Consider the system shown in the figure below:

$The transfer function \frac{C}{R} of this system is$

G_{4} + \frac{G_{1} G_{2} G_{3}}{1 + G_{2} H_{1} - G_{1} G_{2} H_{1} + G_{3} H_{2}}

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G_{4} + \frac{G_{1} G_{2} G_{3}}{1 + G_{1} H_{1} - G_{2} H_{1} + G_{2} G_{3} H_{2}}

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G_{4} + \frac{G_{1} G_{2} G_{3}}{1 + G_{2} H_{1} (1 - G_{1}) + G_{2} G_{3} H_{2}}

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G_{4} + \frac{G_{1} G_{2} G_{3}}{1 + G_{2} H_{1} (1 + G_{1}) + G_{2} G_{3} H_{2}}

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Solution

The correct option is C $G_{4} + \frac{G_{1} G_{2} G_{3}}{1 + G_{2} H_{1} (1 - G_{1}) + G_{2} G_{3} H_{2}}$
The signal flow graph corresponding to the given block diagram can be drawn as shown below.

Forward path gains:
$P_{1} = G_{1} G_{2} G_{3}$
$P_{2} = G_{4}$
Individual loop gains:
$L_{1} = - G_{2} H_{1}$
$L_{2} = G_{1} G_{2} H_{1}$
$L_{3} = - G_{2} G_{3} H_{2}$
The is no pair of two non-touching loops.
$S o, Δ = 1 + G_{2} H_{1} - G_{1} G_{2} H_{1} + G_{2} G_{3} H_{2}$
$Δ_{1} = 1$
$Δ_{2} = Δ$
So, $\frac{C}{R} = G_{4} + \frac{G_{1} G_{2} G_{3}}{1 + G_{2} H_{1} (1 - G_{1}) + G_{2} G_{3} H_{2}}$

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