I have the following closed loop transfer function $$T(s)=\frac{k}{(s+10)^8+k}$$. I want to find the value for k so that the system is critically stable.
I can do this easily on Matlab but this supposed to be a question from an exam sheet so I am looking for an answer without the use of a computer . I can draw the root locus by hand. This is how it is given in Matlab.
So I'm looking for the certain k for which those 2 poles in the picure lie on the imaginary axis.
How can I solve for k without many calculations? Is there a trick or something? I usually use the Ruth-Hurwitz criterion but it seems too long to be what the teacher wants for a control theory exam.
The question is basically asking you to solve
$$ (i\,\omega + 10)^8 + k = 0, \tag{1} $$
with $i$ the imaginary unit, and $\omega$ and $k$ real numbers. The number within the brackets can be written in polar form as follows
$$ i\,\omega + 10 = \sqrt{\omega^2 + 100}\, e^{i\,\tan^{-1}\left(\frac{\omega}{10}\right)}. \tag{2} $$
Raising this to the eighth power is then trivial, namely
$$ (i\,\omega + 10)^8 = (\omega^2 + 100)^4\, e^{8\,i\,\tan^{-1}\left(\frac{\omega}{10}\right)}. \tag{3} $$
There can only be a real value for $k$ which solves equation $(1)$ if the term raised to the eighth power is also real. This is only the case when the argument is a multiple of $\pi$, so
$$ 8\,\tan^{-1}\left(\frac{\omega}{10}\right) = n\,\pi. \tag{4} $$
Solving this for $\omega$ gives
$$ \omega = 10\,\tan\left(\frac{n\,\pi}{8}\right). \tag{5} $$
Now we know that for these values for $\omega$ equation $(3)$ should be real, such that solving equation $(1)$ for $k$ becomes trivial. Substituting equation $(5)$ with equation $(3)$ into equation $(1)$ and solving for $k$ yields
$$ k = -10^8\left(\left(\tan\left(\frac{n\,\pi}{8}\right)\right)^2 + 1\right)^4\, \cos(n\,\pi). \tag{6} $$
It can be noted that both equation $(5)$ and $(6)$ are periodic, namely both with period 8. However equation $(5)$ is an odd function and equation $(6)$ is an even function. This means that the same value for $k$ is obtained if the negated value of $n$ is used, but the negated $\omega$ also satisfies equation $(1)$. So for $\omega\neq0$ you get a complex conjugate pole pairs on the imaginary axis. Since equation $(6)$ is even and periodic then one will only get unique values for $k$ for $n=0,\,1,\,2,\,3,\,4$. The resulting frequencies and gains can be seen in the table below.
$$ \begin{array}{|c|ccccc|} \hline n & 0 & 1 & 2 & 3 & 4\\ \hline \omega & 0 & 4.142 & 10 & 24.142 & 1.633\cdot10^{17}\\ k & -10^8 & 1.884\cdot10^8 & -1.6\cdot10^9 & 2.174\cdot10^{11} & -5.06\cdot10^{137} \\ \hline \end{array} $$
Equation $(6)$ only gives a gain for which the system has poles on the imaginary axis, but it does not say anything about the other poles. So in order to say anything about for which $k$ the system is critically stable one can use that for $k=0$ the closedloop poles are equal to the openloop poles and for $k\to\infty$ the closedloop poles go to the openloop zeros and if there are not enough zeros then the modulus of poles go to infinity. In this case there are no zeros, so all poles should go to infinity for $k\to\infty$. This would imply that when $|k|$ is small that the system is stable. Critically stable indicates a boundary between stable and unstable, so the values for $k$ obtained from $n=0$ and $n=1$ should give bounds between which the system is stable. Therefore at $n=0$ and $n=1$ the closedloop should be critically stable.