Exercise 18 (a).   Construct a filter using the zeros   [Graphics:Images/ZTransformFilterModHome_gr_2105.gif]   for "zeroing out"   [Graphics:Images/ZTransformFilterModHome_gr_2106.gif],  [Graphics:Images/ZTransformFilterModHome_gr_2107.gif],  and  [Graphics:Images/ZTransformFilterModHome_gr_2108.gif].

Solution 18 (a).

See text and/or instructor's solution manual.

Answer.   Compute the product  

                    [Graphics:../Images/ZTransformFilterModHome_gr_2112.gif].  

The desired filter is  

                    [Graphics:../Images/ZTransformFilterModHome_gr_2113.gif].  

Solution.   Use the conjugate pair of zeros   [Graphics:../Images/ZTransformFilterModHome_gr_2114.gif]   and   [Graphics:../Images/ZTransformFilterModHome_gr_2115.gif],  

and the "zero out factors"   [Graphics:../Images/ZTransformFilterModHome_gr_2116.gif]   and   [Graphics:../Images/ZTransformFilterModHome_gr_2117.gif].  

Then calculate  

                    [Graphics:../Images/ZTransformFilterModHome_gr_2118.gif]  

This is a basic filter and we can use Property (i) Zeroing Out Filter.  

The transfer function for part (a) has the form  

                    [Graphics:../Images/ZTransformFilterModHome_gr_2119.gif]  

and corresponds to the filter

                    [Graphics:../Images/ZTransformFilterModHome_gr_2120.gif].  

        For this exercise, we use   [Graphics:../Images/ZTransformFilterModHome_gr_2121.gif]   and   [Graphics:../Images/ZTransformFilterModHome_gr_2122.gif]   in these equations to get the desired recursive formula  

                    [Graphics:../Images/ZTransformFilterModHome_gr_2123.gif].  

Therefore, the desired filter for part (a) is

                    [Graphics:../Images/ZTransformFilterModHome_gr_2124.gif].  

 

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ZTransformFilterModHome_gr_2125.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2126.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_2127.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2128.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_2129.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2130.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_2131.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2132.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_2133.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2134.gif]

Aside.  The Maple commands are similar  

[Graphics:../Images/ZTransformFilterModHome_gr_2135.gif]  

                                                            [Graphics:../Images/ZTransformFilterModHome_gr_2136.gif]  


[Graphics:../Images/ZTransformFilterModHome_gr_2137.gif]  

                                                            [Graphics:../Images/ZTransformFilterModHome_gr_2138.gif]  

 

We are really done.   

Aside.  We can graph the amplitude response for the filter   [Graphics:../Images/ZTransformFilterModHome_gr_2139.gif].  

 

                    [Graphics:../Images/ZTransformFilterModHome_gr_2140.gif]          [Graphics:../Images/ZTransformFilterModHome_gr_2141.gif]

                    Amplitude response   [Graphics:../Images/ZTransformFilterModHome_gr_2142.gif],   and zero-pole plot of   [Graphics:../Images/ZTransformFilterModHome_gr_2143.gif],  

                    for the filter   [Graphics:../Images/ZTransformFilterModHome_gr_2144.gif].

                    We can see that some of the high-range frequencies are attenuated, and   [Graphics:../Images/ZTransformFilterModHome_gr_2145.gif]   for   [Graphics:../Images/ZTransformFilterModHome_gr_2146.gif].  

 

Aside.  In Exercise 12 (b) we investigated the filter   [Graphics:../Images/ZTransformFilterModHome_gr_2147.gif].  

 

We are really really done.   

Aside.  For illustration, we can graph the causal input sequence   [Graphics:../Images/ZTransformFilterModHome_gr_2148.gif],  

and the corresponding causal output sequence   [Graphics:../Images/ZTransformFilterModHome_gr_2149.gif].  

The signal component  [Graphics:../Images/ZTransformFilterModHome_gr_2150.gif]  will be amplified  by the factor  [Graphics:../Images/ZTransformFilterModHome_gr_2151.gif],  and  

The signal component  [Graphics:../Images/ZTransformFilterModHome_gr_2152.gif]  will be attenuated by the factor  [Graphics:../Images/ZTransformFilterModHome_gr_2153.gif].  

[Graphics:../Images/ZTransformFilterModHome_gr_2154.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2155.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_2156.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2157.gif]


[Graphics:../Images/ZTransformFilterModHome_gr_2158.gif]

[Graphics:../Images/ZTransformFilterModHome_gr_2159.gif]

                    [Graphics:../Images/ZTransformFilterModHome_gr_2160.gif]          [Graphics:../Images/ZTransformFilterModHome_gr_2161.gif]

                    The causal input sequence   [Graphics:../Images/ZTransformFilterModHome_gr_2162.gif]   and the corresponding causal output sequence.

                    Here we have  [Graphics:../Images/ZTransformFilterModHome_gr_2163.gif]   and   [Graphics:../Images/ZTransformFilterModHome_gr_2164.gif]
                    
                    This filter reduces the proportion of the signal component   [Graphics:../Images/ZTransformFilterModHome_gr_2165.gif]   by a factor of   [Graphics:../Images/ZTransformFilterModHome_gr_2166.gif].

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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