The transfer function of the open loop system G(s) which is represented by the singal flow graph shown in the figure below is

\frac{G_{1} G_{2}}{1 + G_{1} H_{1} + G_{2} H_{2}}

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

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

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

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Solution

The correct option is A $\frac{G_{1} G_{2}}{1 + G_{1} H_{1} + G_{2} H_{2}}$
Forward path gain
$P_{1} = G_{1} G_{2}$

Loops
$L_{1} = - G_{1} H_{1}$
$L_{2} = - G_{2} H_{2}$

No non-touching loop
$Δ = 1 - (L_{1} + L_{2})$
$= 1 + G_{1} H_{1} + G_{2} H_{2}$

Therefore according to Mason gain formula
$T (s) = \frac{C (s)}{R (s)} = \frac{P_{1} Δ_{1}}{Δ}$

$= \frac{G_{1} G_{2}}{1 + G_{1} H_{1} + G_{2} H_{2}}$

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