Control Systems Test 5

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Control Systems Test 5

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Weekly Quiz Competition

Question 1
1 / -0

In the formation of Routh-Hurwitz array for a polynomial, all the elements of a row have zero values. This premature termination of the array indicates the presence of
1. a pair of real roots with opposite sign
2. complex conjugate roots on the imaginary axis
3. a pair of complex conjugate roots with opposite real parts
Which of the above statements are correct?

A
Only 2

B
2 and 3

C
Only 3

D
1, 2 and 3
Solution
Routh-Hurwitz Stability Criterion: It is used to test the stability of an LTI system.
The characteristic equation for a given open loop transfer function G(s) is
1 + G(s) H(s) = 0
According to the Routh tabulation method,
The system is said to be stable if there are no sign changes in the first column of Routh array
The number of poles lie on the right half of s plane = number of sign changes
A row of zeros in a Routh table:
This situation occurs when the characteristic equation has
- a pair of real roots with opposite sign (±a)
- complex conjugate roots on the imaginary axis (± jω)
- a pair of complex conjugate roots with opposite real parts (-a ± jb, a ± jb)
The procedure to overcome this as follows:
- Form the auxiliary equation from the preceding row to the row of zeros
- Complete Routh array by replacing the zero row with the coefficients obtained by differentiating the auxiliary equation.
- Roots of the auxiliary equation are also the roots of the characteristic equation.
- The roots of the auxiliary equation occur in pairs and are of opposite sign of each other.
- The auxiliary equation is always even order.
Question 2
1 / -0

Which of the following results is/are true regarding the polynomial
s⁴ + 3s³ + 6s² + 12s + 8

A
2 roots on the imaginary axis

B
2 roots in the right half complex plane

C
2 roots in the left half complex plane

D
The frequency of oscillation is 2√2 rad/sec

Solution

Concept:
The characteristic equation for a given open loop transfer function G(s) is
1 + G(s) H(s) = 0
According to the Routh tabulation method,
The system is said to be stable if there are no sign changes in the first column of Routh array
The number of poles lie on the right half of s plane = number of sign changes
Calculation:
s⁴ + 3s³ + 6s² + 12s + 8 = 0
\(\begin{array}{*{20}{c}} {\left. {\begin{array}{*{20}{c}} {{s^4}}\\ {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}} \right|}&{\begin{array}{*{20}{c}} 1&6&8\\ 3&{12}&0\\ 2&8&{}\\ 0&0&{}\\ {}&{}&{} \end{array}} \end{array}\)
As the raw of s¹ is zero, 2 roots lie on the imaginary axis.
As there are no sign changes, there are no roots on the right half of the s-plane.
Number of roots on the left half of s-plane = 2
Auxiliary equation: 2s² + 8 = 0
⇒ s = ±j2 rad / sec
Frequency of oscillations, = ±2 rad / sec
Question 3
1 / -0

Consider the following statement about the system gain k of a closed loop system G(s)?
a) k = 0 at both pole and zero
b) k = 0 at pole and k = ∞ at zero
c) k = ∞ at pole and k = 0 at zero
d) k = ∞ at pole and k = ∞ at zero
Which of the above statement is true

A
a)

B
b)

C
c)

D
d)

Solution

Using the Magnitude condition, as both Pole and zero lie on root locus.
Let \(G\left( s \right) = \frac{{k\left( {s + 1} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)
Zero at s = -1
Pole at s = -2 & s = -3
\(\left| {\frac{{k\left( {s + 1} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}} \right| = 1\)
At zero s = -1
\(\left| {\frac{{k\left( 0 \right)}}{{\left( 1 \right)\left( 2 \right)}}} \right| = 1\)
\(k = \frac{2}{0} = \infty\)
At Pole s = -2
\(\left| {\frac{{k\left( { - 1} \right)}}{{\left( 0 \right)\left( 1 \right)}}} \right| = 1\)
K = 1 × 0
k = 0
Question 4
1 / -0

The root locus plot of a system having open loop transfer function will have angle of asymptotes are,
\(G\left( s \right)H\left( s \right) = \frac{{k\left( {s + 10} \right)\left( {s + 70} \right)}}{{{s^3}\left( {s + 100} \right)\left( {s + 200} \right)}}\)

A
60°, 100°

B
60°, 180°, 300°

C
60°, 120°, 180°

D
60°, 90°, 120°

Solution

Angle of asymptotes (θ_L):
\({\theta _L} = \frac{{\left( {2L + 1} \right)\pi }}{{P - z}}L = 0,1,2,\;...\left| {P - z} \right| - 1\)
From the transfer function
Number of poles (P) = 5
Number of zeros (Z) = 2
(P – Z ) = 5 – 2 = 3
\({\theta _L} = \frac{{\left( {2l + 1} \right)\pi }}{3}\) L = 0, 1, 2
\(\begin{array}{l}{\rm{For\;}}l = 0,{\rm{\;}}{\theta _0} = \frac{\pi }{3} = 60^\circ \\l = 1,\;{\theta _1} = \pi = 180^\circ \\l = 2,\;{\theta _2} = \frac{{5\pi }}{3} = 300^\circ \end{array}\)
Question 5
1 / -0

A system has the characteristic equation: s⁴ + Ks³ + 2s² + (K + 1) s + 10 = 0
The range of K so that the system is stable is ________

A
0.33 < K < 1

B
K > 1

C
System is stable for all the positive values of K

D
System is unstable for any value of K

Solution

Concept:
The characteristic equation for a given open-loop transfer function G(s) is
1 + G(s) H(s) = 0
According to the Routh tabulation method,
The system is said to be stable if there are no sign changes in the first column of Routh array
The number of poles lies on the right half of s plane = number of sign changes
Calculation:
By applying Routh tabulation method,
\(\begin{array}{*{20}{c}}{{s^4}}\\{{s^3}}\\{{s^2}}\\{{s^1}}\\{{s^0}}\end{array}\left| {\begin{array}{*{20}{c}}1&2&{10}\\K&{K + 1}&0\\{\frac{{K - 1}}{K}}&{10}&{}\\{\frac{{ - 9{K^2} - 1}}{{K - 1}}}&0&{}\\{10}&{}&{}\end{array}} \right.\)
The system to be stable,
From the first three rows, K > 0 and K > 1
⇒ K > 1
From the fourth row, \(\frac{{ - 9{K^2} - 1}}{{K - 1}} > 0\)
For K > 1, the above fraction always gives a negative value.
Therefore, the system is unstable for all values of K.
Question 6
1 / -0

The following system given by its characteristic equation
1 + G(s) H(s) = s⁵ + 5s⁴ + 11s³ + 23 s² + 28s + 12 = 0 has

A
two complex poles on right half of the s plane

B
complex conjugate roots on the imaginary axis

C
a pair of real roots with opposite sign

D
a pair of complex conjugate roots with opposite real parts
Solution
Concept:
The characteristic equation for a given open loop transfer function G(s) is
1 + G(s) H(s) = 0
According to the Routh tabulation method,
The system is said to be stable if there are no sign changes in the first column of Routh array
The number of poles lies on the right half of s plane = number of sign changes
A row of zeros in a Routh table:
This situation occurs when the characteristic equation has
- a pair of real roots with opposite sign (±a)
- complex conjugate roots on the imaginary axis (± jω)
- a pair of complex conjugate roots with opposite real parts (-a ± jb, a ± jb)
The procedure to overcome this as follows:
- Form the auxiliary equation from the preceding row to the row of zeros
- Complete Routh array by replacing the zero row with the coefficients obtained by differentiating the auxiliary equation.
- Roots of the auxiliary equation are also the roots of the characteristic equation.
- The roots of the auxiliary equation occur in pairs and are of the opposite sign of each other.
- The auxiliary equation is always even ordered.
Application:
By applying Routh tabulation method,
\(\left. {\begin{array}{*{20}{c}}{{s^5}}\\{{s^4}}\\{{s^3}}\\{{s^2}}\\{{s^1}}\\{{s^0}}\end{array}} \right|\begin{array}{*{20}{c}}1&{11}&{28}\\5&{23}&{12}\\{6.4}&{25.6}&0\\3&{12}&0\\0&0&0\\{}&{}&{}\end{array}\)
The auxiliary equation is, 3s² + 12 = 0
By differentiating with respect to s,
⇒ 6s = 0
Now the table becomes,
\(\left. {\begin{array}{*{20}{c}}{{s^5}}\\{{s^4}}\\{{s^3}}\\{{s^2}}\\{{s^1}}\\{{s^0}}\end{array}} \right|\begin{array}{*{20}{c}}1&{11}&{28}\\5&{23}&{12}\\{6.4}&{25.6}&0\\3&{12}&0\\6&0&{}\\6&{}&{}\end{array}\)
The roots of the auxiliary equation are:
3s² + 12 = 0
⇒ s = ± j2
Therefore, the system has complex conjugate roots on the imaginary axis.
Note: The system is marginally stable as the system has complex conjugate roots on the imaginary axis.
Question 7
1 / -0

A system has the characteristic equation: s⁴ + 25 s³ + 15 s² + 20 s + K = 0
The frequency of oscillations of the system is ___________ (in rad/sec, rounded off up to two decimal places)

A
0.87-0.92

B
0.874-0.921

C
0.877-0.926

D
0.879-0.922
Solution
Concept:
The characteristic equation for a given open-loop transfer function G(s) is
1 + G(s) H(s) = 0
According to the Routh tabulation method,
The system is said to be stable if there are no sign changes in the first column of Routh array
The number of poles lies on the right half of s plane = number of sign changes
A row of zeros in a Routh table:
This situation occurs when the characteristic equation has
- a pair of real roots with opposite sign (±a)
- complex conjugate roots on the imaginary axis (± jω)
- a pair of complex conjugate roots with opposite real parts (-a ± jb, a ± jb)
The procedure to overcome this as follows:
- Form the auxiliary equation from the preceding row to the row of zeros
- Complete Routh array by replacing the zero row with the coefficients obtained by differentiating the auxiliary equation.
- Roots of the auxiliary equation are also the roots of the characteristic equation.
- The roots of the auxiliary equation occur in pairs and are of the opposite sign of each other.
- The auxiliary equation is always even ordered.
The system is marginally stable if the s¹ becomes a row of zero and the roots of the auxiliary equation are, s = ± jω
Where ω is the frequency of oscillations in rad/sec
Application:
By applying Routh tabulation method,
\(\left. {\begin{array}{*{20}{c}}{{s^4}}\\{{s^3}}\\{{s^2}}\\{{s^1}}\\{{s^0}}\end{array}} \right|\begin{array}{*{20}{c}}1&{15}&K\\{2.5}&{20}&0\\{14.2}&K&0\\{20 - 1.76K}&0&0\\K&{}&{}\end{array}\)
The condition for the system to be marginally stable is,
20 – 1.76 K = 0
⇒ K = 11.36
The auxiliary equation is, 14.2 s² + K = 0
⇒ 14.2 s² + 11.36 = 0
⇒ s = ± j0.894
Therefore, the system has complex conjugate roots on the imaginary axis.
The frequency of oscillations = 0.894 rad/sec

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Control Systems Test 5

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Question 1

Only 2

2 and 3

Only 3

1, 2 and 3

Solution

Question 2

2 roots on the imaginary axis

2 roots in the right half complex plane

2 roots in the left half complex plane

The frequency of oscillation is 2√2 rad/sec

Solution

Question 3

a)

b)

c)

d)

Solution

Question 4

60°, 100°

60°, 180°, 300°

60°, 120°, 180°

60°, 90°, 120°

Solution

Question 5

0.33 < K < 1

K > 1

System is stable for all the positive values of K

System is unstable for any value of K

Solution

Question 6

two complex poles on right half of the s plane

complex conjugate roots on the imaginary axis

a pair of real roots with opposite sign

a pair of complex conjugate roots with opposite real parts

Solution

Question 7

0.87-0.92

0.874-0.921

0.877-0.926

0.879-0.922

Solution

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