The transfer function between $y_{2}$ and $y_{1}$ in figure

a + b

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(a + b) c

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\frac{a + b}{1 - c}

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\frac{a + b}{1 + c}

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Solution

The correct option is C $\frac{a + b}{1 - c}$
Between $y_{1}$ and $y_{2}$
forward paths,
$P_{1} = a, P_{2} = b$
Loop $\Rightarrow L_{1} = c$

$T . F . = \frac{P_{1} Δ_{1} + P_{2} Δ_{2}}{Δ}$

$= \frac{a (1 - 0) + b (1 - 0)}{1 - c} = \frac{a + b}{1 - c}$

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The transfer function between y2 and y1 in figure

The correct option is C a+b1−cBetween y1 and y2 forward paths, P1=a,P2=b Loop ⇒L1=c T.F.=P1Δ1+P2Δ2Δ =a(1−0)+b(1−0)1−c=a+b1−c

The transfer function between $y_{2}$ and $y_{1}$ in figure

The correct option is C $\frac{a + b}{1 - c}$
Between $y_{1}$ and $y_{2}$
forward paths,
$P_{1} = a, P_{2} = b$
Loop $\Rightarrow L_{1} = c$

$T . F . = \frac{P_{1} Δ_{1} + P_{2} Δ_{2}}{Δ}$

$= \frac{a (1 - 0) + b (1 - 0)}{1 - c} = \frac{a + b}{1 - c}$