Exercise 2 (b).  Solve      with   .   
                          Hint.  Get  .  

Solution 2 (b).

See text and/or instructor's solution manual.

Answer.   .  

Alternative Answer.   .  

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

Solution.   

Method 1.  The characteristic equation  

                      

has complex roots  .  

The general solution is  

                    .  

Solve the linear system  

                    

and get    and  .

Therefore,  

                    .

Method 2.  Take the z-transform of both sides and use the initial conditions   :  

                    ,  

and then get    

                    .  

New solve for    and obtain:    

                    .  

Using Tables.   Using Table 9.1 and the formula      we get:  

                      

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

Therefore,  

                    .

We are done.   

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

and the substitution    and    ,     and get

                      

Aside.  We can verify the substitution.

We are done.   

Using Residues.   Calculate residues of    at the poles   .  

                      

At the conjugate pole we can use the computation

                    .  

Thus,  

                      

Therefore,

                    .  

We are done.   

Aside.  We can let Mathematica double check our work.

Aside.  The Maple commands are similar  

  

                                                            


  

                                                            


  

                                                            

                                                            
   

                                                            


  

                                                            

We are really done.   

Aside.  We can use Mathematica's Rsolve subroutine.

Aside.  The Maple command is similar  

  

                                                            

For curiosities sake, we can compute some of the terms in the sequence with the various formulae.

We leave it for the reader to prove that the four forms of the answer are equivalent.

We are really really done.   

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

                    The sequence   .  

We are really really really done.   

The Details for the Partial Fractions.   

Aside.  How can we expand      into the proper partial fractions?

Method (i).   Use the following algebra steps  

                      

Now we have the desired partial fraction form:

                    .

Method (ii).   Find the linear combination of   ,  

                    .  

Equate the numerators   ,  

and solve the linear system  

                      

and get   .   

Therefore, the desired partial fraction form is  

                    .  

Aside.   The Mathematica commands for Method (ii)  are

Method (iii).   The substitution      does not apply when there are complex roots.

We are really really really really done.   

Method 2 Revisited.  Take the z-transform of both sides and use the initial conditions   :  

                    ,  

and then get    

                    .  

New solve for    and obtain:    

                    .  

Using Residues.   Calculate residues of    at the poles   .  

                      

At the conjugate pole we can use the computation

                    .  

Thus,  

                      

Therefore,  

                    .  

This solution is complements of the authors.

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