/ZTransformDEModHome_gr_939.gif]  

                                                            [Graphics:../Images/ZTransformDEModHome_gr_940.gif]


[Graphics:../Images/ZTransformDEModHome_gr_941.gif]  

                                                            [Graphics:../Images/ZTransformDEModHome_gr_942.gif]


[Graphics:../Images/ZTransformDEModHome_gr_943.gif]  

                                                            [Graphics:../Images/ZTransformDEModHome_gr_944.gif]


[Graphics:../Images/ZTransformDEModHome_gr_945.gif]  

                                                            [Graphics:../Images/ZTransformDEModHome_gr_946.gif]


[Graphics:../Images/ZTransformDEModHome_gr_947.gif]  

                                                            [Graphics:../Images/ZTransformDEModHome_gr_948.gif]

We are really really really done.   

Aside.  We can use Mathematica's Rsolve subroutine.

[Graphics:../Images/ZTransformDEModHome_gr_949.gif]

[Graphics:../Images/ZTransformDEModHome_gr_950.gif]

Aside.  The Maple command is similar  

[Graphics:../Images/ZTransformDEModHome_gr_951.gif]  

                                                            [Graphics:../Images/ZTransformDEModHome_gr_952.gif]

We are really really really really done.   

 

Aside.  We can graph some of the terms in the sequence.

 

          [Graphics:../Images/ZTransformDEModHome_gr_953.gif]     [Graphics:../Images/ZTransformDEModHome_gr_954.gif]     [Graphics:../Images/ZTransformDEModHome_gr_955.gif]

                    The sequence   [Graphics:../Images/ZTransformDEModHome_gr_956.gif].  

 

We are really really really really really done.   

The Details for the Partial Fractions.   

Aside.  How can we expand   [Graphics:../Images/ZTransformDEModHome_gr_957.gif]   into the proper partial fractions?

It is natural to use the standard partial fraction expansion and the command:

[Graphics:../Images/ZTransformDEModHome_gr_958.gif]

[Graphics:../Images/ZTransformDEModHome_gr_959.gif]

However, as we have seen, this will produce a solution involving the  [Graphics:../Images/ZTransformDEModHome_gr_960.gif]  functions.   

This can be overcome if we use a special partial fraction expansion that is easier to use with Table 9.1.

Method (i).   Use the following algebra steps  

                    [Graphics:../Images/ZTransformDEModHome_gr_961.gif]  

Now we have the desired partial fraction form:

                    [Graphics:../Images/ZTransformDEModHome_gr_962.gif].  

Method (ii).   Find the linear combination of   [Graphics:../Images/ZTransformDEModHome_gr_963.gif],  

                    [Graphics:../Images/ZTransformDEModHome_gr_964.gif].  

Equate the numerators   [Graphics:../Images/ZTransformDEModHome_gr_965.gif],  

and solve the linear system  

                    [Graphics:../Images/ZTransformDEModHome_gr_966.gif]

and get   [Graphics:../Images/ZTransformDEModHome_gr_967.gif].   

Therefore, the desired partial fraction form is  

                    [Graphics:../Images/ZTransformDEModHome_gr_968.gif].  

Aside.   The Mathematica commands for Method (ii)  are

[Graphics:../Images/ZTransformDEModHome_gr_969.gif]

[Graphics:../Images/ZTransformDEModHome_gr_970.gif]


[Graphics:../Images/ZTransformDEModHome_gr_972.gif]

[Graphics:../Images/ZTransformDEModHome_gr_973.gif]

Method (iii).   The substitution   [Graphics:../Images/ZTransformDEModHome_gr_974.gif]   does not apply when there are multiple roots.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell