According to the zero pole plot we can clearly see

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Unformatted text preview: cos(! ). Therefore if there exists poles which are close to the unit circle and have positive real part, the coefficient a1 corresponding to it would become nearly 2 in magnitude. According to the zero pole plot, we can clearly see that the pair 3 biquad’s poles have positive real part, therefore this may cause problem. We need to a1 of this biquad by first dividing it by 2 and then do the multiplication, and at last multiply the result by 2. Q4. Answer: i): As ! ! = ! !! ! !! ! !! ! !!!"# ! , ! = 0,1, … , ! − 1. then: ! ! −! = 2 ! !! ! [! ]! (! ! !! ! ! ) ! !!! ! !! ! ! ! !!"# ∗ ! = ! !! !!!"# ! ! !! As when n is odd, x[n]=0, we only need to worry about when n is even. As when n is even, ! !!"# = 1. Then we can get: ! ! −! = 2 ! !! !!! !!!"# ! = ! ∗ [! ] ! !! Therefore X[k] is conjugate symmetric about N/4. ii): As ! ! = ! !! ! !! ! ! !! !!!"# ! , ! = 0,1, … , ! ...
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