Find stability without solving every root
Keep a=2 and b=3. Compare c=4, 6 and 8. Why does positivity of every coefficient fail to guarantee stability?
First, picture it
This bench restricts a and b to positive values and c to nonnegative values. For p=s³+as²+bs+c, put [1,b] in the s³ row and [a,c] in the s² row. Elimination gives [(ab−c)/a,0] in s¹, then [c,0] in s⁰. Away from zero rows, the number of sign changes in the first column equals the number of right-half-plane roots.
What the model says
Start with a=2, b=3, c=4: the first column is 1,2,1,4 and all poles lie on the left. Increase c to 8. All coefficients remain positive, but the first column becomes 1,2,−1,8: two sign changes reveal two unstable roots. The exact stable range in this family is 0<c<ab. Pole locations are plotted as a separate numerical check.
p(s)=s³+as²+bs+c · first column: 1, a, (ab−c)/a, cAt c=0, p=s(s²+as+b): one pole is exactly zero. The other two are left-half-plane for positive a,b, but the full polynomial is not strictly stable. General polynomials may also have a zero first entry without an all-zero row, requiring a symbolic epsilon treatment. That case is outside the allowed parameter range here; this tool is not a general polynomial stability solver. Routh determines stability, not settling time or robustness.
A coefficient test for a cubic
This bench restricts a and b to positive values and c to nonnegative values. For p=s³+as²+bs+c, put [1,b] in the s³ row and [a,c] in the s² row. Elimination gives [(ab−c)/a,0] in s¹, then [c,0] in s⁰. Away from zero rows, the number of sign changes in the first column equals the number of right-half-plane roots.
Positive coefficients are not enough
Start with a=2, b=3, c=4: the first column is 1,2,1,4 and all poles lie on the left. Increase c to 8. All coefficients remain positive, but the first column becomes 1,2,−1,8: two sign changes reveal two unstable roots. The exact stable range in this family is 0<c<ab. Pole locations are plotted as a separate numerical check.
Treat the zero row explicitly
At c=ab, the s¹ row is entirely zero. Do not divide by zero or claim strict stability. Factor p=(s+a)(s²+b): poles are −a and ±j√b. The auxiliary polynomial from the s² row is as²+c; its derivative is 2as, which can replace the zero row in the general method. This bench displays the factorization directly. Imaginary-axis poles fail strict asymptotic stability; a minimal transfer function with these poles is not BIBO stable.
An origin pole and the scope of the result
At c=0, p=s(s²+as+b): one pole is exactly zero. The other two are left-half-plane for positive a,b, but the full polynomial is not strictly stable. General polynomials may also have a zero first entry without an all-zero row, requiring a symbolic epsilon treatment. That case is outside the allowed parameter range here; this tool is not a general polynomial stability solver. Routh determines stability, not settling time or robustness.
Make it concrete
Keep a=2 and b=3. Compare c=4, 6 and 8. Why does positivity of every coefficient fail to guarantee stability?
Open the concept benchAt c=0, p=s(s²+as+b): one pole is exactly zero. The other two are left-half-plane for positive a,b, but the full polynomial is not strictly stable. General polynomials may also have a zero first entry without an all-zero row, requiring a symbolic epsilon treatment. That case is outside the allowed parameter range here; this tool is not a general polynomial stability solver. Routh determines stability, not settling time or robustness.
Check your understanding+
Keep a=2 and b=3. Compare c=4, 6 and 8. Why does positivity of every coefficient fail to guarantee stability?
At c=ab, the s¹ row is entirely zero. Do not divide by zero or claim strict stability. Factor p=(s+a)(s²+b): poles are −a and ±j√b. The auxiliary polynomial from the s² row is as²+c; its derivative is 2as, which can replace the zero row in the general method. This bench displays the factorization directly. Imaginary-axis poles fail strict asymptotic stability; a minimal transfer function with these poles is not BIBO stable.