The number of sign changes in the first column of Routh array reveals.
the number of roots of the characteristic polynomial in right half plane.
What the Routh–Hurwitz test operates on. The array is constructed from the coefficients of the characteristic polynomial — the denominator of the closed-loop transfer function set to zero:
\(1+G(s)H(s)=0\)
The roots of this equation are the closed-loop poles, and it is their location that decides stability: any root with a positive real part produces a term \(e^{\sigma t}\) with σ > 0 in the response, which grows without bound.
The theorem itself. After building the array, examine the first column. Then:
1. The number of sign changes in the first column equals the number of characteristic-equation roots in the right half of the s-plane.
2. The system is stable if and only if there are no sign changes, i.e. all first-column entries have the same sign.
Small worked illustration. For \(s^{3}+s^{2}+2s+8=0\) the first column comes out as 1, 1, −6, 8 — two sign changes (+ to − and − to +), so two roots lie in the right half plane and the system is unstable. Factorising confirms roots at s = −2 and \(s = 0.5 \pm j1.94\).
Why the other options are wrong. The Routh array is built only from the characteristic polynomial, so it says nothing about zeros — zeros affect the shape of the response and any non-minimum-phase behaviour, but never stability. Nor does it report open-loop pole locations: an open-loop unstable plant can perfectly well be stabilised by feedback, and the array reflects the closed-loop result. (Counting open-loop right-half-plane poles is what the Nyquist criterion needs, via \(Z = N + P\) — a different test.)
Special cases worth remembering. A zero in the first column is replaced by a small ε and the limit taken; an entire row of zeros signals symmetrically placed roots (often a pair on the jω axis, giving marginal stability) and is handled by forming the auxiliary polynomial from the row above.
Hence, the number of sign changes reveals the number of roots of the characteristic polynomial in the right half plane.
Similar Questions
Routh-Hurwitz criterion applicable for
A. absolute stability
B. the number of roots lying on the right half of the S-plane
C. the gain margin and phase margin
D. the phase crossover frequency
E. the gain crossover frequency
Choose the most appropriate answer from the options given below :
Consider a sixth order system with characteristic equation s6 + 2s5 + 8s4 + 12s3 + 20s2 + 16s + 16 = 0, The control system is:
Assertion (A) : According to Routh-Hurwitz criterion, the system represented by characteristic equation F(S) will be unstable, if the first column of the array contains no sign change.
Reason (R) : If any row of the Routh's table is multiplied or divided by a positive integer, then the system stability will not be affected.
Select your answer using the codes given below.
Assertion (A) : The roots of the characteristic equation must lie in the left hand s-plane for the system to exhibit a stable time response.
Reason (R) : Routh-Hurwitz criterion is a necessary and sufficient criterion for stability.
Select your answer using the codes given below.
A fourth order system is characterised by the equation S4 + 8S3 + 18S2 + 16S + 5 = 0. This is a :
According to Routh's stability criterion the necessary conditions for the system to be stable are defined as :
(a) All the coefficients of the characteristics equation should be missing.
(b) None of the coefficients should be real and should have different sign.
(c) None of the coefficients of the characteristics equation should be missing or zero.
(d) All the coefficients should be real and should have the same sign.Which of the above statements are correct ?
What is the relation between k and T such that a unity feedback control system whose open loop transfer function below is stable :
\(G(s)=\dfrac{k}{s\left[s(s+10)+T\right]}\)
Assertion (A) : Routh Hurwitz criterion gives both absolute as well as relative stability using characteristic equation.
Reason (R) : For a system to be stable all roots of characteristic equation must lie in left half of s-plane.
Select your answer using the codes given below :
Which of the following statements are correct in respect of Routh's stability criterion ?
(A) This stability orientation applies to polynomials with only finite number of terms
(B) When this criterion is applied to a control system, the information about absolute stability can be obtained directly from the coefficients of the characteristic equations
(C) The Routh's stability criterion tells whether or not there are unstable roots in polynomial equation without actually solving for them
(D) Routh's criterion is not related to system stabilityChoose the most appropriate answer from the options given below :
Important Questions from Routh-Hurwitz Stability Criteria
Match List I with List II:
List I
(Coefficients of s 2+ a 1s + a 2= 0)
List II
(Nature of Roots)
(A) a \(_1^2\) > 4a 2 (I) Negative real and equal (B) a \(_1^2\) = 4a 2 (II) Conjugate Imaginary (C) a \(_1^2\) < 4a 2 (III) Negative Real and Unequal (D) a 1= 0
a 2≠ 0
(IV) Conjugate Complex (Real part negative) Choose the correct answer from the options given below:
A closed-loop control system has a characteristic equation given by s 3 + 2.4s 2 + 1.8s + 0.5 = 0. Find out the value of a, b, c, and d using the Routh Hurwitz criterion.
s 3
1
1.8
s 2
2.4
0.5
s 1
a
c
s 0
b
d
Determine the stability of system:
S 3+ S 2+ S + 4
Which of the following is NOT the advantage of Routh-Hurwitz criterion of control systems?
The number of roots of s 3+ 5s 2+ 7s + 3 = 0 in the left half of the s-plane is
