In the below figure, output C₁ due to R₁ and R₂ is given by:

Question

Question

This question was previously asked in

MPPKVVCL Indore JE Electrical 21 August 2018 Official Paper

$\frac{G_{1} R_{1} - G_{1} G_{3} G_{4} R_{2}}{1 - G_{1} G_{2} G_{3} G_{4}}$
$\frac{G_{1} R_{1} - G_{2} G_{3} G_{4} R_{2}}{1 - G_{1} G_{2} G_{3} G_{4}}$
$\frac{G_{1} R_{1} - G_{1} G_{3} G_{2} R_{2}}{1 - G_{1} G_{2} G_{3} G_{4}}$
$\frac{G_{1} R_{1} - G_{1} G_{2} G_{3} G_{4} R_{2}}{1 - G_{1} G_{2} G_{3} G_{4}}$

Answer 1

Concept:

According to Mason’s gain formula, the transfer function is given by

$T F = \frac{\sum_{k = 1}^{n} M_{k} Δ_{k}}{Δ}$

Where, n = no of forward paths

Mk = kth forward path gain

Δk = 1 - Sum of the loop that exists after removal of the kth forward path + sum of the gain product of two non-touching loops

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Calculation

While calculating output C1 due to R1 and R2 , keep C₂ = 0

The forward path from R₁ to C₁ is:

$M_{1} = G_{1} R_{1}$

The forward path from R₂ to C1 is:

$M_{2} = G_{1} G_{3} G_{4} R_{2}$

Self-loop gain:

$Δ = 1 - G_{1} G_{2} G_{3} G_{4}$

$T (s) = \frac{G_{1} R_{1} - G_{1} G_{3} G_{4} R_{2}}{1 - G_{1} G_{2} G_{3} G_{4}}$

Hence, the correct answer is option 1.