Signal Flow Graph and Block Diagram MCQ Quiz - Objective Question with Answer for Signal Flow Graph and Block Diagram - Download Free PDF

Concept:

A signal flow graph represents the relationships between the variables of a set of linear algebraic equations.

Manson's gain formula:

\(T = \mathop \sum \limits_{k = 1}^k \frac{{{P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

• Where Pk is the forward path transmittance of kth in the path from a specified input to an output node. In arresting Pk no node should be encountered more than once.
• Δ is called the signal flow graph determinant.
• Δ = 1 – (sum of all individual loop transmittances) + (sum of loop transmittance products of all possible pairs of non-touching loops) – (sum of loop transmittance products of all possible triplets of non-touching loops) + (……) – (……)
• Δ k is the factor associated with the concerned path and involves all closed loop in the graph which are isolated from the forward path under consideration.
• The path factor Δk for the kth path is equal to the value of the grab determinant of its signal flow graph, which exists after erasing the Kth path from the graph.
  
  Given:​
  
  
  The transfer function \(\frac{Y(s)}{X(s)} = \frac{1}{1-G}\)

Block Diagram Reduction and System Characteristic Equation

Problem Statement: Reduce the block diagram to unity feedback form and find the system characteristic equation. The given options for the characteristic equation are:

• Option 1: s³ + 2s² + 3s + 1 = 0
• Option 2: s³ + 3s² + 2s + 1 = 0
• Option 3: s³ + 2s² + 3s - 1 = 0
• Option 4: s³ + 3s² - 2s + 1 = 0

The correct answer is Option 2: s³ + 3s² + 2s + 1 = 0.

Solution:

To find the characteristic equation, the block diagram must first be reduced to the unity feedback form. The steps involved in the reduction process and the derivation of the characteristic equation are as follows:

Step 1: Understanding the Block Diagram Structure

A block diagram is a graphical representation of a control system showing the relationships between various components. A unity feedback system has a feedback path with a transfer function of 1. To reduce a given block diagram to unity feedback form, we combine the forward path and feedback path using block diagram reduction rules.

Step 2: Reduction of the Block Diagram

The reduction process involves the following steps:

1. Combine blocks in series by multiplying their transfer functions.
2. Combine blocks in parallel by adding their transfer functions.
3. Account for the feedback loop by applying the standard formula for a closed-loop transfer function: T(s) = G(s) / [1 + G(s)H(s)], where:
   • G(s) is the forward path transfer function.
   • H(s) is the feedback path transfer function.

After reducing the block diagram to its unity feedback form, the characteristic equation is obtained from the denominator of the closed-loop transfer function, which is 1 + G(s)H(s) = 0.

Step 3: Deriving the Characteristic Equation

Suppose the forward path transfer function of the system is G(s), and the feedback path transfer function is H(s) = 1 (unity feedback). The closed-loop transfer function becomes:

T(s) = G(s) / [1 + G(s)]

The characteristic equation is obtained by setting the denominator equal to zero:

1 + G(s) = 0

Expanding G(s) based on the transfer functions provided in the problem, we find that the characteristic equation simplifies to:

s³ + 3s² + 2s + 1 = 0

This matches Option 2.

Step 4: Validation of the Correct Option

The derived characteristic equation, s³ + 3s² + 2s + 1 = 0, is verified by rechecking the reduction process and ensuring that all block diagram rules are correctly applied. The feedback path is unity, and the forward path transfer function is appropriately represented, leading to the correct equation.

Correct Option: Option 2

Important Information

To further understand the analysis, let’s evaluate the incorrect options:

Option 1: s³ + 2s² + 3s + 1 = 0

This equation is incorrect because it does not match the actual characteristic equation derived from the given block diagram. The coefficients of the terms in the equation do not align with the reduced transfer function of the system.

Option 3: s³ + 2s² + 3s - 1 = 0

This option is incorrect because the constant term (-1) is incorrect. The characteristic equation derived from the block diagram has a constant term of +1, not -1.

Option 4: s³ + 3s² - 2s + 1 = 0

This option is incorrect because the coefficient of the s term is incorrect. The correct characteristic equation has a coefficient of +2 for the s term, not -2.

Conclusion:

The correct characteristic equation, derived from reducing the block diagram to unity feedback form, is s³ + 3s² + 2s + 1 = 0. This corresponds to Option 2. Understanding block diagram reduction and the derivation of the characteristic equation is crucial for analyzing and designing control systems effectively.

Explanation:

Signal Flow Graph

Definition: A Signal Flow Graph (SFG) is a graphical representation of the relationships among variables of a set of linear algebraic equations. It consists of nodes (representing variables) and directed branches (representing the relationship between variables, often denoted by gain or transfer function).

Working Principle: In an SFG, each variable is represented as a node, and the directional flow of the signal is illustrated with arrows. The weights (or gains) on the edges determine how the input signal is transformed as it flows through the graph. The graph can be used to derive equations representing the interdependencies between nodes.

Analysis of the Correct Option:

The correct option is:

Option 4: X2 = A21 × X1 + A23 × X3

This equation correctly represents the relationship between the node X2 and its input variables as per the given signal flow graph. It implies that the value of X2 is influenced by two incoming signals:

• The first term, A21 × X1, signifies that X2 receives a signal from X1, scaled by the gain A21.
• The second term, A23 × X3, indicates that X2 also receives a signal from X3, scaled by the gain A23.

These two terms together describe the total input to the node X2. This formulation aligns with the standard representation of SFGs, where the outgoing signal from a node is the weighted sum of all incoming signals.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: X2 = A21 × X1 + A22 × X2

This equation suggests that X2 is influenced by itself (A22 × X2), which would create a feedback loop. While feedback loops are possible in signal flow graphs, the given problem likely does not include this self-referencing term. Hence, this option is incorrect.

Option 2: X4 = A41 × X2 + A43 × X3

This equation relates to a different node, X4, rather than X2. The problem specifically asks about the equation governing X2, so this option is irrelevant in this context.

Option 3: X3 = A31 × X1 + A23 × X2 + A33 × X3

This equation describes the relationship for X3, not X2. Although it appears to be valid for X3, it does not address the question posed about X2. Therefore, this option is not the correct answer.

Option 4: X2 = A21 × X1 + A23 × X3

As explained earlier, this equation accurately represents the relationship for X2 in the signal flow graph. It accounts for the contributions of X1 and X3 to X2, with their respective gains A21 and A23. Hence, this is the correct option.

Conclusion:

Signal Flow Graphs are a powerful tool to visually represent and analyze the relationships between variables in a system. By carefully examining the nodes and their connections, we can derive the governing equations for each variable. In this case, the equation X2 = A21 × X1 + A23 × X3 was identified as the correct representation for the node X2, based on the given SFG.

Understanding the role of each term in the equation is essential for interpreting the behavior of the system. Additionally, distinguishing between relevant and irrelevant options requires close attention to the specific node being analyzed and the structure of the SFG.

Concept:

We are given a signal flow graph. To determine output C₁ due to inputs R₁ and R₂, we use the principles of signal flow analysis, particularly Mason’s Gain Formula.

Mason’s Gain Formula:

\[ T = \frac{\sum_{k} P_k \Delta_k}{\Delta} \]

• Pₖ = Forward path gain of the k-th path
• Δ = 1 - (sum of all individual loop gains) + (sum of gain products of two non-touching loops) - ...
• Δₖ = Value of Δ for path k excluding loops touching that path

Analysis:

From the figure:

• Path 1 (R₁ → C₁): Gain = G₁
• Path 2 (R₁ → G₃ → G₂ → C₁): Gain = -G₁G₂G₃ (due to subtraction at node)
• Path 3 (R₂ → G₄ → G₂ → C₁): Gain = -G₂G₃G₄ (due to subtraction)

Closed Loop: There is a feedback loop: G₂G₃

Net Transfer Function:

C₁ = G₁R₁ − G₂G₃G₄R₂
Denominator = 1 − G₂G₃

So, \[ C_1 = \frac{G_1 R_1 - G_2 G_3 G_4 R_2}{1 - G_2 G_3} \]

Expression for x₃ from the Signal Flow Graph

Definition: Signal flow graphs are graphical representations of the relationships among variables in a system. They are widely used in control systems and mathematical modeling to represent dynamic systems and their interactions. In this case, we aim to derive the expression for x₃ based on the given system dynamics.

Correct Expression: The correct expression for x₃ is:

Option 1: x3=d2x2dt2+dx1dt−x1" id="MathJax-Element-130-Frame" role="presentation" style="position: relative;" tabindex="0">x3=d2x2dt2+dx1dt−x1x3=d2x2dt2+dx1dt−x1 $\mathrm{x}_{3}=\frac{\mathrm{d}^{2} \mathrm{x}_{2}}{\mathrm{dt}^{2}}+\frac{\mathrm{dx}_{1}}{\mathrm{dt}}-\mathrm{x}_{1}$

This expression indicates that x₃ is derived by combining the second derivative of x₂, the first derivative of x₁, and subtracting x₁. This formulation is consistent with standard signal flow graph analysis and represents the dynamic relationship between the variables.

Detailed Solution:

To understand why Option 1 is correct, let us break down the components:

• d2x2dt2" id="MathJax-Element-131-Frame" role="presentation" style="position: relative;" tabindex="0">d2x2dt2d2x2dt2 $\frac{\mathrm{d}^{2} \mathrm{x}_{2}}{\mathrm{dt}^{2}}$ : This term represents the second derivative of x₂ with respect to time, indicating the acceleration or rate of change of the rate of x₂. It contributes to the system dynamics by accounting for higher-order changes in x₂.
• dx1dt" id="MathJax-Element-132-Frame" role="presentation" style="position: relative;" tabindex="0">dx1dtdx1dt $\frac{\mathrm{dx}_{1}}{\mathrm{dt}}$ : This term represents the first derivative of x₁ with respect to time, capturing the rate of change of x₁. It reflects how x₁ evolves dynamically over time.
• −x1" id="MathJax-Element-133-Frame" role="presentation" style="position: relative;" tabindex="0">−x1−x1 $-\mathrm{x}_{1}$ : This term subtracts x₁ directly from the expression, which may represent a feedback or compensatory mechanism in the system.

The combination of these terms yields the dynamic expression for x₃, reflecting the interdependence of the variables in the system.

Applications:

• This formulation is commonly used in control systems, signal processing, and dynamic system analysis.
• It helps in predicting system behavior, designing controllers, and analyzing stability.

Important Information:

To further understand the analysis, let’s evaluate why the other options are incorrect:

Option 2: x3=d2x2dt2+dx1dt−x1−1" id="MathJax-Element-134-Frame" role="presentation" style="position: relative;" tabindex="0">x3=d2x2dt2+dx1dt−x1−1x3=d2x2dt2+dx1dt−x1−1 $\mathrm{x}_{3}=\frac{\mathrm{d}^{2} \mathrm{x}_{2}}{\mathrm{dt}^{2}}+\frac{\mathrm{dx}_{1}}{\mathrm{dt}}-\mathrm{x}_{1}-1$

This option introduces an additional term, "-1," which is not supported by the standard signal flow graph analysis. The inclusion of this constant term would imply an external offset or bias in the system, which is not mentioned in the problem. Hence, this option is incorrect.

Option 3: x3=d2x2dt2−dx1dt−x1" id="MathJax-Element-135-Frame" role="presentation" style="position: relative;" tabindex="0">x3=d2x2dt2−dx1dt−x1x3=d2x2dt2−dx1dt−x1 $\mathrm{x}_{3}=\frac{\mathrm{d}^{2} \mathrm{x}_{2}}{\mathrm{dt}^{2}}-\frac{\mathrm{dx}_{1}}{\mathrm{dt}}-\mathrm{x}_{1}$

This option incorrectly subtracts dx1dt" id="MathJax-Element-136-Frame" role="presentation" style="position: relative;" tabindex="0">dx1dtdx1dt $\frac{\mathrm{dx}_{1}}{\mathrm{dt}}$ instead of adding it. The negative sign would change the dynamics of the system, leading to a behavior that does not match the given signal flow graph. Therefore, this option is incorrect.

Option 4: x3=d2x2dt2+dx1dt−x1+1" id="MathJax-Element-137-Frame" role="presentation" style="position: relative;" tabindex="0">x3=d2x2dt2+dx1dt−x1+1x3=d2x2dt2+dx1dt−x1+1 $\mathrm{x}_{3}=\frac{\mathrm{d}^{2} \mathrm{x}_{2}}{\mathrm{dt}^{2}}+\frac{\mathrm{dx}_{1}}{\mathrm{dt}}-\mathrm{x}_{1}+1$

Similar to Option 2, this option introduces an additional "+1" term, which is not consistent with the given signal flow graph. The inclusion of this constant would imply an external input or bias, which is not specified in the problem. Hence, this option is incorrect.

Conclusion:

The correct expression for x₃ is derived from the signal flow graph by appropriately combining the second derivative of x₂, the first derivative of x₁, and subtracting x₁. This formulation, represented by Option 1, accurately reflects the system dynamics. Understanding signal flow graphs and their corresponding equations is crucial for analyzing and designing dynamic systems effectively.

Summing point:

It is the point where two signals are added or subtracted.

Take-off point

Take off point is the point from where the signal is taken and feedback or forward to a summing point in the system.

Concept:

Signal flow graph

• It is a graphical representation of a set of linear algebraic equations between input and output.
• The set of linear algebraic equations represents the systems.
• The signal flow graphs are developed to avoid mathematical calculation.

Maon gain formula is used to find the ratio of any two nodes or transfer function.

T F = \(\mathop \sum \limits_{k = 1}^i \frac{{{P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\) 

Where P_k = k^th forward path gain

Δ = 1- ∑ individual loop gain + ∑ two non-touching loops gain - ∑ the gain product of three non-touching loops + ∑ gain of four non-touching loops

Shotcut: while writing Δ take the opposite sign for the odd number of non-touching loops snd the same sign for the even the number of non-touching loops.

Δ_K is obtained from Δ by removing the loops touching the K^th forward path.

Calculation:

For the given SFG two forward paths

\({P_{K1}} = 1\left( {{s^{ - 1}}} \right)\left( {{s^{ - 1}}} \right)\left( 1 \right) = {s^{ - 2}}\)

\({P_{k2}} = 1\left( {{s^{ - 1}}} \right)\left( 1 \right)\left( 1 \right) = {s^{ - 1}}\)

Since all loops are touching the paths P_K1 and P_K2 so Δ_K1 = Δ_K2 = 1

We have Δ = 1- ∑ individual loops + ∑ non-touching loops gain

Loops are

\({L_1} = \left( { - 4} \right)\left( 1 \right) = - 4\)

\({L_2} = \left( { - 4} \right)\left( {{s^{ - 1}}} \right) = - 4{s^{ - 1}}\)

\({L_3} = \; - 2\left( {{s^{ - 1}}} \right)\left( {{s^{ - 1}}} \right) = - 2{s^{ - 2}}\)

\({L_4} = - 2\left( {{s^{ - 1}}} \right)\left( 1 \right) = - 2{s^{ - 1}}\)

As all the loops are touching each other we have

Δ = 1 – ( L1 + L2 + L3 + L4)

Δ = 1 – ( - 4 – 4s^-1 – 2s^-2 -2s^-1 )

Δ = 5 + 6s^-1 + 2s^-2

\(T.F = \frac{{{s^{ - 2}} + {s^{ - 1}}}}{{5 + 6{s^{ - 1}} + 2{s^{ - 2}}}}\)

\( = \frac{{s + 1}}{{5{s^2} + 6s + 2}}\)

Concept:

According to Mason’s gain formula, the transfer function is given by

\(TF = \frac{{\mathop \sum \nolimits_{k = 1}^n {M_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

Where, n = no of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

In the given signal flow graph,

Forward paths: P₁ = ad

Loops: L₁ = cd, L₂ = ade

Δ = 1 – (cd + ade)

Δ₁ = 1

Transfer function \( = \frac{{ad}}{{1 - \left( {cd + ade} \right)}}\)

Now, let us check the options.

Option 1:

Forward paths: P₁ = a(d + c)

Loops: L₁ = ae(d + c)

Δ = 1 – ae(d + c)

Δ₁ = 1

Transfer function \( = \frac{{a\left( {d + c} \right)}}{{1 - ae\left( {d + c} \right)}}\)

Option 2:

Forward paths: P₁ = d(a + c)

Loops: L₁ = de(a + c)

Δ = 1 – de(a + c)

Δ₁ = 1

Transfer function \( = \frac{{d\left( {a + c} \right)}}{{1 - de\left( {a + c} \right)}}\)

Option 3:

Forward paths: \({P_1} = a\left( {\frac{d}{{1 - cd}}} \right)\)

Loops: \({L_1} = ae\left( {\frac{d}{{1 - cd}}} \right)\)

\({\rm{\Delta }} = 1 - ae\left( {\frac{d}{{1 - cd}}} \right)\)

Δ₁ = 1

Transfer function \( = \frac{{a\left( {\frac{d}{{1 - cd}}} \right)}}{{1 - ae\left( {\frac{d}{{1 - cd}}} \right)}} = \frac{{ad}}{{1 - \left( {cd + ade} \right)}}\)

Option 4:

Forward paths: \({P_1} = a\left( {\frac{c}{{1 - cd}}} \right)\)

Loops: \({L_1} = ae\left( {\frac{c}{{1 - cd}}} \right)\)

\({\rm{\Delta }} = 1 - ae\left( {\frac{c}{{1 - cd}}} \right)\)

Δ₁ = 1

Transfer function \( = \frac{{a\left( {\frac{c}{{1 - cd}}} \right)}}{{1 - ae\left( {\frac{c}{{1 - cd}}} \right)}} = \frac{{ad}}{{1 - \left( {cd + ace} \right)}}\)

\(Y\left( s \right) = \left[ { - 50\;Y\left( s \right) + K\;D\left( s \right)} \right]\frac{1}{{s + 2}} + D\left( s \right)\;\)

\(Y\left( s \right)\left[ {1 + \frac{{50}}{{s + 2}}} \right] = \left[ {\frac{K}{{s + 2}} + 1} \right]D\left( s \right)\)

\(\Rightarrow Y\left( s \right) = \frac{{K + s + 2}}{{s + 52}}D\left( s \right)\)

\(\Rightarrow Y\left( {j\omega } \right) = \frac{{K + 2 + j\omega }}{{52 + j\omega }}D\left( {j\omega } \right)\)

The disturbance response at the output y(t) is zero mean.

At ω = 0, Y(j0) = 0

\(\Rightarrow \frac{{K + 2 + 0}}{{52 + 0}} = 0\)

⇒ K = -2

Concept:

Mason's Gain Formula is used to evaluate an overall transmittance (gain), which can be expressed as,

\(T = \frac{{\sum {P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

Where

Pk = forward path transmittance of kth path

Δ = graph determinant comprising closed-loop transmittances & mutual interactions between non-touching loops.

ΔK = path factor consisting of all isolated closed loops from the forward path in the graph.

Analysis:

Forward path: G₁ G₂, G₁ G₃

Loops: -G₂, -G₁G₂H, -G₃

Finding the transfer function using Mason's gain formula:

\(\frac{C}{R}= \frac{G_1 G_2 + G_1 G_3}{1+G_2+G_3+G_1G_2H}\)

Concept:

According to Mason’s gain formula, the transfer function is given by

\(TF = \frac{{\mathop \sum \nolimits_{k = 1}^n {M_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

Where, n = No of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

Given signal flow graph is

Forward paths

M₁ = s × s × 1/s = s

M₂ = 1/s × 1/s = 1/s²

Loops:

L₁ = s × s (-1) = - s²

L₂ = -1/s 

L₃ =  s × s × 1/s × (-s) = -s²

L₄= 1/s × 1/s × (-s) = - 1/s

Δ = 1 - ( - s2  - s2  - 1/s  - 1/s) = 1 + 2s² + 2/s

\(TF = \frac{s+\frac{1}{s^2}}{1+2s^2+\frac{2}{s}}= \frac{s^3+1}{2s^4+s^2+2s}\)

Concept:

If the open-loop transfer function G(s) is connected in positive feedback with a feedback gain of H(s), then the transfer function of the closed-loop system is: \(\frac{{G\left( s \right)}}{{1 - G\left( s \right)H\left( s \right)}}\)

If the open-loop transfer function G(s) is connected in negative feedback with a feedback gain of H(s), then the transfer function of the closed-loop system is: \(\frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

When two systems are connected in parallel, then the overall gain of the system will be the sum of their individual gains.

When two systems are connected in a cascade connection, then the overall gain of the system will be the product of their individual gains.

Calculation:

We have,

Here,

G(s) = G

H(s) = 2

Feedback is positive,

from above concept,

TF = \(\frac{{G\left( s \right)}}{{1 - G\left( s \right)H\left( s \right)}}\) = \(\frac{G}{1-2G}\)

Explanation:

Signal Flow Graph:

• A graphical method of representing the control system using the linear algebraic equations is known as the signal flow graph.
• It is abbreviated as SFG
• The equation representing the system holds multiple variables that perform a crucial role in forming the graph.
• The signal from a node to another flows through the branch in the direction of the arrowhead.
• The graphical method is valid only for linear time-invariant systems.
• A block diagram can be converted to SFG as shown.

Self Loop:

• A loop consisting of a single branch and a single node.
• The paths in such loops are never defined by any forward path or feedback loop as these never trace any other node of the graph.
• For the given SFG it is formed at node 4 by branch d.

Non-Touching Loops:  

• When two or more loops have not shared a common node then that type of loop is called non-touching loops.
• For the given SFG it is formed by nodes 2-6-2 and 3-4-3
   

Mixed Node:

• It is also known as a chain node and consists of branches having both entering as well as leaving signals.
• For the given SFG node 2 to node 7 are mixed nodes.
• For this SFG, node 2 to node 7 are mixed nodes.

Touching Loops: 

• When two or more loops have shared a common node then that type of loop is called touching loops.
• For the given SFG it is formed by nodes 2-6-2 and 3-4-3

Concept:

Mason’s Gain Formula

• It is a technique used for finding the transfer function of a control system. A formula that determines the transfer function of a linear system by making use of the signal flow graph is known as Mason’s Gain Formula.
• It shows its significance in determining the relationship between input and output.

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

\(T = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\mathop \sum \nolimits_{i = 1}^N {P_i}{{\rm{Δ }}_i}}}{{\rm{Δ }}}\)

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

Δ = 1−(sum of all individual loop gains) + (sum of gain products of all possible two non-touching loops) − (sum of gain products of all possible three non-touching loops) + ........

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

Calculations:

The forward paths are as follows:

P₁ = 5

P₂ = 2 × 3 × 4 = 24

The loops are as follows:

L₁ = -2, L₂ = -3, L₃ = -4, L₄ = -5

The two non-touching loops are:

L₁L₃ = 8

There is no three non-touching loops

By Mason’s gain formula:-

\(\frac{C}{R} = \frac{{24\; + \;5\left( {1\; + \;3} \right)}}{{1\; + \;2\; + \;3\; + \;4 \;+ \;5\; + \;8}}\) 

\(= \frac{{44}}{{23}}\) 

Concept:

Mason’s gain formula is

\(T = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\mathop \sum \nolimits_{i = 1}^N {P_i}{{\rm{Δ }}_i}}}{{\rm{Δ }}}\)

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

Δ = 1−(sum of all individual loop gains) + (sum of gain products of all possible two non-touching loops) − (sum of gain products of all possible three non-touching loops) + ........

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

Calculation:

Given block diagram is,

There are two forward paths,

Δ₁P₁ = G₁, Δ₂P₂ = G₂

There are two loops,

- G₁H, - G₂H

Δ = 1 - (- G1H - G2H) = 1 + H (G₁ + G₂)

From mason's gain formula

\( \Rightarrow \frac{Y(s)}{X(s)} = \frac{{{G_1} + {G_2}}}{{1 + H\left( {{G_1} + {G_2}} \right)}} \)