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

• 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

$TF=\frac{{\sum }_{k=1}^n{M}_k{\Delta }_k}{\Delta }$

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:

$\mathrm{\Delta }=1-G_1{G}_2{G}_3{G}_4$

$T(s)=\frac{{\mathrm{G}}_1{\mathrm{R}}_1-{\mathrm{G}}_1{\mathrm{G}}_3{\mathrm{G}}_4{\mathrm{R}}_2}{1-{\mathrm{G}}_1{\mathrm{G}}_2{\mathrm{G}}_3{\mathrm{G}}_4}$

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 = $\frac{{\omega }_n^2}{S^2+2\zeta {\omega }_{n}S+{\omega }_n^2}$ .... (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) = $e^{(\frac{-\zeta \pi }{\sqrt{1-{\zeta }^2}})}$

We have,

Now,

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

Here, $\varphi =-\frac{\pi }{2}-ta{n}^{-1}\frac{\omega }{K+1}$

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 = $\frac{{\omega }_n^2}{S^2+2\zeta {\omega }_{n}S+{\omega }_n^2}$ .... (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) = $e^{(\frac{-\zeta \pi }{\sqrt{1-{\zeta }^2}})}$

We have,

Now,

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

Here, $\varphi =-\frac{\pi }{2}-ta{n}^{-1}\frac{\omega }{K+1}$

At: ω → 0 ⇒ ϕ = -(π/2)

At: ω → ∞ ⇒ ϕ = -(π)

Hence, K will only affct the Peak overshoot.

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:

G₁ and H₁ are connected in positive feedback.

The equivalent for this combination is $=\frac{G_1}{1-G_1{H}_1}$

Now, this block and H₂ are in negative feedback.

The equivalent transfer function is,

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{\frac{G_1}{1-G_1{H}_1}}{1+\frac{G_1{H}_2}{1-G_1{H}_1}}$

$=\frac{G_1}{1-G_1{H}_1+G_1{H}_2}$

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

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

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

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

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

$T=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\sum }_{i=1}^N{P}_i{\Delta }_i}{\Delta }$

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

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:

$TF=\frac{1}{\mathrm{\Delta }}\sum _K^{}{P_k\mathrm{\Delta }}_k$

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

 $TF=\frac{1}{\mathrm{\Delta }}\sum _K^{}{P_k\mathrm{\Delta }}_k$

$TF=\frac{1}{s^2}+\frac{1}{s}+1$

$TF=\frac{s^2+s+1}{s^2}$

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:

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

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

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 = $\frac{{\omega }_n^2}{S^2+2\zeta {\omega }_{n}S+{\omega }_n^2}$ .... (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) = $e^{(\frac{-\zeta \pi }{\sqrt{1-{\zeta }^2}})}$

We have,

Now,

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

Here, $\varphi =-\frac{\pi }{2}-ta{n}^{-1}\frac{\omega }{K+1}$

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

$TF=\frac{{\sum }_{k=1}^n{M}_k{\Delta }_k}{\Delta }$

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:

$\mathrm{\Delta }=1-G_1{G}_2{G}_3{G}_4$

$T(s)=\frac{{\mathrm{G}}_1{\mathrm{R}}_1-{\mathrm{G}}_1{\mathrm{G}}_3{\mathrm{G}}_4{\mathrm{R}}_2}{1-{\mathrm{G}}_1{\mathrm{G}}_2{\mathrm{G}}_3{\mathrm{G}}_4}$

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

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{\frac{G_1{G}_2}{1+G_2{H}_1}}{1+\frac{G_1{G}_2{H}_2}{1+G_2{H}_1}}$

$=\frac{G_1{G}_2}{1+G_2{H}_1+G_1{G}_2{H}_2}$

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

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

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

Thus, only $\frac{{\mathrm{G}}_1\left(\mathrm{s}\right)}{{\mathrm{G}}_2\left(\mathrm{s}\right)}$ 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: