Exercise 7.  The single-pole low-pass filter is   ,   where constant K is between  .  

7 (b).  Use   and find   ,   ,   ,   ,   and   .  

Solution 7 (b).

See text and/or instructor's solution manual.

Answer.   ,   ,   ,   ,   .  

Solution.   Substitute    and get   ,  

the transfer function (9-27) is   

                    ,  

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

                    .  

Calculate  

                    ,  

                    ,  

                    ,  

                      

The higher frequencies are attenuated and      when     .

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 attenuated and      when      and      when    .  

We are really really done.   

Remark.   The transfer function is    and has a zero at the origin, and a pole at  .  

The argument of the pole    is    and points toward the to the low frequencies that are boosted up.

For a single-pole low-pass filter, when the pole is closer to    the higher frequencies are more severely attenuated.

We are really really really done.   

Aside.  Let us investigate how well the filter works to retain low frequencies.

        For illustration purposes we will explore the casual input signal   .

The signal component    will be  slightly attenuated by the factor  ,  and

The signal component    will be attenuated  more by the factor  .  

This filter reduces the proportion of the signal component      by a factor of   .

                    The input signal      and output signal   .  

                    Here the amplitude response is      and   .  

                    This filter reduces the proportion of the signal component      by a factor of   .

This solution is complements of the authors.

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