Chapter 5 Part 2

Alrafidain University College

Assistant Lecturer / Hussain Kassim Ahmad 1

Control Engineering Fundamentals

Chapter 5 Part 2

Signal Flow Graph and Mason’s Gain Formula

Mason's Gain Formula for Signal Flow Graphs:

In many practical cases, we wish to determine the relationship between an

input variable and an output variable of the signal flow graph. The transmittance

between an input node and an output node is the overall gain, or overall

transmittance, between these two nodes. Mason's gain formula, which is applicable

to the overall gain, is given by

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

∑ 𝑃𝐾∆𝐾𝑁𝑘=1

∆

Where

PK = path gain or transmittance of Kth forward path.

Δ = determinant of graph.

= 1 - (sum of all individual loop gains) + (sum of gain products of all

possible combinations of two non-touching loops) - (sum of gain

products

of all possible combinations of three non-touching loops) + . . .

∆ = 1 − ∑ 𝐿𝑎

𝑎

+ ∑ 𝐿𝑏𝐿𝑐

𝑏,𝑐

− ∑ 𝐿𝑑𝐿𝑒𝐿𝑓

𝑑,𝑒,𝑓

∑ 𝐿𝑎

𝑎

= 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠

∑ 𝐿𝑏𝐿𝑐

𝑏,𝑐

= 𝑠𝑢𝑚 𝑜𝑓 𝑔𝑎𝑖𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡𝑤𝑜

Nontouching loops.



∑ 𝐿𝑑𝐿𝑒𝐿𝑓

𝑑,𝑒,𝑓

= 𝑠𝑢𝑚 𝑜𝑓 𝑔𝑎𝑖𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓

𝑡ℎ𝑟𝑒𝑒 Nontouching loops.

ΔK = Same as Δ but formed by loops not touching the (PK) kth forwatd path.

(cofactor of the kth forward path determinant of the graph with the

loops

touching the kth forward path removed, that is, the cofactor ΔK, is

obtained from Δ by removing the loops that touch path Pk).

(Note that the summations are taken over all possible paths from input to

output.)

In the following, we shall illustrate the use of Mason's gain formula by means

of these examples.



Example (5.1): Consider the system shown in Figure (6-3). Obtain the closed-loop

transfer function C(s) /R(s) by use of Mason's gain formula.

Figure (6-3)

Solution:

In this system, there are three forward paths between the input R(s) and the

output C(s).

The forward path gains are

𝑃1 = 𝐺1𝐺2𝐺3𝐺4𝐺5

𝑃2 = 𝐺1𝐺6𝐺4𝐺5

𝑃3 = 𝐺1𝐺2𝐺7

There are four individual loops, the gains of these loops are

𝐿1 = −𝐺4𝐻1

𝐿2 = −𝐺2𝐺7𝐻2

𝐿3 = −𝐺6𝐺4𝐺5𝐻2

𝐿4 = −𝐺2𝐺3𝐺4𝐺5𝐻2

Loop L1 does not touch loop L2. Hence, the determinant A is given by

∆= 1 − ( 𝐿1 + 𝐿2 + 𝐿3 + 𝐿4) + 𝐿1𝐿2

∆= 1 − (−𝐺4𝐻1 − 𝐺2𝐺7𝐻2 − 𝐺6𝐺4𝐺5𝐻2 − 𝐺2𝐺3𝐺4𝐺5𝐻2) + 𝐺4𝐻1𝐺2𝐺7𝐻2

The cofactor Δ1 is obtained from Δ by removing the loops that touch path P1.

Therefore, by removing L1, L2, L3, L4, and L1, L2 from Equation of Δ, we obtain

∆1= 1



Similarly, the cofactor Δ2 is

∆2= 1

The cofactor Δ3 is obtained by removing L2, L3, L4, and L1, L2 from Equation

of Δ, giving

∆3= 1 − 𝐿1

The closed-loop transfer function C(s) /R(s) is then

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1∆1 + 𝑃2∆2 + 𝑃3∆3

∆

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1(1) + 𝑃2(1) + 𝑃3(1 − 𝐿1)

1 − ( 𝐿1 + 𝐿2 + 𝐿3 + 𝐿4) + 𝐿1𝐿2

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝐺1𝐺2𝐺3𝐺4𝐺5 + 𝐺1𝐺6𝐺4𝐺5 + 𝐺1𝐺2𝐺7(1 + 𝐺4𝐻1)

1 − (−𝐺4𝐻1 − 𝐺2𝐺7𝐻2 − 𝐺6𝐺4𝐺5𝐻2 − 𝐺2𝐺3𝐺4𝐺5𝐻2) + 𝐺4𝐻1𝐺2𝐺7𝐻2

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝐺1𝐺2𝐺3𝐺4𝐺5 + 𝐺1𝐺6𝐺4𝐺5 + 𝐺1𝐺2𝐺7 + 𝐺1𝐺2𝐺7𝐺4𝐻1)

1 + 𝐺4𝐻1 + 𝐺2𝐺7𝐻2 + 𝐺6𝐺4𝐺5𝐻2 + 𝐺2𝐺3𝐺4𝐺5𝐻2 + 𝐺4𝐻1𝐺2𝐺7𝐻2



Example (5-2): Consider the system shown in Figure (6-4). Obtain the closed-loop

transfer function C(s) /R(s) by use of Mason's gain formula.

Figure (6-4)

Solution:

First we draw a signal flow graph for this system.

In this system, there are three forward paths between the input R(s) and the

output C(s).


𝑃1 = 𝐺1𝐺2𝐺3

From Figure for A signal flow graph, we see that there are three individual

loops. The gains of these loops are

𝐿1 = 𝐺1𝐺2𝐻1

𝐿2 = −𝐺2𝐺3𝐻2

𝐿3 = −𝐺1𝐺2𝐺3

Note that since all three loops have a common branch, there are no

nontouching loops. Hence, the determinant A is given by



∆ = 1 − ( 𝐿1 + 𝐿2 + 𝐿3)

∆ = 1 – 𝐺1𝐺2𝐻1 + 𝐺2𝐺3𝐻2 + 𝐺1𝐺2𝐺3

The cofactor Δ1 of the determinant along the forward path connecting the

input node and output node is obtained from Δ by removing the loops that touch

this path. Since path Pl touches all three loops, we obtain

∆1= 1

Therefore, the overall gain between the input R(s) and the output C(s), or the

closed-loop transfer function, is given by

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1∆1

∆

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1(1)

1 − ( 𝐿1 + 𝐿2 + 𝐿3)

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝐺1𝐺2𝐺3

1 − 𝐺1𝐺2𝐻1 + 𝐺2𝐺3𝐻2 + 𝐺1𝐺2𝐺3

This is the same as the closed-loop transfer function obtained by block

diagram reduction. Mason's gain formula thus gives the overall gain C(s)/R(s)

without a reduction of the graph.



Example (5.3): Find the transfer function for the system given by signal flow

graph of Figure below by using of Mason's gain formula.

Solution:

In this system, there are three forward paths between the input R(s) and the output C(s).


𝑃1 =1

𝑆3𝑏3 =

𝑏3

𝑆3

𝑃2 =1

𝑆2𝑏2 =

𝑏2

𝑆2

𝑃3 =1

𝑆𝑏1 =

𝑏1

𝑆

There are three individual loops, the gains of these loops are

𝐿1 = −𝑎1

𝑆

𝐿2 = −𝑎2

𝑆2

𝐿3 = −𝑎3

𝑆3

The determinant ∆ is given by

∆= 1 − ( 𝐿1 + 𝐿2 + 𝐿3)

∆= 1 − (−𝑎1

𝑆−

𝑎2

𝑆2−

𝑎3

𝑆3)

∆= 1 +𝑎1

𝑆+

𝑎2

𝑆2+

𝑎3

𝑆3

The cofactor Δ1, Δ2, Δ3 is obtained

∆1= 1

∆2= 1

∆3= 1

The closed-loop transfer function C(s) /R(s) is

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1∆1 + 𝑃2∆2 + 𝑃3∆3

∆

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑏3𝑆3+

𝑏2𝑆2+

𝑏1𝑆

1+𝑎1𝑆

+𝑎2𝑆2+

𝑎3𝑆3



Example (5.4): Find the transfer function for the system given by signal flow

graph of Figure (6.5) by using of Mason's gain formula.

Figure (6.5)


𝑃1 = 𝐺1𝐺2𝐺3𝐺4𝐺5

𝑃2 = 𝐺6

There are four individual loops, the gains of these loops are

𝐿1 = −𝐺2𝐻1

𝐿2 = −𝐺4𝐻2

𝐿3 = −𝐺6𝐻3

𝐿4 = −𝐺1𝐺2𝐺3𝐺4𝐺5𝐻3

∆= 1 − ( 𝐿1 + 𝐿2 + 𝐿3 + 𝐿4) + (𝐿1𝐿2 + 𝐿1𝐿3 + 𝐿2𝐿3) + (𝐿1𝐿2𝐿3)

∆= 1 − ( −𝐺2𝐻1 − 𝐺4𝐻2 − 𝐺6𝐻3−𝐺1𝐺2𝐺3𝐺4𝐺5𝐻3)

+ ( 𝐺2𝐻1𝐺4𝐻2 + 𝐺2𝐻1𝐺6𝐻3 + 𝐺4𝐻2𝐺6𝐻3 ) − (−𝐺2𝐻1𝐺4𝐻2𝐺6𝐻3 )

∆= 1 + 𝐺2𝐻1 + 𝐺4𝐻2 + 𝐺6𝐻3+𝐺1𝐺2𝐺3𝐺4𝐺5𝐻3 + 𝐺2𝐻1𝐺4𝐻2 + 𝐺2𝐻1𝐺6𝐻3

+ 𝐺4𝐻2𝐺6𝐻3 + 𝐺2𝐻1𝐺4𝐻2𝐺6𝐻3

The cofactor Δ1 is obtained from Δ by removing the loops that touch path P1.

Therefore, by removing L1, L2, L3, L4, we obtain

∆1= 1



Similarly, the cofactor Δ2 is obtained from Δ by removing the loops that touch

path P2. Therefore, by removing L3, L4, we obtain

∆2= 1 − (𝐿1 + 𝐿1) + (𝐿1 𝐿2)

∆2= 1 − ( −𝐺2𝐻1 − 𝐺4𝐻2) + ( 𝐺2𝐻1𝐺4𝐻2)

∆2= 1 + 𝐺2𝐻1 + 𝐺4𝐻2 + 𝐺2𝐻1𝐺4𝐻2

The closed-loop transfer function C(s) /R(s) is then

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1∆1 + 𝑃2∆2

∆

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝑃1(1) + 𝑃2[1 − (𝐿1 + 𝐿1) + (𝐿1 𝐿2)]

1 − ( 𝐿1 + 𝐿2 + 𝐿3 + 𝐿4) + (𝐿1𝐿2 + 𝐿1𝐿3 + 𝐿2𝐿3) + (𝐿1𝐿2𝐿3)

𝑇. 𝐹. =𝐶(𝑠)

𝑅(𝑠)=

𝐺1𝐺2𝐺3𝐺4𝐺5 + 𝐺6(1 + 𝐺2𝐻1 + 𝐺4𝐻2 + 𝐺2𝐻1𝐺4𝐻2)1 + 𝐺2𝐻1 + 𝐺4𝐻2 + 𝐺6𝐻3+𝐺1𝐺2𝐺3𝐺4𝐺5𝐻3 + 𝐺2𝐺4𝐻1𝐻2 + 𝐺2𝐺6𝐻1𝐻3 + 𝐺4𝐺6𝐻2𝐻3 + 𝐺2𝐺4𝐺6𝐻1𝐻2𝐻3

Chapter 5 Part 2

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