Exercise 16 (c).  Construct the combination filter using the zeros       and  poles     

for attenuating high frequencies, and boosting up low frequencies.

Solution 16 (c).

See text and/or instructor's solution manual.

Answer.   Compute the quotient  

                    .  

The desired filter is  

                    .  

Solution.   The zeroing out portion of this filter design is similar to the filter in Example 9.26.  

Use the conjugate pair of zeros      and   ,  

and the "zero out factors"      and   .  

Then calculate

                        

The numerator of the transfer function has the form   

                      

and we see that    .

        For the boosting up portion of this filter design we choose the one developed in Example 9.26:  

Use the conjugate pair of poles  ,  

and the "boost up factors"   .  

Then calculate  

                    

The denominator of the transfer function has the form   

                      

and we see that    .  

This is a basic filter and we can use Property (iii) Combination Filter.  

The transfer function has the form  

                    ,
and corresponds to the filter

                    .  

        For this exercise, we use   ,  
        
and      in these equations to get the desired recursive 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   .

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

We are really really done.   

Aside.  For illustration, we can graph the causal input sequence   ,  

and the corresponding causal output sequence   .  

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.

                    Here we have     and   .
                    
                    This filter reduces the proportion of the signal component      by a factor of   .

This solution is complements of the authors.

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