Block Diagram Reduction Technique MCQ Quiz in বাংলা - Objective Question with Answer for Block Diagram Reduction Technique - বিনামূল্যে ডাউনলোড করুন [PDF]

The given block diagram can be reduced as shown below.

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

At D(s) = 0, the block diagram can be redrawn as shown below.

By applying Mason’s gain formula

Forward paths = G₁G₂G₃, G₄G₂G₃

Here, we are having only one loop (G₄ is connected to input R(s). So, it cannot form a loop)

Loops = -G₁G₂G₃ H₁

Δ = 1 + G₁G₂G₃ H₁

Δ₁ = 1, Δ₂ = 1

Transfer function \(= \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{{G_1}{G_2}{G_3} + {G_2}{G_3}{G_4}}}{{1 + {G_1}{G_2}{G_3}\;{H_1}}}\)

Assume N = 0

The Transfer function

\(\frac{{{V_o}}}{{{V_i}}} = \frac{{{A_1}{A_2}}}{{1 + \beta {A_1}{A_2}}}\)         ........(1)

Assume

\(\frac{{{V_0}}}{N} = \frac{{{A_2}}}{{1 + \beta {A_1} {A_2}}}\)           .......(2)

\(\frac{{{V_0}}}{N} = \frac{{{A_2}}}{{1 + \beta {A_1.A_2}}} = 0\)

\(\frac{{{V_0}}}{N} = \frac{{{1}}}{{\frac{{{1}}}{A_2} + \beta {A_1}}} = 0\)

Now, to reduce the effect of noise:

1/A₂ + βA₁ should tend to ∞.

For this to happen, either:

1) A₂ should tend to zero, or

2) β 'or' A₁ should tend to infinity.

But since increasing β to infinity will reduce the effect of V_i from 1^st Transfer function.

So, It is desirable to increase A₁.

Of the given Options only 3^rd is satisfying this requirement.

\(\frac{C}{R} = \frac{{KG}}{{1 + KGH}}\)

Here, If KGH > > 1,

1 + KGH ≈ KGH

\(\Rightarrow \frac{C}{R} = \frac{{KG}}{{KGH}} = \frac{1}{H}\)

The transfer function given is

\(\frac{C}{R} = \frac{1}{H}\times\frac{{GH}}{{1 - GH}} = \frac{G}{{1 - GH}}\)

Hence option (c) is wrong

Considering option (b), \(\frac{C}{R} = H\times\frac{G}{{1 - G}} = \frac{{GH}}{{1 - GH}}\)

Considering option (a), \(\frac{C}{R} = \frac{H}{{1 - GH}}\)

None of the options are correct.

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:

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).

Calculation:

Forward path: G₃

Loop: -G₃ H₁

The transfer function is,

\(\frac{C}{{{R_1}}} = \frac{{{G_3}}}{{1 + {G_3}{H_1}}}\)

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:

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,

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

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

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

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

From block diagram B

First Case: Z = XG – Y

Second Case: Z = (X – Y/G)G = XG – Y

From block diagram C

First Case: Z = (X – Y)G

Second Case: Z = (XG- YG) = (X – Y)G

From block diagram A,

First Case: Z = (XG₁ – Y)G₂ = XG₁G₂ – YG₂

Second Case: Z = (XG₂ – Y)G₁ = XG₁G₂ – YG₁