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The Routh array counts sign changes in its first column to determine the number of continuous-system poles in the right half of the s-plane without solving for roots explicitly.
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Formula
s in s−1; polynomial coefficients carry powers consistent with the chosen normalization; Routh entries are sign-test values, not measured quantities
| Symbol | Meaning | Unit |
|---|---|---|
| Real characteristic polynomial of a continuous-time LTI system | polynomial in s | |
| Laplace variable or pole; Re(s)>0 is the right half-plane | s−1 | |
| Coefficients of the normalized cubic characteristic polynomial | matching powers of s | |
| Entries in the first column of the cubic Routh array | sign value |
p(s)=s^3+a_2s^2+a_1s+a_0;\quad R_1=1,\ R_2=a_2,\ R_3=\frac{a_2a_1-a_0}{a_2},\ R_4=a_0Assume a real characteristic polynomial of a continuous-time linear time-invariant system; asymptotic stability requires every pole strictly in the left half-plane. For s3+a2s2+a1s+a0, this is equivalent to a2>0, a1>0, a0>0 and a2a1>a0. In general, sign changes in the first Routh column count right-half-plane roots. A zero pivot or an all-zero row is a special case requiring the auxiliary-polynomial procedure; imaginary-axis poles are not asymptotically stable. Routh is a coefficient-based stability test. It is distinct from the Nyquist sampling theorem fs>2fmax and also distinct from the Nyquist plot, another control-stability method.
Real characteristic polynomial of a continuous-time LTI system. Asymptotic stability requires all poles strictly in the left half-plane. Handle zero pivots, zero rows and axis poles separately.
The Routh array uses its first column to count right-half-plane poles.
Cubic stability condition
For real coefficients; axis poles preclude asymptotic stability.
Third first-column entry
Directly valid only for a2≠0; otherwise use special-case procedure.
Test p(s)=s3+2s2+3s+4 with Routh–Hurwitz.
First column 1, 2, (6−4)/2=1, 4.
No sign changes means no right-half-plane poles; the cubic stability conditions hold.
The system is asymptotically stable.
Interpret the Nyquist sampling rule as a stability test.
Routh checks continuous poles; fs>2fmax concerns signal sampling.
Formula cluster
Coefficients suffice to count unstable poles; numerical root solving is unnecessary.
For p(s)=s3+2s2+3s+4, the first Routh-column entries are 1, 2, (2·3−4)/2=1, 4. They are all positive, so there are no sign changes and no right-half-plane poles: the system is asymptotically stable. Check the cubic condition: 2·3=6>4. For s3+2s2+s+4, the third entry is −1; +,+,−,+ has two changes, hence two right-half-plane poles.
Check asymptotic stability of continuous closed-loop systems from characteristic-polynomial coefficients.
Here is how to work through a typical Routh–Hurwitz stability criterion (For a cubic: a₂>0, a₁>0, a₀>0 and a₂a₁>a₀) task step by step:
Task
Test p(s)=s3+2s2+3s+4 with Routh–Hurwitz.
Solution path
First column 1, 2, (6−4)/2=1, 4.
No sign changes means no right-half-plane poles; the cubic stability conditions hold.
The system is asymptotically stable.
The number of sign changes in the first Routh-array column equals the number of characteristic roots in the right half-plane when special cases are handled correctly. No changes do not prove stability if imaginary-axis poles or an all-zero row were missed. Check the assumptions and full array, not just one polynomial coefficient. Assume real coefficients and a continuous-time LTI system. Asymptotic stability requires poles strictly in the left half-plane; zero pivots, zero rows and axis poles need separate handling. Check the calculation with this case: For p(s)=s3+2s2+3s+4, the first Routh-column entries are 1, 2, (2·3−4)/2=1, 4. They are all positive, so there are no sign changes and no right-half-plane poles: the system is asymptotically stable. Check the cubic condition: 2·3=6>4. For s3+2s2+s+4, the third entry is −1; +,+,−,+ has two changes, hence two right-half-plane poles. Keep the model limits and units in view.
For p(s)=s3+a2s2+a1s+a0, a2, a1 and a0 must be positive and a2a1>a0 must also hold. The first column is 1, a2, (a2a1−a0)/a2, a0. In s3+2s2+3s+4 the third entry is 1; the sequence 1,2,1,4 remains positive, so the system is asymptotically stable. Assume real coefficients and a continuous-time LTI system. Asymptotic stability requires poles strictly in the left half-plane; zero pivots, zero rows and axis poles need separate handling. The corresponding units are s in s−1; polynomial coefficients carry powers consistent with the chosen normalization; Routh entries are sign-test values, not measured quantities. For p(s)=s3+2s2+3s+4, the first Routh-column entries are 1, 2, (2·3−4)/2=1, 4. They are all positive, so there are no sign changes and no right-half-plane poles: the system is asymptotically stable. Check the cubic condition: 2·3=6>4. For s3+2s2+s+4, the third entry is −1; +,+,−,+ has two changes, hence two right-half-plane poles. This checks that the reference quantities are compatible.
No. Routh–Hurwitz checks poles of a continuous closed loop from its characteristic polynomial. The sampling condition fs>2fmax instead concerns reconstructing a band-limited signal after digitization. There is also a Nyquist-plot criterion for feedback stability; it is a different method from Routh and not the sampling condition. Assume real coefficients and a continuous-time LTI system. Asymptotic stability requires poles strictly in the left half-plane; zero pivots, zero rows and axis poles need separate handling. Check the calculation with this case: For p(s)=s3+2s2+3s+4, the first Routh-column entries are 1, 2, (2·3−4)/2=1, 4. They are all positive, so there are no sign changes and no right-half-plane poles: the system is asymptotically stable. Check the cubic condition: 2·3=6>4. For s3+2s2+s+4, the third entry is −1; +,+,−,+ has two changes, hence two right-half-plane poles. Keep the model limits and units in view.
Keep one polynomial normalization and check its coefficients: s in s−1; polynomial coefficients carry powers consistent with the chosen normalization; Routh entries are sign-test values, not measured quantities. Calculate the Routh column consistently. Example: For p(s)=s3+2s2+3s+4, the first Routh-column entries are 1, 2, (2·3−4)/2=1, 4. They are all positive, so there are no sign changes and no right-half-plane poles: the system is asymptotically stable. Check the cubic condition: 2·3=6>4. For s3+2s2+s+4, the third entry is −1; +,+,−,+ has two changes, hence two right-half-plane poles. This verifies the sign changes and stability conclusion; zero-pivot and zero-row cases need separate treatment. Assume real coefficients and a continuous-time LTI system. Asymptotic stability requires poles strictly in the left half-plane; zero pivots, zero rows and axis poles need separate handling.
The background states this starting condition: Assume a real characteristic polynomial of a continuous-time linear time-invariant system; asymptotic stability requires every pole strictly in. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Assume real coefficients and a continuous-time LTI system. Asymptotic stability requires poles strictly in the left half-plane; zero pivots, zero rows and axis poles need separate handling.
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Which formula describes Routh–Hurwitz stability criterion?
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How do you rearrange Routh–Hurwitz stability criterion for Cubic stability condition?
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Which common mistake happens with Routh–Hurwitz stability criterion?
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