[HELP] Coefficients of polynomial

Copied from Control Systems Engineering (Nise/2ed/pg298):

First, if the closed-loop transfer function has only left-half-plane poles, then the factors of the denominator of the closed-loop system transfer function consist of products of terms such as (s+ai), where ai is real and positive, or complex with a positive real part. The product of such terms is a polynomial with all positive coefficients. No term (coefficient) of the polynomial can be missing, since that would imply cancellation between positive and negative coefficients or imaginary axis roots in the factors...

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My questions:
1) What does the highlighted part mean?
2) Would u pls give examples of polynomial for the cases (between positive and negative, or imaginary axis roots) above?

Thanks

Is this question complete??

cedance.

cedance said:

Is this question complete??

cedance.

Click to expand...

My first post have been modified. The text is actually copied from Control System Engineering by Norman S. Nise. It's about the stability of a control system...

Hi,

I got your question now. This is how I interpret.

the first line is pretty obvious. All terms of denominator would be of the form s+ai. But, if by math, if any term of a polynomial is missing, then, it would be impossible to factorise it of the form (s+ai) (s+bi) and so on... coz, in general, (x+a) (x+b).... would have all the powers of the polynomial, the simplest case being, (x+a)² or to the power n. the second line is a work around of the possible condition only when the first line is true, is what, i think the author had tried to mention here. If the polynomial has all powers, then it is definitely factorisable to the form (s+ai).(s+........) and if it contains all positive coefficients here, then a closed loop left half plane pole is a result. I couldnt think of any other significance to this. It helps in may be identifying such systems with closed loop TF by looking at it.

hope this is good enough...

cedance.

Since all coefficients are real, then imaginary roots must appear in complex conjugate pairs. So if you have a root -a+jb then you must also have a root -a-jb so that the product of the roots would give real coefficients for this product (s-(-a+jb))*(s-(-a-jb))

If you have a system with denominator \[s^2+1\] it is clear that you will have roots on the imaginary axis at \[\pm j1\].

Best regards,
v_c