This section of Control Systems Multiple Choice Questions and Answers (MCQs) focuses on ” Routh Hurwitz Criterion for stability analysis”.It will help you to prepare for various competitive exams.
CONTROL SYSTEM -ROUTH HURWITZ CRITERION MCQs
1. A linear time-invariant system is said to be stable if it satisfies the condition:
Explanation:- A linear time-invariant system is said to be stable if it satisfies the condition (i)For a bounded input, the output must be bounded. (ii)In the absence of input, the output of the system tends to zero irrespective of the initial conditions(This concept of stability is known as Asymptotic stability.)
2.Investigate the stability of a closed-loop system having characteristic equation given as s4+2s3+3s2+4s+5=0
Asymptotic Stable
Explanation:-By using the Routh Hurwitz criterion we observe that number of sign changes in the first column is 2, so 2 roots are in the right half of the s-plane. Hence the system is unstable.
3. The number of sign changes in the first column of Routh’s array indicates:
Explanation:- The number of sign changes in the first column of Routh’s array indicates that the number of roots of the characteristics equation is in the Right half of the s plane if any roots are in the right half of the s-plane then the system is unstable.
4. If the system having complex poles on Jw-axis(Imaginary axis)then the system will be:
Explanation:- if the system having complex poles on Jw-axis(Imaginary axis) then the corresponding time response would be a sustained oscillation of constant amplitude. So the system will be marginally stable.
5. The open-loop transfer function of a feedback control is given as G(s)H(s)=K/s(s+4)(s2+2s+2).Find the value of K for stability.
Explanation:- For stability, all the elements of the first column of Routh’s array should be of the same sign, hence(i) K>0 (ii) 8-36K/52 >0 i.e K<11.56 so the range is 0<K<11.56
6. When all the elements in any row of Routh’s array are zero then what would be the possibility of the location of roots of the characteristics equation in the s-plane?
Explanation:- When anyone row of the Routh table is zero, it indicates that the equation has at least one pair of roots which lie radially opposite to each other and equidistant from origin. so all the options given in the question are possible.
- b0 , b1 , b2 , b3 > 0 and b1 b2 –
b0 b3 < 0
b0 , b1 , b2 , b3 > 0 and b1 b2 – b0 b3 > 0
b0 , b1 , b2 , b3 > 0 and b1 b2 – b0 b3 = 0
b0 , b2 > 0 and b3 , b1 < 0
Explanation:-The necessary condition for a system to be stable is (i) all the coefficients of the equation should have the same sign. (ii)There should be no missing term. if these two conditions are satisfied and all the elements of the first column of the Routh table are of the same sign then the system is stable. So option (B) best suited the above conditions.
8. A system is said to be unstable for which of the following condition:
Explanation:- A system is said to be unstable if there are repeated roots on the jw axis and if any roots of the characteristic equation have a positive real part, or any roots lie in the right half of the s-plane.
9. The polynomial whose coefficients are the elements of the row just above the rows of zeros in the Routh Array is called the Auxiliary equation.
Explanation:- The polynomial whose coefficients are the elements of the row just above the rows of zeros in the Routh Array is called the Auxiliary equation. It has the following properties: (i)It gives the number and location of root pairs of the characteristic equation which are symmetrically located in the s-plane. (ii)The order of the auxiliary equation is always EVEN.
10. The characteristic equation of a feedback control system is s3+ ks2 + 5s + 10 = 0.For the system to be critically stable, the value of K should be:
Explanation:-For a critically stable system, all the elements of the one row of the Routh Table of the given characteristic equation should be zero. so (10k-20)/k =0 i.e k=2.
More Related MCQ’s