Exercise 17 (c).   Construct a filter using the zeros and poles in part (a) and (b).  

From part (a) use the zeros     for "zeroing out"   ,  ,  and  .  

From part (b) use the poles      for boosting up signals near    and    and low frequency signals.  

Solution 17 (c).

See text and/or instructor's solution manual.

Answer.   Compute the quotient  

                    .  

The desired filter is  

                    .  

Solution.   From part 17 (a), use the conjugate pair of zeros      and   ,  

and the "zero out factors"      and   .  

Then calculate the numerator  

                      

From part 17 (b),  the conjugate pair of poles      and   ,  

and the "boost up factors"      and   .  

Then calculate the denominator  

                      

        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  

                    .  

The transfer function for this exercise is  
    
                    .  

        From part (a) use      and    ,   and from part (b) use   .  

and get the desired recursive formula  

                    .  

Therefore, the desired filter for part (c) is

                    .  

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   ,  

                    We can see that some of 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