Question

3) Compute the Z-Transforms of the following time series: (a) x(n)k2"u(n) (b) + 1) x(n) = u(-n (c) x(n) -k2"u-1) (d) x(n) 0.5%1(n) + 3"11(-n) (e) x(n) = 4-nu(n) + 5-nu(n + 1) In the above, u(n) stand for the unit step signal in the discrete time domain. Also, if you can in each case determine the region of convergence of the Z-Transform you obtain.

Answer 1

(a). x(n) = k 2^n u(n) x(z) = Sigma^infinity_n = -infinity x(n) z^- n = Sigma^infinity_n = 0 k 2^n z^-n because u(n) = 0 for n < 0 = 1 for n greaterthanorequalto 0 = k Sigma^infinity_n = 0 (z/2)^- n = k[1 + 2/z + (2/z)^2 + ...] x(z) = k 1/1 - 2 z^- 1, Vertical-bar 2/z Vertical-bar < 1 Rightarrow Vertical-bar z Vertical-bar > 2 where we used the binomial identity, 1/1 - x = 1 + x + x^2 + ...., Vertical-bar x Vertical-bar < 1 \r\n (b). x(z) = Sigma^infinity_n = -infinity u(- n + 1) z^- n but, u(- n + 1) = {1, - n + 1 greaterthanorequalto 0 Rightarrow 1 greaterthanorequalto n or n lessthanorequalto 1 0, - n + 1 < 0 Rightarrow n > 1 x(z) = Sigma^1_n = -infinity z^- n put n = -m, we get x(z) = Sigma^infinity_m = -1 z^m = z^- 1 + (1 + z + z^2 + ...) = z^- 1 + 1/1 - z, Vertical-bar z Vertical-bar < 1