Find the total transfer function of the system based on the block diagram of a multi-loop control system given in the figure.
The given block diagram is:
Moving the take off point between ${G}_{3}\mathrm{and}{G}_{4}$ to the right of ${G}_{4}$. So the modified diagram is
Solving the feedback loop (1)
$T{F}_{1}=\frac{{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}}$
The modified block diagram is:
Solving the feedback loop (2)
$\begin{array}{rcl}T{F}_{2}& =& \frac{\frac{{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}}}{1-\frac{{G}_{3}{G}_{4}{H}_{3}}{1+{G}_{3}{G}_{4}{H}_{2}}}\\ & =& \frac{{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}}\end{array}$
The modified block diagram is:
Solving the feedback loop (3)
$\begin{array}{rcl}T{F}_{3}& =& \frac{\frac{{G}_{2}{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}}}{1-\frac{{G}_{2}{G}_{3}{G}_{4}\left({\displaystyle \raisebox{1ex}{${H}_{4}$}\!\left/ \!\raisebox{-1ex}{${G}_{4}$}\right.}\right)}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}}}\\ & =& \frac{{G}_{2}{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}-{G}_{2}{G}_{3}{H}_{4}}\end{array}$
The modified block diagram is:
Solving the feedback loop (3)
$\begin{array}{rcl}T{F}_{4}& =& \frac{\frac{{G}_{1}{G}_{2}{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}-{G}_{2}{G}_{3}{H}_{4}}}{1+\frac{{G}_{1}{G}_{2}{G}_{3}{G}_{4}{H}_{1}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}-{G}_{2}{G}_{3}{H}_{4}}}\\ & =& \frac{{G}_{1}{G}_{2}{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}-{G}_{2}{G}_{3}{H}_{4}+{G}_{1}{G}_{2}{G}_{3}{G}_{4}{H}_{1}}\end{array}$
Therefore, the total transfer function of the block diagram is
$\begin{array}{rcl}TF& =& \frac{C\left(s\right)}{R\left(s\right)}=\frac{{G}_{1}{G}_{2}{G}_{3}{G}_{4}}{1+{G}_{3}{G}_{4}{H}_{2}-{G}_{3}{G}_{4}{H}_{3}-{G}_{2}{G}_{3}{H}_{4}+{G}_{1}{G}_{2}{G}_{3}{G}_{4}{H}_{1}}\end{array}$