Exercise 2.  Given the recursion formula   .  

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

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

Solution 2.

See text and/or instructor's solution manual.

Answer.   ,   ,   ,   ,   .  

The signal component       is amplified by the factor   .  

The signal component      is attenuated quite a bit by the factor   .  

Solution 2 (a).  Given the filter   ,  

the transfer function (9-27) is   

                    ,  

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

                    .  

Now calculate  

                    ,  

                    ,  

                    ,  

                    .  

Solution 2 (b).  From the above calculations we expect that component      of the signal is amplified by the factor     

and the component      attenuated quite a bit by the factor   .  

Hence the filter almost eliminates the signal component      which is close to the "zero-out" frequency   .  

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 higher frequencies are slightly attenuated and      when   .    

We are really really done.   

Remark.   The transfer function is  

                    ,  

and has conjugate zeros at   .  

The argument of      is   ,   and there is a zero amplitude response at   .  

Since  ,   we can expect that   .  

Also, the amplitude response is decreasing for values of     in the interval   .

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 higher frequencies are attenuated and   .  

                    We can see that the signal component    is practically eliminated.

Remark.  In Exercise 4 we will see what happens when we change the sign of the term   .  

This solution is complements of the authors.

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