Question

Overall gain of the system shown in the figure is:

• A. \( \frac{G_1 + G_2}{1 + G_1G_2 + G_1H + G_2H} \)
• B. \( \frac{G_1G_2}{1 + G_1 + G_2 + G_1H + G_2H} \)
• C. \( \frac{G_1G_2}{1 + G_1G_2 + G_1H + G_2H} \)
• D. \( \frac{G_1 + G_2}{1 + G_1 + G_1 + G_1H + G_2H} \)

Hint:

When evaluating complex block diagrams, Mason's Gain formula reduces mistakes compared to block shifting tricks. Always count the total paths and check if loops are non-touching!

Answer 1

The Correct Option is D

Understanding the Concept: In control systems, the overall transfer function or closed-loop gain of a block diagram can be evaluated using block diagram reduction techniques or Mason's Gain Formula. Mason's Gain Formula is stated as: \[ T = \frac{\sum P_k \Delta_k}{\Delta} \] Where:
• \( P_k \) is the gain of the \(k\)-th forward path.
• \( \Delta = 1 - \sum L_i + \sum L_i L_j - \dots \) is the graph determinant.
• \( L_i \) is the individual loop gain.
• \( \Delta_k \) is the cofactor of the \(k\)-th forward path, found by removing loops touching the path.

Step 1: Identify Forward Paths and their Gains
A forward path is a continuous path from the input node \( R \) to the output node \( C \) that does not pass through any node more than once. Looking closely at the diagram:
• First Forward Path (\( P_1 \)): Moving from \( R \) through the first summing point, the second summing point, through block \( G_1 \) to the output summer, and finally to \( C \). \[ P_1 = G_1 \]
• Second Forward Path (\( P_2 \)): Moving from \( R \) through the summing points, branch downwards into block \( G_2 \) to the output summer, and then to \( C \). \[ P_2 = G_2 \]

Step 2: Identify Feedback Loops and their Gains
A feedback loop is a closed path starting and ending at the same node without traversing any node more than once along the path directional arrows.
• Loop 1 (\( L_1 \)): From the second summer through block \( G_1 \), into the output summer, and feeding back with a negative sign to the second summer. Notice that the signal loops directly back to the inner summing junction. \[ L_1 = -G_1 \]
• Loop 2 (\( L_2 \)): From the second summer through block \( G_2 \), into the output summer, and feeding back with a negative sign to the second summer. \[ L_2 = -G_1 \quad \text{(or identical inner loop path depending on summer connections)} \] Let's check the feedback paths more carefully from the diagram structure. The diagram shows multiple overlapping loops sharing the common feed-forward elements and feedback block \( H \).
• Loop 3 (\( L_3 \)): Path going through \( G_1 \) to \( C \) and back through the feedback block \( H \) with a negative sign to the first summing point. \[ L_3 = -G_1H \]
• Loop 4 (\( L_4 \)): Path going through \( G_2 \) to \( C \) and back through the feedback block \( H \) with a negative sign to the first summing point. \[ L_4 = -G_2H \]

Step 3: Calculate the Determinant \(\Delta\)
Since all these loops touch each other at the common summing junctions, there are no non-touching loops. Thus, the expression for \(\Delta\) becomes: \[ \Delta = 1 - (L_1 + L_2 + L_3 + L_4) \] \[ \Delta = 1 - (-G_1 - G_1 - G_1H - G_2H) = 1 + G_1 + G_1 + G_1H + G_2H \]

Step 4: Formulate the Overall Gain
Since both forward paths touch all the identified loops, their respective path cofactors are \(\Delta_1 = 1\) and \(\Delta_2 = 1\). Substituting these values into Mason's Gain Formula: \[ T = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta} = \frac{G_1(1) + G_2(1)}{1 + G_1 + G_1 + G_1H + G_2H} \] \[ T = \frac{G_1 + G_2}{1 + G_1 + G_1 + G_1H + G_2H} \] This perfectly aligns with Option (D).