Exercise 15 (b).  Construct a filter using the zeros   .   What signals are "zeroed out" ?

Solution 15 (b).

See text and/or instructor's solution manual.

Answer.   Compute the product  

                    .  

The desired filter is  

                    .  

Solution.   Use the conjugate pairs of zeros    and    and calculate  

                       

        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   ,    and      in these equations to get the desired recursive formula  

                    .  

Therefore, the desired filter for (b) is

                    .  

for "zeroing out" the signals   .

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   .

                    We can see that the high-range frequencies are attenuated, and      for   .  

Remark.  In part 15 (a) we saw what happens when we change the sign of the terms   .  

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 amplified by the factor  .  

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.

                    Since we have     and   ,
                    
                    this filter reduces the proportion of the signal component      by a factor of   .
                    
                    Since we have     and   ,
                    
                    this filter reduces the proportion of the signal component      by a factor of   .

We are really really really done.   

Aside.  Let us investigate how well the filter works to eliminate the signal   which is close to a "zero-out" frequency.

and retains the signal   .  

Remark.  In part (a) we saw how to eliminate the latter.

        For illustration purposes we will explore the casual input signal   .

The signal component    will be amplified by the factor  .  

The signal component    will be amplified  by the factor  ,  and  

The signal component    will be attenuated by the factor  .  

                    The causal input sequence      and the corresponding causal output sequence.

                    Since we have     and   ,
                    
                    this filter reduces the proportion of the signal component      only little by the factor of   .

                    Since we have     and   ,
                    
                    this filter reduces the proportion of the signal component      a lot by the factor of   .

This solution is complements of the authors.

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