13.5. a. Find the controllable-form realization for the transfer function in (13.22). b. Show that the realization and the state-variable equation in (13.24) are equivalent. Can you conclude that the transfer function of (13.24) equals (13.22)? (Hint: The similarity transformation is T -T Example 13.2.4 Consider C(s) = 2(s 2)/(s2 + 4s + 3). Then Z 2 2 T C(z) = T(5)|s=(2, 13/7 1 + 4 + 3 T T (13.22) 2T(z 1 2T) 22 + (4T 2)2 + (312 4T + 1) is a digital compensator obtained using the forward-difference method. If T = 0.5, then (13.22) becomes 2.0.5(z 1 - 1) C(z) 22 + (4 0.5 2) + (3 0.25 - 2 + 1) (13.23) 2 2 22 0.25 (z + 0.5)(2 0.5) This is the digital compensator. If we realize C(s) as 4 3 X(t) X(t) + e(t) 1 Z Z = 1 -:] u(1) [2-4]X(t) Then x(k + 1) ['i 4T T -31 1 x(k) + [1] elk) (13.24a) u(k) [2 -4]x(k) (13.24b) 13.2 DIGITAL IMPLEMENTATIONS OF ANALOG COMPENSATORS 519 is the digital compensator. It can be shown that the transfer function of (13.24) equals (13.22). See Problem 13.5. 13.5. a. Find the controllable-form realization for the transfer function in (13.22). b. Show that the realization and the state-variable equation in (13.24) are equivalent. Can you conclude that the transfer function of (13.24) equals (13.22)? (Hint: The similarity transformation is T -T Example 13.2.4 Consider C(s) = 2(s 2)/(s2 + 4s + 3). Then Z 2 2 T C(z) = T(5)|s=(2, 13/7 1 + 4 + 3 T T (13.22) 2T(z 1 2T) 22 + (4T 2)2 + (312 4T + 1) is a digital compensator obtained using the forward-difference method. If T = 0.5, then (13.22) becomes 2.0.5(z 1 - 1) C(z) 22 + (4 0.5 2) + (3 0.25 - 2 + 1) (13.23) 2 2 22 0.25 (z + 0.5)(2 0.5) This is the digital compensator. If we realize C(s) as 4 3 X(t) X(t) + e(t) 1 Z Z = 1 -:] u(1) [2-4]X(t) Then x(k + 1) ['i 4T T -31 1 x(k) + [1] elk) (13.24a) u(k) [2 -4]x(k) (13.24b) 13.2 DIGITAL IMPLEMENTATIONS OF ANALOG COMPENSATORS 519 is the digital compensator. It can be shown that the transfer function of (13.24) equals (13.22). See Problem 13.5