Mason's Gain Formula



Let us now discuss the Mason’s Gain Formula. Suppose there are ‘N’ forward paths in a signal flow graph. The gain between the input and the output nodes of a signal flow graph is nothing but the transfer function of the system. It can be calculated by using Mason’s gain formula.

Mason’s gain formula is

$$T=\frac{C(s)}{R(s)}=\frac{\Sigma ^N _{i=1}P_i\Delta _i}{\Delta}$$

Where,

  • C(s) is the output node

  • R(s) is the input node

  • T is the transfer function or gain between $R(s)$ and $C(s)$

  • Pi is the ith forward path gain

$\Delta =1-(sum \: of \: all \: individual \: loop \: gains)$

$+(sum \: of \: gain \: products \: of \: all \: possible \: two \:nontouching \: loops)$

$$-(sum \: of \: gain \: products \: of \: all \: possible \: three \: nontouching \: loops)+...$$

Δi is obtained from Δ by removing the loops which are touching the ith forward path.

Consider the following signal flow graph in order to understand the basic terminology involved here.

Mason Formula Basic

Path

It is a traversal of branches from one node to any other node in the direction of branch arrows. It should not traverse any node more than once.

Examples − $y_2 \rightarrow y_3 \rightarrow y_4 \rightarrow y_5$ and $y_5 \rightarrow y_3 \rightarrow y_2$

Forward Path

The path that exists from the input node to the output node is known as forward path.

Examples − $y_1 \rightarrow y_2 \rightarrow y_3 \rightarrow y_4 \rightarrow y_5 \rightarrow y_6$ and $y_1 \rightarrow y_2 \rightarrow y_3 \rightarrow y_5 \rightarrow y_6$.

Forward Path Gain

It is obtained by calculating the product of all branch gains of the forward path.

Examples − $abcde$ is the forward path gain of $y_1 \rightarrow y_2 \rightarrow y_3 \rightarrow y_4 \rightarrow y_5 \rightarrow y_6$ and abge is the forward path gain of $y_1 \rightarrow y_2 \rightarrow y_3 \rightarrow y_5 \rightarrow y_6$.

Loop

The path that starts from one node and ends at the same node is known as loop. Hence, it is a closed path.

Examples − $y_2 \rightarrow y_3 \rightarrow y_2$ and $y_3 \rightarrow y_5 \rightarrow y_3$.

Loop Gain

It is obtained by calculating the product of all branch gains of a loop.

Examples − $b_j$ is the loop gain of $y_2 \rightarrow y_3 \rightarrow y_2$ and $g_h$ is the loop gain of $y_3 \rightarrow y_5 \rightarrow y_3$.

Non-touching Loops

These are the loops, which should not have any common node.

Examples − The loops, $y_2 \rightarrow y_3 \rightarrow y_2$ and $y_4 \rightarrow y_5 \rightarrow y_4$ are non-touching.

Calculation of Transfer Function using Mason’s Gain Formula

Let us consider the same signal flow graph for finding transfer function.

Mason Formula Basic
  • Number of forward paths, N = 2.

  • First forward path is - $y_1 \rightarrow y_2 \rightarrow y_3 \rightarrow y_4 \rightarrow y_5 \rightarrow y_6$.

  • First forward path gain, $p_1 = abcde$.

  • Second forward path is - $y_1 \rightarrow y_2 \rightarrow y_3 \rightarrow y_5 \rightarrow y_6$.

  • Second forward path gain, $p_2 = abge$.

  • Number of individual loops, L = 5.

  • Loops are - $y_2 \rightarrow y_3 \rightarrow y_2$, $y_3 \rightarrow y_5 \rightarrow y_3$, $y_3 \rightarrow y_4 \rightarrow y_5 \rightarrow y_3$, $y_4 \rightarrow y_5 \rightarrow y_4$ and $y_5 \rightarrow y_5$.

  • Loop gains are - $l_1 = bj$, $l_2 = gh$, $l_3 = cdh$, $l_4 = di$ and $l_5 = f$.

  • Number of two non-touching loops = 2.

  • First non-touching loops pair is - $y_2 \rightarrow y_3 \rightarrow y_2$, $y_4 \rightarrow y_5 \rightarrow y_4$.

  • Gain product of first non-touching loops pair, $l_1l_4 = bjdi$

  • Second non-touching loops pair is - $y_2 \rightarrow y_3 \rightarrow y_2$, $y_5 \rightarrow y_5$.

  • Gain product of second non-touching loops pair is - $l_1l_5 = bjf$

Higher number of (more than two) non-touching loops are not present in this signal flow graph.

We know,

$\Delta =1-(sum \: of \: all \: individual \: loop \: gains)$

$+(sum \: of \: gain \: products \: of \: all \: possible \: two \:nontouching \: loops)$

$$-(sum \: of \: gain \: products \: of \: all \: possible \: three \: nontouching \: loops)+...$$

Substitute the values in the above equation,

$\Delta =1-(bj+gh+cdh+di+f)+(bjdi+bjf)-(0)$

$\Rightarrow \Delta=1-(bj+gh+cdh+di+f)+bjdi+bjf$

There is no loop which is non-touching to the first forward path.

So, $\Delta_1=1$.

Similarly, $\Delta_2=1$. Since, no loop which is non-touching to the second forward path.

Substitute, N = 2 in Mason’s gain formula

$$T=\frac{C(s)}{R(s)}=\frac{\Sigma ^2 _{i=1}P_i\Delta _i}{\Delta}$$

$$T=\frac{C(s)}{R(s)}=\frac{P_1\Delta_1+P_2\Delta_2}{\Delta}$$

Substitute all the necessary values in the above equation.

$$T=\frac{C(s)}{R(s)}=\frac{(abcde)1+(abge)1}{1-(bj+gh+cdh+di+f)+bjdi+bjf}$$

$$\Rightarrow T=\frac{C(s)}{R(s)}=\frac{(abcde)+(abge)}{1-(bj+gh+cdh+di+f)+bjdi+bjf}$$

Therefore, the transfer function is -

$$T=\frac{C(s)}{R(s)}=\frac{(abcde)+(abge)}{1-(bj+gh+cdh+di+f)+bjdi+bjf}$$

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