Block Diagram Reduction Technique MCQ Quiz - Objective Question with Answer for Block Diagram Reduction Technique - Download Free PDF

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.

• When blocks are connected in parallel, the overall transfer function is obtained by adding the transfer functions of the individual blocks.
• In a block diagram, when blocks are connected in parallel, it means that they all receive the same input and provide outputs that contribute to the overall output.
• If you have two or more blocks connected in parallel with transfer functions \(G_1(S), G_2(s),G_3(S)\)then overall transfer function with G(s) for the parallel configuration is given by:
  \(G(S) = G_1(S) + G_2(s) + G_3(S)\)
• The transfer function of the whole system will be the addition of the transfer function of each block.

Concept:

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

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

Where, n = no of forward paths

Mk = kth forward path gain

Δk = 1 - Sum of the loop that exists after removal of the kth forward path + sum of the gain product of two non-touching loops

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)

Calculation

While calculating output C1 due to R1 and R2 , keep C₂ = 0

The forward path from R₁ to C₁ is:

\(M_1={G_1}{R_1}\)

The forward path from R₂ to C1 is:

\(M_2={G_1}{G_3}{G_4}{R_2}\)

Self-loop gain:

\(\Delta={1-G_1 G_2 G_3 G_4}\)

\(T(s)=\rm\frac{G_1 R_1−G_1 G_3 G_4 R_2}{1−G_1 G_2 G_3 G_4}\)

Hence, the correct answer is option 1.

Given Block Diagram:

Given block diagram can be reduced by,

Further, It can be reduced by,

Hence,

\(\frac{Y(S)}{R(S)}=\frac{1}{S^2+S(K+1)+1}\) .... (1)

The standard equation of 2nd order system can be written as,

TF = \(\frac{ω_n^2}{S^2+2ζω_nS+ω_n^2}\) .... (2)

From equation (1) & (2),

ω_n = ±1

2ζω_n = (K + 1)

or, 2ζ = (K + 1)

or, \(\zeta=\frac{K+1}{2}\)

Peak over shoot (M_p) = \({\large e}^{(\frac{-\zeta π}{\sqrt{1-\zeta^2}})}\)

We have,

Now,

\(\frac{Y(S)}{U(S)}=\frac{1}{S(S+1+K)}\)

Here, \(ϕ = -\frac{π}{2}-tan^{-1}\frac{ω}{K+1}\)

At: ω → 0 ⇒ ϕ = -(π/2)

At: ω → ∞ ⇒ ϕ = -(π)

Hence, K will only affct the Peak overshoot.

Given Block Diagram:

Given block diagram can be reduced by,

Further, It can be reduced by,

Hence,

\(\frac{Y(S)}{R(S)}=\frac{1}{S^2+S(K+1)+1}\) .... (1)

The standard equation of 2nd order system can be written as,

TF = \(\frac{ω_n^2}{S^2+2ζω_nS+ω_n^2}\) .... (2)

From equation (1) & (2),

ω_n = ±1

2ζω_n = (K + 1)

or, 2ζ = (K + 1)

or, \(\zeta=\frac{K+1}{2}\)

Peak over shoot (M_p) = \({\large e}^{(\frac{-\zeta π}{\sqrt{1-\zeta^2}})}\)

We have,

Now,

\(\frac{Y(S)}{U(S)}=\frac{1}{S(S+1+K)}\)

Here, \(ϕ = -\frac{π}{2}-tan^{-1}\frac{ω}{K+1}\)

At: ω → 0 ⇒ ϕ = -(π/2)

At: ω → ∞ ⇒ ϕ = -(π)

Hence, K will only affct the Peak overshoot.

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

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

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

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

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

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

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

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

Concept:

Mason’s gain formula to find the transfer function is given by:

\(TF = \frac{1}{{\rm{\Delta }}}\mathop \sum \limits_K^ {}{{P_k\rm{\Delta }}_k}\)

Pk = Path Gain of kth forward path

Δ = 1 – (sum of loop gains of all individual loops) + (sum of the gain product of all possible combinations of two Non-toucing loops) ….

Δk  = Value of Δ obtained by removing all the loops touching kth forward path.

Calculation:

signal flow graph of the given block diagram

There are three forward paths having gain

P₁ = 1 / s², P₂ = 1 / s, P₃ = 1

and here Δ = 1 (∵ there is no loop)

Δ1 = Δ2 = Δ3 = 1

Transfer function using masson gain formula

 \(TF = \frac{1}{{\rm{\Delta }}}\mathop \sum \limits_K^ {}{{P_k\rm{\Delta }}_k}\)

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

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

The transfer function of the left side block \( = \frac{{{G_1}}}{{1 + {G_1}{H_1}}}\)

The transfer function of the right-side block \( = \frac{{{G_2}}}{{1 + {G_2}{H_2}}}\)

These two blocks are in cascade connection. Therefore, the overall transfer function will multiplication of their individual transfer functions.

\(TF = \frac{{{G_1}{G_2}}}{{\left( {1 + {G_1}{H_1}} \right)\left( {1 + {G_2}{H_2}} \right)}} = \frac{{{G_1}{G_2}}}{{1 + {G_1}{H_1} + {G_2}{H_2} + {G_1}{G_2}{H_1}{H_2}}}\)

Solving the loop:

\(\frac{{s + \frac{1}{s}}}{{1 + \left( {s + \frac{1}{s}} \right)}} = \frac{{{s^2} + 1}}{{{s^2} + s + 1}}\)

Solving the second loop, we get:

\(\frac{{\frac{1}{s}\left( {\frac{{{s^2} + 1}}{{{s^2} + s + 1}}} \right)}}{{1 + \frac{{{s^2} + 1}}{{s\left( {{s^2} + s + 1} \right)}}}}\)

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

Given Block Diagram:

Given block diagram can be reduced by,

Further, It can be reduced by,

Hence,

\(\frac{Y(S)}{R(S)}=\frac{1}{S^2+S(K+1)+1}\) .... (1)

The standard equation of 2nd order system can be written as,

TF = \(\frac{ω_n^2}{S^2+2ζω_nS+ω_n^2}\) .... (2)

From equation (1) & (2),

ω_n = ±1

2ζω_n = (K + 1)

or, 2ζ = (K + 1)

or, \(\zeta=\frac{K+1}{2}\)

Peak over shoot (M_p) = \({\large e}^{(\frac{-\zeta π}{\sqrt{1-\zeta^2}})}\)

We have,

Now,

\(\frac{Y(S)}{U(S)}=\frac{1}{S(S+1+K)}\)

Here, \(ϕ = -\frac{π}{2}-tan^{-1}\frac{ω}{K+1}\)

At: ω → 0 ⇒ ϕ = -(π/2)

At: ω → ∞ ⇒ ϕ = -(π)

Hence, K will only affct the Peak overshoot.

Concept:

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

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

Where, n = no of forward paths

Mk = kth forward path gain

Δk = 1 - Sum of the loop that exists after removal of the kth forward path + sum of the gain product of two non-touching loops

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)

Calculation

While calculating output C1 due to R1 and R2 , keep C₂ = 0

The forward path from R₁ to C₁ is:

\(M_1={G_1}{R_1}\)

The forward path from R₂ to C1 is:

\(M_2={G_1}{G_3}{G_4}{R_2}\)

Self-loop gain:

\(\Delta={1-G_1 G_2 G_3 G_4}\)

\(T(s)=\rm\frac{G_1 R_1−G_1 G_3 G_4 R_2}{1−G_1 G_2 G_3 G_4}\)

Hence, the correct answer is option 1.

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 cascade connection, then the overall gain of the system will be the product of their individual gains.

Calculation:

The transfer function of the right part of the given block diagram is \(\frac{{{G_a}{G_b}}}{{1 + {G_a}{G_b}{H_b}}}\)

Both are connected in cascade connection, therefore the overall transfer function is

\(=\frac{{{G_a}{H_b}}}{{{H_a}\left( {1 + {G_a}{G_b}{H_b}} \right)}}\)

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 cascade connection, then the overall gain of the system will be the product of their individual gains.

Calculation:

G₂ and H₁ are connected in a feedback loop.

The simplified expression for this block is \(\frac{{{G_2}}}{{1 + {G_2}{H_1}}}\)

Now, the block diagram can be simplified as shown below.

Now, the overall transfer function is

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{{G_1}{G_2}}}{{1 + {G_2}{H_1}}}}}{{1 + \frac{{{G_1}{G_2}{H_2}}}{{1 + {G_2}{H_1}}}}}\)

\( = \frac{{{G_1}{G_2}}}{{1 + {G_2}{H_1} + {G_1}{G_2}{H_2}}}\)

\({{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right)\) can be realized directly by cascading the systems.

\({{\rm{G}}_1}\left( {\rm{s}} \right)\left( {\frac{1}{{{{\rm{G}}_1}\left( {\rm{s}} \right)}} + {{\rm{G}}_2}\left( {\rm{s}} \right)} \right) = 1 + {{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right)\) can be realised by providing a unity feed forward path around the cascaded arrangement.

\({{\rm{G}}_1}\left( {\rm{s}} \right)\left( {\frac{1}{{{{\rm{G}}_1}\left( {\rm{s}} \right)}} - {{\rm{G}}_2}\left( {\rm{s}} \right)} \right) = 1 - {{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right)\)  can be realised by providing a unity feed forward path around the cascaded arrangement and multiplying -1 to the cascaded arrangement of \({{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right)\).

Thus, only \(\frac{{{{\rm{G}}_1}\left( {\rm{s}} \right)}}{{{{\rm{G}}_2}\left( {\rm{s}} \right)}}{\rm{}}\) cannot be realised using the above systems.

Additional Information

Block diagram reduction rules:

Rule 1: Moving the branch point ahead of the block:

Rule 2: Moving the branch point before the block:

Rule 3: Moving the summing point ahead of the block

Rule 4: Moving the summing point before the block

Rule 5: Interchanging the summing points

Application: