Question

Control Systems

3. Y(s) Derive the transfer function G(s) = rule. U(S) of the following system using Mason's gain (18 marks) G9 G Gs Gi G2 G3 G4 GS G6 UO юү Hi H2 H3

Answer 1

G9 Gs 54 G4 UO G1 X2 G2 X3 G X1 OY X4 GS X6 G6 X7 X5 Hi H2 H3

From the figure,

X7 *1 = Y; X 6G6 + X 5 G8 + X1 G9 = X7 X5 G5 = X6; X 4G4 = X 5 = X4G4G5 = X6 U *1 = X1; X1G1 + X3H1 + X6H3 = X2 X2G2 = X3; X3 G3 + X1G7+ X5 H2 = X 4

simplify

→ G4 G5X4G6 + G4X4G8 + UG9 X7 =Y → (G4G5 G6 + 64G3)X4 = Y - UG,

= X4 X3 G3 + UG7 + X 4G4H2 (1 – G4H2) G7 - U- X4 G3 G3 X3

X3 → UG1 + X3H1 + X4G4 G5 H3 = X2 G2 H1)X3 = UG1 + X4G4G5 H3 G2 G1 G2 G4G5 H3G2 → X3 = U + X4 1 - H1G2 1 - HG2

equate X3 from both.

$\tiny X_4\frac{(1-G_4H_2)}{G_3 } - U\frac{G_7}{G_3 }= U \frac{G_1G_2}{1-H_1G_2} +X_4 \frac{G_4G_5H_3G_2} {1-H_1G_2} \\ \\ \\ \\ X_4[\frac{(1-G_4H_2)}{G_3 }- \frac{G_4G_5H_3G_2} {1-H_1G_2} ]= U[ \frac{G_1G_2}{1-H_1G_2} +\frac{G_7}{G_3 }]$

$\tiny \Rightarrow X_4[\frac{(1-G_4H_2)(1-H_1G_2)- G_2G_3G_4G_5H_3}{G_3(1-H_1G_2) } ]= U[ \frac{G_1G_2G_3+G_7(1-H_1G_2)}{G_3(1-H_1G_2)} ]$$\tiny \Rightarrow X_4= U[ \frac{G_1G_2G_3+G_7(1-H_1G_2)}{(1-G_4H_2)(1-H_1G_2)- G_2G_3G_4G_5H_3} ]$

We know:

$\tiny (G_4G_5 G_6+G_4 G_8)X_4= Y-UG_9$

$\tiny \Rightarrow (G_4G_5 G_6+G_4 G_8)U[ \frac{G_1G_2G_3+G_7(1-H_1G_2)}{(1-G_4H_2)(1-H_1G_2)- G_2G_3G_4G_5H_3} ]= Y-UG_9$$\tiny \Rightarrow [ \frac{ (G_4G_5 G_6+G_4 G_8)(G_1G_2G_3)+G_7 (G_4G_5 G_6+G_4 G_8)(1-H_1G_2)}{(1-G_4H_2)(1-H_1G_2)- G_2G_3G_4G_5H_3} + G_9]U= Y$$\tiny Y = [ \frac{ (G_4G_5 G_6+G_4 G_8)(G_1G_2G_3)+G_7 (G_4G_5 G_6+G_4 G_8)(1-H_1G_2) + (1-G_4H_2)(1-H_1G_2) G_9- G_2G_3G_4G_5 G_9H_3}{(1-G_4H_2)(1-H_1G_2) - G_2G_3G_4G_5 H_3} ]U$

$\tiny \Rightarrow G(s) = \frac{Y(s)}{U(s)} = \frac{ (G_4G_5 G_6+G_4 G_8)(G_1G_2G_3)+G_7 (G_4G_5 G_6+G_4 G_8)(1-H_1G_2) + (1-G_4H_2)(1-H_1G_2) G_9- G_2G_3G_4G_5 G_9H_3}{(1-G_4H_2)(1-H_1G_2) - G_2G_3G_4G_5 H_3}$