Question

H4 R(s) C(s) G1 G2 G3 G4 H2 H3 H1 ...

Find the total transfer function of the system based on the block diagram of a multi-loop control system given in the figure.

Transcribed: H4 C(s) R(s) G1 G2 G3 G4 H2 H3 H1

Engineering Electrical Engineering Control Systems

Answer

Step #1 of 6

The given block diagram is:

Step #2 of 6

Moving the take off point between $G_{3} and G_{4}$ to the right of $G_{4}$ . So the modified diagram is

Solving the feedback loop (1)

$T F_{1} = \frac{G_{3} G_{4}}{1 + G_{3} G_{4} H_{2}}$

Step #3 of 6

The modified block diagram is:

Solving the feedback loop (2)

$\begin{array}{rcl} T F_{2} & = & \frac{\frac{G_{3} G_{4}}{1 + G_{3} G_{4} H_{2}}}{1 - \frac{G_{3} G_{4} H_{3}}{1 + G_{3} G_{4} H_{2}}} \\ = & \frac{G_{3} G_{4}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3}} \end{array}$

Step #4 of 6

The modified block diagram is:

Solving the feedback loop (3)

$\begin{array}{rcl} T F_{3} & = & \frac{\frac{G_{2} G_{3} G_{4}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3}}}{1 - \frac{G_{2} G_{3} G_{4} (\frac{H_{4}}{G_{4}})}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3}}} \\ = & \frac{G_{2} G_{3} G_{4}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3} - G_{2} G_{3} H_{4}} \end{array}$

Step #5 of 6

The modified block diagram is:

Solving the feedback loop (3)

$\begin{array}{rcl} T F_{4} & = & \frac{\frac{G_{1} G_{2} G_{3} G_{4}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3} - G_{2} G_{3} H_{4}}}{1 + \frac{G_{1} G_{2} G_{3} G_{4} H_{1}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3} - G_{2} G_{3} H_{4}}} \\ = & \frac{G_{1} G_{2} G_{3} G_{4}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3} - G_{2} G_{3} H_{4} + G_{1} G_{2} G_{3} G_{4} H_{1}} \end{array}$

Step #6 of 6

Therefore, the total transfer function of the block diagram is

$\begin{array}{rcl} T F & = & \frac{C (s)}{R (s)} = \frac{G_{1} G_{2} G_{3} G_{4}}{1 + G_{3} G_{4} H_{2} - G_{3} G_{4} H_{3} - G_{2} G_{3} H_{4} + G_{1} G_{2} G_{3} G_{4} H_{1}} \end{array}$

H4 R(s) C(s) G1 G2 G3 G4 H2 H3 H1 ...

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