Question

3.10 For the system with inner-loop feedback, find an expression for the sampled output Y* ($) (Fig. P3.10).

Answer 1

Assumption - Both switch are closed By using signal flow graph, the system can be drown as graph R(s) c(s) & G(s) y*(s). -H (5) -H₂ (6) By using Mason's yain formula, - Σ In-path Y TF= , ΔΚ KE4 R(s) where, PxKthe path forward gain. A > Determinant of system.

Forward path gain - P. 3 = c(s). G(s). Loop gains - 4 > -->2 - G(). [-He(s)} = - G(3) He(6) L₂ ② 2 3 1 - c(s). G(s). [Hz (s)] = - C(s) G (3) H₂ (8). Az 1- ( sum of all individual loop gains non touching with Pl) D = 1 - (0) = 1 Because both loops are touching 'Pi'. A1- (Sum of all individual loon gains) - 1-[-G (3) Hp (3) - C(s) G (3) H₂ (3)] >> A = 1+G(3) He(s) + c(s) G(3) 42(3). From eq , Y* (6) - PAL ( As there is only & path] R (3) y*(s) = C(s).G(s) (1) R(5) 1 +G (3) H1 (3) + C(s) G (s) Ha (s). vow :: [y*() = (3) 6(5): R (s) It Gl) [H1 (3) +CG).H26)]