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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?

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    \$\begingroup\$ I'm sure you mean "thinking hard", but "thinking hardly" makes one think you mean "hardly thinking", i.e. not thinking much at all! \$\endgroup\$
    – Hearth
    Commented Oct 25, 2018 at 15:02
  • \$\begingroup\$ 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. \$\endgroup\$
    – TimWescott
    Commented Oct 25, 2018 at 15:04

2 Answers 2

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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

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

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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<ad\$; 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}\$

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