Question

Consider the system described above. a. Find the transfer function of the system by reducing the diagram. b. Draw a signal-flow diagram for the given system. c. Using Mason's rule find the transfer function of the system. d. Compare your answers to part (a) and part (c). What do you notice? Explain.

Answer 1

solution H2 R(S) GI 62 613 G14 ccs) HI H2/G4) Step-It shifted the take off point to the right of 64 Rish 62 63 GI 614 Hi Step-It t Here 613 8 614 are connected in series. So overall gain is 61364. Gya H2/64 RCS) G2 1 G3 G4h >e(s) GU HI Step-t!: Eliminating the Feedback loop with #1 H2/64 613 614 1 + 61364 ti CLS) RCS) 62 G3 G4 1+63614 HI GU of cascaded. So are step-IV & G2 and 63 64 It 613 614 H1 overall gan is 62 613 614 it G3614 Hi

H2/614 it 63 64 H1 »e(s) loop with H2/614 RCS) Gu 62 63 614 is G2 413 414 618) It G3 61441 612 613 614 * 64 1 + 672673674 H2 G4+G3 G4 G4 H1 +62613614H2 17613644ti 64. 5 19 Х R(S) GI 612 613674x614 G4 + G3 G4 G4 H1 +42G344 Hz C(s)

Stocks are connected in series step-6. Two cascaded . RCS)+ Or 61 62 63 614 X 614 Cis! G4+63 G4 G4 H1 +62 63 64 H2 step-7? The overall transfer function for unity Feedback system is, Ccs) f162613 614 X 614 RCS) G4 + 4364614 Hit G2 G3 G14 H2 1 + 61 62 63 64 x G4 674 + 61361464 +1 + 612 6136442. → CCS) G1 G2 G3 GAGA - RCS) G4 +63614 G4 HP & G 2 G13 G14 H2 + H1 62 63 64 64 > CCS) 64 (60162 63 64) RCS) 1 +63 64 HJ + G2 G3 Hz + 61 62 63 44 . +41 42 43 444]

ces) 11 R(S) 61 62 63 614 1 + G3 G4 Hit 6 263H2 + 61 62 63 614 (b signal flow Diagram stor -H2. 64 GI 612 13 cos BCS) - HI -1 с Mason's rule

Transfer functions CCS) Mis Ok 11 BCS) Ķ =) D where, 0 - A. Number of forward path ie 1 = C(s) MA BIS) Gain of forward path, Mi= 61 62 63 64 Number of Endividuel loops = 3 Gain of loop -1 és L, = - 42613 H2 Gain of coop-a a is L2= -613 64 H Gain of loop - 3 is L3 = 61 672613614 * Number of two and three non 1-touching Where, A= 1- (s sum of goin of all Endividual loops) +(sum of product of two non-touching loops) - sum of product of three non-touching loops) loops o

→ 4 = 1-(-63614 H1 - 61261342 – 41 42 413 414) +0 -0 A = 1+43614 H1 + 6/2 613 H2 + 61 62 63 64 And, Ah - The value of a, which is not touching the forward path Mi. 41 - 1 .. C(s) R(S) - 61 62 63 64 X 1 1+63674 Hi + 62 63 H2 + 61 61263614 - G71 612 63 64 1+63 64 H1 + G2 G3H2 + 61 612 613 614 CCS) R(S) Lol> The The closed loop transfer function by block diagram reduction & Signal flow using Mason's gain Formula. are same.

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