Exercise 11 (b).  Solve      with   .  
                            Hint.  Get  .  

Solution 11 (b).

See text and/or instructor's solution manual.

Answer.   .  

Remark.   The preferred method to use involves Z-transforms, limits and residues.  

Solution.   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.   

Using Residues.   Calculate residues of    at the poles   .  

                      

At the conjugate pole we can use the computation

                     .

At the other 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.  The solution can be written in the alternative form:  

We are really really done.   

Aside.  We can use Mathematica's Rsolve subroutine.

          Mathematica's answer is quite complicated.  Let's look at Maple's answer.

Aside.  The Maple command is similar  

  

                                                            

We are really really really done.   

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

                    The sequence   .  

We are really 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   ,  

                    ,

where the numerator    is

                      

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.

This solution is complements of the authors.

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