Given a stable polynomial $\phi(s)=a_0+a_1{s}+a_2{s^2}+a_3{s^3}+\cdots+a_n{s^n}=\phi^{e}(s)+s\phi^{o}(s)$ where $\phi^{e}(s)=a_0+a_2{s^2}+a_4{s^4}+\cdots,~\phi^{o}(s)=a_1+a_3{s^2}+a_5{s^4}+\cdots$, how to determine the stability of the polynomial $\phi^{e}(s)+\dfrac{d\phi^{e}(s)}{ds}=a_0+a_2{s^2}+2a_2{s}+a_4{s^4}+4a_4{s^3}+a_6{s^6}+6a_6{s^5}+\cdots$?

By stable polynomial I meant, all the roots of the polynomial are on the open left half plane.

I initially thought of checking interlacing property involving $\phi^{e}(s)$ and $\phi^{o}(s)$ but that didn't take me any further. Next from the characteristic equation $\dfrac{d\phi^{e}(s)}{ds}+\phi^{e}(s)=0$, we get $ln(\phi^{e}(s))=-s+c$,

or $\phi^{e}(s)=ke^{-s}=k(1-s+\dfrac{s^2}{2}-\dfrac{s^3}{6}-\cdots)$. Since the coefficients are not strictly positive or negative, $\phi^{e}(s)$ is not stable and hence $\phi^{e}(s)+\dfrac{d\phi^{e}(s)}{ds}=k(1-s)e^{-s}$ is also unstable.

I was actually supposed to show the stability of the polynomial $\phi^{e}(s)+\dfrac{d\phi^{e}(s)}{ds}$, but I don't know where I went wrong in the analysis. Any suggestions are greatly appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.