Marginal Stability MCQ Quiz - Objective Question with Answer for Marginal Stability - Download Free PDF

Explanation:

To determine the value of \( K \) for the system to be stable, we first need to analyze the open-loop transfer function of the system. The given transfer function of the system is:

\[ G(s) = \frac{K}{s(s+2)(s+4)} \] and the feedback transfer function is \( H(s) = 1 \).

The closed-loop transfer function for a unity feedback system is given by:

\[ T(s) = \frac{G(s)}{1 + G(s)H(s)} = \frac{\frac{K}{s(s+2)(s+4)}}{1 + \frac{K}{s(s+2)(s+4)}} \]

To simplify, multiply the numerator and the denominator by the denominator of \( G(s) \):

\[ T(s) = \frac{K}{s(s+2)(s+4) + K} \]

For the system to be stable, all the poles of the closed-loop transfer function must lie in the left half of the s-plane (i.e., all the poles must have negative real parts). The poles of the closed-loop transfer function are the roots of the characteristic equation:

\[ 1 + G(s)H(s) = 0 \rightarrow 1 + \frac{K}{s(s+2)(s+4)} = 0 \]

Multiplying through by \( s(s+2)(s+4) \):

\[ s(s+2)(s+4) + K = 0 \]

Expanding the polynomial:

\[ s^3 + 6s^2 + 8s + K = 0 \]

To determine the range of \( K \) for stability, we can use the Routh-Hurwitz criterion. The Routh-Hurwitz criterion provides a systematic way to determine the number of roots of the characteristic equation that lie in the right half-plane, left half-plane, and on the imaginary axis. The characteristic equation for the system is:

\[ s^3 + 6s^2 + 8s + K = 0 \]

The Routh array for this characteristic equation is constructed as follows:

For the system to be stable, all the elements in the first column of the Routh array must be positive:

1. The coefficient of \( s^3 \): \( 1 \) (which is positive)
2. The coefficient of \( s^2 \): \( 6 \) (which is positive)
3. The coefficient of \( s^1 \): \(\frac{48 - K}{6} > 0 \rightarrow 48 - K > 0 \rightarrow K < 48\)
4. The coefficient of \( s^0 \): \( K > 0 \)

So, for the system to be stable:

\[ 0 < K < 48 \]

The correct option is therefore option 3: \( K < 48 \).

Additional Information

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

Option 1: \( K > 48 \)

This option is incorrect because if \( K > 48 \), the coefficient of \( s^1 \) would become negative (\(\frac{48 - K}{6} < 0\)), indicating that the system would be unstable.

Option 2: \( K < 24 \)

While \( K < 24 \) is within the stability range, it is not the complete condition for stability. \( K \) must be less than 48 for the system to be stable, so this option is incomplete and not the best representation of the stability condition.

Option 4: \( K > 24 \)

This option is also incorrect because \( K > 24 \) does not guarantee stability. The critical upper limit for stability is \( K < 48 \).

Conclusion:

The stability of a control system is crucial for its correct operation. By applying the Routh-Hurwitz criterion to the characteristic equation, we determined that the system is stable for \( 0 < K < 48 \). This comprehensive analysis helps ensure that the system operates correctly within the specified range of \( K \).

Calculation:

Given, \(G\left( s \right)=\frac{k\left( s+4 \right)}{s\left( s+1 \right)}and~H\left( s \right)=\frac{1}{s+2}\)

The closed-loop transfer function for the given negative feedback will be:

\(CLTF=\frac{G\left( s \right)}{1+G\left( s \right)H\left( s \right)}\)

Putting the respective functions, we get

\(CLTF=\frac{k\left( s+4 \right)}{\left( s+1 \right)\left( s \right)}=\left( 1+\frac{k\left( s+4 \right)}{\left( s+2 \right)\left( s \right)\left( s+1 \right)} \right)\)

\(=\frac{k\left( s+4 \right)\left( s+2 \right)}{s\left( s+1 \right)\left( s+2 \right)+k\left( s+4 \right)}\)

\(=\frac{k\left( s+2 \right)\left( s+4 \right)}{{{s}^{3}}+3{{s}^{2}}+2s+ks+4k}=\frac{k\left( s+2 \right)\left( s+4 \right)}{{{s}^{3}}+3{{s}^{2}}+\left( k+2 \right)s+4k}\)

Characteristic equation of the closed loop system is:

s³ + 3s² + (k + 2)s + 4k = 0

Forming a Routh Array to check for stability:

For the system to be marginally stable the row with s¹ must be 0.

i.e. \(\frac{6-k}{3}=0\)

6 – k = 0

k = 6

Characteristic Equation is

\(\rm {s^3} + 3{s^2} + 4s + 3 + k = 0\)

Routh array-

Frequency oscillations will take place when a entire row is zero. Here, in above Routh array, one element  of \(\rm s^1\)  row is zero. So if 2^nd element of this row (\(\rm s^1\) row) is also zero then oscillations will take place.

i.e \(\rm {{3 \times 4 - 1 \times \left( {3 + k} \right)} \over 3} = 0\)

\(\rm \eqalign{ \rm & \rm 3 \times 4 - 1 \times \left( {3 + k} \right) = 0 \cr & \rm \Rightarrow k{\rm{ }} + {\rm{ }}3 = {\rm{ }}3{\rm{ }} \times {\rm{ }}4 \cr & \rm \Rightarrow k = 9 \cr} \)

Auxiliary Equation is

\(\eqalign{ & \rm 3{s^2} + k + 3 = 0 \cr & \rm 3{s^2} + 12 = 0 \cr &\rm {s^2} = - 4 \cr & \rm s = \pm 2j \cr & \rm \omega = 2\ rad/\sec \cr} \)

Calculation:

Given, \(G\left( s \right)=\frac{k\left( s+4 \right)}{s\left( s+1 \right)}and~H\left( s \right)=\frac{1}{s+2}\)

The closed-loop transfer function for the given negative feedback will be:

\(CLTF=\frac{G\left( s \right)}{1+G\left( s \right)H\left( s \right)}\)

Putting the respective functions, we get

\(CLTF=\frac{k\left( s+4 \right)}{\left( s+1 \right)\left( s \right)}=\left( 1+\frac{k\left( s+4 \right)}{\left( s+2 \right)\left( s \right)\left( s+1 \right)} \right)\)

\(=\frac{k\left( s+4 \right)\left( s+2 \right)}{s\left( s+1 \right)\left( s+2 \right)+k\left( s+4 \right)}\)

\(=\frac{k\left( s+2 \right)\left( s+4 \right)}{{{s}^{3}}+3{{s}^{2}}+2s+ks+4k}=\frac{k\left( s+2 \right)\left( s+4 \right)}{{{s}^{3}}+3{{s}^{2}}+\left( k+2 \right)s+4k}\)

Characteristic equation of the closed loop system is:

s³ + 3s² + (k + 2)s + 4k = 0

Forming a Routh Array to check for stability:

For the system to be marginally stable the row with s¹ must be 0.

i.e. \(\frac{6-k}{3}=0\)

6 – k = 0

k = 6

Explanation:

To determine the value of \( K \) for the system to be stable, we first need to analyze the open-loop transfer function of the system. The given transfer function of the system is:

\[ G(s) = \frac{K}{s(s+2)(s+4)} \] and the feedback transfer function is \( H(s) = 1 \).

The closed-loop transfer function for a unity feedback system is given by:

\[ T(s) = \frac{G(s)}{1 + G(s)H(s)} = \frac{\frac{K}{s(s+2)(s+4)}}{1 + \frac{K}{s(s+2)(s+4)}} \]

To simplify, multiply the numerator and the denominator by the denominator of \( G(s) \):

\[ T(s) = \frac{K}{s(s+2)(s+4) + K} \]

For the system to be stable, all the poles of the closed-loop transfer function must lie in the left half of the s-plane (i.e., all the poles must have negative real parts). The poles of the closed-loop transfer function are the roots of the characteristic equation:

\[ 1 + G(s)H(s) = 0 \rightarrow 1 + \frac{K}{s(s+2)(s+4)} = 0 \]

Multiplying through by \( s(s+2)(s+4) \):

\[ s(s+2)(s+4) + K = 0 \]

Expanding the polynomial:

\[ s^3 + 6s^2 + 8s + K = 0 \]

To determine the range of \( K \) for stability, we can use the Routh-Hurwitz criterion. The Routh-Hurwitz criterion provides a systematic way to determine the number of roots of the characteristic equation that lie in the right half-plane, left half-plane, and on the imaginary axis. The characteristic equation for the system is:

\[ s^3 + 6s^2 + 8s + K = 0 \]

The Routh array for this characteristic equation is constructed as follows:

For the system to be stable, all the elements in the first column of the Routh array must be positive:

1. The coefficient of \( s^3 \): \( 1 \) (which is positive)
2. The coefficient of \( s^2 \): \( 6 \) (which is positive)
3. The coefficient of \( s^1 \): \(\frac{48 - K}{6} > 0 \rightarrow 48 - K > 0 \rightarrow K < 48\)
4. The coefficient of \( s^0 \): \( K > 0 \)

So, for the system to be stable:

\[ 0 < K < 48 \]

The correct option is therefore option 3: \( K < 48 \).

Additional Information

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

Option 1: \( K > 48 \)

This option is incorrect because if \( K > 48 \), the coefficient of \( s^1 \) would become negative (\(\frac{48 - K}{6} < 0\)), indicating that the system would be unstable.

Option 2: \( K < 24 \)

While \( K < 24 \) is within the stability range, it is not the complete condition for stability. \( K \) must be less than 48 for the system to be stable, so this option is incomplete and not the best representation of the stability condition.

Option 4: \( K > 24 \)

This option is also incorrect because \( K > 24 \) does not guarantee stability. The critical upper limit for stability is \( K < 48 \).

Conclusion:

The stability of a control system is crucial for its correct operation. By applying the Routh-Hurwitz criterion to the characteristic equation, we determined that the system is stable for \( 0 < K < 48 \). This comprehensive analysis helps ensure that the system operates correctly within the specified range of \( K \).

Calculation:

Given, \(G\left( s \right)=\frac{k\left( s+4 \right)}{s\left( s+1 \right)}and~H\left( s \right)=\frac{1}{s+2}\)

The closed-loop transfer function for the given negative feedback will be:

\(CLTF=\frac{G\left( s \right)}{1+G\left( s \right)H\left( s \right)}\)

Putting the respective functions, we get

\(CLTF=\frac{k\left( s+4 \right)}{\left( s+1 \right)\left( s \right)}=\left( 1+\frac{k\left( s+4 \right)}{\left( s+2 \right)\left( s \right)\left( s+1 \right)} \right)\)

\(=\frac{k\left( s+4 \right)\left( s+2 \right)}{s\left( s+1 \right)\left( s+2 \right)+k\left( s+4 \right)}\)

\(=\frac{k\left( s+2 \right)\left( s+4 \right)}{{{s}^{3}}+3{{s}^{2}}+2s+ks+4k}=\frac{k\left( s+2 \right)\left( s+4 \right)}{{{s}^{3}}+3{{s}^{2}}+\left( k+2 \right)s+4k}\)

Characteristic equation of the closed loop system is:

s³ + 3s² + (k + 2)s + 4k = 0

Forming a Routh Array to check for stability:

For the system to be marginally stable the row with s¹ must be 0.

i.e. \(\frac{6-k}{3}=0\)

6 – k = 0

k = 6

Characteristic equation \(1 + G\left( s \right)H\left( s \right) = 0\)

\(\begin{array}{l} 1 + \frac{K}{{\left( {s + 2} \right)\left( {s + 3} \right)}} = 0\\ {s^2} + 5s + \left( {K + 6} \right) = 0 \end{array}\)

To check w.r.t. \(s = \ – 2\) i.e. shifting origin to \(s = -2\). Hence putting

\(\begin{array}{l} z = s + 2\left( {or} \right)s = z-2\\ {\left( {z - 2} \right)^2} + 5\left( {z - 2} \right) + \left( {K + 6} \right) = 0\\ {z^2} + 4-4z + 5z-10 + K + 6 = 0\\ {z^2} + z + K = 0 \end{array}\)

Forming the routh array.

z²        1     K

z¹          1

z⁰         K

Hence the for the poles to be left of \(z = 0\ \left( {or} \right)s = -2\) axis we should have K > 0.

∴ Any value of K which is greater than zero will give roots left to s = – 2 axis.

∴ the minimum interger value of K = 1.

We know that, \({e^{ - x}} = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}}\)

If x ≪ 1 we can approximate,   \({e^{ - x}} \cong 1-x\)

Hence for low frequencies we can approximate.

\({e^{ - 3s}} = 1-3s\)

Characteristic equations is,   \(\rm q\left( s \right) = 1 + \frac{{K\left( {1 - 3s} \right)}}{{{s^3} + 6{s^2} + 3s + 4}}\)

\(\rm \begin{array}{l} {s^3} + 6{s^2} + 3s + 4 + K-3Ks = 0\\ \rm {s^3} + 6{s^2} + \left( {3 - 3K} \right)s + \left( {K + 4} \right) = 0 \end{array}\)

Obtain the Routh table we get,

s³               1                      3 – 3K

s²               6                      K + 4

s¹               \(\frac{{ - 19K + 14}}{6}\)

s⁰               K + 4

For the system to be stable,      \(\rm K + 4 > 0 \Rightarrow K > - 4\)

\(\rm \frac{{ - 19K + 14}}{6} > 0 \Rightarrow K < \frac{{14}}{{19}}\)

\(\rm \therefore The\ range\ of\ K\ for\ statbility\ is: - 4 < K < \frac{{14}}{{19}}\)

\(\rm \begin{array}{l} a = - 4\ \ and\ b = \frac{{14}}{{19}}\\ \rm \therefore \frac{b}{a} = - \frac{7}{{38}} \end{array}\)

Characteristic Equation is

\(\rm {s^3} + 3{s^2} + 4s + 3 + k = 0\)

Routh array-

Frequency oscillations will take place when a entire row is zero. Here, in above Routh array, one element  of \(\rm s^1\)  row is zero. So if 2^nd element of this row (\(\rm s^1\) row) is also zero then oscillations will take place.

i.e \(\rm {{3 \times 4 - 1 \times \left( {3 + k} \right)} \over 3} = 0\)

\(\rm \eqalign{ \rm & \rm 3 \times 4 - 1 \times \left( {3 + k} \right) = 0 \cr & \rm \Rightarrow k{\rm{ }} + {\rm{ }}3 = {\rm{ }}3{\rm{ }} \times {\rm{ }}4 \cr & \rm \Rightarrow k = 9 \cr} \)

Auxiliary Equation is

\(\eqalign{ & \rm 3{s^2} + k + 3 = 0 \cr & \rm 3{s^2} + 12 = 0 \cr &\rm {s^2} = - 4 \cr & \rm s = \pm 2j \cr & \rm \omega = 2\ rad/\sec \cr} \)

Explanation:

To determine the value of \( K \) for the system to be stable, we first need to analyze the open-loop transfer function of the system. The given transfer function of the system is:

\[ G(s) = \frac{K}{s(s+2)(s+4)} \] and the feedback transfer function is \( H(s) = 1 \).

The closed-loop transfer function for a unity feedback system is given by:

\[ T(s) = \frac{G(s)}{1 + G(s)H(s)} = \frac{\frac{K}{s(s+2)(s+4)}}{1 + \frac{K}{s(s+2)(s+4)}} \]

To simplify, multiply the numerator and the denominator by the denominator of \( G(s) \):

\[ T(s) = \frac{K}{s(s+2)(s+4) + K} \]

For the system to be stable, all the poles of the closed-loop transfer function must lie in the left half of the s-plane (i.e., all the poles must have negative real parts). The poles of the closed-loop transfer function are the roots of the characteristic equation:

\[ 1 + G(s)H(s) = 0 \rightarrow 1 + \frac{K}{s(s+2)(s+4)} = 0 \]

Multiplying through by \( s(s+2)(s+4) \):

\[ s(s+2)(s+4) + K = 0 \]

Expanding the polynomial:

\[ s^3 + 6s^2 + 8s + K = 0 \]

To determine the range of \( K \) for stability, we can use the Routh-Hurwitz criterion. The Routh-Hurwitz criterion provides a systematic way to determine the number of roots of the characteristic equation that lie in the right half-plane, left half-plane, and on the imaginary axis. The characteristic equation for the system is:

\[ s^3 + 6s^2 + 8s + K = 0 \]

The Routh array for this characteristic equation is constructed as follows:

For the system to be stable, all the elements in the first column of the Routh array must be positive:

1. The coefficient of \( s^3 \): \( 1 \) (which is positive)
2. The coefficient of \( s^2 \): \( 6 \) (which is positive)
3. The coefficient of \( s^1 \): \(\frac{48 - K}{6} > 0 \rightarrow 48 - K > 0 \rightarrow K < 48\)
4. The coefficient of \( s^0 \): \( K > 0 \)

So, for the system to be stable:

\[ 0 < K < 48 \]

The correct option is therefore option 3: \( K < 48 \).

Additional Information

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

Option 1: \( K > 48 \)

This option is incorrect because if \( K > 48 \), the coefficient of \( s^1 \) would become negative (\(\frac{48 - K}{6} < 0\)), indicating that the system would be unstable.

Option 2: \( K < 24 \)

While \( K < 24 \) is within the stability range, it is not the complete condition for stability. \( K \) must be less than 48 for the system to be stable, so this option is incomplete and not the best representation of the stability condition.

Option 4: \( K > 24 \)

This option is also incorrect because \( K > 24 \) does not guarantee stability. The critical upper limit for stability is \( K < 48 \).

Conclusion:

The stability of a control system is crucial for its correct operation. By applying the Routh-Hurwitz criterion to the characteristic equation, we determined that the system is stable for \( 0 < K < 48 \). This comprehensive analysis helps ensure that the system operates correctly within the specified range of \( K \).

The characteristic equation of transfer function is

S⁴ + 5s³ + 5s² + 4s + k = 0

Taking Routh array

For system stability k > 0 \(\frac{{84}}{5} - 5k > 0\)

\(\Rightarrow 0 < k < \frac{{84}}{{25}}\) 

⇒ For marginal stability \(k = \frac{{84}}{{25}}\)

Auxiliary polynomial in s² row

\(\begin{array}{l} \frac{{21}}{5}{s^2} + \frac{{84}}{{25}} = 0\\ \Rightarrow s = \sqrt {\frac{{ - 84}}{{25}} \times \frac{5}{{21}}} \\ s = \pm j\sqrt {\frac{4}{5}} = jw \end{array}\) 

Hence absolute value of frequency = 0.894

For certain K, root locus of this system cuts imaginary axis which is K critical.

\(Gain\;Margin = \frac{{{K_{critical}}}}{K}\)

By Routh array \(K_{cri}=\frac{27}{12}\)

\(Gain margin=27/12×4/3\)

gain margin = 3

CE is given by \(\rm 1 + \frac{{K\left( {1 - s} \right)}}{{s\left( {s + 2} \right)}} = 0\)

\(\rm {s^2} + \left( {2 - K} \right)s + K = 0\)

R-H criterion →

\(\rm \left. {\begin{array}{*{20}{c}} {{s^2}}\\ {{s^1}}\\ {{s^2}} \end{array}} \right|\begin{array}{*{20}{r}} 1&\rm K\\ \rm {2 - K}&{}\\ \rm K&{} \end{array}\)

for the intersection with imaginary axis, we should have \(\rm 2-K=0 → K = 2\)

auxiliary equation \(\rm \to {s^2} + K = 0\)

\(\rm \begin{array}{l} {s^2} + 2 = 0\\ \to s = j\pm \sqrt 2 \end{array}\)