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

Concept:

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

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

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:

Number of forward paths = 1

P₁ = G(s)

Number of loops = 3

$L_1=-\frac{G\left(s\right)}{s},L_2=-G\left(s\right),L_3=-sG\left(S\right)$

$\mathrm{\Delta }=1+\frac{G\left(s\right)}{s}+G\left(s\right)+sG\left(s\right)$

$=1+G\left(s\right)\left[\frac{1}{s}+1+s\right]$

Δ = 1

Transfer function $\frac{C\left(s\right)}{R\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)\left[\frac{1}{s}+1+s\right]}$

It is given that, $\frac{C\left(s\right)}{R\left(s\right)}=\frac{s}{s^2+s+2}$

$\Rightarrow \frac{G\left(s\right)}{1+G\left(s\right)\left[\frac{1}{s}+1+s\right]}=\frac{s}{s^2+s+2}$

$\Rightarrow G\left(s\right)\left[s^2+s+2\right]=s+G\left(s\right)\left[1+s+s^2\right]$

⇒ G(s) = s

Concept:

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

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

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:

Number of forward paths = 2

$P_1=\frac{h_0}{s^2},P_2=\frac{h_1}{s}$

Number of loops = 2

$L_1=-\frac{a_1}{s},L_2=-\frac{a_0}{s^2}$

$\mathrm{\Delta }=1+\frac{a_1}{s}+\frac{a_0}{s^2}$

${\mathrm{\Delta }}_1=1,{\mathrm{\Delta }}_2=1+\frac{a_1}{s}$

Transfer function $=\frac{\frac{h_0}{s^2}\left(1\right)+\frac{h_1}{s}\left(1+\frac{a_1}{s}\right)}{1+\frac{a_1}{s}+\frac{a_0}{s^2}}$

$=\frac{h_0+h_1\left(s+a_1\right)}{s^2+a_{1}s+a_0}$

$=\frac{b_0-b_1{a}_1+b_1\left(s+a_1\right)}{s^2+a_{1}s+a_0}$

$G\left(s\right)=\frac{b_{1}s+b_0}{s^2+a_{1}s+a_0}$

Concept:

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

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

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:

Forward paths: P₁ = G₁G₃G₂G₄, P₂ = G₅

Loops: L₁ = -H₁, L₂  -G₃ H₂, L₃ = -G₄H₃

Two no touching loops: + H₁H₂G₃, H₁H₃G₄

Δ = 1 – (-H₁ – G₃H₂ – G₄H₃) + (H₁H₂G₃ + H₁H₃G₄)

= 1 + G₃H₂ + G₄H₃ + H₁ + G₃H₁H₂ + G₄H₁H₃

Δ₁ = 1

Δ₂ = 1 + G₃H₂ + H₁ + G₃H₁H₂ 

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{G_1{G}_2{G}_3{G}_4+G_5\left(1+G_3{H}_2+H_1+G_3{H}_1{H}_2\right)}{\left(1+G_3{H}_2+G_4{H}_3+H_1+G_3{H}_1{H}_2+G_4{H}_1{H}_3\right)}$

Concept:

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

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

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: $L_1=ae\left(\frac{d}{1-cd}\right)$

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

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: $L_1=ae\left(\frac{c}{1-cd}\right)$

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

Δ₁ = 1

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

Hence the signal graph in option (3) is the equivalent representation of the signal flow graph given in the question.

Forward paths: P₁ = G₁G₄G₂, P₂ = G₁G₄G₃

Loops: L₁ = G₁G₄H₁, L₂ = G₁G₄G₂H₂, L₃ = -G₁G₄G₃H₂

There are no nontouching loops.

Δ = 1 – (L₁ + L₂ + L₃)

L₁ = G1G4H1

L₂ =  -G1G2G4H2

L₃ =  -G1G4G3H2

Δ = 1 – G₁G₄H₁ + G₁G₂G₄H₂ + G₁G₄G₃H₂

Δ₁ = 1, Δ₂ = 1

Transfer function, $\frac{C}{R}=\frac{G_1{G}_4\left(G_2+G_3\right)}{1-G_1{G}_4{H}_1+G_1{G}_3{G}_4{H}_2+G_1{G}_2{G}_4{H}_2}$

Concept:

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

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

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: $L_1=ae\left(\frac{d}{1-cd}\right)$

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

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: $L_1=ae\left(\frac{c}{1-cd}\right)$

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

Δ₁ = 1

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

Hence the signal graph in option (3) is the equivalent representation of the signal flow graph given in the question.

No. of forward paths from u₁(s) to Y(s) are:

$P_1=\left(1\right)\left(\frac{1}{L}\right)\left(\frac{1}{S}\right)\left(K_1\right)\left(\frac{1}{J}\right)\left(\frac{1}{S}\right)\left(1\right)=\frac{K_1}{JL{S}^2}$

No. of loops are:

$L_1=\left(\frac{1}{L}\right)\left(\frac{1}{S}\right)\left(-R\right)=\frac{-R}{SL}$

$L_2=\left(\frac{1}{L}\right)\left(\frac{1}{s}\right)\left(K_1\right)\left(\frac{1}{J}\right)\left(\frac{1}{S}\right)\left(-K_2\right)=-\frac{K_1{K}_2}{JL{S}^2}$

$\mathrm{\Delta }=1-\left(L_1+L_2\right)=1+\frac{R}{SL}+\frac{K_1{K}_2}{JL{S}^2}$

From Mason’s gain formula:

Transfer function $=\frac{{\sum }_{K=1}^n{P}_K{\mathrm{\Delta }}_K}{\mathrm{\Delta }}$

$=\frac{\frac{K_1}{JL{S}^2}\left(1\right)}{1+\frac{R}{SL}+\frac{K_1{K}_2}{JL{S}^2}}$

$\frac{Y\left(s\right)}{U_1\left(s\right)}=\frac{K_1}{JL{S}^2+JRS+K_1{K}_2}$

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 signal flow graph,

The number of forward paths from X₂ to Y is 1

P₁Δ₁ = G₂

Number of loops 

L₁ = - G₁G₂, L₂ = G₁

Δ = 1 - (L₁ + L₂) = 1 + G₁ + G₁G₂ 

$TF=\frac{P_1{\Delta }_1}{\Delta }=\frac{G_2}{1+G_1\left[1+G_2\right]}$

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Concept:

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

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

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:

Number of forward paths = 1

P₁ = G(s)

Number of loops = 3

$L_1=-\frac{G\left(s\right)}{s},L_2=-G\left(s\right),L_3=-sG\left(S\right)$

$\mathrm{\Delta }=1+\frac{G\left(s\right)}{s}+G\left(s\right)+sG\left(s\right)$

$=1+G\left(s\right)\left[\frac{1}{s}+1+s\right]$

Δ = 1

Transfer function $\frac{C\left(s\right)}{R\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)\left[\frac{1}{s}+1+s\right]}$

It is given that, $\frac{C\left(s\right)}{R\left(s\right)}=\frac{s}{s^2+s+2}$

$\Rightarrow \frac{G\left(s\right)}{1+G\left(s\right)\left[\frac{1}{s}+1+s\right]}=\frac{s}{s^2+s+2}$

$\Rightarrow G\left(s\right)\left[s^2+s+2\right]=s+G\left(s\right)\left[1+s+s^2\right]$

⇒ G(s) = s

Concept:

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

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

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:

Number of forward paths = 2

$P_1=\frac{h_0}{s^2},P_2=\frac{h_1}{s}$

Number of loops = 2

$L_1=-\frac{a_1}{s},L_2=-\frac{a_0}{s^2}$

$\mathrm{\Delta }=1+\frac{a_1}{s}+\frac{a_0}{s^2}$

${\mathrm{\Delta }}_1=1,{\mathrm{\Delta }}_2=1+\frac{a_1}{s}$

Transfer function $=\frac{\frac{h_0}{s^2}\left(1\right)+\frac{h_1}{s}\left(1+\frac{a_1}{s}\right)}{1+\frac{a_1}{s}+\frac{a_0}{s^2}}$

$=\frac{h_0+h_1\left(s+a_1\right)}{s^2+a_{1}s+a_0}$

$=\frac{b_0-b_1{a}_1+b_1\left(s+a_1\right)}{s^2+a_{1}s+a_0}$

$G\left(s\right)=\frac{b_{1}s+b_0}{s^2+a_{1}s+a_0}$

Manon’s gain Formula for the determination of the overall system gain is given by

$T=\sum _{k=1}^k\frac{P_k{\mathrm{\Delta }}_k}{\mathrm{\Delta }}$

T = Overall gain of the system

P_k = Path gain of the system

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

Δ_k = the value of Δ for the part of the graph not touching with the k^th forward path

Δ₁ = G₁(s) G₃(s)

Δ₂ = G₂(s) G₃(s)

L₁ = G₁(s) H₁(s)

L₂ = G₂(s) H₁(s)

L₃ = G₁(s) G₃(s)

L₄ = G₂(s) G₃(s)

$\frac{\{G_1\left(s\right)+G_2\left(s\right)\}{G}_3\left(s\right)}{1+\left(G_1\left(s\right)+G_2\left(s\right)\right)\left(H_1\left(s\right)+G_3\left(s\right)\right)}$

For N(s) = 0,

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{\frac{2}{s}\cdot \frac{1}{\left(s+3\right)}}{1+\frac{2}{s}\cdot \frac{1}{\left(s+3\right)}}=\frac{2}{s\left(s+3\right)+2}=\frac{2}{\left(s+1\right)\left(s+2\right)}$

For R(s) = 0,

$\frac{C\left(s\right)}{N\left(s\right)}=\frac{\frac{1}{\left(s+3\right)}}{1+\frac{2}{s}\cdot \frac{1}{s+3}}=\frac{s}{\left(s+1\right)\left(s+2\right)}$

Given that, r(t) = u(t), n(t) = δ(t)

$R\left(s\right)=\frac{1}{s},N\left(s\right)=1$

$C\left(s\right)=\frac{2}{\left(s+1\right)\left(s+2\right)}\cdot \frac{1}{s}+\frac{s}{\left(s+1\right)\left(s+2\right)}$

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

$=\frac{1}{s}-\frac{3}{\left(s+1\right)}+\frac{3}{\left(s+2\right)}$

By applying inverse Laplace transform,

c(t) = 1 – 3e^-t + 3e^-2t

Concept:

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

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

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: $L_1=ae\left(\frac{d}{1-cd}\right)$

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

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: $L_1=ae\left(\frac{c}{1-cd}\right)$

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

Δ₁ = 1

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

Hence the signal graph in option (3) is the equivalent representation of the signal flow graph given in the question.