Write and explain the Mason’s gain formula. Determine transfer function of the system from the given signal flow graph. R(s) 1 1 G
1 G
1 C(s) G
-H
-H
-1 If the gain G
is removed, determine the forward path gains, loop gains. Also determine the corresponding transfer function.
The Correct Answer and Explanation is:
Mason’s Gain Formula
Mason’s Gain Formula is a systematic method for finding the overall transfer function (gain) of a linear system represented by a signal flow graph. It expresses the relationship between the input and output variables without requiring the reduction of the graph. The formula is given by:
T = C(s) / R(s) = (1/Δ) * Σ PₖΔₖ
Where:
- T is the overall transfer function of the system.
- C(s) is the output node.
- R(s) is the input node.
- Pₖ is the gain of the k-th forward path from input to output. A forward path is a path that does not pass through any node more than once.
- Δ is the determinant of the graph, calculated as:
Δ = 1 – (Sum of all individual loop gains) + (Sum of gain products of all possible pairs of non-touching loops) – (Sum of gain products of all possible triplets of non-touching loops) + …
(Non-touching loops are loops that do not share any common nodes). - Δₖ is the cofactor of the k-th forward path. It is calculated as the value of Δ for the part of the graph that does not touch the k-th forward path.
Transfer Function of the Given System
To determine the transfer function C(s)/R(s), we will apply Mason’s Gain Formula by identifying all paths and loops.
- Forward Path Gains (Pₖ):
- P₁: The upper path from R(s) to C(s): P₁ = 1 × 1 × G₂ × 1 × G₃ × 1 = G₂G₃
- P₂: The lower path from R(s) to C(s): P₂ = 1 × G₁ × 1 × G₃ × 1 = G₁G₃
- Individual Loop Gains (Lᵢ):
- L₁: The self-loop: 1
- L₂: The loop involving G₂ and -H₁: L₂ = G₂ × (-H₁) = -G₂H₁
- L₃: The loop involving G₃ and -1: L₃ = G₃ × (-1) = -G₃
- L₄: The large loop involving G₂, G₃, and -H₂: L₄ = 1 × G₂ × 1 × G₃ × (-H₂) = -G₂G₃H₂
- L₅: The large loop involving G₁, G₃, and -H₂: L₅ = G₁ × 1 × G₃ × (-H₂) = -G₁G₃H₂
- Pairs of Non-Touching Loops:
- L₁ and L₃ are non-touching. Product: L₁L₃ = (1)(-G₃) = -G₃
- L₁ and L₅ are non-touching. Product: L₁L₅ = (1)(-G₁G₃H₂) = -G₁G₃H₂
- (All other loop pairs share at least one node).
- Triplets of Non-Touching Loops: There are no sets of three mutually non-touching loops.
- Calculate the Determinant (Δ):
Δ = 1 – (L₁ + L₂ + L₃ + L₄ + L₅) + (L₁L₃ + L₁L₅)
Δ = 1 – (1 – G₂H₁ – G₃ – G₂G₃H₂ – G₁G₃H₂) + (-G₃ – G₁G₃H₂)
Δ = 1 – 1 + G₂H₁ + G₃ + G₂G₃H₂ + G₁G₃H₂ – G₃ – G₁G₃H₂
Δ = G₂H₁ + G₂G₃H₂ - Calculate the Cofactors (Δₖ):
- Δ₁ (for P₁): The path P₁ touches all loops. Therefore, no loops remain when P₁ is removed. So, Δ₁ = 1.
- Δ₂ (for P₂): The path P₂ does not touch loop L₁. All other loops (L₂, L₃, L₄, L₅) are touched by P₂. Therefore, Δ₂ = 1 – L₁ = 1 – 1 = 0.
- Calculate the Transfer Function (T):
T = (P₁Δ₁ + P₂Δ₂) / Δ
T = (G₂G₃ × 1 + G₁G₃ × 0) / (G₂H₁ + G₂G₃H₂)
T = G₂G₃ / (G₂H₁ + G₂G₃H₂)
T = G₂G₃ / [G₂(H₁ + G₃H₂)]
T = C(s)/R(s) = G₃ / (H₁ + G₃H₂)
System with Gain G₃ Removed
If the gain G₃ is removed, the branch connecting the last two central nodes is eliminated.
- Forward Path Gains: Both original forward paths, P₁ and P₂, required the gain G₃ to reach the output. With G₃ removed, there are no forward paths from R(s) to C(s).
- P₁ (new) = 0
- P₂ (new) = 0
- Loop Gains: We identify the loops that remain.
- L₁ and L₂ do not involve G₃, so they remain: L₁ = 1 and L₂ = -G₂H₁.
- L₃, L₄, and L₅ all required the gain G₃, so they are eliminated.
- Corresponding Transfer Function:
Since there are no forward paths (all Pₖ = 0), the numerator of Mason’s Gain Formula becomes zero.
T = (Σ PₖΔₖ) / Δ = 0 / Δ = 0
The transfer function becomes zero because removing G₃ breaks the only connections leading towards the output, meaning the input signal R(s) can no longer influence the output C(s).
