Exercise 12 (b).  Construct a filter using the zeros   [Graphics:Images/ZTransformFilterModHome_gr_1101.gif].   What signals are "zeroed out" ?

Solution 12 (b).

See text and/or instructor's solution manual.

Answer.   Compute the product  

                    [Graphics:../Images/ZTransformFilterModHome_gr_1160.gif].  

The desired filter is  

                    [Graphics:../Images/ZTransformFilterModHome_gr_1161.gif].  

Solution.   Use the zeros in part (a) and the additional conjugate pair of zeros   [Graphics:../Images/ZTransformFilterModHome_gr_1162.gif],  

and the "zero out factors"   [Graphics:../Images/ZTransformFilterModHome_gr_1163.gif],    [Graphics:../Images/ZTransformFilterModHome_gr_1164.gif]   and   [Graphics:../Images/ZTransformFilterModHome_gr_1165.gif].  

Then calculate  

                    [Graphics:../Images/ZTransformFilterModHome_gr_1166.gif]   

        Now use the General Filter Equation (9-29) and the corresponding Transfer Function (9-34) in the case  [Graphics:../Images/ZTransformFilterModHome_gr_1167.gif].     

Get the new fact that the transfer function  

                    [Graphics:../Images/ZTransformFilterModHome_gr_1168.gif],  

corresponds to the filter  

                    [Graphics:../Images/ZTransformFilterModHome_gr_1169.gif].  

        For this exercise, we use   [Graphics:../Images/ZTransformFilterModHome_gr_1170.gif], [Graphics:../Images/ZTransformFilterModHome_gr_1171.gif]   and   [Graphics:../Images/ZTransformFilterModHome_gr_1172.gif]   in these equations to get the desired recursive formula  

                    [Graphics:../Images/ZTransformFilterModHome_gr_1173.gif].  

Therefore,  the desired filter for (b) is

                    [Graphics:../Images/ZTransformFilterModHome_gr_1174.gif].  

for "zeroing out" the signals   [Graphics:../Images/ZTransformFilterModHome_gr_1175.gif].

 

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ZTransformFilterModHome_gr_1176.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1177.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1178.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1179.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1180.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1181.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1182.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1183.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1184.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1185.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1186.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1187.gif]

Aside.  The Maple commands are similar  

[Graphics:../Images/ZTransformFilterModHome_gr_1188.gif]  

                                                            [Graphics:../Images/ZTransformFilterModHome_gr_1189.gif]  


[Graphics:../Images/ZTransformFilterModHome_gr_1190.gif]  

                                                            [Graphics:../Images/ZTransformFilterModHome_gr_1191.gif]  

 

We are really done.   

Aside.  We can graph the amplitude response for the filter   [Graphics:../Images/ZTransformFilterModHome_gr_1192.gif].  

 

                    [Graphics:../Images/ZTransformFilterModHome_gr_1193.gif]          [Graphics:../Images/ZTransformFilterModHome_gr_1194.gif]

                    Amplitude response   [Graphics:../Images/ZTransformFilterModHome_gr_1195.gif]   and zero-pole plot of   [Graphics:../Images/ZTransformFilterModHome_gr_1196.gif],  

                    for the filter   [Graphics:../Images/ZTransformFilterModHome_gr_1197.gif].

                    We can see that the mid-range frequencies are attenuated, and   [Graphics:../Images/ZTransformFilterModHome_gr_1198.gif]   for   [Graphics:../Images/ZTransformFilterModHome_gr_1199.gif].

 

Aside.  In Exercise 18 (a) we will investigate the filter   [Graphics:../Images/ZTransformFilterModHome_gr_1200.gif].  

 

We are really really done.   

Aside.  Let us investigate how well the filter works to eliminate signals  [Graphics:../Images/ZTransformFilterModHome_gr_1201.gif] which are close to the "zero-out" frequencies.

        For illustration purposes we will explore the casual input signal   [Graphics:../Images/ZTransformFilterModHome_gr_1202.gif].

The signal component  [Graphics:../Images/ZTransformFilterModHome_gr_1203.gif]  will be amplified by the factor  [Graphics:../Images/ZTransformFilterModHome_gr_1204.gif].  

The signal component  [Graphics:../Images/ZTransformFilterModHome_gr_1205.gif]  will be attenuated by the factor  [Graphics:../Images/ZTransformFilterModHome_gr_1206.gif],  and

The signal component  [Graphics:../Images/ZTransformFilterModHome_gr_1207.gif]  will be attenuated by the factor  [Graphics:../Images/ZTransformFilterModHome_gr_1208.gif],  and

The signal component  [Graphics:../Images/ZTransformFilterModHome_gr_1209.gif]  will be attenuated by the factor  [Graphics:../Images/ZTransformFilterModHome_gr_1210.gif].  

[Graphics:../Images/ZTransformFilterModHome_gr_1211.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1212.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1213.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1214.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1215.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1216.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1217.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1218.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_1219.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_1220.gif]

                    [Graphics:../Images/ZTransformFilterModHome_gr_1221.gif]          [Graphics:../Images/ZTransformFilterModHome_gr_1222.gif]

                    The causal input sequence   [Graphics:../Images/ZTransformFilterModHome_gr_1223.gif]   and the corresponding causal output sequence.

                    We can see that the filter practically eliminates the signals  [Graphics:../Images/ZTransformFilterModHome_gr_1224.gif] which are close to the "zero-out" frequencies.

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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