Problem 4 Note that -transform is linear (a)Given rn]=26{n+3]-6n-2 In this part of the problem, we would like to use one of the most fundamental z-transform pairs 8n-m+ 2, ROC: All z except for 0 if m>0 or oo if m <0 By applying this pair for both terms in the given a[n], we get x(z)=23-z-2 2z5-1 with its ROC is all z except for 0 and oo. Since the unit circle is included in the roC, the Fourier transform of the sequence exists. From the last expression, it can be seen that there are two poles both of which are at 0. There are five zeros and they are the solutions of 2z5-1=0 i.e, 2-5eJjwx where w2 is integer multiples of 2T m To see this, remember that any complex number z can be written in polar form as z=re where r is the length, or radius of the vector a and w is its phase. In our case using this idea we can find the zeros as followsProblem 4 Note that z-transform is linear. (a) Given: x[n] = 2ω[n + 3] − ω[n − 2]. In this part of the problem, we would like to use one of the most fundamental z-transform pairs: ω[n − m] z −m �∩ z , ROC: All z except for 0 if m > 0 or ∗ if m < 0. (6) By applying this pair for both terms in the given x[n], we get: X −2 (z) = 2z3 − z 2z5 − 1 = 2 , z with its ROC is all z except for 0 and ∗. Since the unit circle is included in the ROC, the Fourier transform of the sequence exists. From the last expression, it can be seen that there are two poles both of which are at 0. There are five zeros and they are the solutions of 2z 5 −1 = 0, 1 2� i.e, 2− 5 ej�z where �z is integer multiples of 5 . ≤m ↓e × 2 To see this, remember that any complex number z can be written in polar form as z = rej� where r is the length, or radius of the vector z and � is its phase. In our case using this idea, we can find the zeros as follows: 11