Exercise 6.  Given the recursion formula   .  

6 (a).  Calculate the amplitude response   ,   ,   ,   and   .  

6 (b).  Discuss what happens to the filtered signal for the input   .  

Hint.  This is similar to Example 9.22.  

Solution 6.

See text and/or instructor's solution manual.

Answer.   ,   ,   ,   ,   .  

The signal component      is amplified by the factor   .  

The signal component      is attenuated by the factor   .  

Solution 6 (a).  Given the filter   ,  

the transfer function (9-27) is   

                    ,  

and formula (9-28) for the amplitude response is

                    .  

Now calculate  

                    ,  

                    ,  

                    ,  

                    .  

Solution 6 (b).  From the above calculations we expect that component      of the signal is amplified by the factor   

and the component      attenuated by the factor   .  

Hence the filter "boosted up" the signal component      and attenuates the signal component   .  

Also, the filter reduces the proportion of the signal component      by a factor of   .

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   .  

                    The amplitude response      and zero-pole plot of   ,  
  
                    for the filter   .

                    The frequencies "boosted up" near   .  

                    There is amplification for signals with   ,   and attenuation for signals with   .  

We are really really done.   

Remark.   The transfer function is  

                    ,  

and has conjugate poles at   .  

The argument of      is   ,   and there is a local maximum of the amplitude response near   .  

There is amplification for signals with   ,   and attenuation for signals with   .  

We are really really really done.   

Aside.  We can graph the causal input sequence and the corresponding causal output sequence.  

                    The input signal      and output signal   .  

                    The lower frequencies are "boosted up" near    and   .  

                    The higher frequencies are somewhat attenuated and   .  

Remark.  In Exercise 5 we saw what happens when we divide the coefficient of the term      by   .  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell