Nature of roots of a cubic equation

[pseudockb]

I have derived a transfer function containing 3 poles. All the coefficients are positive but from matlab analysis, there is one LHP real pole and a pair of RHP complex poles. Previously, I have a misconception that a cubic equation having all +ve coeff will yield all poles in the LHP. Could some mathematics experts provide some guidance here? Thanks!

 

[Kral]

pseudockb,
Consider two systems; both have real roots at -1.
One system has complex roots at -SQRT(3) +/-j.5
The other system has complex roots at +SQRT(3) +/-j.5.
Multiply the factors of both systems, i.e.,
(x+1)[x-(SQRT(3)/2 -j/2][x-(SQRT(3)/2 -j/2]
(x+1)[x+(SQRT(3)/2 -j/2][x+(SQRT(3)/2 -j/2]
You will find that both polynomials have all positive coefficients.
Regards,
Kral

 

[pseudockb]

Thanks, I see your point.

By the way, do you happen to know how to determine whether the complex poles reside in the LHP or RHP by looking at the relationship among the coefficients of the cubic equation?

 

[Kral]

pseudockb,
Go to the following website:
h**p://en.wikipedia.org/wiki/Cubic_equation
Scroll down to equation 4.
Let A = the value of u with the + sign under the radical.
Let B = the value of u with the - sign under the radical.
A, B are the principal cube roots.
It can be shown, e.g.,CF Burrington "Handbook of Mathematical Tables and Formulas", available at any decent library, that the solution for the real part of the root is
-(1/2)(A+B).
So if (A+B) is positive, the real part of the root is negative. If (A+B) is neagtive, then the real part of the root is positive.
.
Incidentally, the imaginary parts of the roots are given by +/-[(jSQRT(3)/2)](A-B)
.
This exercise may not be worth the effort, since by the time you calculate A, B, you almost have the complete solution! This is the best that I can do. Hope it helps.
Regards,
Kral