Exercise 10.  Use the transfer function     and show that the moving average filter in Exercise 9

has an alternative formula   .

Solution 10.

See text and/or instructor's solution manual.

Short Solution.   Use the General Filter Equation (9-29) and the corresponding Transfer Function (9-34)

in the case      and     for  .  Then the transfer function  

                    ,  

corresponds to the filter  

                    .  

Now set   ,   and      and obtain the alternative formula for the filter:

                    .

Detailed Solution.   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  

                    .  

Hence, the the moving average filter in Exercise 9 is  

                      

and has the transfer function  

                    .  

Next, we can use the fact that      and write

                    .  

Then make the substitution    and get  

                    .  

It follows that  

                    .  
                    
Therefore, the transfer function for the moving average filter in Exercise 9 can be written as

                    .  

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

Get another new fact that the transfer function  

                    ,  

corresponds to the filter  

                    .  

        For this exercise, we use   ,     for     and      and get  

                    .  

Therefore, the moving average filter in Exercise 9 has the alternative formula   

                    .

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   .    

Remark.   In Exercise 9 we showed that we can multiply the expression for    by    and obtain the transfer function as a product of "zero-out factors"  

                      

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.

Remark.  In Exercise 11 we will investigate the moving average of eight points.  

This solution is complements of the authors.

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