Q1Control System

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Q.1. Find the transfer function of the system shown in fig. using block diagram reduction technique (a multi-loop diagram with forward blocks G1, G2, G3 in cascade from R(s) to C(s), an inner positive feedback via H2 around G2-G3, an outer feedback via H3 from after G2G3 back to before G2, and an outermost negative feedback via H1 around the entire forward path back to the input summing junction).

Answer 1

The block diagram consists of a forward path R(s) → G1 → G2 → G3 → C(s), with three feedback paths: an inner positive feedback H2 taken from after G3 back to before G2 (around G2-G3), a feedback H3 taken from after G2G3 back to a summing point before G2 as well, and an outer negative feedback H1 taken from the output C(s) all the way back to the primary input summing junction (around the entire forward path G1G2G3).

Step 1 — Combine the inner feedback loops around G2-G3: Since both H2 (positive/regenerative, based on the arrow into the summing junction before G2) and H3 are feedback paths around the block G2G3 acting at the same summing point before G2, they combine into an equivalent single feedback path (H2 + H3) or (H2 − H3) around G2G3 depending on their signs at the summer, feeding into the point after G1's output. Taking the diagram's H2 as positive feedback (regenerative) and H3 as negative feedback (as is conventional at that junction), the combined inner loop reduces G2G3 to:

G_{23, e q} (s) = \frac{G _{2} G _{3}}{1 - G _{2} G _{3} H _{2} + G _{2} G _{3} H _{3}} = \frac{G _{2} G _{3}}{1 + G _{2} G _{3} ( H _{3} - H _{2} )}

Step 2 — Combine with G1 in cascade: The equivalent block G23,eq is in cascade with G1, giving a combined forward-path gain:

G_{e q} (s) = G_{1} \cdot G_{23, e q} (s) = \frac{G _{1} G _{2} G _{3}}{1 + G _{2} G _{3} ( H _{3} - H _{2} )}

Step 3 — Reduce the outer feedback loop H1: This combined forward block is now in a standard single negative-feedback loop with H1 around the entire path from R(s) to C(s):

\frac{C ( s )}{R ( s )} = \frac{G _{e q} ( s )}{1 + G _{e q} ( s ) H _{1}} = \frac{\frac{G _{1} G _{2} G _{3}}{1 + G _{2} G _{3} ( H _{3} - H _{2} )}}{1 + \frac{G _{1} G _{2} G _{3} H _{1}}{1 + G _{2} G _{3} ( H _{3} - H _{2} )}}

Multiplying numerator and denominator by [1 + G2G3(H3−H2)] to clear the compound fraction, the final simplified transfer function is:

\frac{C ( s )}{R ( s )} = \frac{G _{1} G _{2} G _{3}}{1 + G _{2} G _{3} ( H _{3} - H _{2} ) + G _{1} G _{2} G _{3} H _{1}}

This result illustrates the systematic block-diagram-reduction technique: (i) collapse cascaded blocks by multiplying their gains, (ii) collapse an inner feedback loop using G/(1∓GH), and (iii) repeat outward until only a single forward block remains in a single feedback loop with the outermost feedback element, at which point the standard closed-loop formula G/(1+GH) yields the final answer.

Alternative check via Mason's Gain Formula. As a cross-check, treating the block diagram as a signal flow graph gives one forward path with gain G1G2G3, and loop gains L1 = G2G3H2 (positive feedback, hence entering Δ with opposite sign), L2 = -G2G3H3, and L3 = -G1G2G3H1 (all sharing common nodes with the forward path and with each other, so no non-touching loop products exist). Substituting into Δ = 1 - ΣLi gives Δ = 1 - G2G3H2 + G2G3H3 + G1G2G3H1, and with Δ1 = 1 (the single forward path touches every loop), Mason's formula directly returns C/R = G1G2G3/[1 + G2G3(H3-H2) + G1G2G3H1], matching the block-diagram-reduction result exactly and confirming the derivation.

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