Exercise 9.  Show that the moving average filter     

is designed to "zero out"  .  

Hint.  This is similar to Example 9.23 (b).

Solution 9.

See text and/or instructor's solution manual.

    Recall that the solutions to      are the sixth roots of unity  

                    .  

and they all lie on the unit circle, and we can write:  

                      

Hence the five roots of      are  

                    .  

We can now multiply the expression for    by    and obtain the transfer function as a product of "zero-out factors"  

                      

        Now use the General Filter Equation (9-29) and the corresponding Transfer Function (9-34) in the case  .     

Get the new fact that the transfer function  

                    ,  

corresponds to the filter  

                    .  

        For this exercise, we use      for      in these equations to get the desired recursive formula  

                    .  

Therefore,  

                    .  

is designed to "zero out"  .  

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 graph the amplitude response for the filter   .  

                    Amplitude response  ,   

                    and zero-pole plot of   ,  

                    for the filter   .

                    The higher frequencies are attenuated and      when   .    

We are really really done.   

Aside.  Let us investigate how well the filter works to eliminate signals    which are close to the "zero-out" frequencies.

        For illustration purposes we will explore the casual input signal   .

The signal component    will be attenuated by the factor  ,  and

The signal component    will be attenuated by the factor  ,  and

The signal component    will be attenuated by the factor  .  

                    The causal input sequence    and the corresponding causal output sequence.

                    We can see that the filter mostly eliminates the signals    which are close to the "zero-out" frequencies.

This solution is complements of the authors.

(c) 20098 John H. Mathews, Russell W. Howell