# Determine the value of a constant for which the system is marginally stable

So I have the following transfer function

$$G(s)=\frac{s^2+0.1s}{10s^3 + 1.1s^2 + 0.01s + 2K}$$

Now I'm trying to determine the value of K so that I have a marginally stable system. I'm not supposed to use the Routh-Hurwitz method. I'm thinking hard but I seem to get to nowhere. I know that for the system to be marginally stable I will need a real pole in the left complex plane and two complex conjugate pure imaginary poles. But how can I determine the exact value of K that will provide me with those 3 specific poles?

• I'm sure you mean "thinking hard", but "thinking hardly" makes one think you mean "hardly thinking", i.e. not thinking much at all! – Hearth Oct 25 '18 at 15:02
• I can't answer without reading your prof's mind. Have they taught you root-locus or Bode plots yet? Either of those could be used. – TimWescott Oct 25 '18 at 15:04

I guess the simplest way of making sure the roots are located correctly is:

A polynomial with one real root and 2 imaginary roots generally looks as follows:

$$A\cdot (s+a)\cdot (s^2 + b)$$

This needs to match the denominator, so

\begin{align} 10\cdot (s + a)\cdot (s^2 + b) &= 10\cdot s^3 + (10a) \cdot s^2 + (10b)\cdot s + (10\cdot a\cdot b) \\ &= 10\cdot s^3 + 1.1 \cdot s^2 + 0.01\cdot s + 2K \end{align}

From this it immediately follows that

\begin{align} a &= \frac{1.1}{10} = 0.11\\ b &= \frac{0.01}{10} = 0.001\\ K &= \frac{10\cdot a\cdot b}{2} = \frac{0.11\cdot 0.001}{2} = 55\cdot 10^{-5} \end{align}

In the case of a 3rd order CLTF denominator: $$\\small as^3+bs^2+cs+d\$$, critical stability is obtained when $$\\small bc=ad\$$. Instability occurs when $$\\small bc; and a stable system has $$\\small bc>ad\$$.

At critical stability the denominator must factorise to $$\\small a(s^2+\omega^2)(s+\alpha )\$$, since there must be a steady (non-decaying) sinusoidal term. By inspection, the frequency of this sinusoid is: $$\ \small \omega= \sqrt{\frac{c}{a}}\:\$$ rad/sec.

For your system, $$\\small 20K=1.1 \times 10^{-2}\$$, giving $$\\small K=5.5 \times 10^{-4}\$$