#20 Control Theory #20 - Routh-Hurwitz Stability Criterion

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Why use Routh–Hurwitz?

   Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.

   For a finite-dimensional continuous-time LTI system with a proper rational transfer function in reduced form, BIBO stability requires every pole to lie strictly in the open left half-plane.

   Routh–Hurwitz counts right-half-plane roots using real polynomial coefficients. It often avoids explicitly solving a high-degree polynomial.

   Imaginary-axis roots and the special cases in the table must also be checked before concluding strict stability.

   Narration transcript

   In the previous lesson, we learned that stability depends on where the poles of the transfer function are located. All poles must be in the left half plane for BIBO stability. But finding the exact roots of a polynomial can be difficult, especially for higher order systems. Is there a way to check stability without actually computing the roots? The answer is yes. The Routh Hurwitz criterion gives us an algebraic test. We only need the coefficients of the characteristic polynomial.

2. 2. Choose the polynomial and the stability question

   Let p(s) = a_n sⁿ + a₍n−1) s⁽n−1) + … + a₁ s + a₀, with real coefficients and a_n ≠ 0. Its degree is n.

   For BIBO stability of a reduced rational transfer function, use the denominator after exact pole-zero cancellation. This conclusion concerns the transfer function; hidden canceled modes can still matter for internal stability.

   Alternatively, to test a finite-dimensional state model for internal asymptotic stability, use its actual characteristic polynomial. Do not silently identify these two questions when cancellations are present.

   Narration transcript

   Every LTI system described by a transfer function has a characteristic polynomial. It is the denominator of the transfer function. We write it as a n times s to the n, plus a n minus one times s to the n minus one, all the way down to a one times s, plus a zero. The roots of this polynomial are the poles. The Routh Hurwitz criterion tells us how many of these roots lie in the right half plane, without solving for them.

3. 3. Check necessary conditions

   Multiply the polynomial by a nonzero constant if needed so that its leading coefficient is positive. This does not move its roots.

   For a real polynomial with all roots strictly in the left half-plane, every coefficient is strictly positive. A missing coefficient or a coefficient with the opposite sign rules out strict Hurwitz stability.

   Positive coefficients alone are not sufficient. Continue with the Routh table, including any zero-pivot or zero-row cases.

   Narration transcript

   Before building the Routh table, there are two quick necessary conditions for stability. First, all coefficients must be present. If any coefficient is zero, the system cannot be stable. Second, all coefficients must have the same sign. If any coefficient differs in sign from the others, there is at least one root in the right half plane. These conditions are necessary but not sufficient. Even if both are satisfied, we still need the Routh table to confirm stability.

4. 4. Build the first two rows

   The table has n+1 rows, labeled sⁿ, s⁽n−1), …, s⁰.

   The first row receives a_n, a₍n−2), a₍n−4), …; the second receives a₍n−1), a₍n−3), a₍n−5), ….

   Pad unused cells with zeros. Use the actual descending powers; the informal labels odd and even depend on whether one means coefficient position or exponent.

   Narration transcript

   Now let us build the Routh table. The table has n plus one rows, where n is the degree of the polynomial. The first two rows come directly from the coefficients. Row s to the n gets the odd indexed coefficients: a n, a n minus two, a n minus four, and so on. Row s to the n minus one gets the even indexed coefficients: a n minus one, a n minus three, a n minus five, and so on. Each row is labeled by descending powers of s.

5. 5. Compute the next row

   Write the two preceding rows as u = [u₀,u₁,…] and v = [v₀,v₁,…], with v immediately above the new row.

   When v₀ is nonzero, the next row has entries w_j = (v₀ u₍j+1) − u₀ v₍j+1))/v₀. Missing entries are zero.

   Repeat down to s⁰. A fixed-width table padded with zeros is convenient; it is not true that every successive row necessarily contains one fewer nonzero entry.

   If the pivot is zero, apply the appropriate special-case procedure before dividing.

   Narration transcript

   The remaining rows are computed using a cross multiplication formula. For each entry, we take the element above and to the left, multiply it by the element diagonally above and to the right, then subtract the reverse diagonal product, and divide by the element above and to the left. This pattern repeats row by row until we reach the s to the zero row. Each row has one fewer element than the previous row, until the last row has a single entry.

6. 6. Count roots, then decide stability

   After proper treatment of special cases, the number of sign changes down the first column equals the number of roots in the open right half-plane, counting multiplicity.

   Zero sign changes means zero right-half-plane roots. For strict Hurwitz stability, also rule out roots on the imaginary axis, including the origin.

   In a regular table with a nonzero first column and no special cases, zero sign changes establishes that all roots lie strictly in the left half-plane.

   Narration transcript

   The Routh Hurwitz criterion states: the number of sign changes in the first column of the Routh table equals the number of roots in the right half plane. If there are zero sign changes, all roots are in the left half plane and the system is stable. If there is one sign change, there is one unstable root. Two sign changes mean two unstable roots, and so on. We only need to look at the first column.

7. 7. Worked example: the corrected table

   Consider p(s) = s³ + 6s² + 11s + 6. The coefficients are all positive, so the necessary coefficient test passes.

   The s³ row is [1,11] and the s² row is [6,6].

   The first entry in the s¹ row is (6×11 − 1×6)/6 = (66−6)/6 = 60/6 = 10.

   The s⁰ entry is (10×6 − 6×0)/10 = 6. The first column is [1,6,10,6].

   There are no sign changes and no zero-row or zero-pivot case. All roots lie in the open left half-plane.

   Independent check: p(s) = (s+1)(s+2)(s+3), so the roots are −1, −2 and −3.

   Source correction: the original video says and displays 9 in the s¹ row. The correct value is 10. The final stability conclusion remains the same, but the arithmetic and first column above are corrected.

   Narration transcript

   Let us work through a third order example. Consider the characteristic polynomial s cubed plus six s squared plus eleven s plus six. First, check the necessary conditions. All coefficients are present and positive. Good. Now build the Routh table. Row s cubed gets one and eleven. Row s squared gets six and six. For the s to the one row: six times eleven minus one times six, all divided by six. That gives sixty minus six over six, which is nine. For the s to the zero row: nine times six minus six times zero, divided by nine, which gives six. The first column is one, six, nine, six. No sign changes. The system is stable.

8. 8. Handle zeros explicitly

   If only the first entry of a row is zero and the rest of that row is not zero, replace that pivot with ε>0, continue the calculation and count signs in the limit ε→0+.

   If an entire row is zero, build the auxiliary polynomial from the row immediately above. Replace the zero row with the coefficients of the derivative of that auxiliary polynomial, then continue.

   The auxiliary factor contains a root set symmetric about the origin. Inspect it for imaginary-axis roots; a zero row must not be read as automatic strict stability.

   For example, q(s) = s³+s²+s+1 = (s+1)(s²+1). The s¹ row is initially all zero. The auxiliary polynomial is s²+1; its derivative gives the replacement row [2,0].

   The completed first column [1,1,2,1] has zero sign changes, but q has roots −1,+j,−j. Thus q is not strictly Hurwitz; the uncancelled transfer 1/q(s) is not BIBO stable.

   Narration transcript

   Two special cases can arise. First, a zero in the first column but the rest of the row is not all zeros. In this case, replace the zero with a small positive number epsilon, continue the table, and then examine the sign changes as epsilon approaches zero. Second, an entire row of zeros. This indicates that the polynomial has symmetric roots about the origin. We form an auxiliary polynomial from the row above, take its derivative, and use the derivative coefficients to replace the zero row. Then continue as normal.

9. 9. Checklist

   Choose the correct real polynomial for the stability question and identify its true degree.

   Check the necessary coefficient conditions, then build the table with zero padding.

   Handle a zero pivot or an entire zero row explicitly.

   Count first-column sign changes for the number of right-half-plane roots; check for imaginary-axis roots before concluding strict stability.

   For the worked example, the corrected first column is [1,6,10,6] and the roots are −1,−2,−3.

   Narration transcript

   Let us summarize. The Routh Hurwitz criterion checks stability using only the polynomial coefficients. Step one: verify that all coefficients are present and have the same sign. Step two: build the Routh table from the coefficients. Step three: count the sign changes in the first column. Zero sign changes means all poles are in the left half plane. The system is stable. In the next lesson, we will study time domain specifications: rise time, settling time, overshoot, and how they relate to pole locations.

Source video: Control Theory #20 - Routh-Hurwitz Stability Criterion (5:28)